a: \(5x^2\left(3x^3-2x^2+x+2\right)\)

\(=15x^5-10x^4+5x^3+10x^2\)

b: \(3x^4\left(-2x^3+5x^2-\dfrac{2}{3}x+\dfrac{1}{3}\right)\)

\(=-6x^7+15x^6-2x^5+x^4\)

1: \(=x^2+1\)

3: \(=\left(x-y-z\right)^2\)

1)  (2x + 1)(3x – 2) = (5x – 8)(2x + 1)

⇔ (2x + 1)(3x – 2) – (5x – 8)(2x + 1) = 0

⇔ (2x + 1).[(3x – 2) – (5x – 8)] = 0

⇔ (2x + 1).(3x – 2 – 5x + 8) = 0

⇔ (2x + 1)(6 – 2x) = 0

⇔\(\left[{}\begin{matrix}2x+1=0\\6-2x=0\end{matrix}\right.\) ⇔ \(\left[{}\begin{matrix}x=\dfrac{-1}{2}\\x=3\end{matrix}\right.\)

Vậy.....

2)  4x² -1 = (2x + 1)(3x - 5)

⇔ (2x-1)(2x+1)-(2x+1)(3x-5)=0

⇔ (2x+1)(2x-1-3x+5)=0

⇔ (2x+1)(4-x)=0

⇔ \(\left[{}\begin{matrix}2x+1=0\\4-x=0\end{matrix}\right.\) ⇔ \(\left[{}\begin{matrix}x=\dfrac{-1}{2}\\x=4\end{matrix}\right.\)

Vậy...

3)  

(x + 1)² = 4(x² – 2x + 1)

⇔ (x + 1)² - 4(x² – 2x + 1) = 0

⇔ x² + 2x +1- 4x² + 8x – 4 = 0

⇔ - 3x² + 10x – 3 = 0

⇔ (- 3x² + 9x) + (x – 3) = 0

⇔ -3x (x – 3)+ ( x- 3) = 0

⇔ ( x- 3) ( - 3x + 1) = 0

⇔\(\left[{}\begin{matrix}x-3=0\\-3x+1=0\end{matrix}\right.\) ⇔\(\left[{}\begin{matrix}x=3\\x=\dfrac{1}{3}\end{matrix}\right.\)

Vậy......

4) 2x³+5x²-3x=0

⇒2x³-x²+6x²-3x=0

⇒(2x³-x²)+(6x²-3x)=0

⇒x²(2x-1)+3x(2x-1)=0

⇒(x²+3x)(2x-1)=0

⇒ hoặc x²+3x=0⇒x(x+3)=0⇒hoặc x=0 hoặc x=-3

hoặc 2x-1=0⇒x=0,5

Vậy ...

5)2x=3x-2

⇒2x-3x=-2

⇒-x=-2

⇒x=2

6) x+15=3x-1

⇒x-3x=-1-15

⇒-2x=-16

⇒x=8

7)2-x=0,5x-4

⇒-x-0,5x=-4-2

⇒-1,5x=-6

⇒x=4

1: Sửa đề: 3x-5

\(=\dfrac{-x^2\left(3x-5\right)-3\left(3x-5\right)}{3x-5}=-x^2-3\)

2: \(=\dfrac{5x^4-5x^3+14x^3-14x^2+12x^2-12x+8x-8}{x-1}\)

=5x^2+14x^2+12x+8

3: \(=\dfrac{5x^3+10x^2+4x^2+8x+4x+8}{x+2}=5x^2+4x+4\)

4: \(=\dfrac{\left(x^2-1\right)\left(x^2+1\right)-2x\left(x^2-1\right)}{x^2-1}=x^2+1-2x\)

5: \(=\dfrac{x^2\left(5-3x\right)+3\left(5-3x\right)}{5-3x}=x^2+3\)

\(1,\)\(\left(2x+3\right)^2=4x^2+12x+9\)

\(2,\)\(\left(3x+2y\right)^2=9x^2+12xy+4x^2\)

\(3,\)\(\left(3a-1\right)^2=9x^2-6x+1\)

\(4,\)\(\left(a-2\right)^2=a^2-4a+4\)

\(5,\)\(\left(1-5a\right)^2=1-10a+25a^2\)

\(6,\)\(\left(x-4\right)^3=x^3-12a^2+48a-64.\)

\(7,\)\(\left(x^2-2y\right)^2=x^4-4x^2y-4y^2\)

\(8,\)\(\left(5x^2-2\right)\left(5x^2+2\right)=25x^4-4\)

\(9,\)\(\left(2a^2-7\right)\left(2a^2+7\right)=4a^4-49\)

\(10,\)\(\left(x-1\right)\left(x^2+x+1\right)=x^3-1\)

\(11,\)\(\left(x^3-2\right)\left(x^6+2x^3+4\right)=x^9-8\)

\(12,\)\(\left(3x+2\right)\left(9x^2-6x+4\right)=27x^3+8\)

\(13,\)\(\left(x^2+3\right)\left(x^4-3x^2+9\right)=x^6+27\)

1, ( 2x + 3 )² = 4x² + 12x + 9

2, ( 3x + 2y )² = 9x² +12xy + 4y²

3 ( 3a - 1 )² = 9a² - 6x + 1

4, ( a - 2 )² = a² - 4a + 4

5, ( 1 - 5a )² = 1 - 10a + 25a²

6,  ( x- 4 )³ = x³ - 12x² + 48x - 64

7, ( x² - 2y )² = x⁴ - 4x²y + 4y²

8, ( 5X² - 2 ).( 5X² + 2 ) = 25X² - 4

9, ( 2a² - 7 ).( 2a² + 7 ) = 4a⁴ - 49

10, ( x - 1 ).( x² + x + 1 ) = x³ - 1

các bạn ơi đừng làm tắt được kg nhờ các bạn đó

\(\left(1-x\right)\left(5x+3\right)=\left(3x-7\right)\left(x-1\right)\)

\(< =>\left(1-x\right)\left(5x+3+3x-7\right)=0\)

\(< =>\left(1-x\right)\left(8x-4\right)=0\)

\(< =>\orbr{\begin{cases}1-x=0\\8x-4=0\end{cases}< =>\orbr{\begin{cases}x=1\\x=\frac{1}{2}\end{cases}}}\)

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

\(< =>\left(x-2\right)\left(x+1\right)=\left(x-2\right)\left(x+2\right)\)

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

\(< =>-1\left(x-2\right)=0\)

\(< =>2-x=0< =>x=2\)

\(2x^3+3x^2-32x=48\)

\(< =>x^2\left(2x+3\right)-16\left(2x+3\right)=0\)

\(< =>\left(x^2-16\right)\left(2x+3\right)=0\)

\(< =>\left(x-4\right)\left(x+4\right)\left(2x+3\right)=0\)

\(< =>\hept{\begin{cases}x=4\\x=-4\\x=-\frac{3}{2}\end{cases}}\)

_a)x=x

_b)x=x^1

a: \(=15x^3-6x^2-3x\)

e: \(=x^2-12x+35\)