Làm bừa thôi nha
ĐK x >=1
VT \(\ge\sqrt{1+3}+\sqrt{1-1}=2\) (1)
VP \(\le2-1^2=1\) (2)
=> PT vô nghiệm
Làm bừa thôi nha
ĐK x >=1
VT \(\ge\sqrt{1+3}+\sqrt{1-1}=2\) (1)
VP \(\le2-1^2=1\) (2)
=> PT vô nghiệm
GPT:
1, \(6x^2+10x-92+\sqrt{\left(x+70\right)\left(2x^2+4x+16\right)}=0\)
2,\(x+3+\sqrt{1-x^2}=3\sqrt{x+1}+\sqrt{1-x}\)
Gpt: \(\sqrt{x+5}+\sqrt{3-x}-2\left(\sqrt{15-2x-x^2}+1\right)=0\)
GPT : \(\sqrt[3]{x+1}+\sqrt[3]{x+2}=1+\sqrt[3]{x^2+3x+2}\)
GPT : \(\frac{1}{\sqrt{x+3}+\sqrt{x+2}}+\frac{1}{\sqrt{x+2}+\sqrt{x+1}}+\frac{1}{\sqrt{x+1}+\sqrt{x}}\)
Gpt: \(1+\frac{2}{3}\sqrt{x-x^2}=\sqrt{x}+\sqrt{1-x}\)
1.Gpt: \(\dfrac{6}{x-3\sqrt{x-2}+7}=\dfrac{1}{\sqrt{x-2}}+\dfrac{\sqrt{3}}{3\sqrt{2\sqrt{x-2}}-3}\)
2.Ghpt: \(\left\{{}\begin{matrix}x^2-y-z=0\\x^3-y^2-z^2+2=0\end{matrix}\right.\)
GPT
\(\sqrt[]{2x+3+\sqrt{x+2}}+\sqrt[]{2x+2-\sqrt{2+x}}=1+2\sqrt{x+2}\)
gpt :
1, \(2\left(2\sqrt{1+x^2}-\sqrt{1-x^2}\right)-\sqrt{1-x^4}=3x^2+1\)
2, \(2\sqrt{1-x}+\sqrt{x+1}+3\sqrt{1-x^2}=3-x\)
GPT: \(\sqrt{2x+3-\sqrt{x+2}}+\sqrt{2x+4+\sqrt{x+2}}=1+2\sqrt{x+2}\)