A) \(...=\left(7y-3\right)^3\)

B) \(...=\left(4y-3\right)^3\)

C) \(...=x^4+2x^2+1-\left(y^2+2y+1\right)\)

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

D) \(...=x^2-6x+9-\left(y^2-10y+25\right)\)

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

cậu có thể giải chi tiết giúp tớ dc ko

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

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

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

phân tích ra nhân tử ak

thế thì la     : X^2-12X+36= : X^2-12X+6^2=(X-6)^2

a: Ta có: \(\left(x+3\right)\left(x+4\right)\left(x+5\right)\left(x+6\right)+1\)

\(=\left(x^2+9x+18\right)\left(x^2+9x+20\right)+1\)

\(=\left(x^2+9x\right)^2+38\left(x^2+9x\right)+360+1\)

\(=\left(x^2+9x\right)^2+2\cdot\left(x^2+9x\right)\cdot19+19^2\)

\(=\left(x^2+9x+19\right)^2\)

b. \(x^2+y^2+2x+2y+2\left(x+1\right)\left(y+1\right)+2\)

\(=\left(x^2+2x+1\right)+2\left(x+1\right)\left(y+1\right)+\left(y^2+2y+1\right)\)

\(=\left(x+1\right)^2+2\left(x+1\right)\left(y+1\right)+\left(y+1\right)^2\)

\(=\left(x+1+y+1\right)^2=\left(x+y+2\right)^2\)

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

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

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

d. \(x^2+2x\left(y+1\right)+y^2+2y+1\)

\(=x^2+2x\left(y+1\right)+\left(y+1\right)^2\)

\(=\left(x+y+1\right)^2\)

Bỏ qua bước gọi,ta có:

\(\frac{a}{b}-\frac{c}{d}=\frac{a^2}{b^2}-\frac{c^2}{d^2}\)

=>\(\frac{a}{b}-\frac{c}{d}=\frac{a}{b}\cdot\frac{a}{b}-\frac{c}{d}\cdot\frac{c}{d}\)

=>\(-\frac{c}{d}+\frac{a}{b}=\frac{a}{b}\cdot\frac{a}{b}-\frac{c}{d}\cdot\frac{c}{d}\)

=>\(0=\frac{a}{b}\cdot\frac{a}{b}-\frac{c}{d}\cdot\frac{c}{d}-\frac{a}{b}+\frac{c}{d}\)

=>\(0=\frac{a}{b}\cdot\left(\frac{a}{b}-1\right)-\frac{c}{d}\cdot\left(\frac{c}{d}-1\right)\)

=>\(\frac{a}{b}\cdot\left(\frac{a}{b}-1\right)=\frac{c}{d}\cdot\left(\frac{c}{d}-1\right)\)

Vì a/b hơn a/b -1 là 1 đơn vị và c/d cũng hơn c/d -1 là 1 đơn vị.

=>Để thỏa mãn đề bài thì a/b = c/d .

Vậy........

Mk còn sót chỗ nào m.n góp ý nha.