chia 2 trường hợp 

mik bt lm nhưng ko đánh đc dấu nên xin lỗi nha

a)<=> 12=!2x-1! 

vô nghiệm VT chẵn VP luôn lẻ

b)

2x-1=x+3=> 2x-x=3+1=4=> x=4

2x-1=-x-2=> 2x+x=-2+1=-1=> vô nghiệm

a: =>|x-5|=3x+12+2x-1=5x+11

TH1: x>=5

=>x-5=5x+11

=>-4x=16

=>x=-4(loại)

TH2: x<5

=>5-x=5x+11

=>-6x=-6

=>x=1(nhận)

b: =>|x+2|-|x-1|=x+5-2x=-x+5

TH1: x<-2

=>-x-2-(1-x)=-x+5

=>-x-2-1+x=-x+5

=>x-3=5

=>x=8(nhận)

Th2: -2<=x<1

=>x+2-1+x=-x+5

=>2x+1=-x+5

=>3x=4

=>x=4/3(loại)

TH3: x>=1

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

=>-x=-2

=>x=2(nhận)

a) ( 2x - 1 )( 2x + 1 ) - ( x - 1 )² = 3x( x - 2 )

<=> 4x² - 1 - ( x² - 2x + 1 ) - 3x( x - 2 ) = 0

<=> 4x² - 1 - x² + 2x - 1 - 3x² + 6x = 0

<=> 8x - 2 = 0

<=> x = 1/4

Vậy phương trình có 1 nghiệm x = 1/4

b) ( 4x - 3 )( 3x + 2 ) = 2( 3x - 1 )( 2x + 5 )

<=> 12x² - x - 6 - 2( 6x² + 13x - 5 ) = 0

<=> 12x² - x - 6 - 12x² - 26x + 10 = 0

<=> -27x + 4 = 0

<=> x = 4/27

Vậy phương trình có 1 nghiệm x = 4/27

c) ( x - 1 )( x² + x + 1 ) - 5( 2x - 3 ) = x( x² - 3 )

<=> x³ - 1 - 10x + 15 - x( x² - 3 ) = 0

<=> x³ + 14 - 10x - x³ + 3x = 0

<=> -7x + 14 = 0

<=> x = 2

Vậy phương trình có nghiệm x = 2

d) \(\frac{3x-2}{4}-\frac{x+4}{3}=\frac{1+x}{12}\)

<=> \(\frac{3x}{4}-\frac{2}{4}-\frac{x}{3}-\frac{4}{3}=\frac{1}{12}+\frac{x}{12}\)

<=> \(\frac{3}{4}x-\frac{1}{3}x-\frac{1}{12}x=\frac{1}{12}+\frac{1}{2}+\frac{4}{3}\)

<=> \(x\left(\frac{3}{4}-\frac{1}{3}-\frac{1}{12}\right)=\frac{23}{12}\)

<=> \(x\cdot\frac{1}{3}=\frac{23}{12}\)

<=> x = 23/4

Vậy phương trình có 1 nghiệm x = 23/4

a)\(2x^2\)+\(3\left(x^2-1\right)\)=\(5x\left(x+1\right)\)

\(2x^2\)+\(3x^2\)\(-3\)=\(5x^2+5x\)

\(5x^2-5x^2-5x=3\)

\(-5x=3\)

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

tự ghi dấu suy ra ở đằng trước nhé

b) Vì \(2x\left(5-3x\right)=2x\left(3x-5\right)-3\left(x-7\right)=3\)

nên chỉ cần giải: \(6x^2-10x-3x+21=3\)

\(\Leftrightarrow6x^2-13x+21=3\)

\(\Leftrightarrow6x^2-13x+18=0\)

\(\Rightarrow\)pt vô nghiệm

(8x−3)(3x+2)−(4x+7)(x+4)=(2x+1)(5x−1)(8x−3)(3x+2)−(4x+7)(x+4)=(2x+1)(5x−1)

