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a: x/2-x/3=1/4
=>1/6x=1/4
hay x=1/4:1/6=3/2
b: \(\dfrac{1}{2}\cdot x:\dfrac{2}{5}=\dfrac{-3}{2}:\dfrac{5}{4}=\dfrac{-3}{2}\cdot\dfrac{4}{5}=\dfrac{-6}{5}\)
\(\Leftrightarrow x\cdot\dfrac{1}{2}\cdot\dfrac{5}{2}=\dfrac{-6}{5}\)
=>5/4x=-6/5
hay x=-24/25
c: \(\dfrac{2}{3}x-\dfrac{1}{3}x=\dfrac{5}{12}\)
nên 1/3x=5/12
=>x=5/4
1: Ta có: \(2x+x\left(x-5\right)=3x^2-x\)
\(\Leftrightarrow2x+x^2-5x-3x^2+x=0\)
\(\Leftrightarrow-2x^2-2x=0\)
\(\Leftrightarrow-2x\left(x+1\right)=0\)
Vì -2≠0
nên \(\left[{}\begin{matrix}x=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)
Vậy: x∈{0;-1}
2) Ta có: \(15-5\left(1-2x\right)=12-x\)
\(\Leftrightarrow15-5+10x-12+x=0\)
\(\Leftrightarrow11x-2=0\)
\(\Leftrightarrow11x=2\)
hay \(x=\frac{2}{11}\)
Vậy: \(x=\frac{2}{11}\)
3) Ta có: \(\frac{2}{3}-\frac{1}{3}\left(x-\frac{3}{2}\right)-\frac{1}{2}\left(2x+1\right)=5\)
\(\Leftrightarrow\frac{2}{3}-\frac{1}{3}x+\frac{1}{2}-x-\frac{1}{2}-5=0\)
\(\Leftrightarrow\frac{-13}{3}-\frac{4}{3}x=0\)
\(\Leftrightarrow\frac{4}{3}x=\frac{-13}{3}\)
hay \(x=\frac{-13}{3}:\frac{4}{3}=\frac{-13}{4}\)
Vậy: \(x=\frac{-13}{4}\)
4) Ta có: \(\left|x-\frac{4}{5}\right|=\frac{3}{5}\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\frac{4}{5}=\frac{3}{5}\\x-\frac{4}{5}=\frac{-3}{5}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{7}{5}\\x=\frac{1}{5}\end{matrix}\right.\)
Vậy: \(x\in\left\{\frac{1}{5};\frac{7}{5}\right\}\)
1. \(2x+x\left(x-5\right)=3x^2-x\)
\(\Leftrightarrow2x+x^2-5x=3x^2-x\)
\(\Leftrightarrow\left(2x-5x+x\right)+\left(x^2-3x^2\right)=0\)
\(\Leftrightarrow-2x-2x^2=0\)
\(\Leftrightarrow-2x\left(1+x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}-2x=0\\1+x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)
2. \(15-5\left(1-2x\right)=12-x\)
\(\Leftrightarrow15-5+10x=12-x\)
\(\Leftrightarrow\left(15-5-12\right)+\left(10x+x\right)=0\)
\(\Leftrightarrow-2+11x=0\)
\(\Leftrightarrow11x=2\Leftrightarrow x=\frac{2}{11}\)
3. \(\frac{2}{3}-\frac{1}{3}\left(x-\frac{3}{2}\right)-\frac{1}{2}\left(2x+1\right)=5\)
\(\Leftrightarrow\frac{2}{3}-\frac{1}{3}x+\frac{1}{2}-x-\frac{1}{2}=5\)
\(\Leftrightarrow\left(\frac{2}{3}+\frac{1}{2}-\frac{1}{2}-5\right)-\left(\frac{1}{3}x+x\right)=0\)
\(\Leftrightarrow-\frac{13}{3}-\frac{4}{3}x=0\)
\(\Leftrightarrow-\frac{4}{3}x=\frac{13}{3}\Leftrightarrow x=-\frac{13}{4}\)
4. \(\left|x-\frac{4}{5}\right|=\frac{3}{5}\)
\(\Rightarrow x-\frac{4}{5}=-\frac{3}{5}\) hoặc \(x-\frac{4}{5}=\frac{3}{5}\)
\(TH1:x-\frac{4}{5}=-\frac{3}{5}\Rightarrow x=\frac{1}{5}\)
\(TH2:x-\frac{4}{5}=\frac{3}{5}\Rightarrow x=\frac{7}{5}\)
a. \(5.\left(x-2\right)+3.\left(x-2\right)=0\)
\(\Rightarrow8.\left(x-2\right)=0\)
\(\Rightarrow x-2=0:8\)
\(\Rightarrow x-2=0\)
\(\Rightarrow x=2\)
Vậy...
