\(\left(x-2\right)^2=9\)
=>\(\left[{}\begin{matrix}x-2=3\\x-2=-3\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=5\\x=-1\end{matrix}\right.\)
\((x-2)^2= 3^2\) hoặc \((x-2)^2=-3^2\)
x-2=3 hoặc x-2=-3
x=3+2 hoặc x=-3+2
x=5 hoặc x=-1
Vậy x=5 hoặc x=-1
\(\left(x-2\right)^2=9\)
=>\(\left[{}\begin{matrix}x-2=3\\x-2=-3\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=5\\x=-1\end{matrix}\right.\)
\((x-2)^2= 3^2\) hoặc \((x-2)^2=-3^2\)
x-2=3 hoặc x-2=-3
x=3+2 hoặc x=-3+2
x=5 hoặc x=-1
Vậy x=5 hoặc x=-1
1)\(3x^3-13^2+30x-4=\sqrt{\left(6x+2\right)\left(3x-4\right)}\)
2)\(2\left(x^2+x+1\right)^2-7\left(x-1\right)^2=13\left(x^3-1\right)\)
3)\(\left(x^2-16\right)\left(x-3\right)^2+9x^2=0\)
4)\(x\left(x^2+9\right)\left(x+9\right)=22\left(x-1\right)^2\)
\(\left(x^2+x+1\right)\left(^3\sqrt{\left(3x-2\right)^2}+^3\sqrt{\left(3x-2\right)}+1\right)=9\)
\(\left\{{}\begin{matrix}\left|x\right|\left(4y+1\right)-2y=-3\\\left|x\left(x^2-12y\right)\right|+4y^2=9\end{matrix}\right.\)
\(9\left(\dfrac{x-2}{x+1}\right)^3+\left(\dfrac{x+2}{x-1}\right)^2-10\left(\dfrac{x^2-4}{x^2-1}\right)=0\)
giải pt \(x\left(x^2+9\right)\left(x+9\right)=22\left(x-1\right)^2\)
\(\left\{{}\begin{matrix}4y^3-12y^2+13y-5=\left(4x+9\right)\sqrt{x+2}\\2\left(x^2-5\left(y-1^2\right)\right)=3\left(y-1\right)\sqrt{x^2-4x-8}\end{matrix}\right.\)
cho x,y,z>0 thỏa mãn \(\left(x^2+y^2\right)\left(y^2+z^2\right)\left(z^2+x^2\right)=8\)
Tìm giá trị nhỏ nhất của S=\(xyz\left(x+y+z\right)^3\)
(có thể dùng BDT \(\left(x+y\right)\left(y+z\right)\left(z+x\right)\ge\dfrac{8}{9}\left(x+y+z\right)\left(xy+yz+zx\right)\))
tks mn<3
\(\left\{{}\begin{matrix}x^3+y^3=xy\sqrt{2\left(x^2+y^2\right)}\\4\sqrt{x\sqrt{x^2-1}}=9\left(y-1\right)\sqrt{2\left(x-1\right)}\end{matrix}\right.\)
Ghpt\(\left\{{}\begin{matrix}\left|x-2\right|+2\left|y-1\right|=9\\x+\left|y-1\right|=-1\end{matrix}\right.\)
\(2\left(x-3\right)^2\left(x+2\right)^2=\left(2x-1\right)^2-9\)