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8 tháng 9 2021

\(a,3\left(2x-3\right)+2\left(2-x\right)=-3\\ \Leftrightarrow6x-9+4-2x=-3\\ \Leftrightarrow4x=2\\ \Leftrightarrow x=\dfrac{1}{2}\\ b,x\left(5-2x\right)+2x\left(x-1\right)=13\\ \Leftrightarrow5x-2x^2+2x^2-2x=13\\ \Leftrightarrow3x=13\\ \Leftrightarrow x=\dfrac{13}{3}\\ c,5x\left(x-1\right)-\left(x+2\right)\left(5x-7\right)=6\\ \Leftrightarrow5x^2-5x-5x^2-3x+14=6\\ \Leftrightarrow-8x=-8\\ \Leftrightarrow x=1\\ d,3x\left(2x+3\right)-\left(2x+5\right)\left(3x-2\right)=8\\ \Leftrightarrow6x^2+9x-6x^2-11x+10=8\\ \Leftrightarrow-2x=-2\\ \Leftrightarrow x=1\)

\(e,2\left(5x-8\right)-3\left(4x-5\right)=4\left(3x-4\right)+11\\ \Leftrightarrow10x-16-12x+15=12x-16+11\\ \Leftrightarrow-14x=-4\\ \Leftrightarrow x=\dfrac{2}{7}\\ f,2x\left(6x-2x^2\right)+3x^2\left(x-4\right)=8\\ \Leftrightarrow12x^2-4x^3+3x^3-12x^2=8\\ \Leftrightarrow-x^3-8=0\\ \Leftrightarrow-\left(x^3+8\right)=0\\ \Leftrightarrow-\left(x+2\right)\left(x^2-2x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-2\\x\in\varnothing\left(x^2-2x+4=\left(x-1\right)^2+3>0\right)\end{matrix}\right.\)

Bài 4:

a: Ta có: \(3\left(2x-3\right)-2\left(x-2\right)=-3\)

\(\Leftrightarrow6x-9-2x+4=-3\)

\(\Leftrightarrow4x=2\)

hay \(x=\dfrac{1}{2}\)

b: Ta có: \(x\left(5-2x\right)+2x\left(x-1\right)=13\)

\(\Leftrightarrow5x-2x^2+2x^2-2x=13\)

\(\Leftrightarrow3x=13\)

hay \(x=\dfrac{13}{3}\)

c: Ta có: \(5x\left(x-1\right)-\left(x+2\right)\left(5x-7\right)=6\)

\(\Leftrightarrow5x^2-5x-5x^2+7x-10x+14=6\)

\(\Leftrightarrow-8x=-8\)

hay x=1

30 tháng 8 2023

a) (x - 5)(x - 3) + 2(x - 5) = 0

(x - 5)(x - 3 + 2) = 0

(x - 5)(x - 1) = 0

x - 5 = 0 hoặc x - 1 = 0

*) x - 5 = 0

x = 5

*) x - 1 = 0

x = 1

Vậy x = 1; x = 5

b) (x - 2)(x² + 2x + 4) - (x + 2)(x² - 2x + 4) = 2(x + 2)

x³ - 8 - x³ - 8 = 2x + 4

2x = -8 - 8 - 4

2x = -20

x = -20 : 2

x = -10

30 tháng 8 2023

a)

\(\left(x-5\right)\left(x-3\right)+2\left(x-5\right)=0\)

\(\left(x-5\right)\left(x-3+2\right)=0\)

\(\left(x-5\right)\left(x-1\right)=0\)

\(x-5=0\) hoặc \(x-1=0\)

+) \(x-5=0\\ \Rightarrow x=5\)

+) \(x-1=0\\ \Rightarrow x=1\)

Vậy \(x=1\) hoặc \(x=5\)

b) \(\left(x-2\right)\left(x^2+2x+4\right)-\left(x+2\right)\left(x^2-2x+4\right)=2\left(x+2\right)\)

\(x^3-8-x^3-8=2x+4\)

\(2x=-8-8-4\)

\(2x=-20\)

 \(x=-20:2\)

 \(x=-10\)

Vậy \(x=-10\)

11 tháng 3 2023

`a,4x-10=0   `

`<=> 4x=10`

`<=>x=10/4`

`<=>x=5/2`

`b, 7-3x=9-x     `

`<=>-3x+x=9-7`

`<=>-2x=2`

`<=>x=-1`

`c, 2x-(3-5x) = 4(x+3)`

`<=>2x-3+5x=4x+12`

`<=>2x+5x-4x=12+3`

`<=>3x=15`

`<=>x=5`

`d, 5-(6-x)=4(3-2x)     `

`<=>5-6+x=12-8x`

`<=>x+8x=12-5+6`

`<=>9x=13`

`<=>x=13/9`

`e, 4(x+3)=-7x+17   `

`<=>4x+12=-7x+17`

`<=>4x+7x=17-12`

`<=>11x=5`

`<=>x=5/11`   

`f, 5(x-3) - 4=2(x-1)+7`

`<=>5x-15-4=2x-2+7`

`<=>5x-2x=15+4-2+7`

`<=>3x=24`

`<=>x=8`

`g, 5(x-3)-4=2(x-1)+7       `

`<=>5x-15-4=2x-2+7`

`<=>5x-2x=15+4-2+7`

`<=>3x=24`

`<=>x=8`

`h,4(3x-2)-3(x-4)=7x+20`

`<=>12x-8-3x+12=7x+20`

`<=>12x-3x-7x=20+8+12`

`<=>2x=40`

`<=>x=20`

6 tháng 12 2019

\(a.3\left(x^2-2x+1\right)-3x^2+15x-2=0\)

