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Điều kiện có 2 nghiệm phân biệt tự làm nha
Theo vi-et ta có:
\(\hept{\begin{cases}x_1+x_2=5\\x_1.x_2=m-2\end{cases}}\)
\(2\left(\frac{1}{\sqrt{x_1}}+\frac{1}{\sqrt{x_2}}\right)=3\)
\(\Leftrightarrow4\left(\frac{1}{x_1}+\frac{1}{x_2}+\frac{2}{\sqrt{x_1.x_2}}\right)=9\)
\(\Leftrightarrow4\left(\frac{5}{m-2}+\frac{2}{\sqrt{m-2}}\right)=9\)
Làm nốt nhé
Câu 1:
M=\(\left(x^2+2xy+y^2\right)+\left(2x+2y\right)+1+\left(4x^2-4x+1\right)+2014\)
=\(\left(\left(x+y\right)^2+2\left(x+y\right)+1\right)+\left(2x-1\right)^2+2014\)
=\(\left(x+y+1\right)^2+\left(2x-1\right)^2+2014\ge2014\)
\(\Rightarrow M\ge2014\Leftrightarrow minM=2014\)
\(\Leftrightarrow\hept{\begin{cases}x+y+1=0\\2x-1=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=0,5\\y=1,5\end{cases}}\)
Đặt \(\left\{{}\begin{matrix}x-2y=a\\\dfrac{1}{2x+3y}=b\end{matrix}\right.\)
hpt trở thành:
\(\left\{{}\begin{matrix}a+b=2\\2a+3b=3\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=3\\b=-1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-2y=3\\\dfrac{1}{2x+3y}=-1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=3+2y\\2x+3y=-1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=3+2y\\2\left(3+2y\right)+3y=-1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=3+2y\\6+4y+3y=-1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=3+2y\\7y=-7\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=3+2.-1\\y=-1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-1\end{matrix}\right.\)
Vậy nghiệm hpt \(\left\{{}\begin{matrix}x=1\\y=-1\end{matrix}\right.\)
Câu 1 là \(\left(8x-4\right)\sqrt{x}-1\) hay là \(\left(8x-4\right)\sqrt{x-1}\)?
Câu 1:ĐK \(x\ge\frac{1}{2}\)
\(4x^2+\left(8x-4\right)\sqrt{x}-1=3x+2\sqrt{2x^2+5x-3}\)
<=> \(\left(4x^2-3x-1\right)+4\left(2x-1\right)\sqrt{x}-2\sqrt{\left(2x-1\right)\left(x+3\right)}\)
<=> \(\left(x-1\right)\left(4x+1\right)+2\sqrt{2x-1}\left(2\sqrt{x\left(2x-1\right)}-\sqrt{x+3}\right)=0\)
<=> \(\left(x-1\right)\left(4x+1\right)+2\sqrt{2x-1}.\frac{8x^2-4x-x-3}{2\sqrt{x\left(2x-1\right)}+\sqrt{x+3}}=0\)
<=>\(\left(x-1\right)\left(4x+1\right)+2\sqrt{2x-1}.\frac{\left(x-1\right)\left(8x+3\right)}{2\sqrt{x\left(2x-1\right)}+\sqrt{x+3}}=0\)
<=> \(\left(x-1\right)\left(4x+1+2\sqrt{2x-1}.\frac{8x+3}{2\sqrt{x\left(2x-1\right)}+\sqrt{x+3}}\right)=0\)
Với \(x\ge\frac{1}{2}\)thì \(4x+1+2\sqrt{2x-1}.\frac{8x-3}{2\sqrt{x\left(2x-1\right)}+\sqrt{x+3}}>0\)
=> \(x=1\)(TM ĐKXĐ)
Vậy x=1
\(2)mx^2-2\left(m-1\right)x+m-1=0\)
Để pt có nghiệm kép \(\Leftrightarrow\left\{{}\begin{matrix}a\ne0\\\Delta=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}m\ne0\\\left[-2\left(m-1\right)\right]^2-4m\left(m-1\right)=0\end{matrix}\right.\)
\(\Leftrightarrow4\left(m^2-2m+1\right)-4m^2+4m=0\)
\(\Leftrightarrow4m^2-8m+4-4m^2+4m=0\)
\(\Leftrightarrow-4m+4=0\)
\(\Leftrightarrow m=1\)
Vậy để pt trên có nghiệm kép thì \(\left\{{}\begin{matrix}m\ne0\\m=1\end{matrix}\right.\)
mn ơi giúp mk vs