Lý thuyết
Dùng quy tắc thực hiện phép tính, quy tắc chuyển vế, quy tắc dấu ngoặc để đưa về các dạng quen thuộc để tìm x:
\(\begin{array}{l}1)x + a = b \Rightarrow x = b - a\\2)x - a = b \Rightarrow x = b + a\\3)a - x = b \Rightarrow x = a - b\\4)a.x = b \Rightarrow x = \dfrac{b}{a}\\5)a:x = b \Rightarrow x = \dfrac{a}{b}\\6)x:a = b \Rightarrow x = a.b\\7)\dfrac{a}{b} = \dfrac{x}{c} \Rightarrow x = \dfrac{{a.c}}{b}\\8){x^2} = {a^2} \Rightarrow \left[ {\begin{array}{*{20}{c}}{x = a}\\{x = - a}\end{array}} \right.\\9){x^3} = {a^3} \Rightarrow x = a\end{array}\)
Bài tập
Bài 1:
Tìm x, biết:
\(\begin{array}{l}a)2x + 3 = 1\dfrac{2}{3}\\b)0,15 - 3x = {( - 10)^0}\\c) - x:\dfrac{2}{5} = 0,8\\d)\dfrac{{3x + 2}}{3} = \dfrac{{ - 4}}{5}\\e)\dfrac{{3x + 2}}{{ - 8}} = \dfrac{{ - 2}}{{3x + 2}}\\f)\left( {x + 1} \right).\left( { - 2x - 3} \right) = 0\end{array}\)
Bài 2:
Tìm \(x\), biết:
a) \(\dfrac{1}{3}x + \dfrac{2}{5}\left( {x - 1} \right) = 0\)
b) \(3 \cdot {\left( {3x - \dfrac{1}{2}} \right)^3} + \dfrac{1}{9} = 0\)
c) \(3 \cdot \left( {1 - \dfrac{1}{2}} \right) - 5 \cdot \left( {x + \dfrac{3}{5}} \right) = {\rm{ \;}} - x + \dfrac{1}{5}\)
d) \(\dfrac{{3 - x}}{{5 - x}} = {\left( {\dfrac{{ - 3}}{5}} \right)^2}\)
e) \(x\;:\;\dfrac{5}{8} = \dfrac{{ - 13}}{{35}} \cdot \dfrac{{15}}{{ - 39}}\)
f) \(\left( {\dfrac{7}{5}\; + \;x} \right):\dfrac{{25}}{{16}} = \dfrac{{ - 4}}{5}\)
g) \( - 4:\left( {x + \dfrac{{ - 2}}{3}} \right) = \dfrac{3}{4}\)
h) \(\left( {\dfrac{{ - 1}}{5} + 2} \right):\left( {x - \dfrac{7}{{10}}} \right) = \dfrac{{ - 1}}{4}\)
Bài 3:
Tìm tập hợp các số nguyên x để: \(\dfrac{5}{6} + \dfrac{{ - 7}}{8} \le \dfrac{x}{{24}} \le \dfrac{{ - 5}}{{12}} + \dfrac{5}{8}\)
Lời giải chi tiết:
Bài 1:
Tìm x, biết:
\(\begin{array}{l}a)2x + 3 = 1\dfrac{2}{3}\\b)0,15 - 3x = {( - 10)^0}\\c) - x:\dfrac{2}{5} = 0,8\\d)\dfrac{{3x + 2}}{3} = \dfrac{{ - 4}}{5}\\e)\dfrac{{3x + 2}}{{ - 8}} = \dfrac{{ - 2}}{{3x + 2}}\\f)\left( {x + 1} \right).\left( { - 2x - 3} \right) = 0\end{array}\)
Phương pháp
Áp dụng quy tắc thực hiện phép tính, quy tắc chuyển vế, quy tắc dấu ngoặc để đưa về các dạng quen thuộc để tìm x.
