có bài 2 nào đâu

Bài 4 ý

Bài 5 ý

2: x+13 là bội của x-1

=>\(x+13⋮x-1\)

=>\(x-1+12⋮x-1\)

=>\(x-1\inƯ\left(12\right)\)

mà x-1>=-1(vì x là số tự nhiên)

nên \(x-1\in\left\{-1;1;2;3;4;6;12\right\}\)

=>\(x\in\left\{0;2;3;4;5;7;13\right\}\)

3: 4x+9 là bội của 2x+1

=>\(4x+9⋮2x+1\)

=>\(4x+2+7⋮2x+1\)

=>\(2x+1\in\left\{1;7\right\}\)

=>\(x\in\left\{0;3\right\}\)

1. Brush your teeth after meals.

2. Don't play soccer in the street

3. Wash and iron your own clothes.

4. Don't eat too much candy.

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1. She is interested in/ fond of/ keen on playing tennis.

2. No house in the street is older than my house.

3. Although the test was difficult, many students did it well. 

4. It's very important to keep the environment clean.

5. What a wonderful holiday they have!

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1. My sister is a fast runner.

2. Hoang plays football badly.

3. Linh is good at swimming.

4. Peter's favourite sport is badminton.

5. Jack is fatter than Mike.

6. Hugo is interested in playing sports in his free time.

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1. There isn't a bookshelf in my room.

2. Ha is harder than Quang.

3. What a lovely baby!

4. Viết lại bằng từ gì nhỉ?

\(b,B=\dfrac{\sqrt{x}+2}{\sqrt{x}-3}+\dfrac{\sqrt{x}-8}{x-5\sqrt{x}+6}\left(x\ge0;x\ne4;x\ne9\right)\\ B=\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)+\sqrt{x}-8}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\\ B=\dfrac{x-4+\sqrt{x}-8}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}=\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}-4\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}=\dfrac{\sqrt{x}-4}{\sqrt{x}-2}\)

\(c,B< A\Leftrightarrow\dfrac{\sqrt{x}-4}{\sqrt{x}-2}< \dfrac{\sqrt{x}+1}{\sqrt{x}-2}\Leftrightarrow\dfrac{\sqrt{x}-4}{\sqrt{x}-2}-\dfrac{\sqrt{x}+1}{\sqrt{x}-2}< 0\\ \Leftrightarrow\dfrac{-5}{\sqrt{x}-2}< 0\Leftrightarrow\sqrt{x}-2>0\left(-5< 0\right)\\ \Leftrightarrow x>4\\ d,P=\dfrac{B}{A}=\dfrac{\sqrt{x}-4}{\sqrt{x}-2}:\dfrac{\sqrt{x}+1}{\sqrt{x}-2}=\dfrac{\sqrt{x}-4}{\sqrt{x}+1}=1-\dfrac{5}{\sqrt{x}+1}\in Z\\ \Leftrightarrow5⋮\sqrt{x}+1\Leftrightarrow\sqrt{x}+1\inƯ\left(5\right)=\left\{-5;-1;1;5\right\}\\ \Leftrightarrow\sqrt{x}\in\left\{-6;-2;0;4\right\}\\ \Leftrightarrow x\in\left\{0;16\right\}\left(\sqrt{x}\ge0\right)\)

\(e,P=1-\dfrac{5}{\sqrt{x}+1}\)

Ta có \(\sqrt{x}+1\ge1,\forall x\Leftrightarrow\dfrac{5}{\sqrt{x}+1}\ge5\Leftrightarrow1-\dfrac{5}{\sqrt{x}+1}\le-4\)

\(P_{max}=-4\Leftrightarrow x=0\)

a, với x > = 0 

\(A=2x-3x+5=-x+5\)

b, Với x>=1 bth có dạng \(B=x-1+3x-2=4x-3\)

Với x<1 bth có dạng \(B=1-x+3x-2=2x-1\)

c, Với x < 0 

\(C=3x+5x+4=8x+4\)

d, Với x =< 2 bth có dạng \(D=2-x-3x+6=-4x+8\)

Với x > 2 bth có dạng \(D=x-2-3x+6=-2x+4\)

dài thế

Tách rời các bài thì mới có người giải nha 

Bài 3: 

a: \(\left(\dfrac{1}{2}-1\right)\left(\dfrac{1}{3}-1\right)\cdot...\cdot\left(\dfrac{1}{1999}-1\right)\)

\(=\dfrac{-1}{2}\cdot\dfrac{-2}{3}\cdot...\cdot\dfrac{-1998}{1999}\)

\(=\dfrac{1}{1999}\)

b: \(\dfrac{5\cdot18-10\cdot27+15\cdot36}{10\cdot36-20\cdot54+30\cdot72}\)

\(=\dfrac{5\cdot9\left(2-2\cdot3+3\cdot4\right)}{10\cdot18\left(2-2\cdot3+3\cdot4\right)}\)

\(=\dfrac{1}{4}\)

c: Ta có: \(\sqrt{x+4\sqrt{x-4}}=5\)

\(\Leftrightarrow\sqrt{x-4}+2=5\)

\(\Leftrightarrow\sqrt{x-4}=3\)

\(\Leftrightarrow x-4=9\)

hay x=13

c: Ta có: 

hay x=13

\(a,ĐK:x\ge-1\\ PT\Leftrightarrow2x^2+2x-4=x^2+2x+1\\ \Leftrightarrow x^2=5\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{5}\left(tm\right)\\x=-\sqrt{5}\left(ktm\right)\end{matrix}\right.\Leftrightarrow x=\sqrt{5}\\ b,ĐK:x\in R\\ PT\Leftrightarrow\left|x-2\right|=2x-5\\ \Leftrightarrow\left[{}\begin{matrix}x-2=2x-5\left(x\ge2\right)\\x-2=5-2x\left(x< 2\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\left(tm\right)\\x=\dfrac{7}{3}\left(ktm\right)\end{matrix}\right.\Leftrightarrow x=3\)

\(c,ĐK:x\ge4\\ PT\Leftrightarrow\sqrt{\left(\sqrt{x-4}+2\right)^2}=5\\ \Leftrightarrow\sqrt{x-4}=3\Leftrightarrow x-4=9\Leftrightarrow x=5\left(tm\right)\)

ta có tam giác ABC vuông cân tại A nên => góc BCA =ABC =45 độ

tương tự ta có tam giác BDC vuông cân tại B nên ta có góc BDC = góc DCB = 45 độ

=> góc BCA = góc DCB (=45 độ)

mà hai góc này nằm ở vị trí so le trong nên => AB // DC => ABDC là hình thang

Mặt khác hình thang ABDC có góc A vuông nên là hình thang vuông