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20(x-2/x+1)^2-5(x+2/x-1)+48(x-2)(x+2)/)(x-1)(x+1)
Đặt x-2/x+1 là a
x+2/x-1 là b
=> Ta có PT: 20a^2-5b+48ab
=20a^2+50ab-2ab-5b
=20a(a+2,5)-2b(a+2,5)
=(20a-2b)(a+2,5)
Xong thay gt a và b vào mà tự tìm.
a,\(\left(\frac{x}{x+1}\right)^2+\left(\frac{x}{x-1}\right)^2=90\)\(\Leftrightarrow\left(\frac{x}{x+1}\right)^2+2.\frac{x}{x+1}.\frac{x}{x-1}+\left(\frac{x}{x-1}\right)^2-\frac{2x^2}{x^2-1}=90\)
\(\Leftrightarrow\left(\frac{x}{x+1}+\frac{x}{x-1}\right)^2-\frac{2x^2}{x^2-1}=90\)\(\Leftrightarrow\left(\frac{x^2-x+x^2+x}{x^2-1}\right)^2-\frac{2x^2}{x^2-1}=90\)
\(\Leftrightarrow\left(\frac{2x^2}{x^2-1}\right)^2-\frac{2x^2}{x^2-1}-90=0\)\(\Leftrightarrow\left(\frac{2x^2}{x^2-1}-10\right)\left(\frac{2x^2}{x^2-1}+9\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}\frac{2x^2}{x^2-1}=10\\\frac{2x^2}{x^2-1}=-9\end{cases}\Leftrightarrow......}\)
b,Đặt \(\frac{x-2}{x+1}=a;\frac{x+2}{x-1}=b\Rightarrow ab=\frac{\left(x-2\right)\left(x+2\right)}{\left(x+1\right)\left(x-1\right)}=\frac{x^2-4}{x^2-1}\)
Từ đó ta có phương trình:\(20a^2-5b^2+48ab=0\Leftrightarrow20a^2-2ab-5b^2+50ab=0\)
\(\Leftrightarrow2a\left(10a-b\right)+5b\left(10a-b\right)=0\Leftrightarrow\left(2a+5b\right)\left(10a-b\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}2a=-5b\\10a=b\end{cases}}\)
TH1:\(2a=-5b\Leftrightarrow\frac{2\left(x-2\right)}{x+1}=\frac{-5\left(x+2\right)}{x-1}\)\(\Rightarrow2\left(x-2\right)\left(x-1\right)=-5\left(x+2\right)\left(x+1\right)\)\(\Leftrightarrow2x^2-6x+4=-5x^2-15x-10\)\(\Leftrightarrow7x^2+9x+14=0\)
\(\Leftrightarrow7\left(x^2+\frac{9}{7}x+2\right)=0\Leftrightarrow7\left(x^2+2.\frac{9}{14}+\frac{81}{196}\right)+\frac{311}{28}=0\)
\(\Leftrightarrow7\left(x+\frac{9}{14}\right)^2+\frac{311}{28}=0\),vô lí
TH2:Tự làm nhé ,tương tự
\(20\left(\frac{x-2}{x+1}\right)^2-5\left(\frac{x+2}{x-1}\right)+48\left(\frac{x^2-4}{x^2-1}\right)\)
ĐK: \(\left|x\right|\ne1\) với \(\left|x\right|=2\Rightarrow V0.N_o\Rightarrow\left|x\right|\ne2\)
Đặt \(\left(\frac{x-2}{x+1}\right)=t\Rightarrow t\ne0\)
\(\Leftrightarrow20t^2-5\left(\frac{1}{t^2}\right)+48=0\Leftrightarrow\left\{\begin{matrix}t^2=y\\y>0\\20y^2+48y-5=0\left(2\right)\end{matrix}\right.\) (I)
\(\left(2\right)\Leftrightarrow y^2+2.\frac{6}{5}y+\left(\frac{6}{5}\right)^{^2}=\left(\frac{13}{10}\right)^2\)
\(\Rightarrow\left\{\begin{matrix}y=\frac{-12-13}{10}=\frac{-5}{2}\left(loai\right)\\y=\frac{-12+13}{10}=\frac{1}{10}\end{matrix}\right.\)
\(\left(I\right)\Rightarrow\left[\begin{matrix}t=-\frac{1}{\sqrt{10}}\\t=\frac{1}{\sqrt{10}}\end{matrix}\right.\) giải tiếp ok!
