Đây là toán lớp 7 mà bạn.

BBạn có trả lời dc ko, ko thì thôi

a) |3x-2|=7

=> 3x-2=7 hoặc 3x-2=-7

=> 3x=9 hoặc 3x=-5

=> x=3 hoặc x=\(-\dfrac{5}{3}\)

a. \(\dfrac{8}{7}-\dfrac{1}{7}:\left(\dfrac{x}{3}-2\right)=-1\)

\(-\dfrac{1}{7}:\left(\dfrac{x}{3}-2\right)=\dfrac{15}{7}\)

\(\dfrac{x}{3}-2=\dfrac{-1}{15}\)

\(\dfrac{x}{3}=\dfrac{29}{15}\)

\(x=5,8\)

b. \(\dfrac{5}{8}+\dfrac{1}{4}\left(2x-1\right)=\dfrac{5}{4}\)

\(\dfrac{1}{4}\left(2x-1\right)=\dfrac{5}{8}\)

\(2x-1=\dfrac{5}{2}\)

\(2x=\dfrac{7}{2}\)

\(x=\dfrac{7}{4}\)

a)

<=> 10x - 35 + 16x - 10 = 5 

<=> 10x + 16x = 5 + 35 + 10

<=> 26x = 50

<=> x = 50/26 = 25/13

\(1,A=\left(3x+7\right)\left(2x+3\right)-\left(2x+3\right)-\left(3x-5\right)\left(2x+11\right)\\ =6x^2+23x+21-2x-3-6x^2-23x+55\\ =73-2x\left(đề.sai\right)\\ B=x^4+x^3-x^2-2x^2-2x+2-x^4-x^3+3x^2+2x\\ =2\\ 2,\\ a,\Leftrightarrow30x^2+18x+3x-30x^2=7\\ \Leftrightarrow21x=7\Leftrightarrow x=\dfrac{1}{3}\\ b,\Leftrightarrow-63x^2+78x-15+63x^2+x-20=44\\ \Leftrightarrow79x=79\Leftrightarrow x=1\\ c,\Leftrightarrow\left(x+5\right)\left(x^2+3x+2\right)-x^3-8x^2=27\\ \Leftrightarrow x^3+3x^2+2x+5x^2+15x+10-x^3-8x^2=27\\ \Leftrightarrow17x=17\Leftrightarrow x=1\)

\(d,\Leftrightarrow7x-2x^2-3+x^2+x-6=-x^2-x+2\\ \Leftrightarrow9x=11\Leftrightarrow x=\dfrac{11}{9}\)

a) \(2x+\frac{3}{15}=\frac{7}{5}\) 

=> \(2x=\frac{7}{5}-\frac{3}{15}=\frac{21}{15}-\frac{3}{15}=\frac{18}{15}\)

=> \(x=\frac{18}{15}:2=\frac{18}{15}\cdot\frac{1}{2}=\frac{9}{15}\cdot\frac{1}{1}=\frac{9}{15}\)

b) \(x-\frac{2}{9}=\frac{8}{3}\)

=> \(x=\frac{8}{3}+\frac{2}{9}\)

=> \(x=\frac{24}{9}+\frac{2}{9}=\frac{26}{9}\)

c) \(\frac{-8}{x}=\frac{-x}{18}\)

=> x(-x) = (-8).18

=> -x² = -144

=> x² = 144(bỏ dấu âm)

=> x = \(\pm\)12

d) \(\frac{2x+3}{6}=\frac{x-2}{5}\)

=> 5(2x + 3) = 6(x - 2)

=> 10x + 15 = 6x - 12

=> 10x + 15 - 6x + 12 = 0

=> 4x + 27 = 0

=> 4x = -27

=> x = -27/4

e) \(\frac{x+1}{22}=\frac{6}{x}\)

=> x(x + 1) = 132

=> x(x + 1) = 11.12

=> x = 11

f) \(\frac{2x-1}{2}=\frac{5}{x}\)

