a: \(\Leftrightarrow\left(x-5\right)\left(x+1\right)\left(x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-1\\x=1\end{matrix}\right.\)

d: \(\Leftrightarrow\left(x+3\right)\left(x^2-4x+5\right)=0\)

\(\Leftrightarrow x+3=0\)

hay x=-3

a/ \(3x(2x-3)=5(3-2x) \Leftrightarrow 3x(2x-3)+5(2x-3)=0 \\\ \Leftrightarrow (2x-3)(3x+5)=0 \)

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

KL: .............

b/ \(\left(x^2+1\right)\left(2x+5\right)=\left(x-1\right)\left(x^2+1\right)\Leftrightarrow\left(x^2+1\right)\left(2x+5\right)-\left(x-1\right)\left(x^2+1\right)=0\)

\(\Leftrightarrow\left(x^2+1\right)\left(2x+5-x+1\right)=0\Leftrightarrow\left[{}\begin{matrix}x^2+1=0\\x+6=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x\in\varnothing\\x=-6\end{matrix}\right.\)

KL: .............

c/ \(3x^3=x^2+3x-1\Leftrightarrow3x^3-x^2-3x+1=0\Leftrightarrow x^2\left(3x-1\right)-\left(3x-1\right)=0\)

\(\Leftrightarrow\left(3x-1\right)\left(x^2-1\right)=0\Leftrightarrow\left[{}\begin{matrix}3x-1=0\\x^2-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{3}\\x=1\\x=-1\end{matrix}\right.\)

KL: ..........

d/ \(x^2-9x+20=0\Leftrightarrow x^2-5x-4x+20=0\Leftrightarrow x\left(x-5\right)-4\left(x-5\right)=0\)

\(\Leftrightarrow\left(x-5\right)\left(x-4\right)=0\Leftrightarrow\left[{}\begin{matrix}x-5=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=4\end{matrix}\right.\)

KL: .............

\(a,3x\left(2x-3\right)=5\left(3-2x\right)\Leftrightarrow6x^2-9x=15-10x\Leftrightarrow6x^2-9x-15+10x=0\Leftrightarrow6x^2+x-15=0\Leftrightarrow\left(3x+5\right)\left(2x-3\right)=0\)

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

\(\left[{}\begin{matrix}x=-\frac{5}{3}\\x=\frac{3}{2}\end{matrix}\right.\)

A= 2x^2 + y^2 - 2xy -2x+3

A= x^2-2xy + y^2 + x^2 - 2x+ 1 +2

A= (x-y)^2 + (x-1)^2 + 2

(x-y)^2> hoặc = 0 với mọi giá trị của x

(x-1)^2 > hoặc =0 với mọi giá trị của x

=> (x-y)^2 + (x-1)^2 > hoặc =0 với mọi giá trị của x

=> (x-y)^2 + (x-1)^2 + 2 > hoặc =2

=> A lớn hơn hoặc bằng 2

=> GTNN của A=2 tại x=y=1

Bài 3

a) x² + 10x + 25

= x² + 2.x.5 + 5²

= (x + 5)²

b) 8x - 16 - x²

= -(x² - 8x + 16)

= -(x² - 2.x.4 + 4²)

= -(x - 4)²

c) x³ + 3x² + 3x + 1

= x³ + 3.x².1 + 3.x.1² + 1³

= (x + 1)³

d) (x + y)² - 9x²

= (x + y)² - (3x)²

= (x + y - 3x)(x + y + 3x)

= (y - 2x)(4x + y)

e) (x + 5)² - (2x - 1)²

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

= (6 - x)(3x + 4)

Bài 4

a) x² - 9 = 0

x² = 9

x = 3 hoặc x = -3

b) (x - 4)² - 36 = 0

(x - 4 - 6)(x - 4 + 6) = 0

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

x - 10 = 0 hoặc x + 2 = 0

*) x - 10 = 0

x = 10

*) x + 2 = 0

x = -2

Vậy x = -2; x = 10

c) x² - 10x = -25

x² - 10x + 25 = 0

(x - 5)² = 0

x - 5 = 0

x = 5

d) x² + 5x + 6 = 0

x² + 2x + 3x + 6 = 0

(x² + 2x) + (3x + 6) = 0

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

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

x + 2 = 0 hoặc x + 3 = 0

*) x + 2 = 0

x = -2

*) x + 3 = 0

x = -3

Vậy x = -3; x = -2

c: \(=\dfrac{x^3+2x^2+x^2+2x-10x-20}{x+2}\)

\(=x^2+x-10\)

Chọn D.

(x − 5)(x + 3) = x(x + 3) – 5( x + 3) =  x 2  + 3x - 5x - 15 =  x 2 − 2x − 15

Bạn chú ý đăng lẻ câu hỏi! 1/

a/ \(=x^3-2x^5\)

b/\(=5x^2+5-x^3-x\)

c/ \(=x^3+3x^2-4x-2x^2-6x+8=x^3=x^2-10x+8\)

d/ \(=x^2-x^3+4x-2x+2x^2-8=3x^2-x^3+2x-8\)

e/ \(=x^4-x^2+2x^3-2x\)

f/ \(=\left(6x^2+x-2\right)\left(3-x\right)=17x^2+5x-6-6x^3\)

Giúp mk vs

a: \(=x^3-2x^5\)

e: \(=x^4+2x^3-x^2-2x\)