Đáp án cần chọn là: B

5 11 .7 12 + 5 11 .7 11 5 12 .7 12 + 9.5 11 .7 11 = 5 11 .7 11 ( 7 + 1 ) 5 11 .7 11 ( 5.7 + 9 ) = 8 44 = 2 11 .

Do đó a = 2,b = 11 nên a + b = 13

\(\left(3x^3y^2-9x^2y^2+15xy^3\right):3xy^2\)

\(=3x^3y^2:3xy^2-9x^2y^2:3xy^2+15xy^3:3xy^2\)

\(=\left(3:3\right)\cdot x^{3-1}\cdot y^{2-2}-\left(9:3\right)\cdot x^{2-1}\cdot y^{2-2}+\left(15:3\right)\cdot x^{1-1}\cdot y^{3-2}\)

\(=x^2-3x+5y\)

TH1: x>=5/3

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

TH2: 3/5<x<5/3

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

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.

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

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

\(C=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{2020}}\\ 2C=1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2019}}\\ 2C-C=\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2019}}\right)-\left(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{2020}}\right)\\ C=1-\dfrac{1}{2^{2020}}=\dfrac{2^{2020}-1}{2^{2020}}\)

Giải:

C=1/2 + 1/2^2 + 1/2^3 + ... + 1/2^2020

2C=1 + 1/2 + 1/2^2 + ... +1/2^2019

2C-C=(1+1/2+1/2^2+...+1/2^2019)-(1/2+1/2^2+1/2^3+...+1/2^2020)

C=1-1/2^2020

Chúc bạn học tốt!

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

a: =8/4-1/4=7/4

b: =3/9=1/3

c: =5/7-4/7=1/7

a,\(\dfrac{7}{4}\)

b,\(\dfrac{1}{3}\)

c,\(\dfrac{1}{7}\)

\(A=\dfrac{x\sqrt{x}+1}{x-1}-\dfrac{x-1}{\sqrt{x}+1}\)

\(A=\dfrac{x\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}-\dfrac{\left(x-1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)

\(A=\dfrac{x\sqrt{x}+1-\left(x-1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)

\(A=\dfrac{x\sqrt{x}+1-\left(x\sqrt{x}-x-\sqrt{x}+1\right)}{x-1}\)

\(A=\dfrac{x\sqrt{x}+1-x\sqrt{x}+x+\sqrt{x}-1}{x-1}\)

\(A=\dfrac{x+\sqrt{x}}{x-1}\)

a: (x+1)(3-x)(x-2)²

\(=\left(3x-x^2+3-x\right)\left(x^2-4x+4\right)\)

\(=\left(-x^2+2x+3\right)\left(x^2-4x+4\right)\)

\(=-x^4+4x^3-4x^2+2x^3-8x^2+8x+3x^2-12x+12\)

\(=-x^4+6x^3-9x^2-4x+12\)

b: \(9x\left(1-x\right)+\left(3x-2\right)\left(3x+2\right)\)

\(=9x-9x^2+\left(3x\right)^2-4\)

\(=9x-9x^2+9x^2-4=9x-4\)