Troi chi oi co ma , o ben phai y .

moi nguoi cho cai

\(2.16\ge2^n>4\)

\(2.2^4\ge2^n>2^2\)

\(2^5\ge2^n>2^2\)

=> \(n\in\left\{3,4,5\right\}\)

Vậy: \(n\in\left\{3,4,5\right\}\)

2 It take more hours to travel by train than to travel

6 The bus run more frequently than the trains

5 That is the farthest distance i have ever run

6 It was the worst mistake I have ever made

5 That's the furthest (or farthest) I've ever run.

6 It is/was the worst mistake I've ever made.

5. That run .... ( mình ko hiểu đề )

6. It make the worst mistake

a)

\(\dfrac{6}{13}\cdot\dfrac{8}{7}\cdot\dfrac{-26}{3}\cdot\dfrac{-7}{8}\)

\(=\dfrac{6}{13}\cdot\dfrac{-26}{3}\cdot\dfrac{8}{7}\cdot\dfrac{-7}{8}\)

\(=-4\cdot\left(-1\right)\\ =4\)

b)

\(\dfrac{6}{11}+\dfrac{11}{3}\cdot\dfrac{3}{22}\)

\(=\dfrac{6}{11}+\dfrac{1}{2}\\ =\dfrac{12}{22}+\dfrac{11}{22}\\ =\dfrac{23}{22}\)

a)=  (6/13*-26/3)* (8/7*-7/8)

= -4*-1=4

b=6/11+1/2=(12+11)/22=23/22

a. \(\dfrac{6.8.\left(-26\right).\left(-7\right)}{13.7.3.8}=\dfrac{2.3.8.\left(-13\right).2.\left(-7\right)}{13.7.3.8}=\dfrac{4}{1}=4\)

\(b.\dfrac{6}{11}+\dfrac{11.3}{3.11.2}\)

\(=\dfrac{6}{11}+\dfrac{1}{2}=\dfrac{23}{22}\)

Bài 5: 

a: Xét (O) có

ΔDMN nội tiếp

MN là đường kính

Do đó: ΔDMN vuông tại D

\(4,=\dfrac{6\left(\sqrt{2}-\sqrt{3}-3\right)}{5-2\sqrt{6}-9}=\dfrac{6\left(\sqrt{2}-\sqrt{3}-3\right)}{-4-2\sqrt{6}}\\ =\dfrac{3\left(3-\sqrt{2}-\sqrt{3}\right)}{2+\sqrt{6}}=\dfrac{\left(9-3\sqrt{2}-3\sqrt{3}\right)\left(\sqrt{6}-2\right)}{2}\\ =\dfrac{9\sqrt{6}-18-6\sqrt{3}+6\sqrt{2}-9\sqrt{2}+6\sqrt{3}}{2}\\ =\dfrac{9\sqrt{6}-3\sqrt{2}-18}{2}\)

\(7,=\dfrac{\sqrt{3}\left(\sqrt{3}+2\right)}{\sqrt{3}}+\dfrac{\sqrt{2}\left(\sqrt{2}+1\right)}{\sqrt{2}+1}-2-\sqrt{3}\\ =\sqrt{3}+2+\sqrt{2}+1-2-\sqrt{3}=1+\sqrt{2}\)

\(10,\dfrac{1}{\sqrt{a}+\sqrt{a+2}}=\dfrac{\sqrt{a}-\sqrt{a+2}}{a-a-2}=\dfrac{\sqrt{a-2}-\sqrt{a}}{2}\)

Do đó \(\dfrac{1}{\sqrt{1}+\sqrt{3}}+\dfrac{1}{\sqrt{3}+\sqrt{5}}+...+\dfrac{1}{\sqrt{47}+\sqrt{49}}\)

\(=\dfrac{\sqrt{3}-\sqrt{1}+\sqrt{5}-\sqrt{3}+...+\sqrt{49}-\sqrt{47}}{2}=\dfrac{-1+\sqrt{49}}{2}=\dfrac{7-1}{2}=3\)

10, \(\dfrac{1}{\sqrt{1}+\sqrt{3}}+\dfrac{1}{\sqrt{3}+\sqrt{5}}+...+\dfrac{1}{\sqrt{17}+\sqrt{19}}=\dfrac{\sqrt{1}-\sqrt{3}}{\left(\sqrt{1}+\sqrt{3}\right)\left(\sqrt{1}-\sqrt{3}\right)}+\dfrac{\sqrt{3}-\sqrt{5}}{\left(\sqrt{3}+\sqrt{5}\right)\left(\sqrt{3}-\sqrt{5}\right)}+...+\dfrac{\sqrt{17}-\sqrt{19}}{\left(\sqrt{17}+\sqrt{19}\right)\left(\sqrt{17}-\sqrt{19}\right)}=\dfrac{1-\sqrt{3}+\sqrt{3}-\sqrt{5}+...+\sqrt{17}-\sqrt{19}}{-2}=-\dfrac{1-\sqrt{19}}{2}\)

Câu 4: 

a: Xét ΔBAE vuông tại A và ΔBHE vuông tại H có

BE chung

\(\widehat{ABE}=\widehat{HBE}\)

Do đó: ΔBAE=ΔBHE

Suy ra: BA=BH và EA=EH

b: Xét ΔAEK vuông tại A và ΔHEC vuông tại H có

EA=EH

\(\widehat{AEK}=\widehat{HEC}\)

Do đó: ΔAEK=ΔHEC

Suy ra: AK=HC

Ta có: BA+AK=BK

BH+HC=BC

mà BA=BH

và AK=HC

nên BK=BC

hay ΔBKC cân tại B

c: \(AC=\sqrt{BC^2-AB^2}=8\left(cm\right)\)

\(\dfrac{8}{3-\sqrt{5}}+\sqrt{9-4\sqrt{5}}=\dfrac{8\left(3+\sqrt{5}\right)}{9-5}+\sqrt{\left(\sqrt{5}-2\right)^2}=\dfrac{24+8\sqrt{5}}{4}+\sqrt{5}-2=6+2\sqrt{5}+\sqrt{5}-2=4+3\sqrt{5}\)

Câu 1: 

a: \(A=\left(4\cdot2\cdot3\right)\cdot x^3y^2z\cdot x^2y^3\cdot xy=24x^6y^6z\)

b: Hệ số là 24

Phần biến là \(x^6;y^6;z\)

Bậc là 13