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ĐKXĐ: \(x\le\dfrac{1}{2}\)
\(4x^2+y^2+2x+y=2-4xy\)
\(\Leftrightarrow\left(4x^2+4xy+y^2\right)+2x+y-2=0\)
\(\Leftrightarrow\left(2x+y\right)^2+2x+y-2=0\)
\(\Rightarrow\left[{}\begin{matrix}2x+y=1\\2x+y=-2\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}1-2x=y\\1-2x=y+3\end{matrix}\right.\)
Thế vào pt dưới:
\(\Rightarrow\left[{}\begin{matrix}8\sqrt{y}+y^2-9=0\\8\sqrt{y+3}+y^2-9=0\end{matrix}\right.\)
\(\Leftrightarrow...\)
\(ĐK:-1\le x\le1\\ PT\Leftrightarrow13\left(1-2x^2\right)\sqrt{\left(1-x^2\right)\left(1+x^2\right)}+9\left(1+2x^2\right)\sqrt{\left(1+x^2\right)\left(1-x^2\right)}=0\\ \Leftrightarrow\sqrt{1-x^4}\left(13-26x^2+9+18x^2\right)=0\\ \Leftrightarrow\sqrt{1-x^4}\left(22-8x^2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}1-x^4=0\\22-8x^2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left(1+x^2\right)\left(1-x\right)\left(1+x\right)=0\\x^2=\dfrac{22}{8}\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x=1\left(tm\right)\\x=-1\left(tm\right)\end{matrix}\right.\\\left[{}\begin{matrix}x=\dfrac{\sqrt{11}}{2}\left(ktm\right)\\x=-\dfrac{\sqrt{11}}{2}\left(ktm\right)\end{matrix}\right.\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)
a) \(2x^2+20x+52=0\Rightarrow x^2+10x+26=0\Rightarrow\left(x+5\right)^2+1=0\)
\(\Rightarrow\) vô nghiệm
b) ĐK: \(x\ne1;-1\)
\(\dfrac{2x-19}{5x^2-5}-\dfrac{17}{x-1}=\dfrac{8}{1-x}\Rightarrow\dfrac{2x-19}{5\left(x-1\right)\left(x+1\right)}-\dfrac{17}{x-1}+\dfrac{8}{x-1}=0\)
\(\Rightarrow\dfrac{2x-19}{5\left(x-1\right)\left(x+1\right)}-\dfrac{9}{x-1}=0\Rightarrow\dfrac{2x-19-45\left(x+1\right)}{5\left(x-1\right)\left(x+1\right)}=0\)
\(\Rightarrow-43x-64=0\Rightarrow x=-\dfrac{64}{43}\)
a) Ta có: \(\Delta'=100-104=-4< 0\)
Vậy phương trình vô nghiệm.
