1 Ann looked at sheself in mirror //sửa lại:herself

2 đúng ko sai

3 He is absent from class today because his sickness //sửa lại:sick

4.There will have a lot of students at tomorrow' s meeting //sửa lại:be

5.When I come home, I'll call to her //sửa lại:call

Tìm và sửa lỗi sai

1.Ann looked at sheself in the mirror   > herself

2.You hadn`t to do that exercise > don't have 

3.He is absent from class today beacause his sickness > sick

4.There will have a lot of students at tomorrow`s meeting > be 

5.When I come home, I `ll call to her  > call 

\(\frac{7x+6}{2x\left(x+7\right)}-\frac{3x+6}{2x^2+14x}\)

\(=\frac{7x+6}{2x^2+14x}-\frac{3x+6}{2x^2+14x}=\frac{7x+6-3x-6}{2x^2+14x}\)

\(=\frac{4x}{2x^2+14x}=\frac{2\cdot2x}{2\left(x^2+7x\right)}=\frac{2x}{x\left(x+7\right)}=\frac{2}{x+7}\)

\(\frac{7x+6}{2x\left(x+7\right)}-\frac{3x+6}{2x^2+14x}\)( ĐKXĐ : \(x\ne0;x\ne-7\))

\(=\frac{7x+6}{2x\left(x+7\right)}-\frac{3x+6}{2x\left(x+7\right)}\)

\(=\frac{7x+6-\left(3x+6\right)}{2x\left(x+7\right)}\)

\(=\frac{7x+6-3x-6}{2x\left(x+7\right)}\)

\(=\frac{4x}{2x\left(x+7\right)}=\frac{2x\cdot2}{2x\left(x+7\right)}=\frac{2}{x+7}\)

\(=\frac{6x-3}{x}.\left(\frac{3x^2}{4x^2-1}\right)\)

\(=\frac{9x}{\left(2x-1\right)\left(2x+1\right)}.\left(\frac{2x-1}{1}\right)\)

\(=\frac{9x}{2x+1}\)

Bài làm 

\(\frac{6x-3}{x}:\frac{4x^2-1}{3x^2}=\frac{6x-3}{x}.\frac{3x^2}{4x^2-1}\)

\(=\frac{18x^3-9x^2}{4x^3-x}\)

1.Nam listened patiently---

2.Lan disd traveling by plane---Lan doesn't travelling by plane

3.You don`t have to go there--- It's is not important to go to there

4.No one helped me do my homework---I have no one helped with my homework

5.To go out alone at night  is dangerous--- It's dangerous to go out at night

\(3^{2x+2}=9^{x+3}\)

\(3^{2x+2}=\left(3^2\right)^{x+3}\)

\(3^{2x+2}=3^{2x+6}\)

\(\Rightarrow2x+2=2x+6\)(Vô Lý)

\(\Rightarrow\)Không có giá trị nào của x

 ko co gia tri nao cua x

Bài làm 

\(\frac{x^2-9}{2x+6}:\frac{3-x}{2}=\frac{x^2-9}{2x+6}.\frac{2}{3-x}\)

\(=\frac{2x^2-18}{6x-2x^2+18-6x}=\frac{2x^2-18}{-2x^2+19}=1\)

\(2A=2+1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2011}}\)

=> \(2A-A=\left(2+1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2011}}\right)-\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2012}}\right)\)

=>  \(A=2-\frac{1}{2^{2012}}=\frac{2^{2013}-1}{2^{2012}}\)

\(A=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2012}}\)

\(2A=2\left(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2012}}\right)\)

\(2A=3+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2011}}\)

\(2A-A=A\)

\(=\left(3+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2011}}\right)-\left(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2012}}\right)\)

\(=3+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2011}}-1-\frac{1}{2}-\frac{1}{2^2}-\frac{1}{2^3}-...-\frac{1}{2^{2012}}\)

\(=2-\frac{1}{2012^2}\)

 \(B=\left(\frac{1}{99}+\frac{12}{999}+\frac{123}{9999}\right)\cdot\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{6}\right)\)

\(B=\left(\frac{1}{99}+\frac{12}{999}+\frac{123}{9999}\right)\cdot\left(\frac{6}{12}-\frac{4}{12}-\frac{2}{12}\right)\)

\(B=\left(\frac{1}{99}+\frac{12}{999}+\frac{123}{9999}\right)\cdot0=0\)

\(\frac{x^2-9}{2x+6}:\frac{3-x}{2}\)

\(=\frac{\left(x-3\right)\left(x+3\right)}{2\left(x+3\right)}.\frac{-2}{x-3}\)

\(=-1\)

(x + 1) + (x + 3) + (x + 5) + ... + (x + 15) = 128

=> x + 1 + x + 3 + x + 5 + ... + x + 15 = 128(bỏ dấu ngoặc)

=> (x + x + ... + x) + (1 + 3 + 5 + ... + 15) = 128 ( chia thành 2 nhóm)

+) Số số hạng là : (15 - 1) : 2 + 1 = 8

Tổng : (1 + 15).8 : 2 = 64

=> 8x + 64 = 128

=> 8x = 64

=> x = 8

Vậy x = 8

(x + 1) + (x + 4) + (x + 7) + ... + (x + 22) = 829

=> x + 1 + x + 4 + x + 7 + ... + x + 22= 829

=> (x + x + ... + x) + (1 + 4 + 7 + ... + 22) = 829

+) Số số hạng là : (22 - 1) : 3 + 1 = 8

Tổng : (1 + 22).8 : 2 = 92

=> 8x + 92 = 829

=> 8x = 737

=> x = 737/8 

a) (x + 1) + (x + 3) + (x + 5) + .... + (x + 15) = 128 (8 cặp số)

=> (x + x + ... + x) + (1 + 3 + 5 + ... + 15) = 128

       8 hạng tử x          8 số hạng

=> 8x + 8.(15 + 1) : 2 = 128

=> 8x + 64 = 128

=> 8x = 64

=> x = 8

b) (x + 1) + (x + 4) + (x + 7) +... + (x + 22) = 829 (8 cặp số)

=> (x + x + ... + x) + (1 + 4 + 7 +... + 22) = 829

    8 hạng tử x                    8 số hạng

=> 8x + 8.(22 + 1):2 = 829

=> 8x + 92 = 829

=> 8x = 737

=> x = 91,125