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a, \(\left|x\right|+0,75=2\Rightarrow\left|x\right|=1,25\Rightarrow x=\pm1,25\)
b, \(\left|x+\dfrac{1}{3}\right|-4=1\Rightarrow\left|x+\dfrac{1}{3}\right|=5\Rightarrow\left|x\right|=\dfrac{14}{3}\Rightarrow x=\pm\dfrac{14}{3}\)
c,
\(\left|x\right|+2,5=0\Rightarrow\left|x\right|=-0,25\) (vô lí)
Vì \(\left|x\right|\ge0\) => x vô nghiệm
a) \(\left|x\right|+0,75=2\)
\(\Leftrightarrow\left|x\right|=2-0,75=1,25\)
\(\Rightarrow\left[{}\begin{matrix}x=-1,25\\x=1,25\end{matrix}\right.\)
b) \(\left|x+\dfrac{1}{3}\right|-4=1\)
\(\Leftrightarrow\left|x+\dfrac{1}{3}\right|=1+4=5\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{3}=5\\x+\dfrac{1}{3}=-5\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{14}{3}\\x=-\dfrac{16}{3}\end{matrix}\right.\)
c) \(\left|x\right|+2,5=0\)
\(\Leftrightarrow\left|x\right|=0-2,5=-2,5\)
Vì \(\left|x\right|\ge0\) mà \(-2,5< 0\) nên \(x\in\varnothing\)
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a) Ta có: \(x-\frac{2}{3}=0,75\)
\(\Leftrightarrow x-\frac{2}{3}=\frac{3}{4}\)
hay \(x=\frac{3}{4}+\frac{2}{3}=\frac{17}{12}\)
Vậy: \(x=\frac{17}{12}\)
b) Ta có: \(\frac{1}{3}+\frac{2}{3}x=-1\)
\(\Leftrightarrow\frac{2}{3}x=-1-\frac{1}{3}=-\frac{4}{3}\)
hay \(x=\frac{-4}{3}:\frac{2}{3}=\frac{-4}{3}\cdot\frac{3}{2}=-2\)
Vậy: x=-2
c) Ta có: \(\left|2x-3\right|-\frac{3}{4}=4,25\)
\(\Leftrightarrow\left|2x-3\right|=\frac{17}{4}+\frac{3}{4}=5\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-3=5\\2x-3=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=8\\2x=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-1\end{matrix}\right.\)
Vậy: x∈{-1;4}
a, \(x-\frac{2}{3}=0.75\)
\(x=0.75-\frac{2}{3}\)
\(x=\frac{1}{12}\)
Vậy...
b, \(\frac{1}{3}+\frac{2}{3}\cdot x=-1\)
\(\frac{2}{3}\cdot x=-1-\frac{1}{3}=-\frac{4}{3}\)
\(x=-\frac{4}{3}:\frac{2}{3}=-2\)
Vậy x = -2
c, \(|2x-3|-\frac{3}{4}=4.25\)
\(|2x-3|=4.25-\frac{3}{4}=\frac{7}{2}\)
=> \(2x-3=\frac{7}{2}hay2x-3=-\frac{7}{2}\)
2x = \(\frac{7}{2}+3\) 2x = \(-\frac{7}{2}+3\)
2x = \(\frac{13}{2}\) 2x = \(-\frac{1}{2}\)
x = \(\frac{12}{2}:2=\frac{13}{4}\) x = \(-\frac{1}{2}:2\) = \(-\frac{1}{4}\)
Vậy...
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\(a,1\dfrac{1}{5}+\dfrac{4}{5}:x=0,75\\ \dfrac{ 4}{5}:x=\dfrac{3}{4}-\dfrac{6}{4}\\ \dfrac{4}{5}:x=-\dfrac{3}{4}\\ x=\dfrac{4}{5}:\left(-\dfrac{3}{4}\right)\\ x=-\dfrac{16}{15}\\ b,x+\dfrac{1}{2}=1-x\\ x+x=1-\dfrac{1}{2}\\ 2x=\dfrac{1}{2}\\ x=\dfrac{1}{2}:2\\ x=\dfrac{1}{4}\)
$ \dfrac{1}{5}+\dfrac{4}{5}: x=0,75$;
b) $x+\dfrac{1}{2}=1-x$.
Em bị trục trặc và nhìn thấy phần câu hỏi bị thế này ạ. Em nhìn em không giải được ạ
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a) Ta có: \(\dfrac{x}{-15}=\dfrac{-60}{x}\)
\(\Leftrightarrow x^2=\left(-15\right)\cdot\left(-60\right)=900\)
hay \(x\in\left\{30;-30\right\}\)
Vậy: \(x\in\left\{30;-30\right\}\)
b) Ta có: \(\left|x\right|+0.573=2\)
\(\Leftrightarrow\left|x\right|=1.427\)
hay \(x\in\left\{1.427;-1.427\right\}\)
Vậy: \(x\in\left\{1.427;-1.427\right\}\)
c) Ta có: \(\left|x+\dfrac{1}{3}\right|-4=-1\)
\(\Leftrightarrow\left|x+\dfrac{1}{3}\right|=3\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{3}=3\\x+\dfrac{1}{3}=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{8}{3}\\x=-\dfrac{10}{3}\end{matrix}\right.\)
Vậy: \(x\in\left\{\dfrac{8}{3};-\dfrac{10}{3}\right\}\)
d) Ta có: \(0.01:2.5=\left(0.75x\right):0.75\)
\(\Leftrightarrow\dfrac{0.75\cdot x}{0.75}=\dfrac{0.01}{2.5}\)
\(\Leftrightarrow x=\dfrac{1}{250}\)
Vậy: \(x=\dfrac{1}{250}\)
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Bài 1 :
\(-\frac{1}{2}-\left|\frac{3}{7}-x\right|=0,75\)
\(\left|\frac{3}{7}-x\right|=-\frac{1}{2}-0,75\)
\(\left|\frac{3}{7}-x\right|=-\frac{5}{4}\)
Vì x > 0
=> Không tõa mãn điều kiện
Bài 2 :
\(\frac{4}{5}+\frac{3}{2}.\left|x+\frac{1}{4}\right|=\frac{1}{2}\)
\(\frac{3}{2}.\left|x+\frac{1}{4}\right|=\frac{1}{2}-\frac{4}{5}\)
\(\frac{3}{2}.\left|x+\frac{1}{4}\right|=-\frac{3}{10}\)
\(\left|x+\frac{1}{4}\right|=-\frac{3}{10}:\frac{3}{2}\)
\(\left|x+\frac{1}{4}\right|=-\frac{3}{10}.\frac{2}{3}\)
\(\left|x+\frac{1}{4}\right|=-\frac{1}{5}\)
Vì x > 0
Vậy không thõa mãn điều kiện
tui giải kiểu lop7
bài1: bỏ tgtđ thì +- nhé
-1/2 -3/7+x =0,75
x= 47/28
-1/2+3/7-x = 0,75
x= -23/28
\(0,75+x=\frac{1}{4}\)
\(\frac{3}{4}+x=\frac{1}{4}\)
\(\Rightarrow x=\frac{1}{4}-\frac{3}{4}=-\frac{1}{2}\)