+\(1\le x\le4\)
+ \(\left(x-1\right)+\left(4-x\right)+2\sqrt{\left(x-1\right)\left(4-x\right)}=8+\frac{29}{5-x}\)
Với \(1\le x\le4\) ta có:
\(VT\le\left(1+1\right)\left(x-1+4-x\right)=6\)
\(VP>8\) vì \(\frac{29}{5-x}>0\)
=> PT vô nghiệm
gpt:
\(3\left(x^2-3x+1\right)+\sqrt{3\left(x^4+x^2+1\right)}=0\)
\(\sqrt[3]{x^3+5x^2}-1=\sqrt{\frac{5x^2-2}{6}}\)
GPT:\(\frac{\left(x+1\right)\left(x+28\right)\left(x+4\right)\left(x-10\right)\left(-5\right)}{\sqrt{x}\left(x-6\right)^{\frac{1}{2}}}\ln\left(x^2-10\right)=0\)
Nhân tài đâu giúp mình với mình tick cho
Cho \(x=\frac{1}{2}\sqrt{\frac{\sqrt{2}-1}{\sqrt{2}+1}}\)
Tính \(A=\left(4x^5+4x^4-x^3+1\right)^{19}+\left(\sqrt{x^5+4x^4-5x^3+5x+3}\right)^3+\left(\frac{1-\sqrt{2x}}{\sqrt{2x^2+2x}}\right)\)
gpt:
\(\sqrt{x}+\sqrt[4]{x\left(1-x\right)}+\sqrt[4]{\left(1-x\right)^3}=\sqrt{1-x}+\sqrt[4]{x^3}+\sqrt[4]{x^2\left(1-x\right)}\)
GPT: \(\left(x+5\right)\sqrt{x+1}+1=\left(3x +4\right)^{\frac{1}{3}}\)
GPT: \(5\left(\sqrt{x+5}+\sqrt{3x+4}\right)=5x^2-11x-1\)
Gpt
a) \(\left(x-3\right)\left(x+1\right)+4\left(x-3\right)\sqrt{\frac{x+1}{x-3}}=-3\)
b)\(\frac{x\left(x^2+1\right)}{\left(x^2-x+1\right)}=2\)
\(\hept{\begin{cases}4\sqrt{1+\frac{1}{x^2}+\frac{1}{\left(x+1\right)^2}}+5y=\left(\sqrt{y}+2\sqrt{y+1}\right)^2\\4\sqrt{1+\frac{1}{y^2}+\frac{1}{\left(y+1\right)^2}}+5x=\left(\sqrt{x}+2\sqrt{x+1}\right)^2\end{cases}}\)
a)Tính giá trị của biểu thức : S=\(\frac{\left(2^4+\frac{1}{4}\right)\left(4^4+\frac{1}{4}\right)\left(6^4+\frac{1}{4}\right)...\left(100^4+\frac{1}{4}\right)}{\left(1^4+\frac{1}{4}\right)\left(3^4+\frac{1}{4}\right)\left(5^4+\frac{1}{4}\right)..\left(99^4+\frac{1}{4}\right)}\)
b) Cho x,y là các số thực dương.Tìm GTNN của biểu thức : P=\(\frac{x+y}{\sqrt{x\left(4x+5y\right)}+\sqrt{y\left(4y+5x\right)}}\)