Lời giải
\(\begin{array}{l}a)2x + 3 = 1\dfrac{2}{3}\\2x + 3 = \dfrac{5}{3}\\2x = \dfrac{5}{3} - 3\\2x = \dfrac{5}{3} - \dfrac{9}{3}\\2x = \dfrac{{ - 4}}{3}\\x = \dfrac{{ - 4}}{3}:2\\x = \dfrac{{ - 4}}{3}.\dfrac{1}{2}\\x = \dfrac{{ - 2}}{3}\end{array}\)
Vậy \(x = \dfrac{{ - 2}}{3}\)
\(\begin{array}{l}b)0,15 - 3x = {( - 10)^0}\\0,15 - 3x = 1\\3x = 0,15 - 1\\3x = 0,85\\3x = \dfrac{{17}}{{20}}\\x = \dfrac{{17}}{{20}}:3\\x = \dfrac{{17}}{{20}}.\dfrac{1}{3}\\x = \dfrac{{17}}{{60}}\end{array}\)
Vậy \(x = \dfrac{{17}}{{60}}\)
\(\begin{array}{l}c) - x:\dfrac{2}{5} = 0,8\\ - x:0.4 = 0,8\\ - x = 0,8.0,4\\ - x = 0,32\\x = - 0,32\end{array}\)
Vậy x = -0,32
\(\begin{array}{l}d)\dfrac{{3x + 2}}{3} = \dfrac{{ - 4}}{5}\\5.(3x + 2) = 3.( - 4)\\15x + 10 = - 12\\15x = - 12 - 10\\15x = - 22\\x = \dfrac{{ - 22}}{{15}}\end{array}\)
Vậy \(x = \dfrac{{ - 22}}{{15}}\)
\(\begin{array}{l}e)\dfrac{{3x + 2}}{{ - 8}} = \dfrac{{ - 2}}{{3x + 2}}\\\left( {3x + 2} \right).\left( {3x + 2} \right) = ( - 8).( - 2)\\{\left( {3x + 2} \right)^2} = 16\\{\left( {3x + 2} \right)^2} = {4^2}\\\left[ {\begin{array}{*{20}{c}}{3x + 2 = 4}\\{3x + 2 = - 4}\end{array}} \right.\\\left[ {\begin{array}{*{20}{c}}{3x = 2}\\{3x = - 6}\end{array}} \right.\\\left[ {\begin{array}{*{20}{c}}{x = \dfrac{2}{3}}\\{x = - 2}\end{array}} \right.\end{array}\)
Vậy \(x \in \left\{ {\dfrac{2}{3}; - 2} \right\}\)
\(\begin{array}{l}f)\left( {x + 1} \right).\left( { - 2x - 3} \right) = 0\\\left[ {\begin{array}{*{20}{c}}{x + 1 = 0}\\{ - 2x - 3 = 0}\end{array}} \right.\\\left[ {\begin{array}{*{20}{c}}{x = - 1}\\{x = \dfrac{{ - 3}}{2}}\end{array}} \right.\end{array}\)
Vậy \(x \in \left\{ { - 1;\dfrac{{ - 3}}{2}} \right\}\)
Bài 2:
Tìm \(x\), biết:
a) \(\dfrac{1}{3}x + \dfrac{2}{5}\left( {x - 1} \right) = 0\)
b) \(3 \cdot {\left( {3x - \dfrac{1}{2}} \right)^3} + \dfrac{1}{9} = 0\)
c) \(3 \cdot \left( {1 - \dfrac{1}{2}} \right) - 5 \cdot \left( {x + \dfrac{3}{5}} \right) = {\rm{ \;}} - x + \dfrac{1}{5}\)
d) \(\dfrac{{3 - x}}{{5 - x}} = {\left( {\dfrac{{ - 3}}{5}} \right)^2}\)
e) \(x\;:\;\dfrac{5}{8} = \dfrac{{ - 13}}{{35}} \cdot \dfrac{{15}}{{ - 39}}\)
f) \(\left( {\dfrac{7}{5}\; + \;x} \right):\dfrac{{25}}{{16}} = \dfrac{{ - 4}}{5}\)
g) \( - 4:\left( {x + \dfrac{{ - 2}}{3}} \right) = \dfrac{3}{4}\)
h) \(\left( {\dfrac{{ - 1}}{5} + 2} \right):\left( {x - \dfrac{7}{{10}}} \right) = \dfrac{{ - 1}}{4}\)
Phương pháp
Áp dụng các qui tắc cộng, trừ, nhân, chia phân số, qui tắc tính giá trị của biểu thức.
