1. ( 3x + 2)² - 4

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

= 3x(3x+4)

2. 4x² - 25y²

= (2x-5y)(2x+5y)

3. 4x²- 49

=(2x-7)(2x+7)

4. 8z³ + 27

=(2z+3)(4x²-6z+9)

5. \(\dfrac{9}{25}x^4-\dfrac{1}{4}\)

= \((\dfrac{3}{5}x^2-\dfrac{1}{2})(\dfrac{3}{5}x^2+\dfrac{1}{2})\)

6. x³²  - 1

=(x¹⁶-1)(x¹⁶+1)

7. 4x² + 4x + 1

=(2x+1)²

8. x² - 20x + 100

=(x-10)²

9. y⁴ -14y² + 49

=(y²-7)²

10.  125x³ - 64y³

= (5x-4y)(25x²+20xy+16y²)

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

2) \(4x^2-25y^2=\left(2x-5y\right)\left(2x+5y\right)\)

3) \(4x^2-49=\left(2x-7\right)\left(2x+7\right)\)

4) \(8z^3+27=\left(2z+3\right)\left(4z^2-6z+9\right)\)

5) \(\dfrac{9}{25}x^4-\dfrac{1}{4}=\left(\dfrac{3}{5}x^2-\dfrac{1}{2}\right)\left(\dfrac{3}{5}x^2+\dfrac{1}{2}\right)\)

6) \(x^{32}-1=\left(x^{16}-1\right)\left(x^{16}+1\right)\)

\(=\left(x^8-1\right)\left(x^8+1\right)\left(x^{16}+1\right)\)

\(=\left(x-1\right)\left(x+1\right)\left(x^2+1\right)\left(x^4+1\right)\left(x^8+1\right)\left(x^{16}+1\right)\)

7) \(4x^2+4x+1=\left(2x+1\right)^2\)

8) \(x^2-20x+100=\left(x-10\right)^2\)

9) \(y^4-14y^2+49=\left(y^2-7\right)^2\)

lam kieu gi

b: 4x^2-20x+25=(x-3)^2

=>(2x-5)^2=(x-3)^2

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

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

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

=>x=8/3 hoặc x=2

c: x+x^2-x^3-x^4=0

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

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

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

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

=>\(x\in\left\{0;1;-1\right\}\)

d: 2x^3+3x^2+2x+3=0

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

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

=>2x+3=0

=>x=-3/2

a: =>x^2(5x-7)-3(5x-7)=0

=>(5x-7)(x^2-3)=0

=>\(x\in\left\{\dfrac{7}{5};\sqrt{3};-\sqrt{3}\right\}\)

\(a,4x^2-4y^2-20x+20y=4\left(x^2-y^2\right)-\left(20x-20y\right)=4\left(x-y\right)\left(x+y\right)-20\left(x-y\right)=\left(x-y\right)\left(4x+4y-20\right)=4\left(x-y\right)\left(x+y-5\right)\\ b,16x^2-25+\left(4x-5\right)=\left(4x-5\right)\left(4x+5\right)+\left(4x-5\right)=\left(4x-5\right)\left(4x+5+1\right)=\left(4x-5\right)\left(4x+6\right)=2\left(4x-5\right)\left(2x+3\right)\)

\(c,\left(x+5y\right)^3=x^3+15x^2y+75xy^2+125y^3\\ e,x^2-4x+4-y^2=\left(x-2\right)^2-y^2=\left(x-y-2\right)\left(x+y-2\right)\\ g,x^2-3x-4=\left(x^2-4x\right)+\left(x-4\right)=x\left(x-4\right)+\left(x-4\right)=\left(x+1\right)\left(x-4\right)\)

kh hiểu bn ơi

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

Đáp án cần chọn là: B

đầu bài là gì vậy? có phải là chứng minh phương trình vô nghiệm không? nếu phải thì đây là lời giải:

a) \(x^2-2x+2\)

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

\(=\left(x-1\right)^2+1\ge0\)

vậy phương trình vô nghiệm.

b) \(9x^2-6x+5\)

\(=\left(3x\right)^2-6x+1+4\)

\(=\left(3x+1\right)^2+4\ge0\)

vậy phương trình vô nghiệm.

a)

\(x^2-4\sqrt{15}x+19=0\\ < =>x^2-4\sqrt{15}x+60-41=0\\ < =>\left(x-2\sqrt{15}\right)^2-41=0\\ < =>\left(x-2\sqrt{15}-\sqrt{41}\right)\left(x-2\sqrt{15}+\sqrt{41}\right)=0\\ < =>\left[{}\begin{matrix}x-2\sqrt{15}-\sqrt{41}=0\\x-2\sqrt{15}+\sqrt{41}=0\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x=2\sqrt{15}+\sqrt{41}\\x=2\sqrt{15}-\sqrt{41}\end{matrix}\right.\)

b)

\(4x^2+4\sqrt{5}x+5=0\\ < =>\left(2x+\sqrt{5}\right)^2=0\\ < =>2x+\sqrt{5}=0\\ < =>2x=-\sqrt{5}\\ < =>-\dfrac{\sqrt{5}}{2}\)

a: Δ=(4căn 15)^2-4*1*19=164>0

Phương trình có hai nghiệm phân biệt là:

\(\left\{{}\begin{matrix}x=\dfrac{4\sqrt{5}-2\sqrt{41}}{2}=2\sqrt{5}-\sqrt{41}\\x_2=2\sqrt{5}+\sqrt{41}\end{matrix}\right.\)

b: \(\Leftrightarrow\left(2x\right)^2+2\cdot2x\cdot\sqrt{5}+5=0\)

=>(2x+căn 5)^2=0

=>2x+căn 5=0

=>x=-1/2*căn 5

1) Ta có: \(x^2-4x+4=0\)

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

\(\Leftrightarrow x-2=0\)

hay x=2

Vậy: S={2}