Mày nhìn cái chóa j

ĐKXĐ: \(x\ne\left\{0;-1;-2;-3;-4;-5;-6;-7\right\}\)

\(\frac{1}{x}+\frac{1}{x+2}+\frac{1}{x+5}+\frac{1}{x+7}=\frac{1}{x+1}+\frac{1}{x+3}+\frac{1}{x+4}+\frac{1}{x+6}\)

\(\Rightarrow\frac{1}{x}+\frac{1}{x+7}+\frac{1}{x+2}+\frac{1}{x+5}=\frac{1}{x+1}+\frac{1}{x+6}+\frac{1}{x+3}+\frac{1}{x+4}\)

\(\Rightarrow\frac{x+7+x}{x\left(x+7\right)}+\frac{x+5+x+2}{\left(x+2\right)\left(x+5\right)}=\frac{x+6+x+1}{\left(x+1\right)\left(x+6\right)}+\frac{x+4+x+3}{\left(x+3\right)\left(x+4\right)}\)

\(\Rightarrow\frac{2x+7}{x^2+7x}+\frac{2x+7}{x^2+7x+10}=\frac{2x+7}{x^2+7x+6}+\frac{2x+7}{x^2+7x+12}\)

\(\Rightarrow\left(2x+7\right)\left(\frac{1}{x^2+7x}+\frac{1}{x^2+7x+10}-\frac{1}{x^2+7x+6}-\frac{1}{x^2+7x+12}\right)=0\)

mà \(\frac{1}{x^2+7x}+\frac{1}{x^2+7x+10}-\frac{1}{x^2+7x+6}-\frac{1}{x^2+7x+12}\ne0\)

=> 2x + 7 = 0 => x = -7/2 

                                                                              Vậy x = -7/2

Bài làm:

PT:

đkxđ: \(x\ne0;x\ne2\)

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

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

\(\Rightarrow x^2+2x=2+x-2\)

\(\Leftrightarrow x^2+x=0\)

\(\Leftrightarrow x\left(x+1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\left(vl\right)\\x+1=0\end{cases}}\Rightarrow x=-1\)

BPT:

Ta có: \(\frac{x+1}{2}-x\le\frac{1}{2}\)

\(\Leftrightarrow\frac{x+1}{2}-x-\frac{1}{2}\le0\)

\(\Leftrightarrow\frac{x+1-2x-1}{2}\le0\)

\(\Leftrightarrow\frac{-x}{2}\le0\)

\(\Rightarrow-x\le0\)

\(\Rightarrow x\ge0\)

a) \(ĐKXĐ:\hept{\begin{cases}x\ne0\\x\ne2\end{cases}}\)

\(\frac{x+2}{x-2}=\frac{2}{x^2-2x}+\frac{1}{x}\)

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

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

\(\Leftrightarrow-x^2-x=0\)

\(\Leftrightarrow-x\left(x+1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x+1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\left(ktm\right)\\x=-1\left(tm\right)\end{cases}}}\)

Vậy \(S=\left\{-1\right\}\)

b) \(\frac{x+1}{2}-x\le\frac{1}{2}\)

\(\Leftrightarrow x+1-2x-1\le0\)

\(\Leftrightarrow-x\le0\)

\(\Leftrightarrow x\ge0\)

Vậy \(x\ge0\)

ĐKXĐ : \(x\ne0;2\)

\(\frac{x+2}{x-2}=\frac{2}{x^2-2x}+\frac{1}{x}\)

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

\(\Leftrightarrow x^2+2x=2+x-2\)

\(\Leftrightarrow x^2+x=0\Leftrightarrow x\left(x+1\right)=0\Leftrightarrow\orbr{\begin{cases}x=0\left(ktm\right)\\x=-1\left(tm\right)\end{cases}}\)

Điều kiện \(x\ne0\) nhận thấy 

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

Do đó phương trình tương đương với 

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

\(\Leftrightarrow2^{\frac{1-x^2}{x^2}}+\frac{1}{2}.\frac{1-x^2}{x^2}=2^{\frac{1-2x}{x^2}}+\frac{1}{2}.\frac{1-2x}{x^2}\)

Mặt khác \(f\left(t\right)=2^t+\frac{t}{2}\) là hàm đồng biến trên R

Do đó từ : \(f\left(\frac{1-x^2}{x^2}\right)=f\left(\frac{1-2x}{x^2}\right)\)

Suy ra 

\(\frac{1-x^2}{x^2}=\frac{1-2x}{x^2}\)