\(3x^2-6x+3-3x^2+15x-2=0\)

\(9x+1=0\)

\(x=-\frac{1}{9}\)

\(b.4x^2-12x+9=0\)

\(4x^2-6x-6x+9=0\)

\(2x\left(x-3\right)-3\left(x-3\right)=0\)

\(\left(2x-3\right)\left(x-3\right)=0\)

\(\Rightarrow\orbr{\begin{cases}2x-3=0\\x-3=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{3}{2}\\x=3\end{cases}}\)

6 tháng 12 2019

\(c.\left(2x-3\right)^2-\left(x+5\right)^2=0\)

\(\left(2x-3-x-5\right)\left(2x-3+x+5\right)=0\)

\(\left(x-8\right)\left(3x+2\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-8=0\\3x+2=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=8\\x=-\frac{2}{3}\end{cases}}\)

14 tháng 8 2019

a) (x - 2)(x + 3) = 6

=> x2 + 3x - 2x - 6 = 6

=> x2 + x - 6 - 6 = 0

=> x2 + x - 12 = 0

=> x2 + 4x - 3x - 12 = 0

=> x(x + 4) - 3(x + 4) = 0

=> (x - 3)(x + 4) = 0

=> \(\orbr{\begin{cases}x-3=0\\x+4=0\end{cases}}\)

=> \(\orbr{\begin{cases}x=3\\x=-4\end{cases}}\)

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

=> 2x2 + 4x - 3x - 6 = 4

=> 2x2 + x - 6 - 4 = 0

=> 2x2 + x - 10 = 0

=> 2x2 + 5x - 4x - 10 = 0

=> x(2x + 5) - 2(2x + 5) = 0

=> (x - 2)(2x + 5) = 0

=> \(\orbr{\begin{cases}x-2=0\\2x+5=0\end{cases}}\)

=> \(\orbr{\begin{cases}x=2\\x=-\frac{5}{2}\end{cases}}\)

2 tháng 10 2020

a) 2x( x - 7 ) - ( x + 3 )( x - 2 ) - ( x + 4 )( x - 4 )

= 2x2 - 14x - ( x2 + x - 6 ) - ( x2 - 16 )

= 2x2 - 14x - x2 - x + 6 - x2 + 16

= 22 - 15x

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

= 2x2 + x - 10 - 3( x2 + 4x + 4 ) + x2 + 2x + 1

= 3x2 + 3x - 9 - 3x2 - 12x - 12

= -9x - 21

c) ( x + 3 )( x - 3 ) - ( x + 5 )( x - 1 ) - ( x - 4 )2

= x2 - 9 - ( x2 + 4x - 5 ) - ( x2 - 8x + 16 )

= x2 - 9 - x2 - 4x + 5 - x2 + 8x - 16

= -x2 + 4x - 20

d) 2x( x + 1 )2 - ( x - 1 )3 - ( x - 2 )( x2 + 2x + 4 ) 

= 2x( x2 + 2x + 1 ) - ( x3 - 3x2 + 3x - 1 ) - ( x3 - 8 )

= 2x3 + 4x2 + 2x - x3 + 3x2 - 3x + 1 - x3 + 8

= 7x2 - x + 9

e) ( x + 5 )( x - 5 )( x + 2 ) - ( x + 2 )3

= ( x2 - 25 )( x + 2 ) - ( x3 + 6x2 + 12x + 8 )

= x3 + 2x2 - 25x - 50 - x3 - 6x2 - 12x - 8

= -4x2 - 37x - 58

a) Ta có: \(\left(x+5\right)\left(2x-1\right)=\left(2x-3\right)\left(x+1\right)\)

\(\Leftrightarrow\left(x+5\right)\left(2x-1\right)-\left(2x-3\right)\left(x+1\right)=0\)

\(\Leftrightarrow2x^2-x+10x-5-\left(2x^2+2x-3x-3\right)=0\)

\(\Leftrightarrow2x^2+9x-5-2x^2+x+3=0\)

\(\Leftrightarrow10x-2=0\)

hay 10x=2

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

Vậy: \(x=\frac{1}{5}\)

b) Ta có: \(\left(x+1\right)\left(x+9\right)=\left(x+3\right)\left(x+5\right)\)

\(\Leftrightarrow x^2+9x+x+9=x^2+5x+3x+15\)

\(\Leftrightarrow x^2+10x+9-x^2-8x-15=0\)

