Question

giải giúp em với ạ 

Answer 1

16*:

a) Ta có:
\(VP=x^4-y^4\\ =\left(x^2-y^2\right)\left(x^2+y^2\right)\\ =\left(x-y\right)\left(x+y\right)\left(x^2+y^2\right)\\ =\left(x-y\right)\left[x\left(x^2+y^2\right)+y\left(x^2+y^2\right)\right]\\ =\left(x-y\right)\left(x^3+xy^2+x^2y+y^3\right)=VT\) 

b) Ta có: 

\(VP=x^5+y^5\\ =\left(x^5+x^4y\right)+\left(xy^4+y^5\right)-\left(x^4y+xy^4\right)\\ =x^4\left(x+y\right)+y^4\left(x+y\right)-xy\left(x^3+y^3\right)\\ =\left(x+y\right)\left(x^4+y^4\right)-xy\left(x+y\right)\left(x^2-xy+y^2\right)\\ =\left(x+y\right)\left[\left(x^4+y^4\right)-xy\left(x^2-xy+y^2\right)\right]\\ =\left(x+y\right)\left(x^4-x^3y+x^2y^2-xy^3+y^4\right)=VT\)

Answer 2

14:

a: \(4x\left(x-5\right)-\left(x-1\right)\left(4x-3\right)=5\)

=>\(4x^2-20x-\left(4x^2-7x+3\right)=5\)

=>\(4x^2-20x-4x^2+7x-3=5\)

=>-13x=8

=>\(x=-\dfrac{8}{13}\)

b: \(\left(x+3\right)\left(4-x\right)+\left(x-1\right)\left(x+1\right)=10\)

=>\(4x-x^2+12-3x+x^2-1=10\)

=>x+11=10

=>x=-1

c: \(-4x^2\left(x-7\right)+4x\left(x^2-5\right)=28x^2-13\)

=>\(28x^2-13=-4x^3+28x^2+4x^3-20x=28x^2-20x\)

=>20x=13

=>\(x=\dfrac{13}{20}\)

15:

a: \(A=\left(x-5\right)\left(2x+3\right)-2x\left(x-3\right)+x+7\)

\(=2x^2+3x-10x-15-2x^2+6x+x+7\)

=-8

b: \(B=4\left(y-6\right)-y^2\left(3y+2\right)+y\left(5y-4\right)+3y^2\left(y-1\right)\)

\(=4y-24-3y^3-2y^2+5y^2-4y+3y^3-3y^2\)

=-24