Từ đó dễ dàng tìm ra được x=2 là nghiệm duy nhất của phương trình

pt <=> 1/(x+2).(x+3) + 1/(x+3).(x+4) + 1/(x+4).(x+5) + 1/(x+5).(x+6) = 1/8

<=> 1/x+2 - 1/x+3 + 1/x+3 - 1/x+4 + 1/x+4 - 1/x+5 + 1/x+5 - 1/x+6 = 1/8

<=> 1/x+2 - 1/x+6 = 1/8

<=> (x+6-x-2)/(x+2).(x+6) = 1/8

<=> 4/(x+2).(x+6) = 1/8

<=>(x+2).(x+6) = 4 : 1/8 = 32

<=>x^2 + 8x + 12 = 32

<=> x^2+8x+12-32=0

<=>x^2+8x-20=0

<=>(x-2).(x+10)=0

<=> x-2 =0 hoặc x+10 = 0

<=> x=2 hoặc x=-10

giang sinh an lanh $%###Xuyen gam cu chuoi###%$

ĐKXĐ \(x\ne0,-1,-2,...,-100\)

\(\frac{1}{x^2+x}+\frac{1}{x^2+3x+2}+...+\frac{1}{x^2+199x+9900}=\frac{25}{51}\)

\(\Leftrightarrow\frac{1}{x\left(x+1\right)}+\frac{1}{x^2+x+2x+2}+...+\frac{1}{x^2+99x+100x+9900}=\frac{25}{51}\)

\(\Leftrightarrow\frac{1}{x\left(x+1\right)}+\frac{1}{x\left(x+1\right)+2\left(x+1\right)}+....+\frac{1}{x\left(x+99\right)+100\left(x+99\right)}=\frac{25}{51}\)

\(\Leftrightarrow\frac{1}{x\left(x+1\right)}+\frac{1}{\left(x+1\right)\left(x+2\right)}+...+\frac{1}{\left(x+99\right)\left(x+100\right)}=\frac{25}{21}\)

\(\Leftrightarrow\frac{1}{x}-\frac{1}{x+1}+\frac{1}{x+1}-\frac{1}{x+2}+...+\frac{1}{x+99}-\frac{1}{x+100}=\frac{25}{21}\)

\(\Leftrightarrow\frac{1}{x}-\frac{1}{x+100}=\frac{25}{21}\)

\(\Leftrightarrow\frac{x+100-x}{x\left(x+100\right)}=\frac{25}{21}\)

\(\Leftrightarrow\frac{100}{x\left(x+100\right)}=\frac{25}{21}\)

\(\Leftrightarrow25x^2+2500x=2100\)

\(\Leftrightarrow x^2+100x-84=0\)

\(\Leftrightarrow x^2+2.x.50+50^2-50^2-84=0\)

\(\Leftrightarrow\left(x+50\right)^2-2584=0\)

\(\Leftrightarrow\left(x+50-2\sqrt{646}\right)\left(x+50+2\sqrt{646}\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=-50+2\sqrt{646}\\x=-50-2\sqrt{646}\end{cases}}\)

Vậy ...

Lê Tài Bảo Châu Vậy ý bạn là \(x^2+4x+3=\left(x+2\right)\left(x+3\right)\)?????

Ban đầu mik cũng có ý tưởng như bạn nhưng thấy nó k đúng với hạng tử thứ 3, xong mới đăng lên đây tìm lời giải khác á~

p/s: nhưng cũng có thể xảy ra trường hợp đề bài sai :((

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

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

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

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

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

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

\(\frac{4x+6}{3x+5}=\frac{6}{3x-1}\)

\(\Rightarrow\left(4x+6\right)\left(3x-1\right)=6\left(3x+5\right)\)

\(\Rightarrow12x^2-4x+18x-6=18x+30\)

\(\Rightarrow12x^2-4x+18x-18x=30+6\)

\(\Rightarrow12x^2-4x-36=0\)

\(\Rightarrow3x^2-x-9=0\)

\(\Rightarrow x^2-\frac{1}{3}x-3=0\)

\(\Rightarrow x^2-2.\frac{1}{6}x+\frac{1}{36}-\frac{1}{36}-3=0\)

\(\Rightarrow\left(x-\frac{1}{6}\right)^2-\frac{109}{36}=0\)

\(\Rightarrow\left(x-\frac{1}{6}-\frac{\sqrt{109}}{6}\right)\left(x-\frac{1}{6}+\frac{\sqrt{109}}{6}\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{1+\sqrt{109}}{6}\\x=\frac{1-\sqrt{109}}{6}\end{cases}}\)

làm lại nhé, chỗ kia quy đồng sai 

lần này làm theo cách khác

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

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

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

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

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

\(\Rightarrow3x+6=-2x-3\)

\(\Rightarrow3x+2x=-3-6\)

\(\Rightarrow5x=-9\)

\(\Rightarrow x=\frac{-9}{5}\)

vậy \(x=\frac{-9}{5}\)