Ta có : \(\frac{3}{2x+1}+\frac{10}{4x+2}-\frac{6}{6x+3}=\frac{12}{26}\)

\(\Rightarrow\frac{3}{2x+1}+\frac{5.2}{2\left(2x+1\right)}-\frac{3.2}{3\left(2x+1\right)}=\frac{6}{13}\)

=> \(\frac{3}{2x+1}+\frac{5}{2x+1}-\frac{2}{2x+1}=\frac{6}{13}\)

=> \(\frac{3+5-2}{2x+1}=\frac{6}{13}\)

=> \(\frac{6}{2x+1}=\frac{6}{13}\)

=> 2x + 1 = 13

=> 2x = 12

=> x = 6

Vậy x = 6

\(\frac{3}{2x+1}+\frac{10}{2\left(2x+1\right)}-\frac{6}{3\left(2x+1\right)}=\frac{6}{13}\)                

\(\frac{3}{2x+1}+\frac{5}{2x+1}-\frac{2}{2x+1}=\frac{6}{13}\)                

\(\frac{6}{2x+1}=\frac{6}{13}\)  

\(\Rightarrow2x+1=13\left(6=6\right)\)         

\(2x=12\)    

\(x=6\)     

We usually play volleyball together on Sunday

We usually play volleyball together on Sunday

1 giờ vòi chảy dc:

1:10=1/10(bể)

7 giờ chảy dc:

7:10=7/10(bể)

8 giờ chảy dc:

8:10=8/10(bể)

đáp số:............ 

tự ghi nha bạn

1 giờ vòi nước chảy được số phần bể là

\(1\div10=\frac{1}{10}\) bể 

7 giờ vòi nước chảy được số phần bể là

\(7\div10=\frac{7}{10}\) bể

8 giờ vòi nước chảy được số phần bể là

\(8\div10=\frac{8}{10}\) bể

Hok tốt !!!!!!!!!!!!

1.is looking

2.starts

3.learned hoặc learnt

Put the verbs in brackets in the correct verb tense :

1. Be careful. The teacher [look ] _is looking______ at you.

2. Our school performance [ start ] __started________ late last Sunday because of the heavy rain.

3. The boy [ learn] _has learnt______ for three years, but he can't understand this letter.

4. Would you some coffee? I [ just / make ] __have just made__________ some.

5.[ she/ feed] _Have she fed_______ the cat yet ?

6. [ your dog/ ever bite ] _Has your dog ever biten__________ anyone ?

7.[ you/ ever be ] ____Have_you ever been_____ to Ha Noi ?

8.I [not be ] __was't_______very happy yesterday.

9. The people in the cafe' [ not be] __wasn't_________ friendly when I was there yesterday .

10. I [ leave] __have left________ my school bag at scool this morning .

11. It [be] __was_____ a great flim in 2001.

12.Our teacher [ tell] __told________ us to be quiet yesterday. 

13.I went to the shop but I [ not have] ___didn't have______ any money . 

14.Susan [ not know] __didb't know_________ about the exam and she did very badly.

15.I [ buy] ___bought______ a ket for the football match yesterday.

Bài làm:

Ta có: \(\frac{12^6.5^2-2^{10}.3^5.5^2}{12^6.10-2^{12}.3^8}\)

\(=\frac{2^{12}.3^6.5^2-2^{10}.3^5.5^2}{2^{13}.3^6.5-2^{12}.3^8}\)

\(=\frac{2^{10}.3^5.5^2.\left(2^2.3-1\right)}{2^{12}.3^6.\left(2.5-3^2\right)}\)

\(=\frac{5^2.11}{2^2.3}=\frac{275}{12}\)

Tổng lúc đầu của hai số là :

          244 412 - (8576 + 1268) = 234 568

Số lớn là :

          (234 568 + 16384) : 2 = 125 476

Số bé là :

          125 476 - 16 384 = 109 092

Thử lại : 125 476 + 109 092 = 234 568(đúng)

                                             Đáp số : Số lớn = 125 476

                                                             Số bé = 109 092

Nếu đúng thì bạn Ti ck cho mình nha!

Cám ơn rất nhiều !

