\(\frac{2}{3}\times\frac{4}{5}+\frac{4}{5}\times\frac{1}{2}+\frac{4}{5}\times\frac{1}{6}\)

\(=\left(\frac{2}{3}+\frac{1}{2}+\frac{1}{6}\right)\times\frac{4}{5}\)

\(=\frac{4}{3}\times\frac{4}{5}\)

\(=\frac{16}{15}\)

22,

1, Đặt √(3-√5) = A

=> √2A=√(6-2√5)

=> √2A=√(5-2√5+1)

=> √2A=|√5 -1|

=> A=\(\dfrac{\sqrt{5}-1}{\text{√2}}\)

=> A= \(\dfrac{\sqrt{10}-\sqrt{2}}{2}\)

2, Đặt √(7+3√5) = B

=> √2B=√(14+6√5)

 => √2B=√(9+2√45+5)

=> √2B=|3+√5|

=> B= \(\dfrac{3+\sqrt{5}}{\sqrt{2}}\)

=> B= \(\dfrac{3\sqrt{2}+\sqrt{10}}{2}\)

3, 

Đặt √(9+√17) - √(9-√17) -\(\sqrt{2}\)=C

=> √2C=√(18+2√17) - √(18-2√17) -\(2\)

=> √2C=√(17+2√17+1) - √(17-2√17+1) -\(2\)

=> √2C=√17+1- √17+1 -\(2\)

=> √2C=0

=> C=0

26,

|3-2x|=2\(\sqrt{5}\)

TH1: 3-2x ≥ 0 ⇔ x≤\(\dfrac{-3}{2}\)

3-2x=2\(\sqrt{5}\)

-2x=2\(\sqrt{5}\) -3

x=\(\dfrac{3-2\sqrt{5}}{2}\) (KTMĐK)

TH2: 3-2x < 0 ⇔ x>\(\dfrac{-3}{2}\)

3-2x=-2\(\sqrt{5}\)

-2x=-2√5 -3

x=\(\dfrac{3+2\sqrt{5}}{2}\) (TMĐK)

Vậy x=\(\dfrac{3+2\sqrt{5}}{2}\)

2, \(\sqrt{x^2}\)=12 ⇔ |x|=12 ⇔ x=12, -12

3, \(\sqrt{x^2-2x+1}\)=7

⇔ |x-1|=7 

TH1: x-1≥0 ⇔ x≥1

x-1=7 ⇔ x=8 (TMĐK)

TH2: x-1<0 ⇔ x<1

x-1=-7 ⇔ x=-6 (TMĐK)

Vậy x=8, -6

4, \(\sqrt{\left(x-1\right)^2}\)=x+3

⇔ |x-1|=x+3

TH1: x-1≥0 ⇔ x≥1

x-1=x+3 ⇔ 0x=4 (KTM)

TH2: x-1<0 ⇔ x<1

x-1=-x-3 ⇔ 2x=-2 ⇔x=-1 (TMĐK)

Vậy x=-1

A = 4/1.3 + 4/3.5 + 4/5.7 + ... + 4/11.13

A = 2.(2/1.3 + 2/3.5 + 2/5.7 + ... + 2/11.13)

A = 2.(1 - 1/3 + 1/3 - 1/5 + 1/5 - 1/7 + ... + 1/11 - 1/13)

A = 2.(1 - 1/13)

A = 2.12/13

A = 24/13

24/13 đúng 100%

bài 4:so sánh

5/2 lớn hơn 3/7

4/3 lớn hơn,3/2 lớn hơn 

bài 6:rút gọn các phân số sau:

3/9=1/3      9/12=3/4          8/18=4/9         60/36=10/6         17/34=1/2              17/51=1/3           35/100=7/20           25/100=1/4                  8/1000=1/125                 24/30=4/5           18/54=1/3           72/42=12/7

đay nhé mk chưa làm hết đc bn viết liền quá mk nhìn khó mà mk hỏi bài 7 là nhân hay cộng vậy?

4 phần 5 trừ 11 phần 5 =

1 + 1 = 2

2 + 2 = 4

3 + 3 = 6

4 + 4 = 8

5 + 5 = 10

\(\dfrac{-1}{9}.\dfrac{-3}{5}+\dfrac{5}{-6}.\dfrac{-3}{5}-\dfrac{7}{2}.\dfrac{3}{5}\)

\(=\dfrac{3}{5}.\left(\dfrac{1}{9}+\dfrac{5}{6}-\dfrac{7}{2}\right)\)

\(=\dfrac{3}{5}.\left(\dfrac{2}{18}+\dfrac{15}{18}-\dfrac{63}{18}\right)\)

\(=\dfrac{3}{5}.\left(-\dfrac{23}{9}\right)\)

\(=-\dfrac{69}{45}\)

a )  \(-\frac{3}{7}.\frac{3}{11}+-\frac{3}{7}.\frac{8}{11}+1\frac{3}{7}\)

\(=-\frac{3}{7}.\left(\frac{3}{11}+\frac{8}{11}\right)+\frac{10}{7}\)

\(=-\frac{3}{7}.\frac{11}{11}+\frac{10}{7}\)

\(=-\frac{3}{7}.1+\frac{10}{7}\)

\(=\frac{10}{7}\)

b )   \(75\%.10,5=\frac{3}{4}.10,5=7,875\)

c )  \(5-3.\left(\left|-4\right|-30:15\right)\)

\(=5-3.\left(4-2\right)\)

\(=5-3.2\)

\(=5-6\)

\(=-1\)

d )  \(-\frac{5}{7}.\frac{2}{11}+-\frac{5}{7}.\frac{9}{11}+1\frac{5}{7}\)

\(=-\frac{5}{7}.\left(\frac{2}{11}+\frac{9}{11}\right)+\frac{12}{7}\)

\(=-\frac{5}{7}.1+\frac{12}{7}\)

\(=\frac{7}{7}\)

\(=1\)

Chúc bạn học tốt !!! 

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.

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

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