:V \(\left(a+b\right)^{100}\)cái này thì tính siêu to khổng lổ à

\(\left(a+b\right)^{100}\)làm gì có cái bài nào nó áp dụng cái hằng đẳng thức này à

vẫn là 66 bạn ơi vì 66 hs trả lời câu a 54 hs trả lời câu b và 18 hs trả lời được câu a,b cho nên 66 hs trả lời được câu a

\(a,a^2y^2+b^2x^2-2abxy\\ =\left(ay\right)^2-2abxy+\left(bx\right)^2\\ =\left(ay-bx\right)^2=\left(bx-ay\right)^2\\ ---\\ b,100-\left(3x-y\right)^2\\ =10^2-\left(3x-y\right)^2\\ =\left(10-3x+y\right)\left(10+3x-y\right)\)

a) \(=\left(ay\right)^2-2abxy+\left(bx\right)^2\)

\(=\left(ay-bx\right)^2\)

b) \(100-\left(3x-y\right)^2\)

\(=10^2-\left(3x-y\right)^2\)

\(=\left(10-3x+y\right)\left(10+3x-y\right)\)

(a+b)³-(a-b)³=a³+3a²b+3ab²+b³-(a³-3a²b+3ab²-b³)

                    =a³+3a²b+3ab²+b³-a³+3a²b-3ab²+b³

                        =6a²b+2b³

Áp dụng hđt a³-b³=(a-b)(a²+ab+b²) ấy

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

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

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

\(7,=\left(\sqrt{x}\right)^2+2\cdot2\sqrt{x}+2^2=\left(\sqrt{x}+2\right)^2\\ 8,=\left(\sqrt{x}\right)^2-2\cdot3\sqrt{x}+3^2=x-6\sqrt{x}+9\\ 9,=\sqrt{x^3}+\sqrt{y^3}=\left(\sqrt{x}+\sqrt{y}\right)\left(x-\sqrt{xy}+y\right)\\ 10,=\sqrt{x^3}-\sqrt{y^3}=\left(\sqrt{x}-\sqrt{y}\right)\left(x+\sqrt{xy}+y\right)\\ 11,=\sqrt{x^3}+1^3=\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)\\ 12,=\sqrt{x^3}-2^3=\left(\sqrt{x}-2\right)\left(x+2\sqrt{x}+4\right)\)

7: \(x+4\sqrt{x}+4=\left(\sqrt{x}+2\right)^2\)

8: \(\left(\sqrt{x}-3\right)^2=x-6\sqrt{x}+9\)

9: \(x\sqrt{x}+y\sqrt{y}=\left(\sqrt{x}+\sqrt{y}\right)\left(x-\sqrt{xy}+y\right)\)

7) \(x+4\sqrt{x}+4=\left(\sqrt{x}\right)^2+2\sqrt{x}.2+2^2=\left(\sqrt{x}+2\right)^2\)

8) \(\left(\sqrt{x}-3\right)^2=\left(\sqrt{x}\right)^2-2.\sqrt{x}.3+3^2=x-6\sqrt{x}+9\)

9) \(x\sqrt{x}+y\sqrt{y}=\sqrt{x^3}+\sqrt{y^3}=\left(\sqrt{x}+\sqrt{y}\right)\left(x-\sqrt{xy}+y\right)\)

10) \(x\sqrt{x}-y\sqrt{y}=\sqrt{x^3}-\sqrt{y^3}=\left(\sqrt{x}-\sqrt{y}\right)\left(x+\sqrt{xy}+y\right)\)

11) \(x\sqrt{x}+1=\sqrt{x^3}+1^3=\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)\)

12) \(x\sqrt{x}-8=\sqrt{x^3}-2^3=\left(\sqrt{x}-2\right)\left(x+2\sqrt{x}+4\right)\)

7. x + \(4\sqrt{x}+4\)

= \(\left(\sqrt{x}\right)^2+2.2.\sqrt{x}+2^2\)

= \(\left(\sqrt{x}+2\right)^2\)

9) \(x-1=\left(\sqrt{x}\right)^2-1=\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)\)

10) \(x\sqrt{x}-1=\sqrt{x^3}-1=\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)\)

11) \(x-2\sqrt{x}-63=\left(x-2\sqrt{x}+1\right)-64=\left(\sqrt{x}-1\right)^2-8^2=\left(\sqrt{x}-1-8\right)\left(\sqrt{x}-1+8\right)=\left(\sqrt{x}-9\right)\left(\sqrt{x}+7\right)\)

sao k làm 12 nữa