=> x(2x - 1) = 10

=> 2x² - x = 10

=> 2x² - x - 10 = 0

tới đây tự làm đi nhé

g) \(\frac{2x-1}{21}=\frac{3}{2x+1}\)

=> (2x - 1)(2x + 1) = 63

=> 4x² - 1 = 63

=> 4x² = 64

=> x² = 16

=> x = \(\pm\)4

h) Tương tự

a) \(\frac{2x+3}{15}=\frac{7}{5}\Leftrightarrow10x+15=105\Leftrightarrow10x=90\Rightarrow x=9\)

b) \(\frac{x-2}{9}=\frac{8}{3}\Leftrightarrow3x-6=72\Leftrightarrow3x=78\Rightarrow x=26\)

c) \(\frac{-8}{x}=\frac{-x}{18}\Leftrightarrow x^2=144\Leftrightarrow\orbr{\begin{cases}x=12\\x=-12\end{cases}}\)

d) \(\frac{2x+3}{6}=\frac{x-2}{5}\Leftrightarrow10x+15=12x-12\Leftrightarrow2x=27\Rightarrow x=\frac{27}{2}\)

e) \(\frac{x+1}{22}=\frac{6}{x}\Leftrightarrow x^2+x-132=0\Leftrightarrow\left(x-11\right)\left(x+12\right)=0\Leftrightarrow\orbr{\begin{cases}x=11\\x=-12\end{cases}}\)

f) \(\frac{2x-1}{2}=\frac{5}{x}\Leftrightarrow2x^2-x-10=0\Leftrightarrow\left(x-2\right)\left(2x+5\right)=0\Leftrightarrow\orbr{\begin{cases}x=2\\x=-\frac{5}{2}\end{cases}}\)

g) \(\frac{2x-1}{21}=\frac{3}{2x+1}\Leftrightarrow4x^2=64\Leftrightarrow x^2=16\Rightarrow\orbr{\begin{cases}x=4\\x=-4\end{cases}}\)

h) \(\frac{10x+5}{6}=\frac{5}{x+1}\Leftrightarrow10x^2+15x-25=0\Leftrightarrow5\left(x-1\right)\left(2x+5\right)=0\Leftrightarrow\orbr{\begin{cases}x=1\\x=-\frac{5}{2}\end{cases}}\)

a, \(\left(x+1\right)^2=3\left(x+1\right)\)

\(\Rightarrow\left(x+1\right)^2-3\left(x+1\right)=0\)

\(\Rightarrow\left(x+1\right).\left(x+1-3\right)=0\)

\(\Rightarrow\left(x+1\right).\left(x-2\right)=0\)

\(\Rightarrow\left\{{}\begin{matrix}x+1=0\\x-2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-1\\x=2\end{matrix}\right.\)

Vậy \(x\in\left\{-1;2\right\}\)

b, \(\left(2x-7\right)^3=8\left(7-2x\right)^2\)

\(\Rightarrow\left(2x-7\right)^3-8\left(2x-7\right)^2=0\) (do \(\left[A\left(x\right)\right]^2=\left[-A\left(x\right)\right]^2\))

\(\Rightarrow\left(2x-7\right)^2.\left(2x-7-8\right)=0\)

\(\Rightarrow\left(2x-7\right)^3.\left(2x-15\right)=0\)

\(\Rightarrow\left\{{}\begin{matrix}\left(2x-7\right)^3=0\\2x-15=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}2x-7=0\\2x=15\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}2x=7\\x=7,5\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=3,5\\x=7,5\end{matrix}\right.\)

Vậy \(x\in\left\{3,5;7,5\right\}\)

Chúc bạn học tốt!!!

Tìm x biết :a) ( 2x - 3 ).( x +1 ) > 0b) ( x + 5 ).(x-7) < 0c) | 2x - 3 | + 8 = 10d) ( 2x + 5 ) . | x -8 | . ( x2 + 1 ) = 0