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

\(x^2-10x+25+4x^2+12x+9-5x^2-10x+3x+6=0\)

\(-5x+40=0\)

\(-5\left(x-8\right)=0\)

\(x-8=0\)

\(x=8\)

a) x^2+4x+3=x^2+x+3x+3=x(x+1)+3(x+1)=(x+1)(x+3)

b) 4x^2+4x-3=4x^2+4x+1-4=(2x+1)^2-4=(2x+1-2)(2x+1+2)=(2x-1)(2x+3)

c) x^2-x-12=x^2-4x+3x-12=x(x-4)+3(x-4)=(x-4)(x+3)

d) 4x^4+4x^2y^2-8y^4=4(x^4+x^2y^2-2y^4)=4(x^4-x^2y^2+2x^2y^2-2y^4)=4(x^2-y^2)(x^2+2y^2)=4(x-y)(x+y)(x^2+2y^2)

a) \(x^2+4x+3\)

\(=x^2+x+3x+3\)

\(=\left(x^2+x\right)+\left(3x+3\right)\)

\(=x\left(x+1\right)+3\left(x+1\right)\)

\(=\left(x+1\right)\left(x+3\right)\)

c) \(x^2-x-12\)

\(=x^2-4x+3x-12\)

\(=\left(x^2-4x\right)+\left(3x-12\right)\)

\(=x\left(x-4\right)+3\left(x-4\right)\)

\(=\left(x-4\right)\left(x+3\right)\)

\(a+\frac{1}{b}=1\)\(\Leftrightarrow\left(a+\frac{1}{b}\right)^2=1\)\(\Leftrightarrow a^2+\frac{1}{b^2}+\frac{2a}{b}=1\)\(\Leftrightarrow\frac{a}{b}=-1\)

\(a^2+\frac{1}{b^2}=3\)\(\Leftrightarrow\left(a^2+\frac{1}{b^2}\right)^2=9\)\(\Leftrightarrow a^4+\frac{1}{b^4}+\frac{2.a^2}{b^2}=9\)\(\Leftrightarrow a^4+\frac{1}{b^4}=7\)

\(N=\frac{a^4b^4+a^2b^2+1}{b^4}=a^4+\frac{a^2}{b^2}+\frac{1}{b^4}\)

\(\text{Thanks you verry much !!}\)

\(3,\)Nhẩm nghiệm của đa thức trên ta đc : -1

Ta có lược đồ sau :

Phân tích thành nhân tử ta có :\(\left(x+1\right)\left(x^2-4\right)\)

\(a+b=2\Rightarrow\left(a+b\right)^2=4\)

\(\Rightarrow a^2+2ab+b^2=4\)

\(16+2ab=4\)

\(ab=-6\)

\(a^3+b^3=\left(a+b\right)\left(a^2+ab+b^2\right)\)

\(=2\left(16+-6\right)\)

\(=20\)

a + b = 2 => ( a + b) ² = 4

=> a² + 2ab +b² = 4

16 + 2ab = 4

 ab = -6

a³ + b³ = ( a + b).(a² + ai + b²)

= 2 [ 16 + (-6)]

= 20

Có :\(VT=a^3+b^3=\left(a+b\right)\left(a^2+ab+b^2\right)\)

\(=\left(a+b\right)\left(a^2+2ab+b^2-ab\right)\)

\(=\left(a+b\right)^3-ab\left(a+b\right)=VP\)

\(\RightarrowĐPCM\)