a: =>(x-5)(x+5)+(x-5)(3x-15)=0

=>(x-5)(x+5+3x-15)=0

=>(x-5)(4x-10)=0

=>x=5 hoặc x=5/2

c: =>x^3-3x^2+2x^2-6x-8x+24=0

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

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

=>\(x\in\left\{3;-4;2\right\}\)

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

Th1 : \(3x-1=0=>x=\frac{1}{3}\)

Th2 : \(2x-3=0=>x=\frac{3}{2}\)

TH3 : \(x+5=0=>x=-5\)

Mik tl mà chẳng có ai T kì quá z

(3x - 1)(2x - 3)(2x - 3)(x + 5) = 0

=> (3x - 1)(2x - 3)²(x + 5) = 0

=> 3x - 1 = 0 => x = 1/3

hoặc 2x - 3 = 0 => x = 3/2

hoặc x + 5 = 0 => x = -5

                      Vậy x = 1/3 , x = 3/2 , x = -5

c.ơn mn ạ

a: =>6x-3x^2-5=4-3x^2-2

=>6x-5=2

=>6x=7

=>x=7/6

b: =>20x+5-12x^2-3x=6x^2-10x+3x-5

=>-12x^2+17x+5-6x^2+7x+5=0

=>-18x^2+24x+10=0

=>x=5/3 hoặc x=-1/3

mới lớp 1 thôi

Làm tử tế giúp mình đi,,,,,,,

mk ghi kết quả, ko phân tích đc thì ib

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

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

3) \(3x^2-6xy+3y^2-12z^2=3\left(x-y-2z\right)\left(x-y+2z\right)\)

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

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

6)  \(2x^2-18=2\left(x-3\right)\left(x+3\right)\)

7) \(x^2-7xy+10y^2=\left(x-2y\right)\left(x-5y\right)\)

8) \(x^3-2x^2+x-xy^2=x\left(x-y-1\right)\left(x+y-1\right)\)

a) Xét \(x\le\frac{3}{2}\) ta có : \(\left(3-2x\right)-x=\left(2-x\right)\)

\(\Leftrightarrow3-3x=2-x\Leftrightarrow-2x=-1\Rightarrow x=-\frac{1}{2}\left(TM\right)\)

Xét \(\frac{3}{2}\le x\le2\) ta có : \(\left(2x-3\right)-x=2-x\)

\(\Leftrightarrow x-3=2-x\Leftrightarrow2x=5\Rightarrow x=\frac{5}{2}\left(l\right)\)

Xét \(x\ge2\) ta có : \(\left(2x-3\right)-x=x-2\)

\(\Leftrightarrow x-3=x-2\Rightarrow-3=-2\left(l\right)\)

Vậy \(x=-\frac{5}{2}\)

b) \(VT=\left|x+3\right|+\left|x+1\right|\ge0\forall x\) nên \(VP=3x\ge0\Rightarrow x\ge0\)

\(\Rightarrow\left|x+3\right|+\left|x+1\right|=x+3+x+1=2x+4\)

Ta có \(2x+4=3x\Rightarrow x=4\)

Vậy \(x=4\)

Mơn nha, Đinh Đức Hùng

a) Ta có: \(3x^2\cdot\left(2x^3-x+5\right)\)

\(=6x^5-3x^3+15x^2\)

b) Ta có: \(\left(4xy+3y-5\right)\cdot x^2y\)

\(=4x^3y^2+3x^2y^2-5x^2y\)

c) Ta có: \(\left(3x-2\right)\left(4x+5\right)-6x\left(2x-1\right)\)

\(=12x^2+15x-8x-10-12x^2+6x\)

\(=13x-10\)

d) Ta có: \(\left(3x-5\right)\left(x^2-5x+7\right)\)

\(=3x^3-15x^2+21x-5x^2+25x-35\)

\(=3x^3-20x^2+46x-35\)

\(\text{x}^2+y^2-\text{x}+4y+5=\left(\text{x}^2-\text{x}+\frac{1}{4}\right)+\left(y^2+4y+4\right)+\frac{3}{4}=\left(\text{x}-\frac{1}{2}\right)^2+\left(y+2\right)^2+\frac{3}{4}\) 

\(\ge0+0+\frac{3}{4}=\frac{3}{4}\).Dâu"=" xayr ra khi: 

\(\Leftrightarrow\hept{\begin{cases}\text{x}-\frac{1}{2}=0\\y+2=0\end{cases}}\Leftrightarrow\hept{\begin{cases}\text{x}=\frac{1}{2}\\y=-2\end{cases}}\)