Bài 3 : 

\(D=40^2-28^2+32^2+80.32=40^2+2.40.32+32^2-28^2\)

\(=\left(40+32\right)^2-28^2=72^2-28^2=\left(72+28\right)\left(72-28\right)=46.100=4600\)

4600 nha

Bài 4 : 

\(B=\left(9+1\right)\left(9^2+1\right)...\left(9^{32}+1\right)\)

\(8B=\left(9^2-1\right)\left(9^2+1\right)...\left(9^{32}+1\right)=9^{64}-1\)

\(\Rightarrow B=\frac{9^{64}-1}{8}< 9^{64}-1\Rightarrow B< C\)

bài 5mn giúp em đc k ạ

a, \(A=x\left(3x+1\right)+3x+1=\left(x+1\right)\left(3x+1\right)\)

Thay x = 33 ta được : \(32.100=3200\)

b, \(B=xy+2x+2y+4=x\left(y+2\right)+2\left(y+2\right)=\left(x+2\right)\left(y+2\right)\)

Thay x = 98 ; y = 98 ta được : \(100.100=10000\)

10000 nha

Trả lời :

=> VD : \(\frac{2x+3}{3x^4-x^2+7}\)

\(\frac{2x+3}{3x^4-x^2-7}\)

=> Là phân thức đại số 

* Nguồn : Vietjack *

\(ab\left(x^2+y^2\right)+xy\left(a^2+b^2\right)=abx^2+aby^2+a^2xy+b^2xy\)

\(=ax\left(bx+ay\right)+by\left(ay+bx\right)=\left(ax+by\right)\left(ay+bx\right)\)

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

\(=\left(a+b+c\right)^3-a^3-b^3-c^3\)(\(a=-x+y+z,b=x-y+z,c=x+y-z\))

\(=\left(b+c\right)^3+3a\left(a+b+c\right)\left(b+c\right)-\left[\left(b+c\right)^3-3bc\left(b+c\right)\right]\)

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

\(=3\left(a+b\right)\left(b+c\right)\left(c+a\right)\)

\(=24xyz\)

của bạn đây ạ

help me ctv

help me

a, \(VP=\frac{35xy^2}{65y^2}=\frac{5y^2.7y}{5y^2.13}=\frac{7y}{13}=VT\)( đpcm )

b, \(VP=\frac{x^2-6x-7}{x+1}=\frac{\left(x-7\right)\left(x+1\right)}{x+1}=x-7\)

\(=\frac{\left(x-7\right)\left(x-1\right)}{x-1}=\frac{x^2-8x+7}{x-1}=VT\)( đpcm )