 20x2−16x−34=10x2+3x−120x2−16x−34=10x2+3x−1

 10x2−19x−33=010x2−19x−33=0

 (10x+11)(x−3)=0

chỉ bt lm con b thoy

..army,,,,,,,,,,

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

\(\Leftrightarrow3x^2-12x-2=3x^2-17x+20\)

\(\Leftrightarrow3x^2-12x=3x^2-17x+20+2\)

\(\Leftrightarrow3x^2-12x=3x^2-17x+22\left(3x^2-17x\right)\)

\(\Leftrightarrow5x=22\)

\(\Rightarrow x=\frac{22}{5}\)

b) \(\left(8x-3\right)\left(3x+2\right)-\left(4x+7\right)\left(x+4\right)=\left(2x+1\right)\left(5x-1\right)\)

\(\Leftrightarrow20x^2-16x-34=10x^2+3x+1\)

\(\Leftrightarrow20x^2-16x-33=10x^2+3x\)

\(\Leftrightarrow20x^2-16x-33=10x^2+3x-3x\)

\(\Leftrightarrow20x^2-16x-33=10x^2\)

\(\Leftrightarrow20x^2-16x-33=10x^2-10x^2\)

\(\Leftrightarrow20x^2-16x-33=0\)

\(\Rightarrow\orbr{\begin{cases}x=3\\x=-\frac{11}{10}\end{cases}}\)

a, ( 2x + 3 )(x - 4) + ( x - 5 )( x - 2) =( 3x - 5 ) ( x - 4 )
<-> 2x² - 8x + 3x - 12 + x² - 2x - 5x + 10 = 3x² - 12x - 5x + 20
<-> 3x² - 10x - 2 = 3x² - 15x + 20
<-> 3x² - 3x² - 10x + 15x = 20 + 2
<-> 5x = 22
<-> x = 22/5

Câu 1:

\(\left(x-2\right)\left(x^2+2x+4\right)+25x=x\left(x+5\right)\left(x-5\right)+8\)

\(\Leftrightarrow x^3-8+25x=x\left(x^2-25\right)+8\)

\(\Leftrightarrow x^3-8+25x=x^3-25x+8\)

\(\Leftrightarrow x^3-8+25x-x^3+25x-8=0\)

\(\Leftrightarrow50x-16=0\)

\(\Leftrightarrow50x=16\)

\(\Leftrightarrow x=\dfrac{8}{25}\)

Câu 2 :

\(\dfrac{x+5}{4}+\dfrac{3+2x}{3}=\dfrac{6x-1}{3}-\dfrac{1-2x}{12}\)

<=> \(\dfrac{3\left(x+5\right)}{12}+\dfrac{4\left(3+2x\right)}{12}=\dfrac{4\left(6x-1\right)}{12}-\dfrac{1-2x}{12}\)

<=>\(\dfrac{3x+15+12+8x}{12}=\dfrac{24x-4-1+2x}{12}\)

<=> 3x + 15 + 12 + 8x = 24x - 4 - 1 +2x

<=> 11x+27 = 26x -5

<=> ( 26x - 5 ) - ( 11x + 27 ) = 0

<=> 15x - 32 = 0

<=> 15x = 32

<=> x = \(\dfrac{32}{15}\)

Câu 3:

x - 4/3 - 3x - 1/12 = 3x + 1/4 + 9x - 2/8

<=> 4x - 16 - 3x + 1/12 = 6x + 2 + 9x - 2/8

<=> x - 15/12 = 15x/8

<=> 8x - 120 = 180x

<=> 120 = -172x <=> x = -172/120 = -43/30

2:

a: =>x-1=0 hoặc 3x+1=0

=>x=1 hoặc x=-1/3

b: =>x-5=0 hoặc 7-x=0

=>x=5 hoặc x=7

c: =>\(\left[{}\begin{matrix}x-1=0\\x+5=0\\3x-8=0\end{matrix}\right.\Leftrightarrow x\in\left\{1;-5;\dfrac{8}{3}\right\}\)

d: =>x=0 hoặc x^2-1=0

=>\(x\in\left\{0;1;-1\right\}\)

Bạn tách ra từng câu thoi nhe .

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tớ chịu.

hi hi.