`@` `\text {Ans}`

`\downarrow`

`a)`

`3x(4x-1) - 2x(6x-3) = 30`

`=> 12x^2 - 3x - 12x^2 + 6x = 30`

`=> 3x = 30`

`=> x = 30 \div 3`

`=> x=10`

Vậy, `x=10`

`b)`

`2x(3-2x) + 2x(2x-1) = 15`

`=> 6x- 4x^2 + 4x^2 - 2x = 15`

`=> 4x = 15`

`=> x = 15/4`

Vậy, `x=15/4`

`c)`

`(5x-2)(4x-1) + (10x+3)(2x-1) = 1`

`=> 5x(4x-1) - 2(4x-1) + 10x(2x-1) + 3(2x-1)=1`

`=> 20x^2-5x - 8x + 2 + 20x^2 - 10x +6x - 3 =1`

`=> 40x^2 -17x - 1 = 1`

`d)`

`(x+2)(x+2)-(x-3)(x+1)=9`

`=> x^2 + 2x + 2x + 4 - x^2 - x + 3x + 3=9`

`=> 6x + 7 =9`

`=> 6x = 2`

`=> x=2/6 =1/3`

Vậy, `x=1/3`

`e)`

`(4x+1)(6x-3) = 7 + (3x-2)(8x+9)`

`=> 24x^2 - 12x + 6x - 3 = 7 + (3x-2)(8x+9)`

`=> 24x^2 - 12x + 6x - 3 = 7 + 24x^2 +11x - 18`

`=> 24x^2 - 6x - 3 = 24x^2 + 18x -11`

`=> 24x^2 - 6x - 3 - 24x^2 + 18x + 11 = 0`

`=> 12x +8 = 0`

`=> 12x = -8`

`=> x= -8/12 = -2/3`

Vậy, `x=-2/3`

`g)`

`(10x+2)(4x- 1)- (8x -3)(5x+2) =14`

`=> 40x^2 - 10x + 8x - 2 - 40x^2 - 16x + 15x + 6 = 14`

`=> -3x + 4 =14`

`=> -3x = 10`

`=> x= - 10/3`

Vậy, `x=-10/3`

Hello các bạn còn đó ko?

a: =>|x-3|=4-x

\(\Leftrightarrow\left\{{}\begin{matrix}x< =4\\\left(4-x-x+3\right)\left(4-x+x-3\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x< =4\\\left(7-2x\right)=0\end{matrix}\right.\Leftrightarrow x=\dfrac{7}{2}\)

b: =>|x-5|=3-19x

\(\Leftrightarrow\left\{{}\begin{matrix}x< =\dfrac{3}{19}\\\left(x-5-3+19x\right)\left(x-5+3-19x\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x< =\dfrac{3}{19}\\\left(20x-8\right)\left(-18x-2\right)=0\end{matrix}\right.\Leftrightarrow x\in\left\{-\dfrac{1}{9}\right\}\)

c: =>\(\Leftrightarrow\sqrt{x-3}\left(\sqrt{x+3}+\sqrt{x-3}\right)=0\)

=>căn x-3=0

=>x=3

a) ( 3x + 2 )( x - 1 ) - ( x + 2 )( 3x + 1 ) = 7

<=> 3x² - x - 2 - ( 3x² + 7x + 2 ) = 7

<=> 3x² - x - 2 - 3x² - 7x - 2 = 7

<=> -8x - 4 = 7

<=> -8x = 11

<=> x = -11/8

b) ( 6x + 5 )( 2x + 3 ) - ( 4x + 3 )( 3x - 2 ) = 8

<=> 12x² + 28x + 15 - ( 12x² + x - 6 ) = 8

<=> 12x² + 28x + 15 - 12x² - x + 6 = 8

<=> 27x + 21 = 8

<=> 27x = -13

<=> x = -13/27 

c) 2x( x + 3 ) - ( x + 1 )( 2x + 1 ) - 5 = 9

<=> 2x² + 6x - ( 2x² + 3x + 1 ) - 5 = 9

<=> 2x² + 6x - 2x² - 3x - 1 - 5 = 9

<=> 3x - 6 = 9

<=> 3x = 15

<=> x = 5

d) ( 5x + 3 )( 4x - 7 ) - ( 10x + 9 )( 2x - 3 ) = 10

<=> 20x² - 23x - 21 - ( 20x² - 12x - 27 ) = 10

<=> 20x² - 23x - 21 - 20x² + 12x + 27 = 10

<=> -11x + 6 = 10

<=> -11x = 4

<=> x = -4/11

a, \(\left(3x+2\right)\left(x-1\right)-\left(x+2\right)\left(3x+1\right)=7\Leftrightarrow-8x-4=7\Leftrightarrow x=-\frac{11}{8}\)

b, \(\left(6x+5\right)\left(2x+3\right)-\left(4x+3\right)\left(3x-2\right)=8\Leftrightarrow27x+21=8\Leftrightarrow x=-\frac{13}{27}\)

