a. \(\frac{x+1}{2}=\frac{8}{x+1}\)

\(\Leftrightarrow\left(x+1\right).\left(x+1\right)=8.2\)

\(\Leftrightarrow\left(x+1\right)^2=16\)

\(\Leftrightarrow\left(x+1\right)^2=2^4\)

\(\Leftrightarrow\left(x+1\right)=2^2\)

\(\Leftrightarrow\left(x+1\right)=4\)

\(\Leftrightarrow x=4-1=3\)

b. \(x:\left(9\frac{1}{2}-\frac{3}{2}\right)=\frac{0,4+\frac{2}{9}-\frac{2}{11}}{1,6+\frac{8}{9}-\frac{8}{11}}\)

\(\Leftrightarrow x:\left(\frac{10}{2}-\frac{3}{2}\right)=\frac{0,4+0,2-0,18}{1,6+0,8-0,72}\)

\(\Leftrightarrow x:\frac{7}{2}=\frac{\frac{21}{50}}{\frac{42}{25}}\)

\(\Leftrightarrow x=\frac{\frac{21}{50}}{\frac{42}{25}}.\frac{7}{2}\Leftrightarrow x=\frac{1}{4}.\frac{7}{2}=\frac{7}{8}\)

a )  \(\frac{x+1}{2}=\frac{8}{x+1}\)

\(\Rightarrow\left(x+1\right).\left(x+1\right)=2.8\)

\(\Rightarrow\left(x+1\right)^2=16\)

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

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

\(\Rightarrow\orbr{\begin{cases}x=3\\x=-5\end{cases}}\)

Dấu " \(\orbr{\begin{cases}\\\end{cases}}\)là hoặc nha !!! 

a)\(\frac{x+1}{2}=\frac{8}{x+1}\)

\(\Rightarrow\left(x+1\right)^2=8\times2=16\)

\(\Rightarrow\left(x+1\right)^2=4^2\)

\(\Rightarrow x+1=4\)

\(\Rightarrow x=4-1=3\)

b)tớ ko biết làm . ~ sorry~

~chuk bn hok giỏi~

1, đặt k 2, nhân chéo

Áp dụng tính chất dãy tỉ số bằng nhau 

ta có ; \(\frac{x^2}{4}=\frac{y}{2}=\frac{x^2-3y}{4-\left(3.2\right)}=\frac{-8}{-2}=4.\)

\(\Rightarrow\frac{x^2}{4}=4\Rightarrow x^2=16\Rightarrow x=\pm4\)

\(\Rightarrow\frac{y}{2}=4\Rightarrow y=8\)

b) \(\frac{x+1}{8}=\frac{8}{x+1}\)

\(\Rightarrow\left(x+1\right)^2=64=8^2=\left(-8\right)^2\)

=> x + 1 = 8     => x = 7

x+1 = -8 => x = -9

KL:...

1)

\(2\frac{1}{4}x-9\frac{1}{4}=-7\frac{1}{4}\)

\(2\frac{1}{4}x=\left(-7\frac{1}{4}\right)+9\frac{1}{4}\)

\(2\frac{1}{4}x=2\)

\(x=2:2\frac{1}{4}\)

\(x=\frac{8}{9}\)

Vậy \(x=\frac{8}{9}\)

\(\frac{x+1}{9}+\frac{x+2}{8}+\frac{x+3}{7}+...+\frac{x+9}{1}=-9\)

\(\left(\frac{x+1}{9}+1\right)+\left(\frac{x+2}{8}+1\right)+\left(\frac{x+3}{7}+1\right)+...+\left(\frac{x+9}{1}+1\right)=0\)

\(\frac{x+10}{9}+\frac{x+10}{8}+\frac{x+10}{7}+...+\frac{x+10}{1}=0\)

\(\left(x+10\right).\left(\frac{1}{9}+\frac{1}{8}+\frac{1}{7}+...+1\right)=0\)

vì \(\frac{1}{9}+\frac{1}{8}+\frac{1}{7}+...+1\ne0\)

\(\Rightarrow x+10=0\)

\(\Rightarrow x=-10\)

