Câu 1:

a) Ta có: \(VT=x^4-y^4\)

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

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

\(=\left(x-y\right)\left(x^3+xy^2+x^2y+y^3\right)\)=VP(đpcm)

c) Ta có: \(VT=a\left(b+1\right)+b\left(a+1\right)\)

\(=ab+a+ab+b\)

\(=a+b+2ab\)(1)

Thay ab=1 vào biểu thức (1), ta được:

a+b+2(*)

Ta có: VP=(a+1)(b+1)=ab+a+b+1(2)

Thay ab=1 vào biểu thức (2), ta được:

1+a+b+1=a+b+2(**)

Từ (*) và (**) ta được VT=VP(đpcm)

Câu 2:

Ta có: \(\left(x-3\right)\left(x+x^2\right)+2\left(x-5\right)\left(x+1\right)-x^3=12\)

\(\Leftrightarrow x^2+x^3-3x-3x^2+2\left(x^2+x-5x-5\right)-x^3=12\)

\(\Leftrightarrow x^3-2x^2-3x+2x^2-8x-10-x^3-12=0\)

\(\Leftrightarrow-11x-22=0\)

\(\Leftrightarrow-11x=22\)

hay x=-2

Vậy: x=-2

\(\dfrac{1}{5}\)

b) Ta có:

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

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

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

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

Để f(x) \(⋮\) x² + 2 thì x + 2 \(⋮\) x² + 2

Đến đây tự làm

câu a sai đề k bn

\(A=\dfrac{1}{2}x^2\cdot y\cdot4x^2y^4+3x^2y^3\cdot x^2y^2\)

\(=2x^4y^5+3x^4y^5=5x^4y^5\)

Ta có: \(\left(x-2\right)^{18}+\left|y+1\right|=0\)

=>x-2=0 và y+1=0

=>x=2 và y=-1

\(A=5\cdot2^4\cdot\left(-1\right)^5=-80\)

Ai giúp với =)

Uôn :))

a) \(\left(x-3\right)^2+2\left(x-3\right)\left(x+2\right)+\left(x+2\right)^2\)

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

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

Hằng đẳng thức: \(\left(a+b\right)^2=a^2+2ab+b^2\).

b) \(\left(x+5\right)^2-\left(2x+10\right)\left(x-6\right)+\left(x-6\right)^2\)

\(=\left(x+5\right)^2-2\left(x+5\right)\left(x-6\right)+\left(x-6\right)^2\)

\(=\left[\left(x+5\right)-\left(x-6\right)\right]^2\)

\(=11^2=121\)

Hằng đẳng thức: \(\left(a-b\right)^2=a^2-2ab+b^2\).

a.\(\left(x-3\right)^2+2\left(x-3\right)\left(x+2\right)+\left(x+2\right)^2\)

\(=\left[\left(x-3\right)+\left(x+2\right)\right]^2\)

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

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

b.\(\left(x+5\right)^2-\left(2x+10\right)\left(x-6\right)+\left(x-6\right)^2\)

\(=\left(x+5\right)^2-2\left(x+5\right)\left(x-6\right)+\left(x-6\right)^2\)

\(=\left[\left(x+5\right)-\left(x-6\right)\right]^2\)

\(=\left(x+5-x+6\right)^2\)

Trả lời:

a, ( x - 3 )² + 2 ( x - 3 ) ( x + 2 ) + ( x + 2 )² 

= ( x - 3 + x + 2 )²

= ( 2x - 1 )²

b, ( x + 5 )² - ( 2x + 10 ) ( x - 6 ) + ( x - 6 )²

= ( x + 5 )² - 2 ( x + 5 ) ( x - 6 ) + ( x - 6 )²

= [ ( x + 5 ) - ( x - 6 ) ]²

= ( x + 5 - x + 6 )²

= 11²

= 121