Bài 3:

a) Ta có: \(\left(x+10\right)^2+\left(x-10\right)^2\)

\(=x^2+20x+100+x^2-20x+100\)

\(=2x^2+200\)

b) Ta có: \(\left(x-12\right)^2+\left(x+12\right)^2\)

\(=x^2-24x+144+x^2+24x+144\)

\(=2x^2+288\)

c) Ta có: \(\left(x+7\right)^2-\left(x-7\right)^2\)

\(=\left(x+7-x+7\right)\left(x+7+x-7\right)\)

\(=14\cdot2x\)

=28x

Bài 1:

a) Ta có: \(\left(a+12\right)^2\)

\(=a^2+2\cdot a\cdot12+12^2\)

\(=a^2+24a+144\)

b) Ta có: \(\left(3a+\dfrac{1}{3}\right)^2\)

\(=\left(3a\right)^2+2\cdot3a\cdot\dfrac{1}{3}+\left(\dfrac{1}{3}\right)^2\)

\(=9a^2+2a+\dfrac{1}{9}\)

c) Ta có: \(\left(5a^2+6\right)^2\)

\(=\left(5a^2\right)^2+2\cdot5a^2\cdot6+6^2\)

\(=25a^4+60a^2+36\)

d) Ta có: \(\left(\dfrac{1}{2}+4b\right)^2\)

\(=\left(\dfrac{1}{2}\right)^2+2\cdot\dfrac{1}{2}\cdot4b+\left(4b\right)^2\)

\(=\dfrac{1}{4}+4b+16b^2\)

e) Ta có: \(\left(a^m+b^n\right)^2\)

\(=\left(a^m\right)^2+2\cdot a^m\cdot b^n+\left(b^n\right)^2\)

\(=a^{2m}+2a^mb^n+b^{2n}\)

Bài 2: 

a) Ta có: \(\left(x-0.3\right)^2\)

\(=x^2-2\cdot x\cdot0.3+0.3^2\)

\(=x^2-0.6x+0.9\)

b) Ta có: \(\left(6x-3y\right)^2\)

\(=\left(6x\right)^2-2\cdot6x\cdot3y+\left(3y\right)^2\)

\(=36x^2-36xy+9y^2\)

c) Ta có: \(\left(5-2xy\right)^2\)

\(=5^2-2\cdot5\cdot2xy+\left(2xy\right)^2\)

\(=4x^2y^2-20xy+25\)

d) Ta có: \(\left(x^4-1\right)^2\)

\(=\left(x^4\right)^2-2\cdot x^4\cdot1+1^2\)

\(=x^8-2x^4+1\)

e) Ta có: \(\left(x^5-y^3\right)^2\)

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

\(=x^{10}-2x^5y^3+y^6\)

\(5x\left(3x^2y-2xy^2+1\right)-3xy\left(5x^2-3xy\right)+x^2y^2-10=0\)

\(\Leftrightarrow15x^3y-10x^2y^2+5x-15x^3y+9x^2y^2+x^2y^2-10=0\)

\(\Leftrightarrow5x-10=0\Leftrightarrow x-2=0\Leftrightarrow x=2\)

Em chụp xoay ảnh đúng chiều khi đăng nhé!

Bài 3

2 was

3 loved

4 cared

5 died

B

1 decided

2 got

3 was flying - were playing

5 asked

6 was enjoying 

7 covered

Bài 4

1 used to

2 used to

4 didn't use to

5 didn't use to

Bài 5

2 live

3 watch

4 going

5 ride

6 washing

a) Ta có: \(Q=\dfrac{3x+\sqrt{9x}-3}{x+\sqrt{x}-2}-\dfrac{\sqrt{x}+1}{\sqrt{x}+2}+\dfrac{\sqrt{x}-2}{1-\sqrt{x}}\)

\(=\dfrac{3x+3\sqrt{x}-3-\left(x-1\right)-\left(x-4\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)

\(=\dfrac{3x+3\sqrt{x}-3-x+1-x+4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)

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

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

b) Thay \(x=4+2\sqrt{3}\) vào Q, ta được:

\(Q=\dfrac{\sqrt{3}+1+1}{\sqrt{3}+1-1}=\dfrac{2+\sqrt{3}}{\sqrt{3}}=\dfrac{2\sqrt{3}+3}{3}\)

c) Để Q=3 thì \(\sqrt{x}+1=3\sqrt{x}-3\)

\(\Leftrightarrow\sqrt{x}-3\sqrt{x}=-3-1\)

\(\Leftrightarrow2\sqrt{x}=4\)

hay x=4

d) Để \(Q>\dfrac{1}{2}\) thì \(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}-\dfrac{1}{2}>0\)

\(\Leftrightarrow\dfrac{2\sqrt{x}+2-\sqrt{x}+1}{2\left(\sqrt{x}-1\right)}>0\)

\(\Leftrightarrow\sqrt{x}-1>0\)

\(\Leftrightarrow x>1\)

Kết hợp ĐKXĐ, ta được: x>1

e) Để Q nguyên thì \(\sqrt{x}+1⋮\sqrt{x}-1\)

\(\Leftrightarrow2⋮\sqrt{x}-1\)

\(\Leftrightarrow\sqrt{x}-1\in\left\{-1;1;2\right\}\)

\(\Leftrightarrow\sqrt{x}\in\left\{0;2;3\right\}\)

hay \(x\in\left\{0;4;9\right\}\)

Cơ chế hình thành cây có kiểu gen Aaa là do rồi loạn giảm phân, diễn ra ở kì sau của giảm phân 1. 

Sơ đồ lai: 

P:       Aa          x             Aa

GP: Aa ; 0  ;           A   ;    a

F1:  Aaa             ;            a

Sao dài quá z?

a: =38-21-58-69=-20-90=-110