a, \(A=\left(100+50\right)^2=22500\)

b, \(B=\left(127+73\right)^2=40000\)

c, \(C=-6x+25\)Thay x = 100 ta có : 

\(C=-6.100+25=-600+25=-575\)

\(A=100^2+200.50+50^2\)

\(=100^2+2.100.5+50^2\)

\(=\left(100+50\right)^2=150^2\)

\(B=127^2+146.127+73^2\)

\(=127^2+2.73.127+73^2\)

\(=\left(127+73\right)^2=200^2\)

\(a,\left(x^2+2\right)\left(x^4-2x^2+4\right)=\left(x^2\right)^3+8=x^6+8\)

\(b,\left(x-\frac{1}{3}\right)\left(x^2+\frac{x}{3}+\frac{1}{9}\right)=x^3-\frac{1}{27}\)

\(c,\left(\frac{1}{2}-x\right)\left(\frac{1}{4}+\frac{1}{2}x+x^2\right)=\frac{1}{8}-x^3\)

\(d,\left(x^2+3\right)\left(x^4-3x^2+9\right)=x^6+27\)

\(e,\left(2x+1\right)\left(4x^2-2x+1\right)=8x^3+1\)

a)          \(\left(x^2+2\right)\left(x^4-2x^2+4\right)=\left(x^2\right)^3+2^3=x^8+8\)

b)           \(\left(x-\frac{1}{3}\right)\left(x^2+\frac{x}{3}+\frac{1}{9}\right)=[x^3-\left(\frac{1}{3}\right)^3]=x^3-\frac{1}{9}\)

c)          \(\left(\frac{1}{2}-x\right)\left(\frac{1}{4}+\frac{1}{2}x+x^2\right)=[\left(\frac{1}{2}\right)^3-x^3]=\frac{1}{8}-x^3\)

d)          \(\left(x^2+3\right)\left(x^4-3x^2+9\right)=\left(x^2\right)^3+3^3=x^8+27\)

e)           \(\left(2x+1\right)\left(4x^2-2x+1\right)=\left(2x\right)^3+1^3=8x^3+1\)

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

Vậy nó \(=\left(x+y-2\right)^2\)

ĐỀ BÀI BỊ THIẾU HAY SAO ZẬY ??

\(\left(x+y\right)^2-4\left(x+y\right)+4\)

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

\(=x^2+xy+xy+y^2-4x-4y+4\)

\(=2x^2+2xy+y^2-4x-4y+4\)

a) Áp dụng hằng đằng thức hiệu của 2 bình phương ta có

\(x^2-7=x^2-\left(\sqrt{7}\right)^2=\left(x-\sqrt{7}\right)\left(x+\sqrt{7}\right)\)

                                                       Bài giải

\(a,\text{ }x^2-7=x^2-\left(\sqrt{7}\right)^2=\left(x-\sqrt{7}\right)\left(x+\sqrt{7}\right)\)

\(b,\text{ }4x-5=\left(2x\right)^2-\left(\sqrt{5}\right)^2=\left(2x-\sqrt{5}\right)\left(2x+\sqrt{5}\right)\)

A, x(x+y)-(2x+2y)

A=x(x+y)-2(x+y)

A=(x+y)(x-2)

B, 5x(x-2y)+2(y-x)

Đề câu này thấy sai sai á bạn

Check đề hộ mình nha.

                                             Bài giải

\(a,\text{ }x\left(x+y\right)-\left(2x+2y\right)=x\left(x+y\right)-2\left(x+y\right)=\left(x-2\right)\left(x+y\right)\)

\(b,\text{ }5x\left(x-2y\right)+2\left(y-x\right)\)

Đề sai ///

\(\left(\sqrt{2x}-y\right)^2=\left(\sqrt{2x}\right)^2-2\cdot\sqrt{2x}\cdot y+y^2=2x-2\sqrt{2x}\cdot y+y^2\)

\(\left(\sqrt{2x}+\sqrt{8y}\right)^2=\left(\sqrt{2x}\right)^2+2\left(\sqrt{2x}\right)\left(\sqrt{8y}\right)+\left(\sqrt{8y}\right)^2=2x+2\sqrt{16xy}+8y\)

Không chắc nha :)

\8=====D

Giả sử:Đẳng thức trên là đúng

Ta có:\(a^3+b^3+c^3+d^3=\left(a+b\right)\left(a^2-ab+b^2\right)+\left(c+d\right)\left(c^2-cd+d^2\right)\)

\(=\left(c+d\right)\left(c^2-cd+d^2-a^2+ab-b^2\right)=3\left(ab-cd\right)\left(c+d\right)\)

\(\Rightarrow c^2-cd+d^2-a^2+ab-b^2=3\left(ab-cd\right)\)

\(\Rightarrow c^2-cd+d^2-a^2+ab-b^2-3ab+3cd=0\)

\(\Rightarrow c^2+2cd+d^2-a^2-2ab-b^2=0\)

\(\Rightarrow\left(c+d\right)^2-\left(a+b\right)^2=0\)

\(\Rightarrow\left(a+b+c+d\right)\left(c+d-a-b\right)=0\)(Luôn đúng)

Vậy điều giả sử trên là đúng

Suy ra điều phải chứng minh