a) \(99^3=\left(100-1\right)^3=100^3-3.100^2+3.100-1=1000000-30000+300-1=970299\)b) \(91^3+3.91^2.9+3.91.9^2+9^3=\left(91+9\right)^3=100^3=1000000\)

c) \(1001^3=\left(1000+1\right)^3=1000^3+3.1000^2+3.1000+1=1003003001\)d) \(102^3-6.102^2+24.102-8=\left(102-2\right)^3+12.102=100^3+1224=1001224\)

Xét tam giác ABC có:

\(\widehat{A}+\widehat{B}+\widehat{C}=180^0\)( tổng 3 góc trong tam giác)

\(\Rightarrow\widehat{B}+\widehat{C}=180^0-\widehat{A}=180^0-70^0=110^0\)

Xét tam giác ABC có:

\(\left\{{}\begin{matrix}\widehat{B}+\widehat{C}=110^0\\\widehat{B}-\widehat{C}=40^0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\widehat{B}=\left(110^0+40^0\right):2=75^0\\\widehat{C}=\left(110^0-40^0\right):2=35^0\end{matrix}\right.\)

Ta có: \(\widehat{C}< \widehat{A}< \widehat{B}< 90^0\)

Vậy tam giác ABC là tam giác nhọn

\(P=a+a^2+a^3+...+a^{2n}=\left(a+a^2\right)+\left(a^3+a^4\right)+...+\left(a^{2n-1}+a^{2n}\right)=a\left(a+1\right)+a^3\left(a+1\right)+...+a^{2n-1}\left(a+1\right)=\left(a+1\right)\left(a+a^3+...+a^{2n-1}\right)⋮a+1\)

Bài 1:

a) \(\sqrt{3+2\sqrt{2}}=\sqrt{\left(\sqrt{2}+\sqrt{1}\right)^2}=\sqrt{2}+1\)

b) \(\sqrt{28-6\sqrt{3}}=\sqrt{\left(\sqrt{27}-1\right)^2}=\sqrt{27}-1\)

c) \(\sqrt{4\sqrt{5}+9}=\sqrt{\left(\sqrt{5}-2\right)^2}=\sqrt{5}-2\)

d) \(\sqrt{29+4\sqrt{7}}=\sqrt{\left(\sqrt{28}+1\right)^2}=\sqrt{28}+1\)

\(a⋮b\Rightarrow a=b.n\left(n\in Z\right)\left(1\right)\)

\(b⋮c\Rightarrow b=c.m\left(m\in Z\right)\left(2\right)\)

Từ \(\left(1\right),\left(2\right)\Rightarrow a=c.m.n⋮c\)( do \(m,n\in Z\))

a) \(\sqrt{3-2\sqrt{2}}=\sqrt{\left(\sqrt{2}-1\right)^2}=\sqrt{2}-1\)

b) \(\sqrt{12-6\sqrt{3}}=\sqrt{\left(\sqrt{9}-\sqrt{3}\right)^2}=3-\sqrt{3}\)

c) \(\sqrt{4\sqrt{5}+21}=\sqrt{\left(\sqrt{20}+1\right)}=\sqrt{20}+1\)

d) \(\sqrt{11+4\sqrt{7}}=\sqrt{\left(\sqrt{7}+\sqrt{4}\right)^2}=\sqrt{7}+\sqrt{4}\)

\(3x\left(x-2\right)-x+2+5x\left(x-2\right)=3x\left(x-2\right)-\left(x-2\right)+5x\left(x-2\right)=\left(x-2\right)\left(3x=1+5x\right)=\left(x-2\right)\left(8x-1\right)\)

bí mật

\(\left(1-\dfrac{5+\sqrt{5}}{1+\sqrt{5}}\right)\left(\dfrac{5-\sqrt{5}}{1-\sqrt{5}}-1\right)=\left(1-\dfrac{\sqrt{5}\left(1+\sqrt{5}\right)}{1+\sqrt{5}}\right)\left(-\dfrac{\sqrt{5}\left(1-\sqrt{5}\right)}{1-\sqrt{5}}-1\right)=-\left(1-\sqrt{5}\right)\left(1+\sqrt{5}\right)=-\left(1-5\right)=4\)