\(a)\left[\left(1\dfrac{2}{3}\right)^2\right]^3\\ =\left(1\dfrac{2}{3}\right)^{2\cdot3}\\ =\left(\dfrac{5}{3}\right)^6\\ b)\left(-0,18^3\right)^7\\=\left(-0,18\right)^{3\cdot7}\\ =\left(-0,18\right)^{21}\\ c)\left(\dfrac{1}{3^2}\right)^6\\ =\left[\left(\dfrac{1}{3}\right)^2\right]^6\\ =\left(\dfrac{1}{3}\right)^{2\cdot6}\\ =\left(\dfrac{1}{3}\right)^{12}\)

\(\left|x+\dfrac{1}{1\cdot3}\right|+\left|x+\dfrac{1}{3\cdot5}\right|+...+\left|x+\dfrac{1}{197\cdot199}\right|=100x\)

Số lượng số hạng là: \(\left(199-3\right):2+1=99\) (số hạng) 

TH1: \(x\ge-\dfrac{1}{197\cdot199}\)

\(=>\left(x+\dfrac{1}{1\cdot3}\right)+\left(x+\dfrac{1}{3\cdot5}\right)+...+\left(x+\dfrac{1}{197\cdot199}\right)=100x\\ =>\left(x+x+...+x\right)+\left(\dfrac{1}{1\cdot3}+\dfrac{1}{3\cdot5}+...+\dfrac{1}{197\cdot199}\right)\\ =>99x+\dfrac{1}{2}\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{197\cdot199}\right)\\ =>100x-99x=\dfrac{1}{2}\cdot\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{197}-\dfrac{1}{199}\right)\\ =>x=\dfrac{1}{2}\cdot\left(1-\dfrac{1}{199}\right)=\dfrac{1}{2}\cdot\dfrac{198}{199}\\ =>x=\dfrac{99}{199}\left(tm\right)\) 

TH2: \(x\le-\dfrac{1}{1\cdot3}\) 

\(=>-\left(x+\dfrac{1}{1\cdot3}\right)-\left(x+\dfrac{1}{3\cdot5}\right)-...-\left(x+\dfrac{1}{197\cdot199}\right)=100x\\ =>-\left[\left(x+x+...+x\right)+\left(\dfrac{1}{1\cdot3}+\dfrac{1}{3\cdot5}+...+\dfrac{1}{197\cdot199}\right)\right]=100x\\ =>-\left[99x+\dfrac{1}{2}\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{197\cdot199}\right)\right]=100\\ =>-99x-\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{197}-\dfrac{1}{199}\right)=100x\\ =>100x+99x=-\dfrac{1}{2}\left(1-\dfrac{1}{199}\right)\\ =>199x=-\dfrac{1}{2}\cdot\dfrac{198}{199}\\ =>199x=-\dfrac{99}{199}\\ =>x=-\dfrac{99}{199}:199=-\dfrac{99}{39061}\left(ktm\right)\)

Vậy: `x=99/199`

\(\dfrac{4}{\sqrt{5}+\sqrt{3}}-\sqrt{20}\\ =\dfrac{4\left(\sqrt{5}-\sqrt{3}\right)}{\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)}-\sqrt{20}\\ =\dfrac{4\left(\sqrt{5}-\sqrt{3}\right)}{\left(\sqrt{5}\right)^2-\left(\sqrt{3}\right)^2}-\sqrt{2^2\cdot5}\\ =\dfrac{4\left(\sqrt{5}-\sqrt{3}\right)}{5-3}-2\sqrt{5}\\ =\dfrac{4\left(\sqrt{5}-\sqrt{3}\right)}{2}-2\sqrt{5}\\ =2\left(\sqrt{5}-\sqrt{3}\right)-2\sqrt{5}\\ =2\sqrt{5}-2\sqrt{3}-2\sqrt{5}\\ =-2\sqrt{3}\)

1 We are often taken to the zoo by our father at weekends

2 Hundreds of pupils wre greeted by the smell of fresh paint

3 Noise shouldn't be made by you

4 How can our fields be kept in good condition?

5 Was Lan given an English-Vietnamese dictionary by you?

6 By whom was "Gulliver's travels" written?

7 Why were these books bought?

Bài 1

1 am working

2 doesn't rain

3 are you looking

4 is getting

5 am going

6 starts

7 is rising

8 is getting

9 is rising

10 coming

\(\sqrt{5+2\sqrt{6}}-\sqrt{5-2\sqrt{6}}\\ =\sqrt{\left(\sqrt{3}\right)^2+2\cdot\sqrt{3}\cdot\sqrt{2}+\left(\sqrt{2}\right)^2}-\sqrt{\left(\sqrt{3}\right)^2-2\cdot\sqrt{3}\cdot\sqrt{2}+\left(\sqrt{2}\right)^2}\\ =\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^2}-\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}\\ =\sqrt{3}+\sqrt{2}-\sqrt{3}+\sqrt{2}\\ =2\sqrt{2}\)

Bài 3:

