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\(e,\left(5+2\sqrt{6}\right)\left(49-20\sqrt{6}\right)\sqrt{5-2\sqrt{6}}\\ =\left(5+2\sqrt{6}\right)\left(\sqrt{3}-\sqrt{2}\right)\left(5-2\sqrt{6}\right)^2\\ =\left(5-2\sqrt{6}\right)\left(\sqrt{3}-\sqrt{2}\right)\\ =\left(\sqrt{3}-\sqrt{2}\right)^2\left(\sqrt{3}-\sqrt{2}\right)=\left(\sqrt{3}-\sqrt{2}\right)^3\)
\(f,\left(2\sqrt{6}-4\sqrt{3}+5\sqrt{2}-\dfrac{1}{4}\sqrt{8}\right)\cdot3\sqrt{6}\\ =36-36\sqrt{2}+30\sqrt{3}-3\sqrt{3}=36-36\sqrt{2}+27\sqrt{3}\)
\(g,\left(2+\sqrt{3}-\sqrt{2}\right)\left(2-\sqrt{3}-\sqrt{2}\right)\left(3+\sqrt{2}\right)\sqrt{3-2\sqrt{2}}\\ =\left[\left(2-\sqrt{2}\right)^2-\left(\sqrt{3}\right)^2\right]\left(3+\sqrt{2}\right)\sqrt{\left(\sqrt{2}-1\right)^2}\\ =\left(3-4\sqrt{2}\right)\left(3+\sqrt{2}\right)\left(\sqrt{2}-1\right)\\ =\left(1-9\sqrt{2}\right)\left(\sqrt{2}-1\right)\\ =10\sqrt{2}-37\)
\(h,A=\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10+2\sqrt{5}}}\\ A^2=4+\sqrt{10+2\sqrt{5}}+4-\sqrt{10+2\sqrt{5}}+2\sqrt{\left(4+\sqrt{10+2\sqrt{5}}\right)\left(4-\sqrt{10+2\sqrt{5}}\right)}\\ A^2=8+2\sqrt{6-2\sqrt{5}}\\ A^2=8+2\left(\sqrt{5}-1\right)\\ A^2=6+2\sqrt{5}\\ A=\sqrt{6+2\sqrt{5}}=\sqrt{\left(\sqrt{5}+1\right)^2}=\sqrt{5}+1\)

\(b,\sqrt{15-\sqrt{216}}+\sqrt{33-12\sqrt{6}}\\ =\sqrt{15-6\sqrt{6}}+\sqrt{\left(2\sqrt{6}-3\right)^2}\\ =\sqrt{\left(3-\sqrt{6}\right)^2}+2\sqrt{6}-3\\ =3-\sqrt{6}+2\sqrt{6}-3=\sqrt{6}\)
\(c,\sqrt{2-\sqrt{3}}\left(\sqrt{6}+\sqrt{2}\right)\\ =\sqrt{12-6\sqrt{3}}+\sqrt{4-2\sqrt{3}}\\ =\sqrt{\left(3-\sqrt{3}\right)^2}+\sqrt{\left(\sqrt{3}-1\right)^2}\\ =3-\sqrt{3}+\sqrt{3}-1=2\)
c: \(\sqrt{2-\sqrt{3}}\cdot\left(\sqrt{6}+\sqrt{2}\right)\)
\(=\sqrt{4-2\sqrt{3}}\cdot\left(\sqrt{3}+1\right)\)
\(=\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)\)
=3-1
=2

Ta có: \(\dfrac{\sqrt{15}-\sqrt{5}}{\sqrt{3}-1}+\sqrt{\left(2-\sqrt{5}\right)^2}-2\sqrt{5}\)
\(=\sqrt{5}+\sqrt{5}-2-2\sqrt{5}\)
=-2

\(\frac{2^2.3^3.5^7}{2^3.3^4.5^6}=\frac{1.1.5}{2.3.1}=\frac{5}{6}\)
Làm tương tự

\(a,=\dfrac{-\sqrt{a}\left(1-\sqrt{a}\right)}{1-\sqrt{a}}=-\sqrt{a}\\ b,=\dfrac{\sqrt{p}\left(\sqrt{p}-2\right)}{\sqrt{p}-2}=\sqrt{p}\)

`@` `\text {Ans}`
`\downarrow`
\(7\cdot35\cdot7\cdot25\)
`=`\(7^2\cdot7\cdot5\cdot5^2\)
`=`\(7^3\cdot5^3=\left(7\cdot5\right)^3=35^3\)

2.3.4.3.4.5.5.2
=2.2.3.3.4.4.5.5
=22.32.42.52
=26.32.52

\(\dfrac{4x^3+4x^2}{x^2-1}=\dfrac{4x^2\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}=\dfrac{4x^2}{x-1}\)
\(\dfrac{b^2+b}{a+ab}=\dfrac{b\left(b+1\right)}{a\left(b+1\right)}=\dfrac{b}{a}\)
d) Để phân thức \(\dfrac{4x^3+4x^2}{x^2-1}\) có nghĩa thì: \(x^2-1\ne0\Leftrightarrow x\ne\pm1\)
Khi đó: \(\dfrac{4x^3+4x^2}{x^2-1}=\dfrac{4x^2\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}=\dfrac{4x^2}{x-1}\)
e) Để phân thức \(\dfrac{b^2+b}{a+ab}\) có nghĩa thì: \(a+ab\ne0\Leftrightarrow a\ne-ab\)
Khi đó: \(\dfrac{b^2+b}{a+ab}=\dfrac{b\left(b+1\right)}{a\left(1+b\right)}=\dfrac{b}{a}\)
a: Ta có: \(A=2^{100}-2^{99}+2^{98}-2^{97}+2^{96}-\cdots+2^2-2\)
=>\(2A=2^{101}-2^{100}+2^{99}-2^{98}+\cdots+2^3-2^2\)
=>\(2A+A=2^{101}-2^{100}+2^{99}-2^{98}+\cdots+2^3-2^2+2^{100}-2^{99}+2^{98}-\cdots+2^2-2\)
=>\(3A=2^{101}-2\)
=>\(A=\frac{2^{101}-2}{3}\)
b: \(B=3^{100}-3^{99}+3^{98}-3^{97}+\cdots+3^2-3+1\)
=>\(3B=3^{101}-3^{100}+3^{99}-3^{98}+\cdots+3^3-3^2+3\)
=>\(3B+B=3^{101}-3^{100}+3^{99}-3^{98}+\cdots3^3-3^2+3+3^{100}-3^{99}+3^{98}-\cdots+3^2-3+1\)
=>\(4B=3^{101}+1\)
=>\(B=\frac{3^{101}+1}{4}\)