Đáp án là B

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

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

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

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

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

2.\(x^2+3x+2=0\)

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

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

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

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

3.\(2x^2+5x+3=0\)

\(\Leftrightarrow2x^2+2x+3x+3=0\)

\(\Leftrightarrow2x\left(x+1\right)+3\left(x+1\right)=0\)

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

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

4.\(x^3+x^2-12x=0\)

\(\Leftrightarrow x\left(x^2+x-12\right)=0\)

\(\Leftrightarrow x\left(x+4\right)\left(x-3\right)=0\)

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

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

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

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

=>x=-2 hoặc x=1

b: =>(x+1)(x+2)=0

=>x=-1 hoặc x=-2

c: =>(2x+3)(x+1)=0

=>x=-1 hoặc x=-3/2

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

hay \(x\in\left\{0;-4;3\right\}\)

a: =>x^3(x-2)+10x(x-2)=0

=>(x-2)(x^3+10x)=0

=>x(x-2)(x^2+10)=0

=>x(x-2)=0

=>x=0 hoặc x=2

b: =>x^2*(x-3)-16(x-3)=0

=>(x-3)(x^2-16)=0

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

=>\(x\in\left\{3;4;-4\right\}\)

a: Ta có: \(\left(x^2+2\right)\left(x-4\right)-\left(x+2\right)^3=-16\)

\(\Leftrightarrow x^3-4x^2+2x-8-x^3-6x^2-12x-8=-16\)

\(\Leftrightarrow-10x^2-10x=0\)

\(\Leftrightarrow-10x\left(x+1\right)=0\)

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

c: Ta có: \(x^3+3x^2+3x+28=0\)

\(\Leftrightarrow\left(x+1\right)^3=-27\)

\(\Leftrightarrow x+1=-3\)

hay x=-4

Bài 1.

Bạn xem đã viết đúng đề chưa vậy?

a: Sửa đề: \(\left(6x^3+x^2\right):2x-3x+2=0\)

ĐKXĐ: x<>0

Ta có: \(\left(6x^3+x^2\right):2x-3x+2=0\)

=>\(\frac{6x^3}{2x}+\frac{x^2}{2x}-3x+2=0\)

=>\(3x+\frac12x-3x+2=0\)

=>\(\frac12x+2=0\)

=>\(\frac12x=-2\)

=>x=-4(nhận)

b: Sửa đề: \(\left(5x^4-3x^2\right):x^2-x\left(5x+6\right)=0\)

ĐKxĐ: x<>0

Ta có: \(\left(5x^4-3x^2\right):x^2-x\left(5x+6\right)=0\)

=>\(5x^2-3-5x^2-6x=0\)

=>-6x-3=0

=>6x+3=0

=>3(2x+1)=0

=>2x+1=0

=>2x=-1

=>\(x=-\frac12\) (nhận)

Lời giải:

a.

PT $\Leftrightarrow 3x^2+\frac{x}{2}-3x^2+3x+2=0$
$\Leftrightarrow \frac{7}{2}x+2=0$
$\Leftrightarrow \frac{7}{2}x=-2$
$\Leftrightarrow x=-2: \frac{7}{2}=\frac{-4}{7}$
b.

PT $\Leftrightarrow 5x^2-3-5x^2-6x=0$

$\Leftrightarrow -3-6x=0$

$\Leftrightarrow 6x=-3$

$\Leftrightarrow x=\frac{-3}{6}=\frac{-1}{2}$

a.

\(\Leftrightarrow\left(x-1\right)^3=10^3\)

\(\Leftrightarrow x-1=10\)

\(\Rightarrow x=11\)

b.

\(\Leftrightarrow x^2-4x+4=25\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x-2=5\\x-2=-5\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=7\\x=-3\end{matrix}\right.\)

a) x³-3x²+3x-1=1000

⇒(x-1)³=1000

⇒x-1=10

⇒x=11

b) x²-4x-21=0

⇒ x²-7x+3x-21=0

⇒x(x-7)+3(x-7)=0

⇒(x+3)(x-7)=0

⇒ hoặc x+3 = 0 ⇒ x=-3

hoặc x-7=0⇒x=7

vậy x={-3; 7}

a) 3x² - 6x = 0

3x(x - 2) = 0

3x = 0 hoặc x - 2 = 0

*) 3x = 0

x = 0

*) x - 2 = 0

x = 2

Vậy x = 0; x = 2

b) x² + 10x + 25 = 64

x² + 2.x.5 + 5² = 64

(x + 5)² = 64

x + 5 = -8 hoặc x + 5 = 8

*) x + 5 = -8

x = -8 - 5

x = -13

*) x + 5 = 8

x = 8 - 5

x = 3

Vậy x = -13; x = 3

c) (2 - 3x)(x + 4) + 3x - 2 = 0

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

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

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

2 - 3x = 0 hoặc x + 3 = 0

*) 2 - 3x = 0

3x = 2

*) x + 3 = 0

x = 0 - 3

x = -3

Vậy: