help me 

please !!!

giúp tớ với

cái này trong sách í

a) \(\frac{3x-2}{2}=\frac{1-2x}{3}\)

\(\Leftrightarrow3\left(3x-2\right)=2\left(1-2x\right)\)

\(\Leftrightarrow9x-6=2-4x\)

\(\Leftrightarrow13x-8=0\)

\(\Leftrightarrow x=\frac{8}{13}\)

b) \(\frac{x-1}{3}+2=3-\frac{2x+5}{4}\)

\(\Leftrightarrow\frac{4\left(x-1\right)+24}{12}=\frac{36-3\left(2x+5\right)}{12}\)

\(\Leftrightarrow4x-4+24=36-6x-15\)

\(\Leftrightarrow4x+20=-6x+21\)

\(\Leftrightarrow10x=1\)

\(\Leftrightarrow x=\frac{1}{10}\)

c) \(\frac{x-1}{5}+x=\frac{x+1}{7}\)

\(\Leftrightarrow\frac{7\left(x-1\right)+35x}{35}=\frac{5\left(x+1\right)}{7}\)

\(\Leftrightarrow7x-7+35x=5x+5\)

\(\Leftrightarrow37x=12\)

\(\Leftrightarrow x=\frac{12}{37}\)

d) \(2.\left(x-2.5\right)=0,25+\frac{4x-3}{8}\)

\(\Leftrightarrow8\left(x-2,5\right)=2+4x-3\)

\(\Leftrightarrow8x-20=4x-1\)

\(\Leftrightarrow4x=19\)

\(\Leftrightarrow x=\frac{4}{19}\)

a: \(\Leftrightarrow\left|x+2\right|=6x+1\)

\(\Leftrightarrow\left\{{}\begin{matrix}x>=-\dfrac{1}{6}\\\left(6x+1-x-2\right)\left(6x+1+x+2\right)=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge-\dfrac{1}{6}\\\left(5x-1\right)\left(7x+3\right)=0\end{matrix}\right.\)

\(\Leftrightarrow x=\dfrac{1}{5}\)

b: Trường hợp 1: x<2

Pt sẽ là 3-x+2-x=7

=>5-2x=7

=>2x=-2

hay x=-1(nhận)

Trường hợp 2: 2<=x<3

Pt sẽ là 3-x+x-2=7

=>1=7(vô lý)

Trường hợp 3: x>=3

Pt sẽ là x-3+x-2=7

=>2x-5=7

=>x=6(nhận)

d: \(\Leftrightarrow4^x\cdot\left(1+4^3\right)=4160\)

\(\Leftrightarrow4^x=64\)

hay x=3

Mình chỉ hướng dẫn giải thôi nhá chứ nhiều bài quá

a) Đặt \(\frac{x}{5}=\frac{y}{7}=k\Rightarrow x=5k;y=7k\)

Thay x.y=315 => 5k.7k=315 <=> 35k²=315 => k²=9 => k=3

x=5.3=15 ; y=7.3=21

b) 5x=9y<=> \(\frac{x}{9}=\frac{y}{5}\)

Theo TCDTSBN ta có : \(\frac{x}{9}=\frac{y}{5}=\frac{2x+3y}{2.9+3.5}=\frac{-33}{33}=-1\)

x/9=-1=>x=-9 ; y/5=-1=>y=-5

các bài còn lại tương tự b 

\(\frac{x+2015}{5}+\frac{x+2016}{4}=\frac{x+2017}{3}+\frac{x+2018}{2}\)

\(\Leftrightarrow\left(\frac{x+2015}{5}+1\right)+\left(\frac{x+2016}{4}+1\right)=\left(\frac{x+2017}{3}+1\right)+\left(\frac{x+2018}{2}+1\right)\)

\(\Leftrightarrow\frac{x+2020}{5}+\frac{x+2020}{4}-\frac{x+2020}{3}-\frac{x+2020}{2}=0\)

\(\Leftrightarrow\left(x+2020\right)\left(\frac{1}{5}+\frac{1}{4}-\frac{1}{3}-\frac{1}{2}\right)=0\)

\(\Leftrightarrow x+2020=0\)vì \(\frac{1}{5}+\frac{1}{4}+\frac{1}{3}+\frac{1}{2}\ne0\)

\(\Leftrightarrow x=-2020\)

khó lắm

bây h thì bạn giải đc chưa

Cảm ơn bạn rất nhiều mình đã hiểu rồi 

Chúc bạn học tốt nhé

a: \(\Leftrightarrow\left(x-\dfrac{2}{5}\right):\dfrac{3}{2}=-\dfrac{5}{4}+\dfrac{5}{2}=\dfrac{5}{4}\)

\(\Leftrightarrow x-\dfrac{2}{5}=\dfrac{5}{4}\cdot\dfrac{3}{2}=\dfrac{15}{8}\)

hay x=91/40

b: \(\Leftrightarrow\left(2.5x-3.6\right)=-1\cdot\dfrac{12}{7}=\dfrac{-12}{7}\)

=>2,5x=66/35

hay x=132/175

c: \(\Leftrightarrow\left(\dfrac{15}{4}-2x\right)=\dfrac{19}{9}:\dfrac{4}{3}=\dfrac{19}{9}\cdot\dfrac{3}{4}=\dfrac{19}{12}\)

