\(\sqrt{2x+1}-\sqrt{5-x}+x-6=0\)

\(\Leftrightarrow\left(\sqrt{2x+1}-3\right)+\left(1-\sqrt{5-x}\right)+x-4=0\)

\(\Leftrightarrow\frac{2\left(x-4\right)}{\sqrt{2x+1}+3}+\frac{x-4}{\sqrt{5-x}+1}+x-4=0\)

\(\Leftrightarrow\left(x-4\right)\left(\frac{2}{\sqrt{2x+1}+3}+\frac{1}{\sqrt{5-x}+1}+1\right)=0\)

\(\Leftrightarrow x=4\)

trả lời nhanh hộ t nhé cc :)

\(\frac{5\left(\sqrt{6}-1\right)\left(\sqrt{6}-1\right)}{\left(\sqrt{6}+1\right)\left(\sqrt{6}-1\right)}+\frac{\left(\sqrt{2}-\sqrt{3}\right)\left(\sqrt{2}-\sqrt{3}\right)}{\left(\sqrt{2}+\sqrt{3}\right)\left(\sqrt{2}-\sqrt{3}\right)}+\sqrt{\left(\sqrt{2}\right)^2-2\sqrt{2}+1}\)

\(=\frac{5\left(\sqrt{6}-1\right)^2}{5}-\frac{\left(\sqrt{2}-\sqrt{3}\right)^2}{1}+\sqrt{\left(\sqrt{2}-1\right)^2}\)

\(=\left(\sqrt{6}-1\right)^2-\left(\sqrt{2}-\sqrt{3}\right)^2+\left(\sqrt{2}-1\right)\)

\(=6-2\sqrt{6}+1-2+2\sqrt{6}-3+\sqrt{2}-1=\sqrt{2}\)

Ta có

\(x=\frac{\sqrt{4+2\sqrt{3}}-\sqrt{3}}{\left(\sqrt{5}+2\right)\sqrt[3]{17\sqrt{5}-38}-2}\)

\(=\frac{\sqrt{3+2\sqrt{3}+1}-\sqrt{3}}{\left(\sqrt{5}+2\right)\sqrt[3]{5\sqrt{5}-3.5.2+3.4.\sqrt{5}-8}-2}\)

\(=\frac{\sqrt{3}+1-\sqrt{3}}{\left(\sqrt{5}+2\right)\left(\sqrt{5}-2\right)-2}=\frac{1}{5-4-2}=-1\)

Thế vào ta được

\(P=\left(x^2+x+1\right)^{2013}+\left(x^2+x-1\right)^{2013}\)

\(=\left(1-1+1\right)^{2013}+\left(1-1-1\right)^{2013}=1-1=0\)

\(x=\frac{\sqrt{\left(\sqrt{5}+1\right)^2}+\sqrt{\left(\sqrt{5-1}\right)^2}}{\sqrt{20}}=\frac{2\sqrt{5}}{\sqrt{20}}=1\)

=>P=(1+1-1)²⁰¹⁶=1

a)  \(A=\sqrt{10+\sqrt{99}}=\sqrt{10+3\sqrt{11}}=\frac{1}{\sqrt{2}}.\sqrt{20+6\sqrt{11}}\)

\(=\frac{1}{\sqrt{2}}.\sqrt{\left(3+\sqrt{11}\right)^2}=\frac{3+\sqrt{11}}{2}\)

b)  \(B=\sqrt{21+6\sqrt{6}}-\sqrt{21-6\sqrt{6}}=\sqrt{\left(3\sqrt{2}+\sqrt{3}\right)^2}-\sqrt{\left(3\sqrt{2}-\sqrt{3}\right)^2}\)

\(=3\sqrt{2}+\sqrt{3}-3\sqrt{2}+\sqrt{3}=2\sqrt{3}\)

c) bn ktra lại đề

d) ĐK:  \(x\ge0\)

 \(\sqrt{x+1+2\sqrt{x}}=\sqrt{\left(\sqrt{x}+1\right)^2}=\sqrt{x}+1\)

e) đk:  \(x\ge-1\)

 \(\sqrt{2x+3+2\sqrt{x^2+3x+2}}=\sqrt{x+1+2\sqrt{\left(x+1\right)\left(x+2\right)}+x+2}\)

\(=\sqrt{\left(\sqrt{x+1}+\sqrt{x+2}\right)^2}=\sqrt{x+1}+\sqrt{x+2}\)

Đặt \(\hept{\begin{cases}\sqrt[3]{x-16}=a\\\sqrt[3]{x+13}=b\end{cases}}\)

\(\Rightarrow b^3-a^3=29\)

Từ đó ta có hệ \(\hept{\begin{cases}1+a=b\\b^3-a^3=29\end{cases}}\)

Thế pt đầu vào pt sau ta được

\(a^3+3a^2+3a+1-a^3=29\)

\(\Leftrightarrow3a^2+3a-28=0\)

\(\Leftrightarrow\orbr{\begin{cases}a=\frac{-3+\sqrt{345}}{6}\\a=\frac{-3-\sqrt{345}}{6}\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}b=\frac{3+\sqrt{345}}{6}\\b=\frac{3-\sqrt{345}}{6}\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{31\sqrt{345}+27}{18}\\x=\frac{-31\sqrt{345}+27}{18}\end{cases}}\)