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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 ...
\(x\ne\pm2\)
Đặt \(\left\{{}\begin{matrix}\frac{x+3}{x-2}=a\\\frac{x-3}{x+2}=b\end{matrix}\right.\) phương trình trở thành:
\(a^2+6b^2=7ab\)
\(\Leftrightarrow a^2-7ab+6b^2=0\)
\(\Leftrightarrow a^2-ab-6ab+6b^2=0\)
\(\Leftrightarrow a\left(a-b\right)-6b\left(a-b\right)=0\)
\(\Leftrightarrow\left(a-6b\right)\left(a-b\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=b\\a=6b\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\frac{x+3}{x-2}=\frac{x-3}{x+2}\\\frac{x+3}{x-2}=\frac{6\left(x-3\right)}{x+2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(x+3\right)\left(x+2\right)=\left(x-3\right)\left(x-2\right)\\\left(x+3\right)\left(x+2\right)=6\left(x-3\right)\left(x-2\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}5x=-5x\\x^2-7x+6=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=0\\x=1\\x=6\end{matrix}\right.\)
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
.
Sr bạn nha, nhưng điều kiện là \(x\in R\backslash\left\{-5,-4,-3,-2,-1\right\}\). (Xét thiếu :>)
Chúc bạn học tốt nha.
Cái này bạn đặt x+3/x-2 = a
x-3/x+2 = b
=> x^2-9/x^2-4 = ab
Ta có : a^2 - 7ab + 6b^2 = 0
<=> a^2 - 6ab - ab + 6b^2 = 0
PT đa thức thành nhân tử là xong :D
\(48x\left(x+1\right)\left(x^3-4\right)=\left(x^4+8x+12\right)^2\)
\(\Leftrightarrow4\left(12x+12\right)\left(x^4-4x\right)=\left(x^4+8x+12\right)^2\)
Đặt \(\left\{{}\begin{matrix}x^4-4x=a\\12x+12=b\end{matrix}\right.\)
\(\Rightarrow4ab=\left(a+b\right)^2\)
\(\Leftrightarrow4ab=a^2+a^2+2ab\)
\(\Leftrightarrow\left(a-b\right)^2=0\)
\(\Leftrightarrow a-b=0\)
\(\Leftrightarrow x^4-16x-12=0\)
\(\Leftrightarrow\left(x^2-2x-2\right)\left(x^2+2x+6\right)=0\)
\(\Leftrightarrow x^2-2x-2=0\)
\(\Rightarrow x=1\pm\sqrt{3}\)
cho e hỏi vs ạ. sao từ \(x^4-16x-12=0\) lại ra \(\left(x^2-2x-2\right)\left(x^2+2x+6\right)=0\) ạ?
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 \(\left\{{}\begin{matrix}\dfrac{x-1}{x+2}=a\\\dfrac{x+2}{x-3}=b\end{matrix}\right.\)
\(\Rightarrow a^2-4b^2+3ab=0\Leftrightarrow\left(a-b\right)\left(a+4b\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a-b=0\\a+4b=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\dfrac{x-1}{x+2}-\dfrac{x+2}{x-3}=0\\\dfrac{x-1}{x+2}+\dfrac{4x+8}{x-3}=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(x-1\right)\left(x-3\right)-\left(x+2\right)^2=0\\\left(x-\right)\left(x-3\right)+4\left(x+2\right)^2=0\end{matrix}\right.\)
\(\Leftrightarrow...\)
a) \(\left(x^2+2x+2\right)\left(x^2+2x+3\right)=0\)
<=> \(\orbr{\begin{cases}x^2+2x+2=0\\x^2+2x+3=0\end{cases}}\)
<=> \(\orbr{\begin{cases}\left(x+1\right)^2+1=0\left(vl\right)\\\left(x+1\right)^2+2=0\left(vl\right)\end{cases}}\)
=> pt vô nghiệm
b) \(\left(x+3\right)\left(x-3\right)\left(x^2-11\right)+3=2\)
<=> \(\left(x^2-9\right)\left(x^2-11\right)+1=0\)
<=> \(\left(x^2-9\right)^2-2\left(x^2-9\right)+1=0\)
<=> \(\left(x^2-9-1\right)^2=0\)
<=> \(x^2-10=0\)
<=> \(x=\pm\sqrt{10}\)
c) \(\left(x+3\right)^4+\left(x+5\right)^4=2\)
<=> \(\left(x+4-1\right)^4+\left(x+4+1\right)^4=2\)
Đặt x + 4 = a
<=> \(\left(a-1\right)^4+\left(a+1\right)^4=2\)
<=> \(a^4-4a^3+6a^2-4a+1+a^4+4a^3+6a^2+4a+1=2\)
<=> \(a^4+12a^2=0\)
<=> \(a^2\left(a^2+12\right)=0\)
<=> a = 0 (vì a2 + 12 > 0)
Vậy S = {0}
\(\left(x+2\right)\left(x-3\right)+3=\left(x-4\right)\left(x+2\right)-7\)
\(\Leftrightarrow x^2-x-6+3=x^2-2x-8-7\)
\(\Leftrightarrow x^2-x-x^2+2x=6-3-8-7\)
\(\Leftrightarrow x=-12\)
Vậy: Phương trình có tập nghiệm \(S=\left\{-12\right\}\)
\(x^2-3x+2x-6+3-x^2-2x+4x+8+7=0\)
\(x+12=0\)
\(x=-12\)