Bài 1: Căn bậc hai

\(=3ab^2\)

\(\sqrt{3^2a^2b^4}\)=3a\(b^2\)

a: ĐKXĐ: \(x\ge\dfrac{1}{2}\)

1 \(\Leftrightarrow\left\{{}\begin{matrix}15x-5y=40\\2x+5y=11\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=3x-8=1\end{matrix}\right.\)

Đề yêu cầu gì?

\(a,ĐK:x>2\\ b,ĐK:x\ge0;x\ne4\)

\(=\dfrac{\sqrt{2}-\sqrt{2-\sqrt{3}}-\sqrt{2}-\sqrt{2+\sqrt{3}}}{\left(\sqrt{2}+\sqrt{2+\sqrt{3}}\right)\left(\sqrt{2}-\sqrt{2-\sqrt{3}}\right)}\\ =\dfrac{-\sqrt{2-\sqrt{3}}-\sqrt{2+\sqrt{3}}}{2-\sqrt{4-2\sqrt{3}}+\sqrt{4+2\sqrt{3}}-2+\sqrt{3}}\\ =\dfrac{-\sqrt{\left(\sqrt{\dfrac{3}{2}}-\sqrt{\dfrac{1}{2}}\right)^2}-\sqrt{\left(\sqrt{\dfrac{3}{2}}+\sqrt{\dfrac{1}{2}}\right)^2}}{\sqrt{3}-\sqrt{3}+1+\sqrt{3}+1}\\ =\dfrac{-\sqrt{\dfrac{3}{2}}+\sqrt{\dfrac{1}{2}}-\sqrt{\dfrac{3}{2}}-\sqrt{\dfrac{1}{2}}}{\sqrt{3}+2}\\ =\dfrac{-\sqrt{6}}{\sqrt{3}+2}=\dfrac{-\sqrt{6}\left(2-\sqrt{3}\right)}{1}=3\sqrt{2}-2\sqrt{6}\)

Lời giải:
\(P=\frac{\sqrt{2}-\sqrt{2+\sqrt{3}}}{2-(2+\sqrt{3})}-\frac{\sqrt{2}+\sqrt{2-\sqrt{3}}}{2-(2-\sqrt{3})}\)

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

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

Xét: \(A=\sqrt{2+\sqrt{3}}-\sqrt{2-\sqrt{3}}\)

\(A^2=4-2\sqrt{(2+\sqrt{3})(2-\sqrt{3})}=2\Rightarrow A=\sqrt{2}\)

\(P=\frac{2\sqrt{2}-A}{-\sqrt{3}}=\frac{2\sqrt{2}-\sqrt{2}}{-\sqrt{3}}=\frac{-\sqrt{2}}{\sqrt{3}}\)

b: 

\(\Leftrightarrow2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}=4\)

\(\Leftrightarrow x-5=4\)

hay x=9

Lời giải:

$(-3)-2\sqrt{3}\approx -6,46$

Cái đề...

C

\(D=2021-\sqrt{\left(x-3\right)^2+1}\le2021-\sqrt{1}=2020\\ D_{max}=2020\left(C\right)\)

\(D=2021-\sqrt{x^2-6x+10}\)

\(\sqrt{x^2-6x+10}=\sqrt{\left(x^2-6x+9\right)+1}=\sqrt{\left(x-3\right)^2+1}\ge\sqrt{0+1}=1\)

\(\Rightarrow D=2021-\sqrt{x^2-6x+10}\le2021-1=2020\)

\(maxD=2020\Leftrightarrow x=3\)

a: ĐKXĐ: \(x\ge0\)

b: ĐKXĐ:\(x\ge7\)

c: ĐKXĐ: \(x\ge-3\)