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a) x=4/7 - 1/3=19/21
b) /x-5/=7 -->x-5=7 hoặc x-5=-7
--> x=12 hoặc x= -2
\(\hept{\begin{cases}\text{|}0,5x\text{|}=0,5x\\\sqrt{\left(0,5x\right)^2}=0,5x\\\left(0,5x\right)^2=\left(0,5x\right)^2\end{cases}}\)
2, tương tự
\(\hept{\begin{cases}\text{|}-\frac{2}{3}x\text{|}=\frac{2}{3}x\\\sqrt{\left(-\frac{2}{3}x\right)^2}=\frac{2}{3}x\\\left(-\frac{2}{3}x\right)^2=\left(\frac{2}{3}x\right)^2\end{cases}}\)
4, tương tự
a,
\(2\frac{2}{3}:x=1\frac{7}{9}:2\frac{2}{3}\)
\(\frac{8}{3}:x=\frac{16}{9}:\frac{8}{3}\)
\(\frac{8}{3}:x=\frac{2}{3}\)
\(\frac{8}{3}:\frac{2}{3}=x\)
\(x=4\)
Vậy x = 4
b,
\(-2^3+0,5x=1,5\)
\(-8+0,5x=1,5\)
\(0,5x=1,5+8\)
\(0,5x=9,5\)
\(x=9,5:0,5\)
\(x=19\)
Vậy x = 19
a) Ta có: \(\frac{3x+2}{5x+7}=\frac{3x-1}{5x+1}\)
\(\Leftrightarrow\left(3x+2\right)\left(5x+1\right)=\left(5x+7\right)\left(3x-1\right)\)
\(\Leftrightarrow3x\left(5x+1\right)+2\left(5x+1\right)=5x\left(3x-1\right)+7\left(3x-1\right)\)
\(\Leftrightarrow15x^2+3x+10x+2=15x^2-5x+21x-7\)
\(\Leftrightarrow15x^2-15x^2+3x+10x+5x-21x=-7-2\)
\(\Leftrightarrow-3x=-9\)
\(\Leftrightarrow x=3\)
Vậy x = 3
b) Ta có: \(\frac{x+1}{2x+1}=\frac{0,5x+2}{x+3}\Leftrightarrow\left(x+1\right)\left(x+3\right)=\left(2x+1\right)\left(0,5x+2\right)\)
\(\Leftrightarrow x\left(x+3\right)+\left(x+3\right)=2x\left(0,5x+2\right)+\left(0,5x+2\right)\)
\(\Leftrightarrow x^2+3x+x+3=x^2+4x+0,5x+2\)
\(\Leftrightarrow x^2-x^2+3x+x-4x-0,5x=2-3\)
\(\Leftrightarrow-0,5x=-1\Leftrightarrow x=2\)
Vậy x = 2
b) \(\left(5x-1\right)\left(2x-\frac{1}{3}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}5x-1=0\\2x-\frac{1}{3}=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}5x=1\\2x=\frac{1}{3}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{5}\\x=\frac{1}{6}\end{matrix}\right.\)
e, \(-\frac{3}{4}-\left|\frac{4}{5}-x\right|=-1\)
\(\Leftrightarrow\left|\frac{4}{5}-x\right|=-\frac{3}{4}-\left(-1\right)\)
\(\Leftrightarrow\left|\frac{4}{5}-x\right|=\frac{1}{4}\)
\(\Leftrightarrow\left[{}\begin{matrix}\frac{4}{5}-x=\frac{1}{4}\\\frac{4}{5}-x=-\frac{1}{4}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{7}{15}\\x=1,05\end{matrix}\right.\)
Vậy ....
Bài 2:
1) \(\frac{x}{12}-\frac{5}{6}=\frac{1}{12}\)
\(\Rightarrow\frac{x}{12}=\frac{1}{12}+\frac{5}{6}\)
\(\Rightarrow\frac{x}{12}=\frac{11}{12}\)
\(\Rightarrow x.12=11.12\)
\(\Rightarrow x.12=132\)
\(\Rightarrow x=132:12\)
\(\Rightarrow x=11\)
Vậy \(x=11.\)
2) \(\frac{2}{3}-1\frac{4}{15}x=\frac{-3}{5}\)
\(\Rightarrow\frac{2}{3}-\frac{19}{15}x=\frac{-3}{5}\)
\(\Rightarrow\frac{19}{15}x=\frac{2}{3}+\frac{3}{5}\)
\(\Rightarrow\frac{19}{15}x=\frac{19}{15}\)
\(\Rightarrow x=\frac{19}{15}:\frac{19}{15}\)
\(\Rightarrow x=1\)
Vậy \(x=1.\)
3) \(\left(-2\right)^3+0,5x=1,5\)
\(\Rightarrow-8+0,5x=1,5\)
\(\Rightarrow0,5x=1,5+8\)
\(\Rightarrow0,5x=9,5\)
\(\Rightarrow x=9,5:0,5\)
\(\Rightarrow x=19\)
Vậy \(x=19.\)
Chúc bạn học tốt!
a, \(-\frac{22}{15}x+\frac{1}{3}=\left|-\frac{2}{3}+\frac{1}{5}\right|=\left|-\frac{7}{15}\right|=\frac{7}{15}\)
\(\Rightarrow\frac{-22}{15}x=\frac{7}{15}-\frac{1}{3}=\frac{2}{15}\)
\(\Rightarrow x=\frac{2}{15}:\frac{-22}{15}=\frac{2}{15}.\frac{15}{-22}=-\frac{1}{11}\)
\(\Rightarrow x.\left(0,5-\frac{2}{3}\right)=\frac{7}{12}\)
\(\Rightarrow x.\left(-\frac{1}{6}\right)=\frac{7}{12}\)
\(\Rightarrow x=\frac{7}{12}:\left(-\frac{1}{6}\right)=-\frac{7}{2}\)
hi cậu qua phần của tớ có hết á