1 Jean advised Bill not to cook the fish for too long

2 I suggested  Brian (should) go and see the Sound of Music

3 Jack warned Tim not to take the A20 in the rush hour

4 Ann insisted that Sophia stay at the night

5 Mary invited Helen to come to lunch on Sunday

6 Clare reminded Andy to buy some milk

7 The customer finally decided to take the brown pair

8 Bob denied taking Sue's calculator

9 Dick's mother told him not to smoke in his room

10 She told me if she knew French well, she would go to France to study

Đề câu cuối có chữ " Gamma*d " lạ nên mình đổi thành " she would " nha, câu điều kiện loại 2=)

1 Jean advised Bill not to cook the fish for too long

2 I suggested Brian go and see the Sound of Music again

3 Jack warned Tim not to take the A20 in the rush hour

4 Ann insisted that Sophia stay the night

5 Mary invited Helen to come to lunch on Sunday

`sqrt{10 + 4sqrt{6}}`

`=sqrt{10 + 2. 2sqrt{6}}`

`=sqrt{sqrt{6}^2 + 2. 2sqrt{6} + 2^2}`

`=sqrt{(sqrt{6}+ 2)^2}`

`= sqrt{6}+ 2`

\(\sqrt{10+4\sqrt{6}}=\sqrt{4.\dfrac{5}{2}+4\sqrt{6}}=2\sqrt{\dfrac{5\sqrt{6}}{2}}\)

Bài 1:

a: a<b

=>a-b<0; b-a>0

2a+3b+1-5b-1

=2a-2b

=2(a-b)<0

=>2a+3b+1<5b+1

b: -5a+7b-10-2b+10=-5a+5b=-5(a-b)>0

=>-5a+7b-10>2b-10

Bài 2:

a:

a>b

=>a-b>0

20a+5b-20b-5a=15a-15b=15(a-b)>0

=>20a+5b>20b+5a

b: -3(a+b)-1+6b+1=-3a-3b+6b=3b-3a=3(b-a)<0

=>-3(a+b)+1>-6b-1

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\(\left(x+2\right)^2-\left(2x+1\right)\left(x+2\right)=0\\ < =>\left(x+2\right)\left[\left(x+2\right)-\left(2x+1\right)\right]=0\\ < =>\left(x+2\right)\left(x+2-2x-1\right)=0\\ < =>\left(x+2\right)\left(1-x\right)=0\\ < =>\left[{}\begin{matrix}x+2=0\\1-x=0\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x=-2\\x=1\end{matrix}\right.\)

Vậy: ... 

\(\left(x+2\right)^2-\left(2x+1\right)\left(x+2\right)=0\)

=>(x+2)(x+2-2x-1)=0

=>(x+2)(-x+1)=0

=>\(\left[{}\begin{matrix}x+2=0\\-x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=1\end{matrix}\right.\)

Ta có: 

\(VT=\sqrt{4-x}+\sqrt{x-2}\le\sqrt{2\left[\left(\sqrt{4-x}\right)^2+\left(\sqrt{x-2}\right)^2\right]}=2\) (1)

\(VP=x^2-6x+11=\left(x^2-6x+9\right)+2=\left(x-3\right)^2+2\ge2\) (2) 

Từ (1) và (2) ta có dấu "=" xảy ra: 

\(\left\{{}\begin{matrix}4-x=x-2\\x-3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x=6\\x=3\end{matrix}\right.\Leftrightarrow x=3\)

Vậy: .. 

VT dùng bất đẳng thức j bạn