https://hoc24.vn/cau-hoi/a-5x-2x-62-50b-5x-x-150-2-3c-6x-x-511-59-31d-5x-3x-36-334-124x-2x-68-219-216.2785429565572

a: \(\Leftrightarrow7x=35\)

hay x=5

b: \(\Leftrightarrow6x=78\)

hay x=13

a: \(\Leftrightarrow6x=30\)

hay x=5

b: \(\Leftrightarrow6x=25+12-1=36\)

hay x=6

A x=5

B x = 6 (36)

a: \(\Leftrightarrow8x=108+12=120\)

hay x=15

b: \(\Leftrightarrow6x=60\)

hay x=10

a) 5x + 3x = 120
    x . (5+3)= 120
    x . 8       = 120
    x            = 120 : 8
    x            = 15

c) 7x - 2x = 40
    x .(7-2)= 40
    x . 5     = 40

    x          = 40 : 5
    x          = 8

`Answer:`

a. \(5x-[2x+1-\left(2x-3\right)-\left(4x+1\right)]\)

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

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

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

\(=9x-3\)

b. \(\left(-3x^2+2x-1\right)+\left(4x^2-2x+3\right)\)

\(=-3x^2+2x-1+4x^2-2x+3\)

\(=\left(-3x^2+4x^2\right)+\left(2x-2x\right)+\left(-1+3\right)\)

\(=x^2+2\)

a) ( x² - 3x )( x² + 7x + 10 ) = 216

<=> x( x - 3 )( x + 2 )( x + 5 ) - 216 = 0

<=> [ x( x + 2 ) ][ ( x - 3 )( x + 5 ) ] - 216 = 0

<=> ( x² + 2x )( x² + 2x - 15 ) - 216 = 0 (1)

Đặt a = x² + 2x

(1) trở thành a( a - 15 ) - 216 = 0 <=> a² - 15a - 216 = 0 <=> ( a - 24 )( a + 9 ) = 0 <=> a = 24 hoặc a = -9

=> \(\orbr{\begin{cases}x^2+2x=24\\x^2+2x=-9\end{cases}}\Leftrightarrow\orbr{\begin{cases}x^2+2x-24=0\\x^2+2x+9=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}\left(x-4\right)\left(x+6\right)=0\\\left(x+1\right)^2+8>0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=4\\x=-6\end{cases}}\)

Vậy S = { 4 ; -6 }

b) ( 2x² - 7x + 3 )( 2x² + x - 3 ) + 9 = 0

<=> ( x - 3 )( 2x - 1 )( x - 1 )( 2x + 3 ) + 9 = 0

<=> [ ( x - 3 )( 2x + 3 ) ][ ( 2x - 1 )( x - 1 ) ] + 9 = 0

<=> ( 2x² - 3x - 9 )( 2x² - 3x + 1 ) + 9 = 0

<=> ( 2x² - 3x - 4 - 5 )( 2x² - 3x - 4 + 5 ) + 9 = 0

<=> ( 2x² - 3x - 4 )² - 16 = 0

<=> x( 2x - 3 )( 2x² - 3x - 8 ) = 0

<=> x = 0 hoặc 2x - 3 = 0 hoặc 2x² - 3x - 8 = 0

<=> x = 0 hoặc x = 3/2 hoặc x = \(\frac{3\pm\sqrt{73}}{4}\)

Vậy S = { 0 ; 3/2 ; \(\frac{3\pm\sqrt{73}}{4}\)}

c) ( 4x + 3 )²( 2x + 1 )( x + 1 ) = 72

<=> ( 16x² + 24x + 9 )( 2x² + 3x + 1 ) - 72 = 0 (1)

Đặt 2x² + 3x + 1 = a

(1) trở thành ( 8a + 1 )a - 72 = 0 <=> 8a² + a - 72 = 0 

Đến đây nghiệm xấu lắm nên thôi dừng nhá:(

\(\left[68-\left(197-216\right)\right]-\left[\left(-43\right)+168-216\right]\\ =\left[68-\left(-19\right)\right]-\left[\left(-43\right)-48\right]\\ =87+91\\ =178\)

[68−(197−216)]−[(−43)+168−216]=[68−(−19)]−[(−43)−48]=87+91=178

b: \(\Leftrightarrow32x^5+1-32x^5+1=2\)

=>2=2(luôn đúng)

a: \(\Leftrightarrow\left[\left(x-3\right)^2-\left(x+3\right)^2\right]\left[\left(x-3\right)^2+\left(x+3\right)^2\right]+24x^3=216\)

\(\Leftrightarrow-12x\left(2x^2+18\right)+24x^3=216\)

=>-216x=216

hay x=-1