a/ (2x² + 3x - 1)² - 4(2x² + 3x + 3) + 20 = 0

Đặt a = 2x² + 3x - 1 , ta đc:

a² - 4.(a + 4) + 20 = 0

=> a² - 4a - 16 + 20 = 0

=> a² - 4a + 4 = 0

=> (a - 2)² = 0 => a = 2

Với a = 2 => 2x² + 3x - 1 = 2 => 2x² + 3x - 3 = 0 

Có : \(\Delta=3^2-4.2.\left(-3\right)=33\Rightarrow\sqrt{\Delta}=\sqrt{33}\)

\(\Rightarrow x_1=\frac{-3+\sqrt{33}}{4};x_2=\frac{-3-\sqrt{33}}{4}\)

Vậy pt có 2 nghiệm như trên 

b, c có 2 cách làm lận, bạn thích cách nào

làm câu b,c nha

bạn vào trang cá nhân tặng coin cho mik rùi mik giải ch

a) Ta có: \(-1\dfrac{1}{3}-x=2.5\)

\(\Leftrightarrow\dfrac{-4}{3}-x=\dfrac{5}{2}\)

hay \(x=\dfrac{-4}{3}-\dfrac{5}{2}=\dfrac{-8}{6}-\dfrac{15}{6}=\dfrac{-23}{6}\)

b) Ta có: \(\dfrac{1}{3}:x-20\%=\dfrac{16}{-32}\)

\(\Leftrightarrow\dfrac{1}{3}:x=\dfrac{-1}{2}+\dfrac{1}{5}=\dfrac{-3}{10}\)

hay \(x=\dfrac{1}{3}:\dfrac{-3}{10}=\dfrac{1}{3}\cdot\dfrac{10}{-3}=\dfrac{-10}{9}\)

c) Ta có: \(\left(\dfrac{3x}{7}+1\right):\left(-4\right)=\dfrac{-1}{28}\)

\(\Leftrightarrow\dfrac{3}{7}x+1=\dfrac{1}{7}\)

\(\Leftrightarrow x\cdot\dfrac{3}{7}=\dfrac{1}{7}-1=\dfrac{-6}{7}\)

hay x=-2

a) Ta có: ⇔−43−x=52⇔−43−x=52

hay 13:x−20%=16−3213:x−20%=16−32

x=13:−310=13⋅10−3=−109x=13:−310=13⋅10−3=−109

c) Ta có: ⇔37x+1=17⇔37x+1=17

11: Ta có: \(\left(x+3\right)^3=125\)

\(\Leftrightarrow x+3=5\)

hay x=2

12: Ta có: \(\left(2x\right)^4=16\)

\(\Leftrightarrow x^4=1\)

hay \(x\in\left\{1;-1\right\}\)

11: Ta có: \left(x+3\right)^3=125(x+3)3=125

$\iff x+3=5$

hay x=2

12: Ta có: \left(2x\right)^4=16(2x)4=16

\Leftrightarrow x^4=1(2x)   = 2   =2  4

           4      

⇔2x=2⇔ x=1

13) 32:(3x-2)=2^3

32:(3x-2)=8

3x-2=32:8=4

3x=4+2=6

x=6:3=2

⇔x=2

mai nah

a, 2x - 3 = 4 - 5x

=> 2x + 5x = 4 + 3

=> 7x = 7

=> x = 7 : 7

=> x = 1

b) 5.(x - 1)³ = 20.(x - 1)

=> 5.x³ - 3x³ + 3x - 1 = 20x - 1

=> x³(5 - 3) + 3x = 20x - 1 + 1

=> 2x³ + 3x = 20x

=> 2x³ = 20x - 3x

=> 2.x³ = 17x

=> 2.x³ - 17x = 0

=> x(2x² - 17) = 0

=> x = 0 hoặc 2x² - 17 = 0

                  => 2x² = 17

                  => x² = 8,5

                  => x = \(\sqrt{8,5}\)

c) 9x² - 16 = 3x - 4

=> 9x² - 3x = -4 + 16

=> x(9x - 3) = 12

=> x[3(3x - 1)] = 12

=> x(3x - 1) = 12 : 3

=> x(3x - 1) = 4  chju nhe

to rat ngu

\(8x-48+4x-12-14=-x+4\)

\(\Leftrightarrow12x-75=-x+4\Leftrightarrow13x=79\Leftrightarrow x=\dfrac{79}{13}\)

\(-7\left(8-x\right)-6\left(x+9\right)=20-x\Leftrightarrow-56+7x-6x-54=20-x\)

\(\Leftrightarrow2x=130\Leftrightarrow x=65\)

\(9x-63-80+60x=-7x+15\Leftrightarrow76x=158\Leftrightarrow x=\dfrac{79}{38}\)

\(-96-16x-60+30x=-40x-16\Leftrightarrow54x=140\Leftrightarrow x=\dfrac{70}{27}\)

\(17x-102-14x-28=4x-24-2x+4\Leftrightarrow x=110\)

1: =(x+y-3x)(x+y+3x)

=(-2x+y)(4x+y)

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

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

=3(x+1)(3x-5)

3: =(2x)^2-(x^2+1)^2

=-[(x^2+1)^2-(2x)^2]

=-(x^2+1-2x)(x^2+1+2x)

=-(x-1)^2(x+1)^2

4: =(2x+1+x-1)(2x+1-x+1)

=3x(x+2)

5: =[(x+1)^2-(x-1)^2][(x+1)^2+(x-1)^2]

=(2x^2+2)*4x

=8x(x^2+1)

6: =(5x-5y)^2-(4x+4y)^2

=(5x-5y-4x-4y)(5x-5y+4x+4y)

=(x-9y)(9x-y)

7: =(x^2+xy+y^2+xy)(x^2+xy-y^2-xy)

=(x^2+2xy+y^2)(x^2-y^2)

=(x+y)^3*(x-y)

8: =(x^2+4y^2-20-4xy+16)(x^2+4y^2-20+4xy-16)

=[(x-2y)^2-4][(x+2y)^2-36]

=(x-2y-2)(x-2y+2)(x+2y-6)(x+2y+6)