cần bài nào?

bài 1 ,2 mỗi đề í

có 4 đề thì mỗi đề chỉ càn làm bài 1 , bài 2 hoi ..

bạn có thể làm cho mình đc hông ạ

a nhờ... thuyền

b bình minh

c qua bao năm tháng 

d bắng... mình 

e chiều chiều , trên bãi thả

f lát sau, để trả thù

g khi... ngỗ

h đêm... quá

nhé

1 Nam said that he came from Thanh Hoa city

2 Mr Cuong asked me if I liked pop music

3 Although the weather was bad, we visited Bat Trang village

4 Because it was raining, they cancelled the trip to Trang An

5 We have to try harder so that our handicrafts can keep up with theirs

6 She grew up in Thanh Hoa city

II

1 Eventhough it was very cold out, we decided to go for a walk

2 Mai studied hard for the exam so that she could have good marks

1. nam said that he  came from Thanh Hoa city.

2.Mr. Cuong asked if I liked pop music

3. although the weather was bad, we visited bat Trang village

(35.5-2.5):x=15

33            :x=15

                x=33:15

                 x=2.2

35,5 : x - 2,5 : x = 15

(35,5-2,5) : x = 15

33 : x = 15

x = 33 : 15

x = \(\frac{33}{15}\)

\(1,ĐK:x\ge2\\ PT\Leftrightarrow\sqrt{3x-6}+x-2-\left(\sqrt{2x-3}-1\right)=0\\ \Leftrightarrow\dfrac{3\left(x-2\right)}{\sqrt{3x-6}}+\left(x-2\right)-\dfrac{2\left(x-2\right)}{\sqrt{2x-3}+1}=0\\ \Leftrightarrow\left(x-2\right)\left(\dfrac{3}{\sqrt{3x-6}}-\dfrac{2}{\sqrt{2x-3}+1}+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2\left(tm\right)\\\dfrac{3}{\sqrt{3x-6}}-\dfrac{2}{\sqrt{2x-3}+1}+1=0\left(1\right)\end{matrix}\right.\)

Với \(x>2\Leftrightarrow-\dfrac{2}{\sqrt{2x-3}+1}>-\dfrac{2}{1+1}=-1\left(3x-6\ne0\right)\)

\(\Leftrightarrow\left(1\right)>0-1+1=0\left(vn\right)\)

Vậy \(x=2\)

\(2,ĐK:x\ge-1\)

Đặt \(\left\{{}\begin{matrix}\sqrt{x+1}=a\\\sqrt{x^2-x+1}=b\end{matrix}\right.\left(a,b\ge0\right)\Leftrightarrow a^2+b^2=x^2+2\)

\(PT\Leftrightarrow2a^2+2b^2-5ab=0\\ \Leftrightarrow\left(a-2b\right)\left(2a-b\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}a=2b\\b=2a\end{matrix}\right.\)

Với \(a=2b\Leftrightarrow x+1=4x^2-4x+4\left(vn\right)\)

Với \(b=2a\Leftrightarrow4x+4=x^2-x+1\Leftrightarrow x^2-5x-3=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5+\sqrt{37}}{2}\left(tm\right)\\x=\dfrac{5-\sqrt{37}}{2}\left(tm\right)\end{matrix}\right.\)

Vậy ...

b \(B=\left|x+2\right|+\left|x-5\right|+\left|x-4\right|+\left|x-1\right|\)

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

Đặt a=|x+2|+|x-4|;b=|-x+5|+|x-1|

Ta có \(\left|x+2\right|\ge0;\left|x-4\right|\ge0với\forall x\)

\(\Rightarrow a=\left|x+2\right|+\left|x-4\right|\ge0với\forall x\left(1\right)\)

\(b=\left|-x+5\right|+\left|x-1\right|\ge-x+5+x-1=4với\forall x\left(2\right)\)

Từ (1) và (2)\(\Rightarrow B=a+b\ge4với\forall x\)

B đạt GTNN \(\Leftrightarrow\hept{\begin{cases}x+2\ge0\\x-4\ge0\end{cases}\Leftrightarrow\hept{\begin{cases}x\ge-2\\x\le4\end{cases}\Leftrightarrow}-2\le x\le4}\)

d \(D=\left|x-2\right|+\left|x-3\right|+\left|x+4\right|+\left|x+5\right|\)

   \(D=\left|-x+2\right|+\left|x-3\right|+\left|-x-4\right|+\left|x+5\right|\)

 Ta có

 \(\left|-x+2\right|+\left|x-3\right|\ge-x+2+x-3=1với\forall\left(1\right)\)

\(\left|-x-4\right|+\left|x+5\right|\ge-x-4+x+5=1với\forall x\left(2\right)\)

Từ(1)và(2)\(\Rightarrow D=\left|-x+2\right|+.....+\left|x+5\right|\ge2\)

D đạt GTNN 

bài này là tìm x hay GTLN(GTNN) của A vậy?

GTNN bạn nhé