Lời giải
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a) \(\dfrac{1}{3}x + \dfrac{2}{5}\left( {x - 1} \right) = 0\) \(\begin{array}{l}\dfrac{1}{3}x + \dfrac{2}{5}x - \dfrac{2}{5} = 0\\\left( {\dfrac{1}{3} + \dfrac{2}{5}} \right)x = \dfrac{2}{5}\\\dfrac{{11}}{{15}}x = \dfrac{2}{5}\end{array}\) \(x = \dfrac{2}{5}:\dfrac{{11}}{{15}}\) \(\begin{array}{l}x = \dfrac{2}{5} \cdot \dfrac{{15}}{{11}}\\x = \dfrac{6}{{11}}\end{array}\) Vậy \(x = \dfrac{6}{{11}} \cdot \) b) \(3.{\left( {3x - \dfrac{1}{2}} \right)^3} + \dfrac{1}{9} = 0\) \(\begin{array}{l}3.{\left( {3x - \dfrac{1}{2}} \right)^3} = - \dfrac{1}{9}\\{\left( {3x - \dfrac{1}{2}} \right)^3} = - \dfrac{1}{9}:3\\{\left( {3x - \dfrac{1}{2}} \right)^3} = - \dfrac{1}{{27}} = \left( {\dfrac{{ - 1}}{3}} \right)\end{array}\) \( \Rightarrow 3x - \dfrac{1}{2} = {\dfrac{{ - 1}}{3}^3}\) \(\begin{array}{l}{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} 3x = \dfrac{{ - 1}}{3} + \dfrac{1}{2}{\kern 1pt} {\kern 1pt} \\{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} 3x = \dfrac{{ - 2}}{6} + \dfrac{3}{6}\\{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} 3x = \dfrac{1}{6}{\kern 1pt} {\kern 1pt} \\{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} x = \dfrac{1}{{18}}\end{array}\) Vậy \(x = \dfrac{1}{{18}} \cdot \)
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c) \(3.\left( {1 - \dfrac{1}{2}} \right) - 5\left( {x + \dfrac{3}{5}} \right) = {\rm{ \;}} - x + \dfrac{1}{5}\) \(\begin{array}{*{20}{l}}{3 - \dfrac{3}{2} - \left( {5x + 5.\dfrac{3}{5}} \right) = {\rm{ \;}} - x + \dfrac{1}{5}}\\{\dfrac{3}{2} - 5x - 3 = {\rm{ \;}} - x + \dfrac{1}{5}}\\{ - 5x + x = \dfrac{1}{5} - \dfrac{3}{2} + 3}\end{array}\) \(\begin{array}{*{20}{l}}{ - 4x = \dfrac{{ - 13}}{{10}} + 3}\\{ - 4x = \dfrac{{17}}{{10}}}\\{x = \dfrac{{17}}{{10}}:\left( { - 4} \right)}\\{x = {\rm{ \;}} - \dfrac{{17}}{{40}}}\end{array}\) Vậy \(x = {\rm{ \;}} - \dfrac{{17}}{{40}} \cdot \) d) \(\dfrac{{3 - x}}{{5 - x}} = {\left( {\dfrac{{ - 3}}{5}} \right)^2}\) Điều kiện: \(5 - x \ne 0 \Leftrightarrow x \ne 5.\) \(\begin{array}{*{20}{l}}{ \Rightarrow \dfrac{{3 - x}}{{5 - x}} = \dfrac{9}{{25}}}\\{ \Rightarrow \left( {3 - x} \right).25 = 9.\left( {5 - x} \right)}\\{{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} 75 - 25x = 45 - 9x{\kern 1pt} }\\{{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} - 25x + 9x = 45 - 75}\\{{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} - 16x = {\rm{ \;}} - 30}\\{{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} x = \dfrac{{ - 30}}{{ - 16}} = \dfrac{{15}}{8}}\end{array}\) Vậy \(x = \dfrac{{15}}{8} \cdot \)
