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\(\frac{2}{3}x+\frac{5}{7}=\frac{3}{10}\)
\(\Rightarrow\frac{2}{3}x=\frac{3}{10}-\frac{5}{7}\)
\(\Rightarrow\frac{2}{3}x=-\frac{29}{70}\)
\(\Rightarrow x=-\frac{29}{70}:\frac{2}{3}\)
\(\Rightarrow x=-\frac{87}{140}\)
tíc mình nha
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a, \(\left|x+\frac{1}{3}\right|=0\Leftrightarrow x=-\frac{1}{3}\)
b, \(\left|\frac{5}{18}-x\right|-\frac{7}{24}=0\)
\(\Leftrightarrow\orbr{\begin{cases}\frac{5}{18}-x=\frac{7}{24}\\\frac{5}{18}-x=-\frac{7}{24}\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-\frac{1}{72}\\x=\frac{41}{72}\end{cases}}\)
c, \(\frac{2}{5}-\left|\frac{1}{2}-x\right|=6\Leftrightarrow\left|\frac{1}{2}-x\right|=-\frac{28}{5}\)vô lí
Vì \(\left|\frac{1}{2}-x\right|\ge0\forall x\)*luôn dương* Mà \(-\frac{28}{5}< 0\)
=> Ko có x thỏa mãn
\(|x+\frac{1}{3}|=0\)
\(< =>x+\frac{1}{3}=0< =>x=-\frac{1}{3}\)
\(|x+\frac{3}{4}|=\frac{1}{2}\)
\(< =>\orbr{\begin{cases}x+\frac{3}{4}=\frac{1}{2}\\x+\frac{3}{4}=-\frac{1}{2}\end{cases}}\)
\(< =>\orbr{\begin{cases}x=-\frac{1}{4}\\x=-\frac{5}{4}\end{cases}}\)
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a ) \(1\frac{1}{2}+x=\frac{3}{7}-7\)
\(\frac{3}{2}+x=-\frac{46}{7}\)
\(x=-\frac{46}{7}-\frac{3}{2}\)
\(x=-\frac{113}{14}\)
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a)\(\frac{x+3}{x+5}=7\Leftrightarrow x+3=7\left(x+5\right)\)
\(\Leftrightarrow x+3=7x+35\)
\(\Leftrightarrow-6x=32\)
\(\Leftrightarrow x=-\frac{16}{3}\)
b)\(\frac{2x-1}{3x+5}=-\frac{2}{3}\)
\(\Leftrightarrow3\left(2x-1\right)=-2\left(3x+5\right)\)
\(\Leftrightarrow6x-3=-6x-10\)
\(\Leftrightarrow12x=-7\)
\(\Leftrightarrow x=-\frac{7}{12}\)
c)\(\frac{x+1}{4}=\frac{9}{x+1}\Leftrightarrow\left(x+1\right)^2=36\)
\(\Leftrightarrow\left(x+1\right)^2=6^2\)
\(\Leftrightarrow\orbr{\begin{cases}x+1=6\\x+1=-6\end{cases}\Leftrightarrow\orbr{\begin{cases}x=5\\x=-7\end{cases}}}\)
d)\(\frac{6x-1}{2x+3}=\frac{3x}{x+2}\)
\(\Leftrightarrow\left(6x-1\right)\left(x+2\right)=3x\left(2x+3\right)\)
\(\Leftrightarrow6x^2+12x-x-2=6x^2+9x\)
\(\Leftrightarrow2x=2\Leftrightarrow x=1\)
\(a,\left|x-5\right|=7\)\(\Leftrightarrow\orbr{\begin{cases}x-5=7\\x-5=-7\end{cases}\Leftrightarrow\orbr{\begin{cases}x=12\\x=-2\end{cases}}}\)
\(b,\left|x+\frac{7}{5}\right|-\frac{1}{5}=0\)
\(\Leftrightarrow\left|x+\frac{7}{5}\right|=\frac{1}{5}\)
\(\Leftrightarrow\orbr{\begin{cases}x+\frac{7}{5}=\frac{1}{5}\\x+\frac{7}{5}=-\frac{1}{5}\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-\frac{6}{5}\\x=-\frac{8}{5}\end{cases}}}\)
\(c,\left|x\right|=3x-1\)
\(\Leftrightarrow\orbr{\begin{cases}x=3x-1\\x=1-3x\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}-2x=-1\\4x=1\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=\frac{1}{4}\end{cases}}}\)
Bài làm
| X - 5 |=7
=> 2 trường hợp:
TH1: x-5=-7
x. = -7 +5
x. = -2
TH: x-5=7
x. =7+5
x. =12
Vậy x = -2 hoặc x=12
b) | x+7/5|-1/5=0
| x+7/5|. =0+1/5
| x+7/5|. =1/5
=> 2 trường hợp
TH1: x+7/5 =1/5 => x=-6/5
TH2: x+7/5=-1/5 => x =-8/5
Vậy x=-6/5 hoặc x=-8/5
c) | x | =3x-1
=> X = 3x-1
X.x= 3-1
X2 = 2
.......
# Tự làm nốt. Điện thoại hết pin #