Giải phương trình :
a) \(10\left(\frac{x-2}{x+1}\right)^2+\left(\frac{x+2}{x-1}\right)^2-11\cdot\frac{x^2-4}{x^2-1}=0\)
b) \(\left(\frac{x+1}{x-2}\right)^2+\frac{x+1}{x-4}=12\left(\frac{x-2}{x-4}\right)^2\)
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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)
\(A=\frac{1}{2}.\frac{2}{3}\cdot\frac{3}{4}\cdot\cdot\cdot\frac{2013}{2014}\cdot\frac{2014}{2015}\cdot\frac{2015}{2016}\)
\(A=\frac{1}{2016}\)
8 tháng 2 2021
nhìn căng nhể :))
a) ( x - 1 )( x - 3 )( x + 5 )( x + 7 ) - 297 = 0
<=> [ ( x - 1 )( x + 5 ) ][ ( x - 3 )( x + 7 ) ] - 297 = 0
<=> ( x2 + 4x - 5 )( x2 + 4x - 21 ) - 297 = 0
Đặt t = x2 + 4x - 5
pt <=> t( t - 16 ) - 297 = 0
<=> t2 - 16t - 297 = 0
<=> t2 - 27t + 11t - 297 = 0
<=> t( t - 27 ) + 11( t - 27 ) = 0
<=> ( t - 27 )( t + 11 ) = 0
<=> ( x2 + 4x - 5 - 27 )( x2 + 4x - 5 + 11 ) = 0
<=> ( x2 + 4x - 32 )( x2 + 4x + 6 ) = 0
<=> ( x2 - 4x + 8x - 32 )( x2 + 4x + 6 ) = 0
<=> [ x( x - 4 ) + 8( x - 4 ) ]( x2 + 4x + 6 ) = 0
<=> ( x - 4 )( x + 8 )( x2 + 4x + 6 ) = 0
Đến đây dễ rồi :)
\(\text{a) }10\left(\frac{x-2}{x+1}\right)^2+\left(\frac{x+2}{x-1}\right)^2-11\frac{x^2-4}{x^2-1}=0\\ DKXD:x\ne-1;x\ne1\\ \Leftrightarrow10\left(\frac{x-2}{x+1}\right)^2+\left(\frac{x+2}{x-1}\right)^2-11\frac{\left(x+2\right)\left(x-2\right)}{\left(x+1\right)\left(x-1\right)}=0\)
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 \(S=\left\{0;3;\frac{2}{3}\right\}\)
\(\text{b) }\left(\frac{x+1}{x-2}\right)^2+\frac{x+1}{x-4}=12\left(\frac{x-2}{x-4}\right)^2\\ DKXD:x\ne2;x\ne4\\ \Leftrightarrow\left(\frac{x+1}{x-2}\right)^2+\frac{x+1}{x-2}\cdot\frac{x-2}{x-4}-12\left(\frac{x-2}{x-4}\right)^2=0\)
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.