\(a)\left(x+2\right)\left(x+3\right)-\left(x-2\right)\left(x+5\right)=0\\ \Leftrightarrow x^2+3x+2x+6-\left(x^2+5x-2x-10\right)=0\\ \Leftrightarrow x^2+5x+6-x^2-3x+10=0\\ \Leftrightarrow2x+16=0\\ \Leftrightarrow2x=-16\\ \Leftrightarrow x=-\dfrac{16}{2}=-8\\ b)\left(x-3\right)\left(x-2\right)-\left(x+1\right)\left(x-5\right)=0\\ \Leftrightarrow\left(x^2-2x-3x+6\right)-\left(x^2-5x+x-5\right)=0\\ \Leftrightarrow x^2-5x+6-x^2+4x+5=0\\ \Leftrightarrow-x+11=0\\ \Leftrightarrow x=11\\ c)x\left(2x-5\right)-2x\left(x-6\right)=42\\ \Leftrightarrow2x^2-5x-2x^2+12x=42\\ \Leftrightarrow7x=42\\ \Leftrightarrow x=\dfrac{42}{7}\\ \Leftrightarrow x=6\\ d)\left(x-1\right)\left(2x+3\right)-2x\left(x-1\right)=0\\ \Leftrightarrow\left(x-1\right)\left(2x+3-2x\right)=0\\ \Leftrightarrow3\left(x-1\right)=0\\ \Leftrightarrow x-1=0\\ \Leftrightarrow x=1\)

Bài 2:

b: 

c:

d:

Bài 4:

a: \(A\left(x\right)=x^7-3x^2-x^5+x^4-x^2+2x-7\)

\(=x^7-x^5+x^4+\left(-3x^2-x^2\right)+2x-7\)

\(=x^7-x^5+x^4-4x^2+2x-7\)

\(B\left(x\right)=x-2x^2+x^4-x^5-x^7-4x^2-1\)

\(=-x^7-x^5+x^4+\left(-2x^2-4x^2\right)+x-1\)

\(=-x^7-x^5+x^4-6x^2+x-1\)

b: A(x)+B(x)

\(=x^7-x^5+x^4-4x^2+2x-7-x^7-x^5+x^4-6x^2+x-1\)

\(=-2x^5-10x^2+3x-8\)

A(x)-B(x)

\(=x^7-x^5+x^4-4x^2+2x-7+x^7+x^5-x^4+6x^2-x+1\)

\(=2x^7+2x^2+x-6\)

c: C(x)=A(x)+B(x

=>\(C\left(x\right)=-2x^5-10x^2+3x-8\)

Thay x=-1 vào C(x), ta được:

\(C\left(-1\right)=-2\cdot\left(-1\right)^5-10\cdot\left(-1\right)^2+3\cdot\left(-1\right)-8\)

=2-10-3-8

=-1-10-8=-19

Bài 1: 

a) Để căn thức có nghĩa thì:

\(x+1\ge0\Leftrightarrow x\ge-1\)

b) Để căn thức có nghĩa thì: 

\(3x-8\ge0\Leftrightarrow3x\ge8\Leftrightarrow x\ge\dfrac{8}{3}\)

c) Để căn thức có nghĩa thì: 

\(2x^2+3>0\)

Mà điều này luôn đúng nên căn thức có nghĩa khi x ∈ R

d) Để căn thức có nghĩa thì: 

\(16-x^2\ge0\Leftrightarrow\left(4-x\right)\left(4+x\right)\ge0\Leftrightarrow-4\le x\le4\)

Câu 1: B

Câu 2: D

Câu 3: ĐKXĐ: 2x+5>=0

=>2x>=-5

=>\(x>=-\dfrac{5}{2}\)

=>Chọn C

Câu 4: ĐKXĐ: 3-4x>=0

=>-4x>=-3

=>4x<=3

=>\(x< =\dfrac{3}{4}\)

=>Chọn D

Câu 5: \(\sqrt{7}-\sqrt{\left(1-\sqrt{7}\right)^2}\)

\(=\sqrt{7}-\left|1-\sqrt{7}\right|\)

\(=\sqrt{7}-\sqrt{7}+1=1\)

=>Chọn B

Câu 6: \(\sqrt{x^2-6x+9}=\sqrt{\left(x-3\right)^2}=\left|x-3\right|\)

x>=3 nên x-3>=0

=>\(\sqrt{x^2-6x+9}=\left|x-3\right|=x-3\)

=>Chọn B

Câu 7: \(\sqrt{25+x^2+10x}=\sqrt{x^2+10x+25}=\sqrt{\left(x+5\right)^2}=\left|x+5\right|\)

x<-6

=>x+6<0

mà x+5<x+6

nên x+5<0

=>\(\sqrt{25+x^2+10x}=-\left(x+5\right)\)

=>Chọn D

Câu 8: \(M=2x-\sqrt{x^4+2x^2+1}\)

\(=2x-\sqrt{\left(x^2+1\right)^2}\)

\(=2x-\left(x^2+1\right)=-\left(x^2-2x+1\right)=-\left(x-1\right)^2\)

Thay x=11 vào M, ta được:

\(M=-\left(11-1\right)^2=-10^2=-100\)

=>Chọn A

Câu 9: \(\sqrt{x-7}=2\)

=>\(x-7=2^2=4\)

=>x=4+7=11

=>Chọn C

Câu 10: \(\sqrt{a^4b^2}=\sqrt{a^4}\cdot\sqrt{b^2}=a^2\cdot\left|b\right|\)

=>Chọn D

Câu 11: \(\dfrac{1}{2+\sqrt{3}}+\dfrac{1}{2-\sqrt{3}}\)

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

\(=\dfrac{4}{4-3}=4\)

=>Chọn C

Câu 12: A

a, Ta có DE vuông AB 

AH vuông AB 

=> DE // AH 

b, Ta có DE // AH => ^BDE = ^ACB ( 2 góc đồng vị ) 

=> ^BDE = ^DCH = 40⁰

c, Ta có DH vuông AC 

AB vuông AC 

=> DH // AB 

Ta có DH // AB; ED//AH ; ^EAH = ^AED = ^AHD = 90⁰

Vậy tứ giác AEDH là hình vuông 

=> DE vuông DH 

Số xe cửa hàng nhập về là:

\(80:32\%=250\) (chiếc)

Số xe cửa hàng còn lại là:
`250 - 80 =170` (chiếc) 

ĐS: ...