b. \(\dfrac{2}{3}+\dfrac{5}{2}:x=\dfrac{2}{4}\)
\(\Rightarrow\dfrac{5}{2}:x=\dfrac{2}{4}-\dfrac{2}{3}\)
\(\Rightarrow\dfrac{5}{2}:x=\dfrac{-1}{6}\)
\(\Rightarrow x=\dfrac{5}{2}:\dfrac{-1}{6}=-15\)
Vậy...
c. \(2.\left(x-\dfrac{1}{7}\right)=0\)
\(\Rightarrow x-\dfrac{1}{7}=0:2\)
\(\Rightarrow x-\dfrac{1}{7}=0\)
\(\Rightarrow x=\dfrac{1}{7}\)
Vậy...
d. \(\dfrac{11}{20}-\left(\dfrac{2}{5}+x\right)=\dfrac{2}{3}\)
\(\Rightarrow\dfrac{2}{5}+x=\dfrac{11}{12}:\dfrac{2}{3}\)
\(\Rightarrow\dfrac{2}{5}+x=\dfrac{1}{4}\)
\(\Rightarrow x=\dfrac{1}{4}-\dfrac{2}{5}=\dfrac{-3}{20}\)
Vậy...
e. \(\dfrac{3}{4}+\dfrac{1}{4}:x=\dfrac{2}{5}\)
\(\Rightarrow\dfrac{1}{4}:x=\dfrac{2}{5}-\dfrac{3}{4}\)
\(\Rightarrow\dfrac{1}{4}:x=\dfrac{-7}{20}\)
\(\Rightarrow x=\dfrac{1}{4}:\dfrac{-7}{20}=\dfrac{-5}{7}\)
Vậy...
g. \(\dfrac{2}{3}x+\dfrac{5}{7}=\dfrac{3}{10}\)
\(\Rightarrow\dfrac{2}{3}x=\dfrac{3}{10}-\dfrac{5}{7}\)
\(\Rightarrow\dfrac{2}{3}x=\dfrac{-29}{70}\)
\(\Rightarrow x=\dfrac{-29}{70}:\dfrac{2}{3}=\dfrac{-87}{140}\)
Vậy...
a ) x + 5/12 = -2/3
=> x = -2/3 - 5/12
=> x = -8/12 - 5/12
=> x = -13/12
b ) 4/5 + 3/4 : x = 1/2
=> 3/4 : x = 1/2 - 4/5
=> 3/4 : x = 5/10 - 8/10
=> 3/4 : x = -3/10
=> x = 3/4 : -3/10
=> x = -5/2
c ) x/2 + x/3 = 1/4
=> 3x/6 + 2x/6 = 1/4
=> ( 3x + 2x )/6 = 1/4
=> 5x/6 = 1/4
=> 20x/24 = 6/24
=> 20x = 6
=> x = 6 : 20
=> x = 0 , 3
Chúc bạn học giỏi !!!
