Đặt \(x-1=t\) pt trở thành:
\(\left(t+1\right)^4+\left(t-1\right)^4=1\)
\(\Leftrightarrow2t^4+12t^2+1=0\)
Vế trái luôn dương nên pt vô nghiệm
Đặt \(x-1=t\) pt trở thành:
\(\left(t+1\right)^4+\left(t-1\right)^4=1\)
\(\Leftrightarrow2t^4+12t^2+1=0\)
Vế trái luôn dương nên pt vô nghiệm
giải pt
a) \(\left|x^2-5x-4\right|=\left|x^2-4\right|\)
b) \(\left|x-1\right|+3\left|x-3\right|=6\)
c) \(\left|\frac{x^2-6x-4}{x^2-4}\right|=1\)
d) \(\left|x-1\right|-2\left|x-2\right|=x^2-x-3\)
giải pt
a) \(x^2+4x-3\left|x+2\right|+4=0\)
b) \(\left(x+2\right)^2-3\left|x+2\right|-4=0\)
c) \(\left(x^2-3\right)^2-6\left|x^2-3\right|+5=0\)
d) \(\frac{x^2-4x+4}{x^2-2x+1}+\frac{\left|2x-4\right|}{x-1}=3\)
e) \(\left|\frac{2x-1}{x+2}\right|-2\left|\frac{x+2}{2x-1}\right|=1\)
f) \(x^2+\frac{1}{x^2}-10=2\left|x-\frac{1}{x}\right|\)
\(\left\{{}\begin{matrix}x^2+y^2=\dfrac{1}{2}\\x^3+3xy^2=\dfrac{1}{2}\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\left(1+x^2\right)^2\left(1+\dfrac{1}{y^4}\right)=8\\\left(1+y^2\right)^2\left(1+\dfrac{1}{x^4}\right)=8\end{matrix}\right.\)
Em cảm ơn ạ !!!
giải hệ phương trình
1 , \(\left\{{}\begin{matrix}\left(x+y\right)\left(x-1\right)=\left(x-y\right)\left(x+1\right)+2xy\\\left(y-x\right)\left(y-1\right)=\left(y+x\right)\left(y-2\right)-2xy\end{matrix}\right.\)
2, \(\left\{{}\begin{matrix}2\left(\frac{1}{x}+\frac{1}{2y}\right)+3\left(\frac{1}{x}-\frac{1}{2y}\right)^2=9\\\left(\frac{1}{x}+\frac{1}{2y}\right)-6\left(\frac{1}{x}-\frac{1}{2y}\right)^2=-3\end{matrix}\right.\)
3 , \(\left\{{}\begin{matrix}\frac{xy}{x+y}=\frac{2}{3}\\\frac{yz}{y+z}=\frac{6}{5}\\\frac{zx}{z+x}=\frac{3}{4}\end{matrix}\right.\)
4 , \(\left\{{}\begin{matrix}2xy-3\frac{x}{y}=15\\xy+\frac{x}{y}=15\end{matrix}\right.\)
5 , \(\left\{{}\begin{matrix}x+y+3xy=5\\x^2+y^2=1\end{matrix}\right.\)
6 , \(\left\{{}\begin{matrix}x+y+xy=11\\x^2+y^2+3\left(x+y\right)=28\end{matrix}\right.\)
7, \(\left\{{}\begin{matrix}x+y+\frac{1}{x}+\frac{1}{y}=4\\x^2+y^2+\frac{1}{x^2}+\frac{1}{y^2}=4\end{matrix}\right.\)
8, \(\left\{{}\begin{matrix}x+y+xy=11\\xy\left(x+y\right)=30\end{matrix}\right.\)
9 , \(\left\{{}\begin{matrix}x^5+y^5=1\\x^9+y^9=x^4+y^4\end{matrix}\right.\)
giải các PT sau :
a) \(\left|2x+3\right|-\left|x\right|+\left|x-1\right|=2x+4\)
b) \(\sqrt{x}-\dfrac{4}{\sqrt{x+2}}+\sqrt{x+2}=0\)
c) \(\sqrt{x+\sqrt{2x-1}}+\sqrt{x-\sqrt{2x-1}}=\sqrt{2}\)
d) \(x+\sqrt{x+\dfrac{1}{2}+\sqrt{x+\dfrac{1}{4}}}=4\)
e) \(\sqrt{4x+3}+\sqrt{2x+1}=6x+\sqrt{8x^2+10x+3}-16\)
f)\(\sqrt[3]{x-2}+\sqrt{x+1}=3\)
GIÚP MÌNH VỚI MÌNH ĐANG CẦN GẤP
1.Giải hpt : a,\(\left\{{}\begin{matrix}\left(x+y+3\right)\sqrt{x-2y}+2y+4=0\\\left(x-y\right)\left(x^2+4\right)=y^2+1\end{matrix}\right.\)
giải hệ phương trình:
1, \(\left\{{}\begin{matrix}\sqrt{3+2x^2y-x^4y^2}+x^2\left(1-2x^2\right)=y^4\\1+\sqrt{1+\left(x-y\right)^2}=-x^2\left(x^4+1-2x^2-2xy^2\right)\end{matrix}\right.\)
2, \(\left\{{}\begin{matrix}\sqrt{x-1}+\sqrt{x}\left(3\sqrt{x}-y\right)+x\sqrt{x}=3y+\sqrt{y-1}\\3xy^2+4=4x^2+2y+x\end{matrix}\right.\)
1)\(\begin{cases}\left(8x-6\right)\sqrt{y}=\left(2+\sqrt{x-2}\right)\left(y+4\sqrt{x-2}+4\right)\\2\sqrt{x^2+3x-y}-\sqrt{y^2+4x}=x+1\end{cases}\)
2)\(\begin{cases}\left(x+\sqrt{x^2+1}\right)\left(y+\sqrt{y^2+1}\right)=1\\x^2+\sqrt{3-x}=2y^2-4\sqrt{2-y}+5\end{cases}\)
Giải các bpt
a) \(\sqrt{x^2-4-12}\le x-4\)
b) \(\sqrt{x^2-8x}\ge2\left(X+1\right)\)
C) \(\left(X-2\right).\sqrt{X^2+4}< X^2-4\)