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1 tháng 12 2021
a,ĐKXĐ:\(x\ge2\)
\(4\sqrt{x-2}+\sqrt{9x-18}-\sqrt{\dfrac{x-2}{4}}=26\\ \Leftrightarrow4\sqrt{x-2}+3\sqrt{x-2}-\dfrac{\sqrt{x-2}}{2}=26\\ \Leftrightarrow8\sqrt{x-2}+6\sqrt{x-2}-\sqrt{x-2}=52\\ \Leftrightarrow13\sqrt{x-2}=52\\ \Leftrightarrow\sqrt{x-2}=4\\ \Leftrightarrow x-2=16\\ \Leftrightarrow x=18\left(tm\right)\)
b,ĐKXĐ:\(x\in R\)
\(3x+\sqrt{4x^2-8x+4}=1\\ \Leftrightarrow2\sqrt{x^2-2x+1}=1-3x\\ \Leftrightarrow\left|x-1\right|=\dfrac{1-3x}{2}\\ \Leftrightarrow\left[{}\begin{matrix}x-1=\dfrac{1-3x}{2}\\x-1=\dfrac{3x-1}{2}\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}2x-2=1-3x\\2x-2=3x-1\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{5}\left(tm\right)\\x=-1\left(tm\right)\end{matrix}\right.\)
c, ĐKXĐ:\(x\ge0\)
\(\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)=7\\ \Leftrightarrow\sqrt{x}\left(2\sqrt{x}+1\right)-2\left(2\sqrt{x}+1\right)=7\\ \Leftrightarrow2x+\sqrt{x}-4\sqrt{x}-2=7\\ \Leftrightarrow2x-3\sqrt{x}-9=0\\ \Leftrightarrow\left(2x+3\sqrt{x}\right)-\left(6\sqrt{x}+9\right)=0\\ \Leftrightarrow\sqrt{x}\left(2\sqrt{x}+3\right)-3\left(2\sqrt{x}+3\right)=0\\ \Leftrightarrow\left(\sqrt{x}-3\right)\left(2\sqrt{x}+3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=3\\2\sqrt{x}=-3\left(vô.lí\right)\end{matrix}\right.\\ \Leftrightarrow x=9\left(tm\right)\)
TN
1
20 tháng 11 2021
\(\Leftrightarrow x+1=3x+9\\ \Leftrightarrow2x=-8\\ \Leftrightarrow x=-4\)
1 tháng 4 2022
(1)-a)Với mọi x, ta luôn có: \(\left(x+1\right)^2+3>0\Leftrightarrow x^2+1+2x+3>0\Leftrightarrow x^2+2x+4>0\)
\(\sqrt{x^2+2x+4}=2\Leftrightarrow x^2+2x+4=2^2=4\)
\(\Leftrightarrow x^2+2x=0\\\Leftrightarrow\left(x+2\right)x=0\\\Leftrightarrow\left[{}\begin{matrix}x+2=0\Leftrightarrow x=-2\\x=0\end{matrix}\right. \)
➤\(x\in\left\{-2;0\right\}\)
b) \(\left\{{}\begin{matrix}x+2y-1=0\\2x+y=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x+2y=1\\4x+2y=10\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2y=1-x\\3x=9\Leftrightarrow x=\dfrac{9}{3}=3\end{matrix}\right.\)
Do \(x=3\Leftrightarrow1-x=1-3=-2\) nên ta có: \(2y=1-x=-2\Leftrightarrow y=\dfrac{-2}{2}=-1\)
➤\(\left\{{}\begin{matrix}x=3\\y=-1\end{matrix}\right.\)
(2): +)ĐK để 2 hàm số cắt nhau là: \(2a\ne1\Leftrightarrow a\ne\dfrac{1}{2}\Leftrightarrow a\ne0,5\)
Ta có hệ phương trình sau: \(\left\{{}\begin{matrix}y=2ax+a+1\\y=x+2\end{matrix}\right.\)
➢Do đó, ta có: \(2ax+a+1=x+2\Leftrightarrow2ax+a-x=2-1=1\)
P
0
3 tháng 4 2022
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 \(\left\{{}\begin{matrix}x=1\\y=-1\end{matrix}\right.\)
PM
0
T
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Đặt x2 = t > 0 ta được
\(2t+1=\dfrac{1}{t}-4\Leftrightarrow2t^2+5t-1=0\\ \Leftrightarrow\left[{}\begin{matrix}t=\dfrac{-5+\sqrt{33}}{4}\\t=\dfrac{-5-\sqrt{33}}{4}\left(loại\right)\end{matrix}\right.\\ \Leftrightarrow x^2=\dfrac{-5+\sqrt{33}}{4}\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{\sqrt{-5+\sqrt{33}}}{2}\\x=\dfrac{\sqrt{-5+\sqrt{33}}}{2}\end{matrix}\right.\)
Vậy pt có 2 nghiệm
\(2x^2+1=\dfrac{1}{x^2}-4\left(1\right)\)
Đặt \(x^2=t\left(t\ge0\right)\)
Khi đó phương trình \(\left(1\right)\) trở thành \(2t+1=\dfrac{1}{t}-4\)
\(\Leftrightarrow2t^2+5t-1=0\)
\(\Leftrightarrow\left[{}\begin{matrix}t=\dfrac{-5+\sqrt{33}}{4}\left(\text{nhận}\right)\\t=\dfrac{-5-\sqrt{33}}{4}\left(\text{loại}\right)\end{matrix}\right.\)
\(\Rightarrow x^2=\dfrac{-5+\sqrt{33}}{4}\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-\sqrt{-5+\sqrt{33}}}{2}\\x=\dfrac{\sqrt{-5+\sqrt{33}}}{2}\end{matrix}\right.\)
Vậy: \(S=\left\{\dfrac{-\sqrt{-5+\sqrt{33}}}{2};\dfrac{\sqrt{-5+\sqrt{33}}}{2}\right\}\)