SKT_NTT làm đúng ùi -_- 100 %

\(\left(\frac{x+1}{9}+1\right)+\left(\frac{x+2}{8}+1\right)+...+\left(\frac{x+9}{1}+1\right)=-9+9\)

\(\Rightarrow\frac{x+10}{9}+\frac{x+10}{8}+...+\frac{x+10}{1}\) =

a) \(\left(x+1\right)+\left(x+2\right)+\left(x+3\right)+...+\left(x+100\right)=5750\)

\(\Rightarrow\left(x+x+x+...+x\right)+\left(1+2+3+..+100\right)=5750\Rightarrow x.100+\left(100+1\right)\cdot100:2=5750\)\

\(\Rightarrow x.100+5050=5750\Rightarrow x.100=700\Rightarrow x=7\)

b) \(\frac{x+1}{2}=\frac{8}{x+1}\Rightarrow\left(x+1\right)\left(x+1\right)=2.8\)

\(\Rightarrow\left(x+1\right)^2=16\Rightarrow\left(x+1\right)^2=4^2\)

\(\Leftrightarrow x+1=4\Rightarrow x=3\)

1.\(\left(x+1\right)+\left(x+2\right)+\left(x+3\right)+...+\left(x+100\right)=5750\)

\(\Leftrightarrow\left(x+x+x+...+x\right)+\left(1+2+3+...+100\right)=5750\)

\(\Leftrightarrow100x+5050=5750\)

\(\Leftrightarrow100x=5750-5050=700\)

\(\Leftrightarrow x=700:100=7\)

2.   \(\frac{x+1}{2}=\frac{8}{x+1}\)

\(\Leftrightarrow\left(x+1\right).\left(x+1\right)=8.2\)

\(\Leftrightarrow\left(x+1\right).\left(x+1\right)=16\)

\(\Leftrightarrow\left(x+1\right)^2=16\)

\(\Leftrightarrow\left(x+1\right)=16:2\)

\(\Leftrightarrow\left(x+1\right)=8\)

\(\Leftrightarrow x=8-1=7\)

Bạn sửa lại chỗ bài 2. Ở chỗ:

\(\left(x+1\right)^2=16\) và các chỗ sau chỗ đó thành:

\(\left(x+1\right)^2=16\)

\(\Leftrightarrow\left(x+1\right)^2=4^2\)

\(\Leftrightarrow x+1=4\)

\(\Leftrightarrow x=4-1=3\)

\(\left(x+\frac{1}{2}\right)+\left(x+\frac{1}{4}\right)+\left(x+\frac{1}{8}\right)+\left(x+\frac{1}{16}\right)=\frac{23}{16}\)

\(4x+\frac{15}{16}=\frac{23}{16}\)

\(4x=\frac{1}{2}\)

\(x=\frac{1}{8}\)

             Vậy \(x=\frac{1}{8}\)

\(\left(x+\frac{1}{2}\right)+\left(x+\frac{1}{4}\right)+\left(x+\frac{1}{8}\right)+\left(x+\frac{1}{16}\right)=\frac{23}{16}\)

\(\Rightarrow\left(x+x+x+x+x\right)+\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}\right)=\frac{23}{16}\)

\(\Rightarrow5x+\frac{15}{32}=\frac{23}{16}\)

\(\Rightarrow5x=\frac{23}{16}-\frac{15}{32}\)

\(\Rightarrow5x=\frac{31}{32}\)

\(\Rightarrow x=\frac{31}{32}.\frac{1}{5}=\frac{31}{160}\)

Có 4x thôi Nguyễn Quang Trung

\(x:\frac{1}{2}+x:\frac{1}{4}+x:\frac{1}{8}+..+x:\frac{1}{512}=x.2+x.4+x.8+...x.512=x\left(2+4+8+...+512\right)=...\)

A\(A=\frac{1}{1.3}+..+\frac{1}{x\left(x+1\right)}\)

\(2A=\frac{1}{1}-\frac{1}{\left(x+1\right)}\)

\(A=\frac{x}{2.\left(x+1\right)}=\frac{8}{17}=\frac{16}{2.17}\)

X=16

17 - 1= 16

= > x = 16

 tk mình nha

17 - 1= 16

= > x = 16

 tk mình nha