0.5 nha bạn

Chúc bạn học tốt! :)

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

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

\(\Rightarrow x^2-4+3x+3=3+x^2-2x+x-2\)

\(\Leftrightarrow x^2-x^2+3x+2x-x=1+4-3\)

\(\Leftrightarrow4x=2\)

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

\(ĐKXĐ:\hept{\begin{cases}x+1\ne0\\x-2\ne0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x\ne-1\\x\ne2\end{cases}}\)

Vậy nên x=1/2 thỏa mãn ĐKXĐ nhé!

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

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

\(x^2-4+3x+3=1+x^2-x\) 

\(x^2+3x-1-1-x^2+x=0\) 

\(4x-2=0\) 

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

Vậy.....

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

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

ĐKXĐ: \(x\ne-1,2\)

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

\(\Leftrightarrow\) \(\left(x^2-4\right)\) \(+3.\left(x+1\right)=\)\(3+\left(x+1\right).\left(x-2\right)\)

\(\Leftrightarrow\) x² - 4 + 3x + 3 = 3 + x² - x - 2

\(\Leftrightarrow\) x² + 3x - x² + x = 4 - 3 + 3 - 2

\(\Leftrightarrow\) 4x = 2

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

Vậy phương trình có nghiệm là: \(x=\frac{1}{2}\)

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

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

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

\(\frac{4x-3}{x^2-x-2}=0\)

\(\Leftrightarrow4x-3=0\Leftrightarrow x=\frac{3}{4}\)

Cái bài đầu giải BPT bn ghi cái dj ak ,mik cx k hỉu nữa

V mik giải bài 2 nghen, sửa lại đề bài đầu rồi mik giải cho

\(3x-3=|2x+1|\)

Điều kiện: \(3x-3\ge0\Leftrightarrow3x\ge3\Leftrightarrow x\ge1\)

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

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

\(\Leftrightarrow\orbr{\begin{cases}-x=-3\\5x=2\end{cases}\Leftrightarrow\orbr{\begin{cases}x=3\left(n\right)\\x=\frac{2}{5}\left(l\right)\end{cases}}}\)

Vậy S={3}

Cài đề câu b ,bn xem lại nhé!

\(\frac{2x-3}{35}+\frac{x\left(x-2\right)}{7}>\frac{x^2}{7}-\frac{2x-3}{5}\)

\(\Leftrightarrow\frac{2x-3}{35}+\frac{5x\left(x-2\right)}{35}-\frac{5x^2}{35}+\frac{7\left(2x-3\right)}{35}>0\)

\(\Leftrightarrow2x-3+5x\left(x-2\right)-5x^2+7\left(2x-3\right)>0\)

\(\Leftrightarrow2x-3+5x^2-10x-5x^2+14x-21>0\)

\(\Leftrightarrow6x-24>0\)

\(\Leftrightarrow x>4\)

VẬY TẬP NGHIỆM CỦA BẤT PHƯƠNG TRÌNH LÀ :  S = {  \(x\text{\x}>4\)}

\(\frac{6x+1}{18}+\frac{x+3}{12}\le\frac{5x+3}{6}+\frac{12-5x}{9}\)

\(\Leftrightarrow\frac{6\left(6x+1\right)}{108}+\frac{9\left(x+3\right)}{108}\le\frac{18\left(5x+3\right)}{108}+\frac{12\left(12-5x\right)}{108}\)

\(\Leftrightarrow36x+6+9x+27\le90x+54+144-60x\)

\(\Leftrightarrow36x+6+9x+27-90x-54-144+60x\le0\)

\(\Leftrightarrow15x-165\le0\)

\(\Leftrightarrow x\le11\)

VẬY TẬP NGHIỆM CỦA BẤT PHƯƠNG trình ..........

tk mk nka !!! chúc bạn học tốt !!!

\(x-\frac{\frac{x}{2}-\frac{3+x}{4}}{2}=3-\frac{\left(1-\frac{6-x}{3}\right).\frac{1}{2}}{2}\)

\(\Leftrightarrow2x-\frac{x}{2}+\frac{3+x}{4}=6-\frac{1}{2}+\frac{6-x}{6}\)

\(\Leftrightarrow24x-6x+9+3x=72-6+12-2x\)

\(\Leftrightarrow23x=69\)

\(\Leftrightarrow x=3\)

Vậy nghiệm của pt x=3

Điều kiện: x khác (-3,-2,1,4)

PT <=> 

\(1+\frac{2}{x-1}+1-\frac{4}{x+2}+1-\frac{6}{x+3}+1+\frac{8}{x-4}=4\)

<=> \(\frac{1}{x-1}-\frac{2}{x+2}-\frac{3}{x+3}+\frac{4}{x-4}=0\)

<=> (x+2)(x+3)(x-4)-2(x-1)(x+3)(x-4)-3(x-1)(x+2)(x-4)+4(x-1)(x+2)(x+3)=0

<=> (x³+x²-14x-24)-2(x³ - 2x²-11x+12) - 3(x³ - 3x²- 6x+8) + 4(x³+4x² + x-6) = 0

<=> x³+x²-14x-24-2x³ + 4x²+22x-24 - 3x³ + 9x²+ 18x-24 + 4x³+16x² + 4x-24 = 0

<=> 30x² + 30x -96=0

<=> 5x² + 5x -16 = 0

Giải ra được: \(\orbr{\begin{cases}x_1=\frac{-5-\sqrt{345}}{10}\\x_2=\frac{-5+\sqrt{345}}{10}\end{cases}}\)

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

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

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

\(\left(x-1\right)\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}\right)\)=0

\(x-1=0\)

\(x=1\)

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 :((