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e) \(x\;:\;\dfrac{5}{8} = \dfrac{{ - 13}}{{35}} \cdot \dfrac{{15}}{{ - 39}}\) \(\begin{array}{*{20}{l}}{x:\dfrac{5}{8} = \dfrac{1}{7}}\\{x\;\;\;\;\; = \dfrac{1}{7} \cdot \dfrac{5}{8}}\\{x\;\;\;\;\; = \dfrac{5}{{56}}.}\end{array}\) Vậy \(x = \dfrac{5}{{56}}\) f) \(\left( {\dfrac{7}{5}\; + \;x} \right):\dfrac{{25}}{{16}} = \dfrac{{ - 4}}{5}\) \(\begin{array}{*{20}{l}}{\dfrac{7}{5}\; + \;x\;\;\;\;\;\;\;\;\;\; = \dfrac{{ - 4}}{5} \cdot \dfrac{{25}}{{16}}}\\{\dfrac{7}{5}\; + \;x\;\;\;\;\;\;\;\;\;\; = \dfrac{{ - 5}}{4}}\\{\;\;{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} x\;\;\;\;\;\;\;\;\;\; = \dfrac{{ - 5}}{4} - \dfrac{7}{5}}\\{\;\;\;\;\;\;x\;\;\;\;\;\;\;\;\;\; = \dfrac{{ - 53}}{{20}}.}\end{array}\) Vậy \(x = \dfrac{{ - 53}}{{20}}\). |
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g) \( - 4:\left( {x + \dfrac{{ - 2}}{3}} \right) = \dfrac{3}{4}\) \(\begin{array}{*{20}{l}}{\;\;\;\;\;\;\;\;\;\;\;x + \dfrac{{ - 2}}{3} = {\rm{ \;}} - 4:\dfrac{3}{4}}\\{\;\;\;\;\;\;\;\;\;\;\;x + \dfrac{{ - 2}}{3} = {\rm{ \;}} - 4.\dfrac{4}{3}}\\{\;\;\;\;\;\;\;\;\;\;\;x + \dfrac{{ - 2}}{3} = \dfrac{{ - 16}}{3}}\\{\;\;\;\;\;\;\;\;\;\;\;x\;\;\;\;\;\;\;\; = \dfrac{{ - 16}}{3} - \left( {\dfrac{{ - 2}}{3}} \right)}\\{\;\;\;\;\;\;\;\;\;\;\;x\;\;\;\;\;\;\;\; = \dfrac{{ - 14}}{3}.}\end{array}\) Vậy \(x = \dfrac{{ - 14}}{3}\). h) \(\left( {\dfrac{{ - 1}}{5} + 2} \right):\left( {x - \dfrac{7}{{10}}} \right) = \dfrac{{ - 1}}{4}\) \(\dfrac{{ - 1 + 10}}{5}:\left( {x - \dfrac{7}{{10}}} \right) = \dfrac{{ - 1}}{4}\) \(\begin{array}{*{20}{l}}{\;\;\;\;\dfrac{9}{5}:\left( {x - \dfrac{7}{{10}}} \right) = \dfrac{{ - 1}}{4}}\\{\;\;\;\;\;\;\;\;\;\;x - \dfrac{7}{{10}} = \dfrac{9}{5}:\dfrac{{ - 1}}{4}}\\{\;\;\;\;\;\;\;\;\;\;x - \dfrac{7}{{10}} = \dfrac{9}{5}.\left( { - 4} \right)}\\{\;\;\;\;\;\;\;\;\;\;x - \dfrac{7}{{10}} = \dfrac{{ - 36}}{5}}\\{\;\;\;\;\;\;\;\;\;\;x\;\;\;\;\;\;\;\; = \dfrac{{ - 36}}{5} + \dfrac{7}{{10}}}\\{\;\;\;\;\;\;\;\;\;\;x\;\;\;\;\;\;\;\; = \dfrac{{ - 13}}{2}.}\end{array}\) Vậy \(x = \dfrac{{ - 13}}{2}\)
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Bài 3:
Tìm tập hợp các số nguyên x để: \(\dfrac{5}{6} + \dfrac{{ - 7}}{8} \le \dfrac{x}{{24}} \le \dfrac{{ - 5}}{{12}} + \dfrac{5}{8}\)
Phương pháp
+ Thực hiện phép cộng các phân số đã biết.
+ Xác định vai trò của số chưa biết trong phép toán rồi kết luận.
Lời giải
\(\begin{array}{*{20}{l}}{\dfrac{5}{6} + \dfrac{{ - 7}}{8} \le \dfrac{x}{{24}} \le \dfrac{{ - 5}}{{12}} + \dfrac{5}{8}}\\{ \Rightarrow \dfrac{{ - 1}}{{24}} \le \dfrac{x}{{24}} \le \dfrac{5}{{24}}}\\{ \Rightarrow {\rm{ \;}} - 1 \le x \le 5}\end{array}\)
Vì \(x \in \mathbb{Z}\) nên \(x \in \left\{ { - 1;0;1;2;3;4;5} \right\}\)
Vậy \(x \in \left\{ { - 1;0;1;2;3;4;5} \right\}\)
