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a: ĐKXĐ: \(\left\{{}\begin{matrix}2x-3>=0\\x-1>0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x>=\dfrac{3}{2}\\x>1\end{matrix}\right.\Leftrightarrow x>=\dfrac{3}{2}\)
\(\dfrac{\sqrt{2x-3}}{\sqrt{x-1}}=2\)
=>\(\sqrt{\dfrac{2x-3}{x-1}}=2\)
=>\(\dfrac{2x-3}{x-1}=4\)
=>4(x-1)=2x-3
=>4x-4=2x-3
=>4x-2x=-3+4
=>2x=1
=>\(x=\dfrac{1}{2}\left(loại\right)\)
b: ĐKXĐ: 2x+15>=0
=>x>=-15/2
\(x+\sqrt{2x+15}=0\)
=>\(\sqrt{2x+5}=-x\)
=>\(\left\{{}\begin{matrix}-x>=0\\\left(-x\right)^2=2x+5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-\dfrac{15}{2}< =x< =0\\x^2-2x-5=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}-\dfrac{15}{2}< =x< =0\\\left(x-1\right)^2=6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-\dfrac{15}{2}< =x< =0\\\left[{}\begin{matrix}x-1=\sqrt{6}\\x-1=-\sqrt{6}\end{matrix}\right.\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}-\dfrac{15}{2}< =x< =0\\\left[{}\begin{matrix}x=\sqrt{6}+1\left(loại\right)\\x=-\sqrt{6}+1\left(nhận\right)\end{matrix}\right.\end{matrix}\right.\)
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a.
\(\Leftrightarrow\left\{{}\begin{matrix}\left(2x-y\right)^2-3\left(2x-y\right)=0\\x+2y=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(2x-y\right)\left(2x-y-3\right)=0\\x+2y=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}2x-y=0\\x+2y=0\end{matrix}\right.\\\left\{{}\begin{matrix}2x-y-3=0\\x+2y=0\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=0\\y=0\end{matrix}\right.\\\left\{{}\begin{matrix}x=\dfrac{6}{5}\\y=-\dfrac{3}{5}\end{matrix}\right.\end{matrix}\right.\)
b.
ĐKXĐ: \(\dfrac{2x-y}{x+y}>0\)
Đặt \(\sqrt{\dfrac{2x-y}{x+y}}=t>0\) pt đầu trở thành:
\(t+\dfrac{1}{t}=2\Leftrightarrow t^2-2t+1=0\)
\(\Leftrightarrow t=1\Leftrightarrow\sqrt{\dfrac{2x-y}{x+y}}=1\)
\(\Leftrightarrow2x-y=x+y\Leftrightarrow x=2y\)
Thay xuống pt dưới:
\(6y+y=14\Rightarrow y=2\)
\(\Rightarrow x=4\)
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Theo hệ thức vi ét x1+x2=2; x1x2=-15
x1-x2= căn (x1-x2)2= căn [(x1+x2)2-4x1x2]
bạn thay vào rồi tính nốt nha
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f(x)g(x)=0<=>f(x)=0 hoặc g(x)=0
ta xét Th (x^3-4x^2-2x-15)/(x^2+x+1)=0
\(\Leftrightarrow\frac{x^3-4x^2-2x-15}{x^2+x+1}=\frac{\left(x-5\right)\left(x^2+x+3\right)}{x^2+x+1}\Rightarrow x=5\)
x2+x+3=0
12-4(1.3=-11
=>pt ko có nghiệm thực
=>x=5 vì (x^3-4x^2-2x-15)/(x^2+x+1)<0
=>\(x\in\left\{-\infty;5\right\}\)
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\(\hept{\begin{cases}2x^2+3xy-2y^2-5\left(2x-y\right)=0\left(1\right)\\x^2-2xy-3y^2+15=0\left(2\right)\end{cases}\left(I\right)}\)
Ta có \(\left(1\right)\Leftrightarrow\left(2x-y\right)\left(x+2y\right)-5\left(2x-y\right)=0\)
\(\Leftrightarrow\left(2x-y\right)\left(x+2y-5\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}y=2x\\x=5-2y\end{cases}}\)
Do đó \(\left(I\right)\Leftrightarrow\hept{\begin{cases}y=2x\\x^2-2x\cdot2x-3\left(2x\right)^2+15=0\end{cases}\left(II\right)}\)hoặc \(\hept{\begin{cases}x=5-2y\\\left(5-2y\right)^2-2\left(5-2y\right)y-3y^2+15=0\end{cases}\left(III\right)}\)
\(\left(II\right)\Leftrightarrow\hept{\begin{cases}y=2x\\-15x^2+15=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=1;y=2\\x=-1;y=-2\end{cases}}}\)
\(\left(III\right)\Leftrightarrow\hept{\begin{cases}x=5-2y\\5y^2-30y+40=0\end{cases}\Leftrightarrow\orbr{\begin{cases}y=2;x=1\\y=4;x=-3\end{cases}}}\)
Vậy hệ phương trình (I) đã cho có nghiệm (x;y)=(1;2);(-1;-2);(-3;4)
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\(3x^2=15\)=> \(x^2=5\)=> \(\orbr{\begin{cases}x=\sqrt{5}\\x=-\sqrt{5}\end{cases}}\)
\(\Leftrightarrow\left(x^2+2x-15\right)\left(x^2+2x+1\right)=0\)
\(\Leftrightarrow\left(x+5\right)\left(x-3\right)\left(x+1\right)^2=0\)
hay \(x\in\left\{-5;3;-1\right\}\)