Cảm ơn bạn Phạm Thị Bích nha

Bài làm:

Ta có: \(2\cdot\left(2-x\right)+\frac{1}{2}\cdot\left(2-x\right)^2=0\)

\(\Leftrightarrow\left(2-x\right)\left[2+\frac{1}{2}\left(2-x\right)\right]=0\)

\(\Leftrightarrow\left(2-x\right)\left(3-\frac{x}{2}\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}2-x=0\\3-\frac{x}{2}=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=2\\x=\frac{3}{2}\end{cases}}\)

2( 2 - x ) + 1/2( 2 - x )²

Đa thức có nghiệm <=> 2( 2 - x ) + 1/2( 2 - x )² = 0

                               <=> ( 2 - x )[ 2 + 1/2( 2 - x ) ] = 0

                               <=> ( 2 - x )[ 2 + 1 - 1/2x ]

                               <=> ( 2 - x )( 3 - 1/2x ) = 0

                               <=> \(\orbr{\begin{cases}2-x=0\\3-\frac{1}{2}x=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=2\\x=6\end{cases}}\)

\(ĐKXĐ:x\ge0\)

Ta có : \(D=\frac{2011x-2\sqrt{x}+1}{\sqrt{x}}=2011\sqrt{x}+\frac{1}{\sqrt{x}}-2\)

Theo BĐT AM - GM ta có :

\(2011\sqrt{x}+\frac{1}{\sqrt{x}}\ge2\sqrt{2011\sqrt{x}\cdot\frac{1}{\sqrt{x}}}=2\sqrt{2011}\)

\(\Rightarrow2011\sqrt{x}+\frac{1}{\sqrt{x}}-2\ge2\left(\sqrt{2011}-1\right)\)

Dấu "=" xảy ra \(\Leftrightarrow x=\frac{1}{2011}\)

Vậy \(D_{min}=2\left(\sqrt{2011}-1\right)\) tại \(x=\frac{1}{2011}\)

\(A=75.(4^{2004}+4^{2003}+...+4^2+4+1)+25\)

Đặt \(B=4^{2004}+4^{2003}+...+4^2+4+1\)

\(4B=4^{2005}+4^{2004}+...+4^3+4^2+4\)

\(4B-B=(4^{2005}+4^{2004}+...+4^3+4^2+4)-\left(4^{2004}+4^{2003}+...+4^2+4+1\right)\)

\(3B=4^{2005}-1\)

\(B=\frac{4^{2005}-1}{3}\)

Thay B vào A ta có

\(A=75.\text{​​}\text{​​}\frac{4^{2005}-1}{3}+25\)

\(A=25.3.(\text{​​}\text{​​}\frac{4^{2005}-1}{3})+25\)

\(A=25.(\text{​​}\text{​​}4^{2005}-1)+25\)

\(A=25.(\text{​​}\text{​​}4^{2005}-1+1)\)

\(A=25.\text{​​}\text{​​}4^{2005}\)

Hok tốt !!!!!!!!!

\(A=75\left(4^{2004}+4^{2003}+4^{2002}+...+4^2+4+1\right)+25\)

\(=75\cdot4^{2004}+75\cdot4^{2003}+75\cdot4^{2002}+...+7\cdot4^2+75\cdot4+\left(75+25\right)\)

\(=3\cdot\left(25\cdot4\right)\cdot4^{2003}+3\cdot\left(25\cdot4\right)\cdot4^{2002}+3\cdot\left(25\cdot4\right)\cdot4^{2001}+...+3\cdot\left(25\cdot4\right)\cdot4+3\cdot\left(25\cdot4\right)+25\cdot4\)

\(=3\cdot100\cdot4^{2003}+3\cdot100\cdot4^{2002}+3\cdot100\cdot4^{2001}+...+3\cdot100\cdot4+3\cdot100+100\)

Mà:

\(3\cdot100\cdot4^{2003}⋮100\)

\(3\cdot100\cdot4^{2002}⋮100\)

\(3\cdot100\cdot4^{2001}⋮100\)

\(...\)

\(3\cdot100⋮100\)

\(100⋮100\)

\(\Rightarrow3\cdot100\cdot4^{2003}+3\cdot100\cdot4^{2002}+3\cdot100\cdot4^{2001}+...+3\cdot100\cdot4+3\cdot100+100⋮100\)

\(\Rightarrow A⋮100\left(đpcm\right)\)