ĐKXĐ: ...

\(\left(\dfrac{x-1}{x+2}\right)^2-4\left(\dfrac{x+2}{x-3}\right)^2+3\left(\dfrac{x-1}{x-3}\right)=0\)

Đặt \(\left\{{}\begin{matrix}\dfrac{x-1}{x+2}=a\\\dfrac{x+2}{x-3}=b\end{matrix}\right.\)

\(\Rightarrow a^2-4b^2+3ab=0\Leftrightarrow\left(a-b\right)\left(a+4b\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}a-b=0\\a+4b=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\dfrac{x-1}{x+2}-\dfrac{x+2}{x-3}=0\\\dfrac{x-1}{x+2}+\dfrac{4x+8}{x-3}=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left(x-1\right)\left(x-3\right)-\left(x+2\right)^2=0\\\left(x-\right)\left(x-3\right)+4\left(x+2\right)^2=0\end{matrix}\right.\)

\(\Leftrightarrow...\)

Giải:

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

ĐKXĐ: \(x\ne\left\{1;2;3;4\right\}\)

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

\(\Rightarrow\left(x-3\right)\left(x-4\right)+\left(x-1\right)\left(x-4\right)=\left(x-1\right)\left(x-2\right)+\left(x-2\right)\left(x-3\right)\)

\(\Leftrightarrow\left(x-4\right)\left[\left(x-3\right)+\left(x-1\right)\right]=\left(x-2\right)\left[\left(x-1\right)+\left(x-3\right)\right]\)

\(\Leftrightarrow x-4=x-2\)

\(\Leftrightarrow0x=2\)

Vậy ...

a); b) Do tích = 0 

=> Từng thừa số = 0 và ta nhận xét: \(x^2+2;x^2+3>0\)

=> a) \(\orbr{\begin{cases}x=1\\x=-\frac{5}{2}\end{cases}}\)

và câu b) \(\orbr{\begin{cases}x=\frac{1}{2}\\x=5\end{cases}}\)

a; *x-1=0 <=>x=1

    *2x+5=0 <=>x=-2,5

    *x²+2=0 <=> ko có x

b; tương tự a

a/ \(\left(x-1\right)\left(2x+5\right)\left(x^2+2\right)=0\)

Vì \(x^2\ge0\Rightarrow x^2+2\ge2>0\)

\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\2x+5=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=1\\x=-\frac{5}{2}\end{cases}}\)

\(a,x^2+2x+1=4.\left(x^2-2x+1\right)\)

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

\(\Leftrightarrow\left(x+1\right)^2-\left(2x-2\right)^2=0\)

\(\Leftrightarrow\left(x+1+2x+2\right).\left(x+1-2x+2\right)=0\)

\(\Leftrightarrow\left(3x+3\right).\left(-x+3\right)=0\)

tự làm tiếp

\(x.\left(x-1\right).\left(x+2\right)-\left(x-5\right).\left(x^2-x+1\right)-7x^2=0\)

\(\Leftrightarrow\left(x^3+x^2-2x\right)-\left(x^3-6x^2+6x-5\right)-7x^2=0\)

\(\Leftrightarrow\left(x^3-6x^2-2x\right)-\left(x^3-6x^2-2x+8x-5\right)=0\)

\(\Leftrightarrow-8x+5=0\)

\(\Leftrightarrow-8x=-5\Rightarrow x=\frac{5}{8}\)

Vậy...

>: sr, t làm lộn

dòng thứ 4\(\left(x+1+2x-2\right).\left(x+1-2x+2\right)=0\)

\(\Leftrightarrow\left(3x-1\right).\left(-x+3\right)=0\)

\(\left(x+1\right)\left(x-1\right)^2-\left(x+1\right)\left(x-2\right)^2=0\)

\(\Leftrightarrow\left(x+1\right)\left[\left(x-1\right)^2-\left(x-2\right)^2\right]=0\)

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

\(\Leftrightarrow\left(x+1\right).1.\left(2x-3\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(2x-3\right)=0\)

\(TH1:x+1=0\)

\(\Leftrightarrow x=-1\)

\(TH2:2x-3=0\)

\(\Leftrightarrow2x=3\)

\(\Leftrightarrow x=\frac{3}{2}\)

Vậy phương trình có tập nghiệm là:\(S=\left\{-1;\frac{3}{2}\right\}\)

\(\left(x+1\right)\left(x-1\right)^2-\left(x+1\right)\left(x-2\right)^2=0\)

\(\Leftrightarrow\left(x+1\right)\left[\left(x-1\right)^2-\left(x-2\right)^2\right]=0\)

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

\(\Leftrightarrow\left(x+1\right)\left[1.\left(2x-3\right)\right]=0\)

\(\Leftrightarrow\left(x+1\right)\left(2x-3\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x+1=0\\2x-3=0\end{cases}}\)

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

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

Vậy................

ĐKXĐ: \(x\ne\left\{2;4\right\}\)

Đặt \(\left\{{}\begin{matrix}\dfrac{x+1}{x-2}=a\\\dfrac{x-2}{x-4}=b\end{matrix}\right.\) \(\Rightarrow\dfrac{x+1}{x-4}=ab\)

Phương trình trở thành:

\(a^2-12b^2+ab=0\)

\(\Leftrightarrow a^2+4ab-3ab-12b^2=0\)

\(\Leftrightarrow a\left(a+4b\right)-3b\left(a+4b\right)=0\)

\(\Leftrightarrow\left(a-3b\right)\left(a+4b\right)=0\Leftrightarrow\left[{}\begin{matrix}a-3b=0\\a+4b=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{x+1}{x-2}-\dfrac{3\left(x-2\right)}{x-4}=0\\\dfrac{x+1}{x-2}+\dfrac{4\left(x-2\right)}{x-4}=0\end{matrix}\right.\)

Bạn tự quy đồng và hoàn thành phần còn lại nhé

Đây không phải toán lớp 1 đâu bạn

Tớ không biết vì tớ mới lớp 5

K mk nha

*Mio*

Tự đăng bài rồi tự làm luôn à bn .

Đây ko pk là Toán lớp nhá 

Học tôt nhé bn

# MissyGirl #

đang giải trên face mà bạn , làm xong chụp gửi nó

ĐK: \(x\in R\backslash\left\{-4,-3,-2,-1\right\}\)

PT ban đầu

\(\Leftrightarrow\frac{x+2-x-1}{\left(x+1\right)\left(x+2\right)}+\frac{x+3-x-2}{\left(x+2\right)\left(x+3\right)}+\frac{x+4-x-3}{\left(x+3\right)\left(x+4\right)}+\frac{x+5-x-4}{\left(x+4\right)\left(x+5\right)}=\frac{1}{x+1}-403\\ \Leftrightarrow\frac{1}{x+1}-\frac{1}{x+2}+\frac{1}{x+2}-\frac{1}{x+3}+\frac{1}{x+3}-\frac{1}{x+4}+\frac{1}{x+4}-\frac{1}{x+5}=\frac{1}{x+1}-403\\ \Leftrightarrow\frac{1}{x+5}=403\\ \Leftrightarrow x+5=\frac{1}{403}\Leftrightarrow x=\frac{-2014}{403}\)

Chúc bạn học tốt nha.