c, \(2x\left(x+3\right)-\left(x+1\right)\left(2x+1\right)-5=9\Leftrightarrow3x-6=9\Leftrightarrow x=5\)

d, \(\left(5x+3\right)\left(4x-7\right)-\left(10x+9\right)\left(2x-3\right)=10\Leftrightarrow-11x+6=10\Leftrightarrow x=-\frac{4}{11}\)

a: \(2x\left(x^2-3x+1\right)=2x^3-6x^2+2x\)

b: \(\left(x+2\right)^2-x^2=4x+4\)

c: \(\left(x+3\right)\left(x^2-3x+9\right)-x^3=27\)

a) Ta có: \(3x\left(x+1\right)-2x\left(x+20\right)=-1-x\)

\(\Leftrightarrow3x^2+3x-2x^2-40x+1+x=0\)

\(\Leftrightarrow x^2-36x+1=0\)

\(\Leftrightarrow x^2-36x+324-323=0\)

\(\Leftrightarrow\left(x-18\right)^2=323\)

\(\Leftrightarrow\left[{}\begin{matrix}x-18=\sqrt{323}\\x-18=-\sqrt{323}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=18+\sqrt{323}\\x=18-\sqrt{323}\end{matrix}\right.\)

Vậy: \(x\in\left\{18+\sqrt{323};18-\sqrt{323}\right\}\)

b) Ta có: \(\left(3x-1\right)\left(2x+7\right)-\left(x+1\right)\left(6x-5\right)=16\)

\(\Leftrightarrow6x^2+21x-2x-7-\left(6x^2-5x+6x-5\right)-16=0\)

\(\Leftrightarrow6x^2+19x-7-\left(6x^2+x-5\right)-16=0\)

\(\Leftrightarrow6x^2+19x-7-6x^2-x+5-16=0\)

\(\Leftrightarrow18x-18=0\)

\(\Leftrightarrow18x=18\)

hay x=1

Vậy: x=1

c) Ta có: \(\left(10x+9\right)\cdot x-\left(5x-1\right)\left(2x+3\right)=8\)

\(\Leftrightarrow10x^2+9x-\left(10x^2+15x-2x-3\right)-8=0\)

\(\Leftrightarrow10x^2+9x-10x^2-13x+3-8=0\)

\(\Leftrightarrow-4x-5=0\)

\(\Leftrightarrow-4x=5\)

hay \(x=\frac{-5}{4}\)

Vậy: \(x=\frac{-5}{4}\)

a) \(x^2-10x+9=x^2-x-9x+9=x\left(x-1\right)-9\left(x-1\right)=\left(x-1\right)\left(x-9\right)\)

b)\(x^2-10x+21=x^2-3x-7x+21=x\left(x-3\right)-7\left(x-3\right)=\left(x-3\right)\left(x-7\right)\) c)\(x^2-2x-3=x^2+x-3x-3=x\left(x+1\right)-3\left(x+1\right)=\left(x+1\right)\left(x-3\right)\)d)\(x^2-10x+16=x^2-2x-8x+16=x\left(x-2\right)-8\left(x-2\right)=\left(x-2\right)\left(x-8\right)\)e)\(x^2-2x-8=x^2+2x-4x-8=x\left(x+2\right)-4\left(x+2\right)=\left(x+2\right)\left(x-4\right)\)f)\(x^2-2x-48=x^2+6x-8x-48=x\left(x+6\right)-8\left(x+6\right)=\left(x+6\right)\left(x-8\right)\)g)\(x^2-10x+24=x^2-4x-6x+24=x\left(x-4\right)-6\left(x-4\right)=\left(x-4\right)\left(x-6\right)\)i)mình nghĩ câu này bị sai nên mình k giải đc. Mình nghĩ đề là \(x^4+3x^2-4\)

j)\(x^2-2x-15=x^2-3x+5x-15=x\left(x-3\right)+5\left(x-3\right)=\left(x-3\right)\left(x+5\right)\)

chúc bạn học tốt........

a,x² - 10x + 9 = x² - x - 9x + 9 = x(x - 1) - 9(x - 1) = (x - 9)(x - 1)

b,x² - 10x + 21 = x² - 3x - 7x + 21 = x(x - 3) - 7(x - 3)

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

d,x² - 10x + 16 = x² - 2x -8x + 16= x(x - 2) - 8(x - 2) = (x - 8)(x - 2)

e,x² - 2x - 8 = x² + 2x - 4x - 8 = x(x + 2) - 4(x + 2) = (x - 4)(x + 2)

f,x² - 2x - 48 = x² - 8x + 6x - 48 = x(x - 8) + 6(x - 8) = (x + 6)(x - 8)

g,x² - 10x + 24 = x² - 4x - 6x + 24 = x(x - 4) - 6(x - 4) = (x - 6)(x - 4)

j,x² - 2x - 15 = x² + 3x - 5x -15 = x(x + 3) - 5(x + 3) = (x - 5)(x + 3)