=>2x=15/4-19/12=45/12-19/12=26/12

=>x=13/12

a)

\((3x-7)^5=0\Rightarrow 3x-7=0\Rightarrow x=\frac{7}{3}\)

b)

\(\frac{1}{4}-(2x-1)^2=0\)

\(\Leftrightarrow (2x-1)^2=\frac{1}{4}=(\frac{1}{2})^2=(-\frac{1}{2})^2\)

\(\Rightarrow \left[\begin{matrix} 2x-1=\frac{1}{2}\\ 2x-1=\frac{-1}{2}\end{matrix}\right.\Rightarrow \Rightarrow \left[\begin{matrix} x=\frac{3}{4}\\ x=\frac{1}{4}\end{matrix}\right.\)

c)

\(\frac{1}{16}-(5-x)^3=\frac{31}{64}\)

\(\Leftrightarrow (5-x)^3=\frac{1}{16}-\frac{31}{64}=\frac{-27}{64}=(\frac{-3}{4})^3\)

\(\Leftrightarrow 5-x=\frac{-3}{4}\)

\(\Leftrightarrow x=\frac{23}{4}\)

d)

\(2x=(3,8)^3:(-3,8)^2=(3,8)^3:(3,8)^2=3,8\)

\(\Rightarrow x=3,8:2=1,9\)

e)

\((\frac{27}{64})^9.x=(\frac{-3}{4})^{32}\)

\(\Leftrightarrow [(\frac{3}{4})^3]^9.x=(\frac{3}{4})^{32}\)

\(\Leftrightarrow (\frac{3}{4})^{27}.x=(\frac{3}{4})^{32}\)

\(\Leftrightarrow x=(\frac{3}{4})^{32}:(\frac{3}{4})^{27}=(\frac{3}{4})^5\)

f)

\(5^{(x+5)(x^2-4)}=1\)

\(\Leftrightarrow (x+5)(x^2-4)=0\)

\(\Leftrightarrow \left[\begin{matrix} x+5=0\\ x^2-4=0\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x+5=0\\ x^2=4=2^2=(-2)^2\end{matrix}\right.\)

\(\Rightarrow \left[\begin{matrix} x=-5\\ x=\pm 2\end{matrix}\right.\)

g)

\((x-2,5)^2=\frac{4}{9}=(\frac{2}{3})^2=(\frac{-2}{3})^2\)

\(\Rightarrow \left[\begin{matrix} x-2,5=\frac{2}{3}\\ x-2,5=\frac{-2}{3}\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=\frac{19}{6}\\ x=\frac{11}{6}\end{matrix}\right.\)

h)

\((2x+\frac{1}{3})^3=\frac{8}{27}=(\frac{2}{3})^3\)

\(\Rightarrow 2x+\frac{1}{3}=\frac{2}{3}\Rightarrow x=\frac{1}{6}\)

a ) \(\left(\frac{2}{5}-x\right):1\frac{1}{3}+\frac{1}{2}=-4\)

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

     \(\left(\frac{2}{5}-x\right):\frac{4}{3}=-4-\frac{1}{2}\)

     \(\left(\frac{2}{5}-x\right):\frac{4}{3}=-\frac{9}{2}\)

        \(\frac{2}{5}-x=-\frac{9}{2}.\frac{4}{3}\)

        \(\frac{2}{5}-x=-3\)

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

                   \(x=\frac{2}{5}+3\)

                   \(x=\frac{3}{5}-\frac{15}{5}\)

                   \(x=-\frac{12}{5}\)

Vay \(x=-\frac{12}{5}\) 

b ) \(\left(-3+\frac{3}{x}-\frac{1}{3}\right):\left(1+\frac{2}{5}+\frac{2}{3}\right)=-\frac{5}{4}\)

     \(\left(-3+\frac{3}{x}-\frac{1}{3}\right):\left(\frac{15}{15}+\frac{6}{15}+\frac{10}{15}\right)=-\frac{5}{4}\)

     \(\left(-3+\frac{3}{x}-\frac{1}{3}\right):\left(\frac{15+6+10}{15}\right)=-\frac{5}{4}\)

     \(\left(-3+\frac{3}{x}-\frac{1}{3}\right):\frac{31}{15}=-\frac{5}{4}\)

     \(\left(-3+\frac{3}{x}-\frac{1}{3}\right)=-\frac{5}{4}.\frac{31}{15}\)

     \(\left(-3+\frac{3}{x}-\frac{1}{3}\right)=-\frac{1}{4}.\frac{31}{3}\)

        \(-3+\frac{3}{x}-\frac{1}{3}=-\frac{31}{12}\)

        \(-3+\frac{3}{x}=-\frac{31}{12}+\frac{1}{2}\)

        \(-3+\frac{3}{x}=-\frac{31}{12}+\frac{6}{12}\)

        \(-3+\frac{3}{x}=\frac{-25}{12}\)

                     \(\frac{3}{x}=\frac{-25}{12}+3\)

                      \(\frac{3}{x}=\frac{-25}{12}+\frac{36}{12}\)

                      \(\frac{3}{x}=\frac{5}{6}\)

                      \(\frac{18}{6x}=\frac{5x}{6x}\)

Đèn dây , bạn tự làm tiếp nhé , de rồi chứ