gpt : a) \(\frac{5x}{\sqrt{4-x^2}}+\frac{8}{x^2}+\frac{2x}{4-x^2}+\frac{5\sqrt{4-x^2}}{x}+4=0\)
b) \(\frac{2x}{\sqrt{8x^2+25}}+\frac{125}{x^2}-14=0\)
c) \(\left(x^3-3x+2\right)\sqrt{3x-2}-2x^3+6x^2-4x=0\)
d) \(\sqrt{x^2-x+6}+\frac{4}{x-1}=x^2+x\)
1. gpt : \(\frac{2x+1}{\sqrt{x^2+2}}+\left(x+1\right)\sqrt{1+\frac{2x+1}{x^2+2}}+x=0\)
2. \(\left\{{}\begin{matrix}x,y,z>0\\x+y+z\le\frac{3}{2}\end{matrix}\right.\) Tìm min \(Q=\frac{x}{y^2z}+\frac{y}{z^2x}+\frac{z}{x^2y}+\frac{x^5}{y}+\frac{y^5}{z}+\frac{z^5}{x}\)
Gpt: a) \(\sqrt[4]{3\left(x+5\right)}-\sqrt[4]{11-x}=\sqrt[4]{13+x}-\sqrt[4]{3\left(3-x\right)}\)
b) \(\frac{1+2\sqrt{x}-x\sqrt{x}}{3-x-\sqrt{2-x}}=2\left(\frac{1+x\sqrt{x}}{1+x}\right)\) c) \(\sqrt{x+1}+\frac{4\left(\sqrt{x+1}+\sqrt{x-2}\right)}{3\left(\sqrt{x-2}+1\right)^2}=3\)
d) \(\sqrt{\frac{x-2}{x+1}}+\frac{x+2}{\left(\sqrt{x+2}+\sqrt{x-2}\right)^2}=1\) e) \(2x+1+x\sqrt{x^2+2}+\left(x+1\right)\sqrt{x^2+2x+2}=0\)
f) \(\sqrt{2x+3}\cdot\sqrt[3]{x+5}=x^2+x-6\)
gpt : a) \(\frac{2+\sqrt{x}}{\sqrt{2}+\sqrt{2+\sqrt{x}}}+\frac{2-\sqrt{x}}{\sqrt{2}-\sqrt{2-\sqrt{x}}}=\sqrt{2}\)
b) \(\sqrt[3]{x+1}+\sqrt[3]{x+2}+\sqrt[3]{x+3}=0\)
c) \(\sqrt[4]{1-x^2}+\sqrt[4]{1+x}+\sqrt[4]{1-x}=3\)
1, gpt
a,\(\sqrt{x^2-3x+2}+\sqrt{x+3}=\sqrt{x-2}+\sqrt{x^2+2x-3}\)
b, \(\left(4x+2\right)\sqrt{x+8}=3x^2+7x+8\)
c,\(\sqrt{x+\sqrt{2x-1}}+\sqrt{x-\sqrt{2x-1}}=\sqrt{2}\)
2/ cho x,y,z thỏa mãn : \(\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right):\frac{1}{x+y+z}=1\)
tính giá trị biểu thức B=\(\left(x^{29}+y^{29}\right)\left(x^{11}+y^{11}\right)\left(x^{2013}+y^{2013}\right)\)
Tìm GTLN:
\(A=\frac{\sqrt{10x-49}}{2020}\\ B=\frac{\sqrt{2x^2-25}}{2020x^2}\\ C=\frac{7x^8+256}{x^7}\left(x>0\right)\\ D=\frac{\sqrt{x}+6\sqrt{x}+34}{\sqrt{x}+3}\\ E=x+\frac{1}{x-1}\left(x>1\right)\)
GPT: \(\frac{2x^2+x}{\sqrt{2x^2+x+10}}+2=\sqrt{2x^2+x+4}\)
Giải phương trình:
a) \(\sqrt{\frac{2x-1}{x+1}}+\sqrt{\frac{x+1}{2x-1}}=2\)
b) \(2\sqrt[3]{\frac{2x-3}{1-x}}+\sqrt[3]{\frac{1-x}{2x-3}}=3\)
c) \(x+\frac{1}{x}+4\left(\sqrt{x}+\frac{1}{\sqrt{x}}\right)+6=0\)
GPT
\(x^2-3\sqrt[3]{3x-2}-12+\frac{1}{\sqrt{x}}=\frac{\sqrt{x}+8}{x}\)