X²=3                              x²=25     

=> X=\(\pm\sqrt{3}\)             => x=5

X²=36                           

=> x=6

2.(x-1)²+5⁰= 9

2.(x-1)²+1= 9

2.(x-1)²= 8

(x-1)² = 8/2

(x-1)² = 4 

(x-1)² = (2)²

x-1=(\(\pm\)2)

TH1: x-1= 2              TH2: x-1=-2

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

        x= 3                          x = -1

Vậy x\(\in\)\(\left\{3;1\right\}\)

a: 5x-20xy

\(=5x\cdot1-5x\cdot4y=5x\left(1-4y\right)\)

b: \(x^2-9=\left(x-3\right)\left(x+3\right)\)

c: \(x^2-2xy+y^2-z^2\)

\(=\left(x-y\right)^2-z^2\)

=(x-y-z)(x-y+z)

d: \(5x\left(x-1\right)-2\left(x-1\right)=\left(x-1\right)\left(5x-2\right)\)

e; \(x^2+4x+3=x^2+x+3x+3\)

=x(x+1)+3(x+1)

=(x+1)(x+3)

f: \(x^3-x+3x^2y+3xy^2+y^3-y\)

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

\(=\left(x+y\right)\left\lbrack\left(x+y\right)^2-1\right\rbrack\)

=(x+y)(x+y-1)(x+y+1)

g: \(x^2-x-y^2-y\)

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

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

=(x+y)(x-y-1)

h: \(16x-5x^2-3\)

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

=-5x(x-3)+(x-3)

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

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

j: \(2x^2-6x=2x\cdot x-2x\cdot3=2x\left(x-3\right)\)

k: \(x^3-3x^2-4x+12\)

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

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

l: \(x^2-y^2-5x+5y\)

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

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

Câu 1

Do x = 2 là nghiệm của A(x)

⇒⇒A(2) = 0

2.2² + a.2 + b = 0

8 + 2a + b = 0

b = -8 - 2a (1)

Do x = 3 là nghiệm của A(x)

⇒ A(3) = 0

2.3² + a.3 + b = 0

18 + 3a + b = 0 (2)

Thay (1) vào (2) ta được:

18 + 3a + (-8 - 2a) = 0

18 + 3a - 8 - 2a = 0

a + 10 = 0

a = -10

Thay a = -10 vào (1) ta được:

b = -8 - 2.(-10)

= 12

Vậy a = -10; b = 12

Đặt \(A\left(x\right)=0\Rightarrow2x^2+ax+b=0\) \(\left(1\right)\)

Thay \(x=2\) vào \(\left(1\right)\Rightarrow2.2^2+2a+b=0\)

\(\Rightarrow2a+b=-8\left(2\right)\)

Thay \(x=3\) vào \(\left(1\right)\Rightarrow2.3^2+3a+b=0\)

\(\Rightarrow3a+b=-18\left(3\right)\)

Từ \(\left(2\right),\left(3\right)\Rightarrow\left\{{}\begin{matrix}a=-10\\b=12\end{matrix}\right.\)

Vậy \(a=-10,b=12\)

Bài 5:

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

=>3x+10-2x=0

=>x+10=0

=>x=-10

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

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

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

=>\(\left[\begin{array}{l}x+1=0\\ x+3=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=-1\\ x=-3\end{array}\right.\)

Bài 6:

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

\(=-6x^5+4x^4+2x^3-2x^2+2x^3-6x^2+2x\)

\(=-6x^5+4x^4+4x^3-8x^2+2x\)

a) Kết quả M = x 4 – 1.

b) Kết quả M =  x 2  – 2x – 3.

a.x^3-1^3

b.x^3-5^3

c)(2x)^3+3^3

d)x^3+1/2^3

c. Ta có h(x) = 0 ⇒ 5x + 1 = 0 ⇒ x = -1/5

Vậy nghiệm của đa thức h(x) là x = -1/5 (1 điểm)

b) \(\Leftrightarrow3x^3+12x-2x^2-8=0\\ \Leftrightarrow\left(3x^3-2x^2\right)+\left(12x-8\right)=0\\ \Leftrightarrow x^2\left(3x-2\right)+4\left(3x-2\right)=0\\ \Leftrightarrow\left(x^2+4\right)\left(3x-2\right)=0\)

Vì \(x^2+4>0\Rightarrow3x-2=0\Rightarrow x=\dfrac{2}{3}\)

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

d) \(\Leftrightarrow x^3-27+x\left(4-x^2\right)=36\\ \Leftrightarrow x^3+4x-x^3=63\\ \Leftrightarrow4x=63\\ \Leftrightarrow x=\dfrac{63}{4}\)

b) 3x(x\(^3\) +12x-2x\(^2\)-8=0

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

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

\(\Leftrightarrow\left[{}\begin{matrix}X^2+4=0\\3X-2=0\end{matrix}\right.\) <=> \(\left[{}\begin{matrix}x\in Z\\X=\dfrac{2}{3}\end{matrix}\right.\)
 

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

x(x+5)=0

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

c)(x-3)(x\(^2\)+3x+9)+x(x+2)(2-x)=36

x\(^3\)-27+x(x+2)(2-x)=36

4x-27=36

4x=36+27

4x=63

x=\(\dfrac{63}{4}\)