3.(x-1/2) -5(x+3/5)=-x+1/5

3x - 3/2 -5x +3 = -x+1/5 

3x-5x+x= 3/2-3+1/5

x.(3-5+1)=15/10 + (-30/10)+2/10

x.(-1)= -13/10

x = -13/10 : (-1)

x=13/10

vậy x=13/10

-5.(x+1/5) -1/2.(x-2/3)=3/2x-5/6

-5x + (-1) -1/2x -1/3=3/2x-5/6

-5x-1/2x-3/2x=1+1/3-5/6

x.(-5-1/2-3/2)= 6/6+2/6+(-5/6)

x.(-10/2+(-1/2)+(-3/2))=3/6

x.6/2=1/2

x=1/2:6/2

x=1/6

Vậy x = 1/6

a) \(x=\frac{9}{10}\)

b) \(x=\frac{-4}{3}\)

c) \(x=\frac{1}{42}\)

d) \(x=\frac{-47}{10}\)

ko có thời gian nên mình chỉ cho đáp án thôi nhé

thông cảm cho mình ngen

đúng thì k đấy

chúc bạn học giỏi

làm chi tiết cho mk nhé

ai làm chi tiết mk k cho nhìu

a) \(\left|\frac{4}{7}-x\right|+\frac{2}{5}=0\)

=> \(\left|\frac{4}{7}-x\right|=-\frac{2}{5}\), vô lí vì \(\left|\frac{4}{7}-x\right|\ge0\)

Vậy không tồn tại giá trị của x thỏa mãn đề bài

b) \(6-\left|\frac{1}{4}x+\frac{2}{5}\right|=0\)

=> \(\left|\frac{1}{4}x+\frac{2}{5}\right|=6-0=6\)

=> \(\left[\begin{array}{nghiempt}\frac{1}{4}x+\frac{2}{5}=6\\\frac{1}{4}x+\frac{2}{5}=-6\end{array}\right.\)=> \(\left[\begin{array}{nghiempt}\frac{1}{4}x=\frac{28}{5}\\\frac{1}{4}x=-\frac{32}{5}\end{array}\right.\)=> \(\left[\begin{array}{nghiempt}x=\frac{112}{5}\\x=-\frac{128}{5}\end{array}\right.\)

Vậy \(\left[\begin{array}{nghiempt}x=\frac{112}{5}\\x=-\frac{128}{5}\end{array}\right.\)

c) \(\left|x-\frac{1}{3}\right|+\left|2-\frac{4}{5}\right|=0\)

=> \(\left|x-\frac{1}{3}\right|+\left|\frac{6}{5}\right|=0\)

=> \(\left|x-\frac{1}{3}\right|+\frac{6}{5}=0\)

=> \(\left|x-\frac{1}{3}\right|=-\frac{6}{5}\), vô lí vì \(\left|x-\frac{1}{3}\right|\ge0\)

Vậy không tồn tại giá trị của x thỏa mãn đề bài

\(\left[\frac{-2}{5}x^3.\left(2x-1\right)^m+\frac{2}{5}x^{m+3}\right]:\left(\frac{-2}{5}x^3\right)\)

\(=\left[\frac{2}{5}x^3\left(2x+1\right)^m+\frac{2}{5}x^3.\left(\frac{2}{5}\right)^m\right]:\left(\frac{-2}{5}x^3\right)\)

\(=\left\{\frac{2}{5}x^3.\left[\left(2x+1\right)^m+\left(\frac{2}{5}\right)^m\right]\right\}:\left(\frac{-2}{5}x^3\right)\)

\(=\left\{\frac{2}{5}x^3.\left[2x+\frac{7}{5}\right]^m\right\}:\frac{-2}{5}x^3\)

\(=-\left(2x+\frac{7}{5}\right)^m\)

đến đây thì mình chịu

\(x\left(\frac{1}{4}+\frac{1}{5}\right)-\left(\frac{1}{7}+\frac{1}{8}\right)=0\)

\(\frac{9}{20}x-\frac{15}{56}=0\)

\(\frac{9}{20}x=\frac{15}{56}\)

\(x=\frac{25}{42}\)

\(x\left[\frac{1}{4}+\frac{1}{5}\right]-\left[\frac{1}{7}+\frac{1}{8}\right]\) = \(0\)

\(\frac{9}{20}x-\frac{15}{56}\)= \(0\)

\(\frac{9}{20}x-\frac{15}{56}\)

\(x=\frac{25}{42}\)

\(VP=\frac{1}{2\left(a+3\right)}+\frac{1}{2\left(a+5\right)}=\frac{2\left(a+5\right)}{2\left(a+3\right)\left(a+5\right)}+\frac{2\left(a+3\right)}{2\left(a+3\right)\left(a+5\right)}\)

\(=\frac{2\left(a+5\right)}{4\left(a+3\right)\left(a+5\right)}+\frac{2\left(a+3\right)}{4\left(a+3\right)\left(a+5\right)}=\frac{2\left(a+5\right)+2\left(a+3\right)}{4\left(a+3\right)\left(a+5\right)}=\frac{2\left[\left(a+3\right)+\left(a+5\right)\right]}{4\left(a+3\right)\left(a+5\right)}=\frac{\left(a+3\right)+\left(a+5\right)}{2\left(a+3\right)\left(a+5\right)}\)

\(=\frac{\left(a+a\right)+\left(3+5\right)}{2\left(a+3\right)\left(a+5\right)}=\frac{2a+8}{2\left(a+3\right)\left(a+5\right)}=\frac{2\left(a+4\right)}{2\left(a+3\right)\left(a+5\right)}=\frac{a+4}{\left(a+3\right)\left(a+5\right)}\)

\(VT=\frac{x-2}{\left(a+3\right)\left(a-5\right)}\)

\(\Rightarrow\frac{x-2}{\left(a+3\right)\left(a-5\right)}=\frac{a+4}{\left(a+3\right)\left(a+5\right)}\)

\(\Rightarrow\frac{x-2}{a+4}=\frac{\left(a+3\right)\left(a-5\right)}{\left(a+3\right)\left(a+5\right)}\Rightarrow\frac{x-2}{a+4}=\frac{a-5}{a+5}\Rightarrow\left(x-2\right)\left(a+5\right)=\left(a-5\right)\left(a+4\right)\)

chịu

\(a\left(x-\frac{1}{2}\right)^3=\frac{1}{27}\\ =>\left(x-\frac{1}{2}\right)=\frac{1}{3}\\ =>x=\frac{1}{3}+\frac{1}{2}\\ =>x=\frac{5}{6}\)

b) \(\left(x+\frac{1}{2}\right)^2=\frac{4}{25}\\ =>\left(x+\frac{1}{2}\right)=\frac{2}{5}\\ =>x=\frac{-1}{10}\)

d) (2x+3)²⁰¹⁶=(2x+3)²⁰¹⁸ khi 2x+3=0 hoặc 1 

Nếu 2x+3=0 

=2x=-3 ( loại ) 

Nếu 2x+3=1

=>2x=-2

=>x=-1 ( thỏa )