Giải phương trình\(\left(\frac{x+1}{x-2}\right)^2+\frac{x+1}{x-4}=12\cdot\left(\frac{x-2}{x-4}\right)^2\)
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29 tháng 7 2019
\(\text{a) }10\left(\frac{x-2}{x+1}\right)^2+\left(\frac{x+2}{x-1}\right)^2-11\frac{x^2-4}{x^2-1}=0\\ DKXD:x\ne-1;x\ne1\\ \Leftrightarrow10\left(\frac{x-2}{x+1}\right)^2+\left(\frac{x+2}{x-1}\right)^2-11\frac{\left(x+2\right)\left(x-2\right)}{\left(x+1\right)\left(x-1\right)}=0\)
Đặt \(\frac{x-2}{x+1}=a;\frac{x+2}{x-1}=b\)
\(Pt\Leftrightarrow10a^2+b^2-11ab=0\\ \Leftrightarrow10a^2-10ab-ab+b^2=0\\ \Leftrightarrow10a\left(a-b\right)-b\left(a-b\right)=0\\ \Leftrightarrow\left(10a-b\right)\left(a-b\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}10a-b=0\\a-b=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}10a=b\\a=b\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\frac{10\left(x-2\right)}{x+1}=\frac{x+2}{x-1}\left(1\right)\\\frac{x-2}{x+1}=\frac{x+2}{x-1}\left(2\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow10\left(x-2\right)\left(x-1\right)=\left(x+1\right)\left(x+2\right)\\ \Leftrightarrow10\left(x^2-3x+2\right)=x^2+3x+2\\ \Leftrightarrow9x^2-33x+18=0\\ \Leftrightarrow9x^2-27x-6x+18=0\\ \Leftrightarrow9x\left(x-3\right)-6\left(x-3\right)=0\\ \Leftrightarrow\left(9x-6\right)\left(x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-3=0\\9x-6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\frac{2}{3}\end{matrix}\right.\left(Tm\right)\)
\(\left(2\right)\Leftrightarrow\left(x-2\right)\left(x-1\right)=\left(x+1\right)\left(x+2\right)\\ \Leftrightarrow x^2-3x+2=x^2+3x+2=0\\ \Leftrightarrow6x=0\\ \Leftrightarrow x=0\left(Tm\right)\)
Vậy pt có tập nghiệm \(S=\left\{0;3;\frac{2}{3}\right\}\)
29 tháng 7 2019
\(\text{b) }\left(\frac{x+1}{x-2}\right)^2+\frac{x+1}{x-4}=12\left(\frac{x-2}{x-4}\right)^2\\ DKXD:x\ne2;x\ne4\\ \Leftrightarrow\left(\frac{x+1}{x-2}\right)^2+\frac{x+1}{x-2}\cdot\frac{x-2}{x-4}-12\left(\frac{x-2}{x-4}\right)^2=0\)
Đặt \(\frac{x+1}{x-2}=a;\frac{x-2}{x-4}=b\)
\(Pt\Leftrightarrow a^2+ab-12b^2=0\\ \Leftrightarrow a^2+4ab-3ab-12b^2=0\\ \Leftrightarrow a\left(a+4b\right)-3b\left(a+4b\right)=0\\ \Leftrightarrow\left(a-3b\right)\left(a+4b\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}a-3b=0\\a+4b=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}a=3b\\a=-4b\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\frac{x+1}{x-2}=\frac{3\left(x-2\right)}{x-4}\left(1\right)\\\frac{x+1}{x-2}=\frac{-4\left(x-2\right)}{x-4}\left(2\right)\end{matrix}\right.\)
Tự giải tiếp nha.
13 tháng 3 2019
\(\Leftrightarrow8\left(x+\frac{1}{x}\right)^2+4\left(x^2+\frac{1}{x^2}\right)\left[\left(x^2+\frac{1}{x^2}\right)-\left(x+\frac{1}{x}\right)^2\right]=\left(x+4\right)^2.ĐKXĐ:x\ne0\)
\(\Leftrightarrow8\left(x+\frac{1}{x}\right)^2+4\left(x^2+\frac{1}{x^2}\right)\left(x^2+\frac{1}{x^2}-x^2-2-\frac{1}{x^2}\right)=\left(x+4\right)^2\)
\(\Leftrightarrow8\left(x+\frac{1}{x}\right)^2-8\left(x^2+\frac{1}{x^2}\right)=\left(x+4\right)^2\)
\(\Leftrightarrow8\left[\left(x+\frac{1}{x}\right)^2-\left(x^2+\frac{1}{x^2}\right)\right]=\left(x+4\right)^2\)
\(\Leftrightarrow8\left(x^2+2+\frac{1}{x^2}-x^2+\frac{1}{x^2}\right)=\left(x+4\right)^2\)
\(\Leftrightarrow16=\left(x+4\right)^2\)
\(\Leftrightarrow x^2+8x+16=16\)
\(\Leftrightarrow x^2+8x=0\)
\(\Leftrightarrow x\left(x+8\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\left(l\right)\\x=-8\left(n\right)\end{cases}}\)
V...\(S=\left\{-8\right\}\)
^^
13 tháng 3 2019
bạn ghi sai đề ở chỗ \(\left(x+\frac{1}{x}\right)^2\)chứ ko phải \(\left(x+\frac{1}{x^2}\right)^2\)nhé
23 tháng 4 2016
b)
\(x-2.\left(\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}+\frac{1}{8\cdot9}\right)=\frac{16}{9}\)
\(x-2\cdot\left(\frac{1}{3}-\frac{1}{9}\right)=\frac{16}{9}\)
\(x-2=\frac{16}{9}:\left(\frac{1}{3}-\frac{1}{9}\right)\)
\(x-2=8\)
=> x = 10
23 tháng 4 2016
a)
\(A=\frac{1}{2}.\frac{2}{3}\cdot\frac{3}{4}\cdot\cdot\cdot\frac{2013}{2014}\cdot\frac{2014}{2015}\cdot\frac{2015}{2016}\)
\(A=\frac{1}{2016}\)
LH
27 tháng 4 2019
ĐK: x khác 0
Đặt \(x+\frac{1}{x}=a\)\(\Rightarrow\left(x+\frac{1}{x}\right)^2=a^2\Leftrightarrow a^2=x^2+\frac{1}{x^2}+2\cdot x\cdot\frac{1}{x}\Leftrightarrow a^2-2=x^2+\frac{1}{x^2}\)
Có:
\(8\left(x+\frac{1}{x}\right)^2+4\left(x^2+\frac{1}{x^2}\right)^2-4\left(x^2+\frac{1}{x^2}\right)\left(x+\frac{1}{x}\right)^2\)
\(=8a^2+4\left(a^2-2\right)^2-4\left(a^2-2\right)a^2\)
\(=8a^2+4\left(a^4-4a^2+4\right)-4\left(a^4-2a^2\right)\)
\(=8a^2+4a^4-16a^2+16-4a^4+8a^2=16\)
Thay \(8\left(x+\frac{1}{x}\right)^2+4\left(x^2+\frac{1}{x^2}\right)^2-4\left(x^2+\frac{1}{x^2}\right)\left(x+\frac{1}{x}\right)^2=16\)
vào phương trình, ta có: \(\left(x-4\right)^2=16\)
\(\Leftrightarrow\orbr{\begin{cases}x-4=-4\\x-4=4\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=8\end{cases}}\)Mà điều kiện x khác 0 nên x=8
Vậy phương trình có nghiệm x=8