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

\(=a^4-2a^2b^2+b^4+4a^2b^2\)

\(=a^4+2a^2b^2+b^4=\left(a^2+b^2\right)^2\)

b: \(\left(ac+bd\right)^2+\left(ad-bc\right)^2\)

\(=a^2c^2+b^2d^2+a^2d^2+b^2c^2\)

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

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

c: \(\left(ax+b\right)^2+\left(a-bx\right)^2+c^2x^2\)

\(=a^2x^2+b^2+a^2+b^2x^2+c^2x^2\)

\(=a^2\left(x^2+1\right)+b^2\left(x^2+1\right)+c^2x^2\)

\(=\left(x^2+1\right)\left(a^2+b^2\right)+c^2x^2\)

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

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

\(\Rightarrowđpcm\)

b)\(a^4+b^4=\left(a^2+b^2\right)^2-2a^2b^2\)

\(VP=a^4+b^4+2a^2b^2-2a^2b^2=a^4+b^4=VT\)\(\Rightarrowđpcm\)

c) ​\(a^6+b^6=\left(a^2+b^2\right)\left[\left(a^2+b^2\right)^2-3a^2b^2\right]\)

\(VP=\left(a^2+b^2\right)\left(a^4-a^2b^2+b^4\right)=a^6+b^6\)

\(VP=VT\Rightarrowđpcm\)

d)\(a^6-b^6=\left(a^2-b^2\right)[\left(a^2+b^2\right)^2-a^2b^2]\)

\(VP=\left(a^2-b^2\right)\left(a^4+a^2b^2+b^4\right)=a^6-b^6=VT\)

\(VP=VT\Rightarrowđpcm\)

a) Ta có : \(a^2+1=a^2+ab+bc+ac=a\left(a+b\right)+c\left(a+b\right)=\left(a+b\right)\left(a+c\right)\)

Tương tự : \(b^2+1=\left(b+a\right)\left(b+c\right)\) ; \(c^2+1=\left(c+a\right)\left(c+b\right)\)

Suy ra \(\left(1+a^2\right)\left(1+b^2\right)\left(1+c^2\right)=\left(a+b\right)^2\left(b+c\right)^2\left(c+a\right)^2\)

Vậy \(A=\frac{\left(a+b\right)^2\left(b+c\right)^2\left(c+a\right)^2}{\left(a+b\right)^2\left(b+c\right)^2\left(c+a\right)^2}=1\)

b) Ta có ; \(a^2+2bc-1=a^2+2bc-\left(ab+bc+ac\right)=a^2-ab+bc-ac=a\left(a-b\right)-c\left(a-b\right)\)

\(=\left(a-b\right)\left(a-c\right)\)

Tương tự : \(b^2+2ac-1=\left(a-b\right)\left(c-b\right)\) ; \(c^2+2ab-1=\left(a-c\right)\left(b-c\right)\)

Suy ra \(\left(a^2+2bc-1\right)\left(b^2+2ac-1\right)\left(c^2+2ab-1\right)=\left(a-b\right)^2.\left(c-a\right)^2.\left[-\left(b-c\right)^2\right]\)

Vậy : \(B=\frac{-\left(a-b\right)^2\left(b-c\right)^2\left(c-a\right)^2}{\left(a-b\right)^2\left(b-c\right)^2\left(c-a\right)}=-1\)

\(a,b)\)Ta có: \(\left(a\pm b\right)^2\)

\(=\left(a\pm b\right)\left(a\pm b\right)\)

\(=a^2\pm ab\pm ab+b^2\)

\(=a^2\pm ab+b^2\)

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

2ab*

a) thay 1=ab+bc+ca vào mẫu và phân tích thành nhân tử .

tính ra 1

b)cũng thay vào tử và cũng tính ra 1

cách làm ra lun ớ

Tìm x nguyên thỏa mãn$x^2\left(x^2-1\right)\left(x^2-5\right)\left(x^2-10\right)<0$x2(x2−1)(x2−5)(x2−10)<0và $\left|x\right|<5$|x|<5Bài này của lớp 6 nhưng lập bảng xét dấu

xin lỗi em mới học lớp 5 

nên ko làm đựơc 

nếu ai cũng vậy thì k cho nhé