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 :)
\(20\left(\frac{x-2}{x+1}\right)^2-5\left(\frac{x+2}{x-1}\right)^2+48\left(\frac{x^2-4}{x^2-1}\right)=0\left(ĐKXĐ:x\ne\pm1\right)\)
Đặt \(\frac{x-2}{x+1}=a,\frac{x+2}{x-1}=b\)thì \(ab=\frac{\left(x-2\right)\left(x+2\right)}{\left(x+1\right)\left(x-1\right)}=\frac{x^2-4}{x^2-1}\). Phương trình trở thành:
\(20a^2-5b^2+48ab=0\).
\(\Leftrightarrow\left(20a^2+50ab\right)-\left(2ab+5b^2\right)=0\).
\(\Leftrightarrow10a\left(2a+5b\right)-b\left(2a+5b\right)=0\).
\(\Leftrightarrow\left(10a-b\right)\left(2a+5b\right)=0\).
\(\Leftrightarrow\orbr{\begin{cases}10a-b=0\\2a+5b=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}10a=b\left(1\right)\\2a=-5b\left(2\right)\end{cases}}\)
Xét phương trình \(\left(1\right)\):
\(\left(1\right)\Leftrightarrow\frac{10\left(x-2\right)}{x+1}=\frac{x+2}{x-1}\).
\(\Leftrightarrow\frac{10\left(x-2\right)\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}=\frac{\left(x+2\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}\).
\(\Rightarrow10\left(x-2\right)\left(x-1\right)=\left(x+2\right)\left(x+1\right)\).
\(\Leftrightarrow10\left(x^2-3x+2\right)=x^2+3x+2\).
\(\Leftrightarrow10x^2-30x+20=x^2+3x+2\).
\(\Leftrightarrow10x^2-30x+20-x^2-3x-2=0\).
\(\Leftrightarrow9x^2-33x+18=0\).
\(\Leftrightarrow3\left(3x^2-11x+6\right)=0\).
\(\Leftrightarrow3x^2-11x+6=0\).
\(\Leftrightarrow\left(3x^2-9x\right)-\left(2x-6\right)=0\).
\(\Leftrightarrow3x\left(x-3\right)-2\left(x-3\right)=0\).
\(\left(3x-2\right)\left(x-3\right)=0\).
\(\Leftrightarrow\orbr{\begin{cases}3x-2=0\\x-3=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{2}{3}\\x=3\end{cases}}\left(TMĐKXĐ\right)\).
Xét phương trình \(\left(2\right)\):
\(\left(2\right)\Leftrightarrow\frac{2\left(x-2\right)}{x+1}=\frac{-5\left(x+2\right)}{x-1}\).
\(\Leftrightarrow\frac{2\left(x-2\right)\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}=\frac{-5\left(x+2\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}\).
\(\Rightarrow2\left(x-2\right)\left(x-1\right)=-5\left(x+2\right)\left(x+1\right)\).
\(\Leftrightarrow2\left(x^2-3x+2\right)=-5\left(x^2+3x+2\right)\).
\(\Leftrightarrow2x^2-6x+4=-5x^2-15x-10\).
\(\Leftrightarrow2x^2-6x+4+5x^2+15x+10=0\).
\(\Leftrightarrow7x^2+9x+14=0\).
\(\Leftrightarrow7\left(x^2+\frac{9}{7}x+2\right)=0\).
\(\Leftrightarrow\left(x^2+2.x.\frac{9}{14}+\frac{81}{196}\right)+\frac{311}{196}=0\).
\(\Leftrightarrow\left(x+\frac{9}{14}\right)^2=\frac{-311}{196}\)(vô nghiệm).
Vậy phương trình có tập nghiệm: \(S=\left\{\frac{2}{3};3\right\}\).
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