\(x^4-2x^3-x^2-2x+1=0\)
\(\Rightarrow\left(x^4+2x^2+1\right)-2x\left(x^2+1\right)+x^2+x^2\)
\(=\left(x^2+1\right)^2-2.x.\left(x^2+1\right)+x^2+x^2\)
\(=\left(x^2-x+1\right)^2+x^2=0\)
\(\Rightarrow\left\{{}\begin{matrix}x^2-x+1=0\\x=0\end{matrix}\right.\)
=> Ko tồn tại giá trị của x
=> PT vô nghiệm.
Tìm x:
(2x2+x)2-4(2x2+x)+3=0
(x+y)2+(1+x)(1+y)=0
(3x-2)(x+1)2(3x+8)=-16
(4x-5)(2x-3)(x-1)=9
Giải phương trình:
1. \(\sqrt{2x^2+4x+7}=x^4+4x^3+3x^2-2x-7\)
2. \(\dfrac{4}{x}+\sqrt{x-\dfrac{1}{x}}=x+\sqrt{2x-\dfrac{5}{x}}\)
3. \(\dfrac{6-2x}{\sqrt{5-x}}+\dfrac{6+2x}{\sqrt{5+x}}=\dfrac{8}{3}\)
4. \(x^2+1-\left(x+1\right)\sqrt{x^2-2x+3}=0\)
5. \(2\sqrt{2x+4}+4\sqrt{2-x}=\sqrt{9x^2+16}\)
6. \(\left(2x+7\right)\sqrt{2x+7}=x^2+9x+7\)
a ) \(x^4+2x^3-4x^2-2x+ 1=0\)
b)\(x^4-3x^3+4x^2-3x+1=0\)
1) \(\dfrac{x-3x^2}{2}+\sqrt{2x^4-x^3+7x^2-3x+3}=2\)
2) \(1+\sqrt{\dfrac{x-2}{1-x}}=\dfrac{2x^2-2x+1}{x^2-2x+2}\)
3) \(x+y+z+\dfrac{3}{x-1}+\dfrac{3}{y-1}+\dfrac{3}{z-1}=2\left(\sqrt{x+2}+\sqrt{y+2}+\sqrt{z+2}\right)\) với x ,y ,z > 1
4) \(\sqrt[3]{x+6}+x^2=7-\sqrt{x-1}\)
5) \(x^4-2x^3+x-\sqrt{2\left(x^2-x\right)}=0\)
Giải các phương trình sau:
a, \(\left(x-3\right)^2+x^4=-y^2+6y-4\)
b, \(\sqrt{2x-3}+\sqrt{5-2x}-x^2+4x-6=0\)
c, \(4+4x-x^2=|x-1|+|x-2|+|2x-3|+|4x-14|\)
d, \(x^2-2x+3=\sqrt{2x^2-x}+\sqrt{1+3x-3x^2}\)
Giải phương trình
\(\left(x^2+x-1\right)^2+2\left(x^2+x-1\right)-3=0\)
\(\left(2x^2-x\right)^2-4\left(2x^2-x\right)+3=0\)
gpt:
1, (x-2)3+(2x-4)3+(7-3x)3=1
2, x4 - 2x3 + x - \(\sqrt{2\left(x^2-x\right)}\) = 0.
Giải phương trình: 1/((x^3(2x^2-x+2))+(x^4+3x^2)>=2
Biết 2x^2+2>x>0