\(\Leftrightarrow2x-6=0\)

hay 2x=6

\(\Leftrightarrow x=3\)

Vậy: x=3

c) Ta có: \(\left(3x+5\right)\left(2x+1\right)=\left(6x-2\right)\left(x-3\right)\)

\(\Leftrightarrow6x^2+3x+10x+5=6x^2-18x-2x+6\)

\(\Leftrightarrow6x^2+13x+5=6x^2-20x+6\)

\(\Leftrightarrow6x^2+13x+5-6x^2+20x-6=0\)

\(\Leftrightarrow33x-1=0\)

\(\Leftrightarrow33x=1\)

hay \(x=\frac{1}{33}\)

Vậy: \(x=\frac{1}{33}\)

d) Ta có: \(\left(x-2\right)\left(3x+5\right)=\left(2x-4\right)\left(x+1\right)\)

\(\Leftrightarrow3x^2+5x-6x-10=2x^2+2x-4x-4\)

\(\Leftrightarrow3x^2-x-10=2x^2-2x-4\)

\(\Leftrightarrow3x^2-x-10-2x^2+2x+4=0\)

\(\Leftrightarrow x^2+x-6=0\)

\(\Leftrightarrow x^2+3x-2x-6=0\)

\(\Leftrightarrow x\left(x+3\right)-2\left(x+3\right)=0\)

\(\Leftrightarrow\left(x+3\right)\left(x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=2\end{matrix}\right.\)

Vậy: \(x\in\left\{-3;2\right\}\)

đ) Ta có: \(9x^2-1=\left(3x+1\right)\left(2x-3\right)\)

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

\(\Leftrightarrow\left(3x+1\right)\left[\left(3x-1\right)-\left(2x-3\right)\right]=0\)

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

\(\Leftrightarrow\left(3x+1\right)\left(x+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}3x+1=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=-1\\x=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\frac{1}{3}\\x=-2\end{matrix}\right.\)

Vậy: \(x\in\left\{-\frac{1}{3};-2\right\}\)

e) Ta có: \(\left(2x+5\right)\left(x-4\right)=\left(x-5\right)\left(4-x\right)\)

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

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

\(\Leftrightarrow\left(x-4\right)\cdot3x=0\)

\(3\ne0\)

nên \(\left[{}\begin{matrix}x-4=0\\x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=0\end{matrix}\right.\)

Vậy: \(x\in\left\{0;4\right\}\)

AH
Akai Haruma
Giáo viên
24 tháng 2 2020

a) $(x+5)(2x-1)=(2x-3)(x+1)$

$\Leftrightarrow 2x^2+9x-5=2x^2-x-3$

$\Leftrightarrow 10x=2\Rightarrow x=\frac{1}{5}$

b)

$(x+1)(x+9)=(x+3)(x+5)$

$\Leftrightarrow x^2+10x+9=x^2+8x+15$

$\Leftrightarrow 2x=6\Rightarrow x=3$

c)

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

$\Leftrightarrow 6x^2+13x+5=6x^2-20x+6$

$\Leftrightarrow 33x=1\Rightarrow x=\frac{1}{33}$

a:=>x^2-1-x=2x-1

=>x^2-x-1=2x-1

=>x^2-3x=0

=>x=0(loại) hoặc x=3(nhận)

b:=>x+2=0 hoặc 5-3x=0

=>x=-2 hoặc x=5/3

c:=>20(1-2x)+6x=9(x-5)-24

=>20-40x+6x=9x-45-24

=>-34x+20=9x-69

=>-43x=-89

=>x=89/43

d: =>x^2+4x+4-x^2-2x+3=2x^2+8x-4x-16-3

=>2x^2+4x-19=-2x+7

=>2x^2+6x-26=0

=>x^2+3x-13=0

=>\(x=\dfrac{-3\pm\sqrt{61}}{2}\)

e: =>(2x-3)(2x-3-x-1)=0

=>(2x-3)(x-4)=0

=>x=4 hoặc x=3/2

a: (2x+1)(3-x)(4-2x)=0

=>(2x+1)(x-3)(x-2)=0

hay \(x\in\left\{-\dfrac{1}{2};3;2\right\}\)

b: 2x(x-3)+5(x-3)=0

=>(x-3)(2x+5)=0

=>x=3 hoặc x=-5/2

c: =>(x-2)(x+2)+(x-2)(2x-3)=0

=>(x-2)(x+2+2x-3)=0

=>(x-2)(3x-1)=0

=>x=2 hoặc x=1/3

d: =>(x-2)(x-3)=0

=>x=2 hoặc x=3

e: =>(2x+5+x+2)(2x+5-x-2)=0

=>(3x+7)(x+3)=0

=>x=-7/3 hoặc x=-3

f: \(\Leftrightarrow2x^3+5x^2-3x=0\)

\(\Leftrightarrow x\left(2x^2+5x-3\right)=0\)

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

hay \(x\in\left\{0;-3;\dfrac{1}{2}\right\}\)

11 tháng 12 2021

a: \(=2x^2-4x-6x+12-2x^2+10x=12\)