\(1+\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+3+...+x}=2\)

\(\Rightarrow1+\frac{1}{2.3}.2+\frac{1}{3.4}.2+...+\frac{1}{x\left(x+1\right)}.2=2\)

=> \(2\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x\left(x+1\right)}\right)=2\)

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

=> \(1-\frac{1}{x+1}=1\)

=> \(\frac{1}{x+1}=0\Rightarrow0\left(x+1\right)=1\Rightarrow x\in\varnothing\)

\(\frac{1}{1.2:2}+\frac{1}{2.3:2}+\frac{1}{3.4:2}+...+\frac{1}{x.\left(x+1\right):2}=2\)

\(\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{x.\left(x+1\right)}=2\)

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

\(1-\frac{1}{x+1}=1\)

\(\frac{1}{x+1}=0\)

Vậy x vô nghiệm.

a) \(x^2+1\ge1\)

\(\Rightarrow x^2+1< 1\)( Vô lí )

=> BPT vô nghiệm 

b) \(x^2+2x< 2x\)

\(\Leftrightarrow x^2+2x-2x< 0\)

\(\Leftrightarrow x^2< 0\)( vô lí )

Vậy BPT vô nghiệm

c) \(x^2-2x+3< -2x+3\)

\(\Leftrightarrow x^2-2x+3+2x-3< 0\)

\(\Leftrightarrow x^2< 0\)

Vậy,,,,,,,,,,,,,,,,,,,

a, \(x^2+1< 1\)(*)

Ta có : \(x^2\ge0< =>x^2+1\ge1\)

Nên không thể bé hơn 1 

Nên (*) vô lí 

b, \(x^2+2x< 2x\)(**)

Ta có : \(x^2\ge0< =>x^2+2x\ge2x\)

Nên không thể bé hơn 2x

Nên (**) vô lí 

c, \(x^2-2x+3< -2x+3\)

\(< =>x^2-2x+2x+3-x< 0\)

\(< =>x^2< 0\)( vô lí )

\(\frac{30303}{80808}=\frac{3\times10101}{8\times10101}=\frac{3}{8}\)

đề bài thiếu : cho tam giác ABC có AB ..........  

Question 1 :

1.How often does Nam have English?

=> Nam has English four times a week.

2.What storybooks can he read in English?

=> He can read Aladdin and the Magic Lamp in English

3.How does he practise speaking?

=> He practises speaking by talking to his foreigh friends

4.How does he practise writing?

=> He practises writing English by sending emails to his friend Hakim in Malaysia.

5.Why does he learn English?

=> Because he wants to watch English cartoons on TV

Question 2 :

My name is Trung. I am a new pupil in class 5B. Today is Tuesday I have five lessons : Maths, Vietnamese, Science, IT and PE.

Tomorrow is Wednesday I'll have four lessons: Maths , Vietnamese, Music and Art 

1.Read and write.

1. Nam have English four times week.

2.He can read Aladdin and the Magic Lamp in English.

3.He practise by talking to his foreigh friends.

4.He practise by sending emails to his friend Hakim in Malaysia.

5.He learn English because his want to watch English cartoons on TV

2.Read and complete

1.Tuseday

2.Lesons

3.Wednesday

4.Four

5.Art

ĐỀ BÀI SAI RỒI !!!!!!

1 SỐ CÓ DẠNG \(\sqrt{x}\ge0\)   VÀ     \(x\ge0\left(đkxđ\right)\)

NHƯNG \(\sqrt{3}< 2\)

=> \(\sqrt{3}-2< 0\)

=> \(\sqrt{\sqrt{3}-2}\)KO TỒN TẠI !!!!!.

Đặt  \(A=\left(\sqrt{6}+\sqrt{2}\right)\left(\sqrt{3}-2\right).\sqrt{\sqrt{3}-2}\)

\(\Rightarrow A^2=\left(\sqrt{6}+\sqrt{2}\right)^2.\left(\sqrt{3}-2\right)^2.\left(\sqrt{3}-2\right)\)

\(\Leftrightarrow A^2=\left(8+4\sqrt{3}\right).\left(-1-4\sqrt{3}\right)\left(\sqrt{3}-2\right)\)

\(\Leftrightarrow A^2=\left(8+4\sqrt{3}\right).\left(-\sqrt{3}+2-12+8\sqrt{3}\right)\)

\(\Leftrightarrow A^2=\left(8+4\sqrt{3}\right)\left(7\sqrt{3}-10\right)\)

\(\Leftrightarrow A^2=56\sqrt{3}-80+84-40\sqrt{3}\)

\(\Leftrightarrow A^2=16\sqrt{3}+4\)

\(\Rightarrow A=\pm\sqrt{16\sqrt{3}+4}\)

    \(\left|x+1\right|+\left|x+2\right|=1\)

\(\Rightarrow\left|x+1+x+2\right|=1\)

\(\Rightarrow\left|2x+3\right|=1\)

\(\Rightarrow\orbr{\begin{cases}2x+3=1\\2x+3=-1\end{cases}\Rightarrow\orbr{\begin{cases}2x=-2\\2x=-4\end{cases}}\Rightarrow\orbr{\begin{cases}x=-1\\x=-2\end{cases}}}\)

     Vậy \(x=-1;x=-2\)

                        Mình nghĩ vậy

Xét \(x< 1\)thì ta có : \(-x-1-x-2=1\)

\(< =>-2x=4< =>x=\frac{4}{-2}=-2\)

Xét \(1\le x< 2\)thì ta có : \(x+1-x-2=1\)

\(< =>-1=1\)(vô lí)

Xét \(x\ge2\)thì ta có : \(x+1+x+2=1\)

\(< =>2x=1-3=-2< =>x=-1\)

Vậy \(x=\left\{-1;-2\right\}\)