b) ĐKXĐ: \(x\ne1;x\ne-1\)
\(\Leftrightarrow\dfrac{2x-19}{5\left(x^2-1\right)}=\dfrac{17}{x-1}-\dfrac{8}{x-1}\)
\(\Leftrightarrow\dfrac{2x-19}{5\left(x-1\right)\left(x+1\right)}=\dfrac{9}{x-1}\)
\(\Leftrightarrow\dfrac{2x-19}{5\left(x-1\right)\left(x+1\right)}=\dfrac{45\left(x+1\right)}{5\left(x-1\right)\left(x+1\right)}\)
\(\Rightarrow2x-19=45x+45\)
\(\Leftrightarrow43x=-64\)
\(\Leftrightarrow x=-\dfrac{64}{43}\)(TM)
Vậy phương trình có nghiệm là: \(x=-\dfrac{64}{43}\)
\(ĐK:x\in R\)
\(\sqrt{x^2+x+4}+\sqrt{x^2+x+1}=\sqrt{2x^2+2x+9}\) (*)
Đặt \(x^2+x+1=a;a\ge0\)
\(\rightarrow\left\{{}\begin{matrix}x^2+x+4=a+3\\2x^2+2x+9=2a+7\end{matrix}\right.\)
(*) \(\Rightarrow\sqrt{a+3}+\sqrt{a}=\sqrt{2a+7}\)
\(\Leftrightarrow\left(\sqrt{a+3}+\sqrt{a}\right)^2=\left(\sqrt{2a+7}\right)^2\)
\(\Leftrightarrow a+3+a+2\sqrt{a\left(a+3\right)}=2a+7\)
\(\Leftrightarrow2\sqrt{a\left(a+3\right)}=4\)
\(\Leftrightarrow\sqrt{a\left(a+3\right)}=2\)
\(\Leftrightarrow a\left(a+3\right)=4\)
\(\Leftrightarrow a^2+3a-4=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=1\left(tm\right)\\a=-4\left(ktm\right)\end{matrix}\right.\)
\(\Rightarrow x^2+x+1=1\)
\(\Leftrightarrow x\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\) \((tm)\)
Vậy \(S=\left\{0;-1\right\}\)
\(\sqrt{\left(2x-1\right)^2}+\sqrt{9\left(2x-1\right)^2}=8\)
\(\left|2x-1\right|+3.\left|2x-1\right|=8\)
\(2x-1+6x-3=8\)
\(8x-4=8\)
\(8x=12\)
\(x=\frac{3}{2}\)
vậy \(x=\frac{3}{2}\)
Lời giải:
Ta có:
\(x^4-2x^3+2x^2+4x-8=0\)
\(\Leftrightarrow x^2(x^2-2)-2x(x^2-2)+4(x^2-2)=0\)
\(\Leftrightarrow (x^2-2)(x^2-2x+4)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-2=0\left(1\right)\\x^2-2x+4=0\left(2\right)\end{matrix}\right.\)
(1) \(\Leftrightarrow x^2-2=0\Leftrightarrow x=\pm \sqrt{2}\)
(2)\(\Leftrightarrow x^2-2x+4=0\Leftrightarrow (x-1)^2+3=0\)
(vô lý vì \((x-1)^2+3\geq 3>0\forall x\in\mathbb{R}\) )
Vậy \(x=\pm \sqrt{2}\)
a) \(\sqrt[]{x^2-4x+4}=x+3\)
\(\Leftrightarrow\sqrt[]{\left(x-2\right)^2}=x+3\)
\(\Leftrightarrow\left|x-2\right|=x+3\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=x+3\\x-2=-\left(x+3\right)\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}0x=5\left(loại\right)\\x-2=-x-3\end{matrix}\right.\)
\(\Leftrightarrow2x=-1\Leftrightarrow x=-\dfrac{1}{2}\)
b) \(2x^2-\sqrt[]{9x^2-6x+1}=5\)
\(\Leftrightarrow2x^2-\sqrt[]{\left(3x-1\right)^2}=5\)
\(\Leftrightarrow2x^2-\left|3x-1\right|=5\)
\(\Leftrightarrow\left|3x-1\right|=2x^2-5\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-1=2x^2-5\\3x-1=-2x^2+5\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}2x^2-3x-4=0\left(1\right)\\2x^2+3x-6=0\left(2\right)\end{matrix}\right.\)
Giải pt (1)
\(\Delta=9+32=41>0\)
Pt \(\left(1\right)\) \(\Leftrightarrow x=\dfrac{3\pm\sqrt[]{41}}{4}\)
Giải pt (2)
\(\Delta=9+48=57>0\)
Pt \(\left(2\right)\) \(\Leftrightarrow x=\dfrac{-3\pm\sqrt[]{57}}{4}\)
Vậy nghiệm pt là \(\left[{}\begin{matrix}x=\dfrac{3\pm\sqrt[]{41}}{4}\\x=\dfrac{-3\pm\sqrt[]{57}}{4}\end{matrix}\right.\)