\(2x^3-50=0\)

\(\Rightarrow2\left(x^3-25\right)=0\)

\(\Rightarrow x^3-25=0\Rightarrow x^3=25\)

\(\Rightarrow x=\sqrt[3]{25}\)

\(x^2-5x=-6\)

\(\Rightarrow x\left(x-5\right)=-6\)

Xét ước

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

\(\Rightarrow4x^2-4x+1-3x-5=0\)

\(\Rightarrow4x^2-4-7x=0\)

\(\Rightarrow4x^2-7x=4\)

\(\Rightarrow x\left(4x-7\right)=4\)

Xét ước

\(4x^2-20x+25=0\)

\(\Rightarrow\left(2x-5\right)^2=0\)

\(\Rightarrow2x=5\Rightarrow x=\dfrac{5}{2}\)

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

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

\(\Rightarrow\left|3x-1\right|=\left|x-2\right|\)

Xét dấu:v

b: ta có: \(x^2-5x=-6\)

\(\Leftrightarrow x^2-5x+6=0\)

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

=>x=2 hoặc x=3

c: Sửa đề:  \(\left(2x-1\right)^2-\left(3x+5\right)^2=0\)

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

\(\Leftrightarrow\left(-x-6\right)\left(5x+4\right)=0\)

=>x=-6 hoặc x=-4/5

d: ta có: \(4x^2-20x+25=0\)

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

=>2x-5=0

hay x=5/2

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

\(\Leftrightarrow\left(2x+1\right)\left(4x-3\right)=0\)

hay \(x\in\left\{-\dfrac{1}{2};\dfrac{3}{4}\right\}\)

a)\(6x^2+5x-6=0\)

\(\Leftrightarrow6x^2-4x+9x-6=0\)

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

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

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

b)\(6x^2-13x+6=0\)

\(\Leftrightarrow6x^2-4x-9x+6=0\)

\(\Leftrightarrow2x\left(3x-2\right)-3\left(3x-2\right)=0\)

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

\(\Leftrightarrow\left[\begin{array}{nghiempt}2x-3=0\\3x-2=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=\frac{3}{2}\\x=\frac{2}{3}\end{array}\right.\)

c)\(10x^2-13x-3=0\)

\(\Leftrightarrow10x^2-15x+2x-3=0\)

\(\Leftrightarrow5x\left(2x-3\right)+\left(2x-3\right)=0\)

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

\(\Leftrightarrow\left[\begin{array}{nghiempt}2x-3=0\\5x+1=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=\frac{3}{2}\\x=-\frac{1}{5}\end{array}\right.\)

d)\(20x^2+19x-3=0\)

\(\Delta=19^2-\left(-4\left(20.3\right)\right)=601\)

\(\Rightarrow x_{1,2}=\frac{-19\pm\sqrt{601}}{40}\)

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

\(\Delta=\left(-1\right)^2-4\left(3.6\right)=-71< 0\)

Suy ra vô nghiệm

bạn tách từng bài ra bn

tl câu hỏi trên cho mik ik

a: =>(2x-5x-1)(2x+5x+1)=0

=>(-3x-1)(7x+1)=0

=>x=-1/3 hoặc x=-1/7

b: =>(5x-5)^2-(x+2)^2=0

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

=>(4x-7)(6x-3)=0

=>x=1/2 hoặc x=7/4

c: =>(x^2+4x-1-x^2+3x-2)(x^2+4x-1+x^2-3x+2)=0

=>(7x-3)(2x^2+x+1)=0

=>7x-3=0

=>x=3/7

kh hiểu bn ơi

`4x=2+xx+1x<=>4x=2+3x<=>4x-3x=2<=>1x=2<=>x=2`

a: \(x\in\left\{0;25\right\}\)

c: \(x\in\left\{0;5\right\}\)

Bài 1:

a) (3x - 2)(4x + 5) = 0

<=> 3x - 2 = 0 hoặc 4x + 5 = 0

<=> 3x = 2 hoặc 4x = -5

<=> x = 2/3 hoặc x = -5/4

b) (2,3x - 6,9)(0,1x + 2) = 0

<=> 2,3x - 6,9 = 0 hoặc 0,1x + 2 = 0

<=> 2,3x = 6,9 hoặc 0,1x = -2

<=> x = 3 hoặc x = -20

c) (4x + 2)(x^2 + 1) = 0

<=> 4x + 2 = 0 hoặc x^2 + 1 # 0

<=> 4x = -2

<=> x = -2/4 = -1/2

d) (2x + 7)(x - 5)(5x + 1) = 0

<=> 2x + 7 = 0 hoặc x - 5 = 0 hoặc 5x + 1 = 0

<=> 2x = -7 hoặc x = 5 hoặc 5x = -1

<=> x = -7/2 hoặc x = 5 hoặc x = -1/5

bài 2:

a, (3x+2)(x^2-1)=(9x^2-4)(x+1)

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

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

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

b, x(x+3)(x-3)-(x-2)(x^2-2x+4)=0

x(x^2-9)-(x^3+8)=0

x^3-9x-x^3-8=0

-9x-8=0

tự tìm x nha