\(a,\frac{1}{3}+\frac{1}{2}:x=\frac{1}{5}\)
\(\Leftrightarrow\frac{1}{2}:x=\frac{1}{5}-\frac{1}{3}\)
\(\Leftrightarrow\frac{1}{2}:x=\frac{3}{15}-\frac{5}{15}\)
\(\Leftrightarrow\frac{1}{2}:x=\frac{-2}{15}\)
\(\Leftrightarrow x=\frac{1}{2}:\frac{-15}{2}=\frac{-15}{4}\)
\(b,\frac{1}{3}x+\frac{2}{5}\left[x+1\right]=0\)
\(\Leftrightarrow\frac{1}{3}x+\frac{2}{5}x+\frac{2}{5}=0\)
\(\Leftrightarrow\frac{11}{15}x=-\frac{2}{5}\Leftrightarrow x=\frac{-2}{5}:\frac{11}{15}=\frac{-2}{5}\cdot\frac{15}{11}=\frac{-2}{1}\cdot\frac{3}{11}=\frac{-6}{11}\)
1, \(5.\left|x+2\right|=10.\left(-2\right)\)
\(\Leftrightarrow5.\left|x+2\right|=-20\)
\(\Leftrightarrow\left|x+2\right|=-\frac{20}{5}=-4\)
Ta thấy : \(\left|x+2\right|\ge0\forall x\) mà \(-4< 0\)
\(\Rightarrow x\in\varnothing\)
Vậy : không có \(x\) thỏa mãn đề.
2, \(-4.\left|x-2\right|=8\)
\(\Leftrightarrow\left|x-2\right|=8:\left(-4\right)=-2\)
Ta thấy : \(\left|x-2\right|\ge0\forall x\) mà \(-2< 0\)
\(\Rightarrow x\in\varnothing\)
Vậy : không có \(x\) thỏa mãn đề.
3, \(2.\left(x-5\right)-3.\left(x+7\right)=12\)
\(\Leftrightarrow2x-10-3x-21=12\)
\(\Leftrightarrow2x-3x=12+10+21\)
\(\Leftrightarrow-x=43\)
\(\Leftrightarrow x=-43\)
Vậy : \(x=-43\)
4, \(7.\left(5-x\right)-2.\left(x+3\right)=15\)
\(\Leftrightarrow35-7x-2x-6=15\)
\(\Leftrightarrow-7x-2x=15-35+6\)
\(\Leftrightarrow-9x=-14\)
\(\Leftrightarrow x=\frac{14}{9}\)
Vậy : \(x=\frac{14}{9}\)
Chúc bạn học tốt !!!
1) \(5.\left|x+2\right|=10.\left(-2\right)\)
=> \(5.\left|x+2\right|=-20\)
=> \(\left|x+2\right|=\left(-20\right):5\)
=> \(\left|x+2\right|=-4\)
Ta luôn có \(\left|x\right|\ge0\) \(\forall x.\)
=> \(\left|x+2\right|>-4\)
=> \(\left|x+2\right|\ne-4\)
Vậy không tồn tại giá trị nào của \(x\) thỏa mãn yêu cầu đề bài.
2) \(-4.\left|x-2\right|=8\)
=> \(\left|x-2\right|=8:\left(-4\right)\)
=> \(\left|x-2\right|=-2\)
Ta luôn có \(\left|x\right|\ge0\) \(\forall x.\)
=> \(\left|x-2\right|>-2\)
=> \(\left|x-2\right|\ne-2\)
Vậy không tồn tại giá trị nào của \(x\) thỏa mã yêu cầu đề bài.
Mình chỉ làm 2 câu này thôi nhé.
Chúc bạn học tốt!
\(=\dfrac{\left(\dfrac{2}{3}\cdot\dfrac{-3}{4}\right)^2\cdot\dfrac{2}{3}\cdot\left(-1\right)}{\left(\dfrac{2}{5}\cdot\dfrac{-5}{12}\right)^2}=\dfrac{\dfrac{1}{4}\cdot\dfrac{2}{3}\cdot\left(-1\right)}{\dfrac{1}{36}}=\dfrac{-2}{12}:\dfrac{1}{36}=-\dfrac{1}{6}\cdot36=-6\)
thank iu bạn nha