Sao ko bảo chị giải đi, kkkk

Haizzzzz

Câu a)

Câu b) Đề bài sai.

Bài 1:
a.

\(\frac{1}{2\sqrt{2}-3\sqrt{3}}=\frac{2\sqrt{2}+3\sqrt{3}}{(2\sqrt{2}-3\sqrt{3})(2\sqrt{2}+3\sqrt{3})}=\frac{2\sqrt{2}+3\sqrt{3}}{(2\sqrt{2})^2-(3\sqrt{3})^2}=\frac{2\sqrt{2}+3\sqrt{3}}{-19}\)

b.

\(=\sqrt{\frac{(3-\sqrt{5})^2}{(3-\sqrt{5})(3+\sqrt{5})}}=\sqrt{\frac{(3-\sqrt{5})^2}{3^2-5}}=\sqrt{\frac{(3-\sqrt{5})^2}{4}}=\sqrt{(\frac{3-\sqrt{5}}{2})^2}=|\frac{3-\sqrt{5}}{2}|=\frac{3-\sqrt{5}}{2}\)

Bài 2.

a. 

\(=\frac{\sqrt{8}(\sqrt{5}+\sqrt{3})}{(\sqrt{5}-\sqrt{3})(\sqrt{5}+\sqrt{3})}=\frac{2\sqrt{2}(\sqrt{5}+\sqrt{3})}{5-3}=\sqrt{2}(\sqrt{5}+\sqrt{3})=\sqrt{10}+\sqrt{6}\)

b.

\(=\sqrt{\frac{(2-\sqrt{3})^2}{(2-\sqrt{3})(2+\sqrt{3})}}=\sqrt{\frac{(2-\sqrt{3})^2}{2^2-3}}=\sqrt{(2-\sqrt{3})^2}=|2-\sqrt{3}|=2-\sqrt{3}\)

Bài 3:

a.

\(M=\left[\frac{15(\sqrt{6}-1)}{(\sqrt{6}+1)(\sqrt{6}-1)}+\frac{4(\sqrt{6}+2)}{(\sqrt{6}-2)(\sqrt{6}+2)}-\frac{12(3+\sqrt{6})}{(3-\sqrt{6})(3+\sqrt{6})}\right](\sqrt{6}+11)\)

\(=\left[\frac{15(\sqrt{6}-1)}{6-1}+\frac{4(\sqrt{6}+2)}{6-2^2}-\frac{12(3+\sqrt{6})}{3^2-6}\right](\sqrt{6}+11)\)

\(=[3(\sqrt{6}-1)+2(\sqrt{6}+2)-4(3+\sqrt{6})](\sqrt{6}+11)=(\sqrt{6}-11)(\sqrt{6}+11)=6-11^2=-115\)

b.

\(N=\left[1-\frac{\sqrt{5}(\sqrt{5}+1)}{\sqrt{5}+1}\right].\left[\frac{\sqrt{5}(\sqrt{5}-1)}{1-\sqrt{5}}-1\right]\)

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

\(=a^2+2a+1-a^2+2a-1-3a^2+3\\ =-3a^2+4a+3\)

\(a,=\sqrt{2}\left(\sqrt{5}+3\right)\sqrt{\left(3-\sqrt{5}\right)^2}=\sqrt{2}\left(\sqrt{5}+3\right)\left(3-\sqrt{5}\right)=4\sqrt{2}\\ b,=\sqrt{\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)}=\sqrt{4}=2\)

a)\(=\left(\sqrt{10}+3\sqrt{2}\right)\sqrt{\left(3-\sqrt{5}\right)^2}=\left(\sqrt{10}+3\sqrt{2}\right)\left(3-\sqrt{5}\right)=3\sqrt{10}-5\sqrt{2}+9\sqrt{2}-3\sqrt{10}=4\sqrt{2}\)

b) \(=\sqrt{\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)}=\sqrt{9-5}=\sqrt{4}=2\)

TH1: x>=5/3

A=3x-5+4x-6=7x-11

TH2: 3/5<x<5/3

A=5-3x+4x-6=x-1

A=(7-2x)(7+2x)+(2x+7)²

    =49-4x²+4x²+28x+49

   = 98+28x

B=(4x-5)²-(2x-1)(8x-5)

  = 16x²-25-((8x(2x-1))-(5(2x-1)))

  = 16x²-25-((16x²+8x)-(10x+5))

  = 16x²-25-(16x²+8x-10x-5)

  = 16x²-25-16x²-8x+10x+5

   = -20+2x

a) Ta có: \(A=\left(7-2x\right)\left(7+2x\right)+\left(2x+7\right)^2\)

\(=7-4x^2+4x^2+28x+49\)

\(=28x+56\)

b) Ta có: \(B=\left(4x-5\right)^2-\left(2x-1\right)\left(8x-5\right)\)

\(=16x^2-40x+25-\left(16x^2-10x-8x+5\right)\)

\(=16x^2-40x+25-16x^2+18x-5\)

\(=-22x+20\)

c) Ta có: \(C=\left(5x-3\right)^2-2\left(5x-3\right)\left(5-5x\right)+\left(5x-5\right)^2\)

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

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

\(=\left(10x-8\right)^2\)

\(=100x^2-160x+64\)

d) Ta có: \(D=\left(2a+3b-c\right)\left(2a-3b+c\right)-\left(4a^2-9b^2-c^2\right)\)

\(=\left[\left(2a+\left(3b-c\right)\right)\left(2a-\left(3b-c\right)\right)\right]-\left(4a^2-9b^2-c^2\right)\)

\(=4a^2-\left(3b-c\right)^2-4a^2+9b^2+c^2\)

\(=-9b^2+6bc-c^2+9b^2+c^2\)

=6bc

\(A=1+2+2^2+...+2^{51}\)

\(2A=2+2^2+2^3+...+2^{52}\)

\(2A-A=\left(2+2^2+2^3+...+2^{52}\right)-\left(1+2+2^2+...+2^{51}\right)\)

\(A=2^{52}-1\)

\(B=5+5^2+5^3+...+5^{100}\)

\(5B=5^2+5^3+5^4+...+5^{101}\)

\(5B-B=\left(5^2+5^3+5^4+...+5^{101}\right)-\left(5+5^2+5^3+...+5^{100}\right)\)

\(4B=5^{101}-5\)

\(B=\frac{5^{101}-5}{4}\)

`a)(5sqrt2-2sqrt5)/(sqrt5-sqrt2)+6/(2-sqrt{10})`

`=(sqrt{10}(sqrt5-sqrt2))/(sqrt5-sqrt2)+(6(2+sqrt{10}))/(4-10)`

`=sqrt{10}-(2+sqrt{10})`

`=-2`

`b)6/(sqrt5-1)+7/(1-sqrt3)-2/(sqrt3-sqrt5)`

`=(6(sqrt5+1))/(5-1)+(7(1+sqrt3))/(1-3)-(2(sqrt3+sqrt5))/(3-5)`

`=(6(sqrt5+1))/4-(7+7sqrt3)/2+sqrt3+sqrt5`

`=(3sqrt5+3)/2-(7+7sqrt3)/2+sqrt3+sqrt5`

`=(3sqrt5+3-7-7sqrt3+2sqrt3+2sqrt5)/2`

`=(5sqrt5-5sqrt3-4)/2`