1. Ta có: \(9< 10\)\(\Rightarrow\sqrt{9}< \sqrt{10}\)\(\Rightarrow3< \sqrt{10}\)\(\Rightarrow3-\sqrt{10}< 0\)(1)

Vì \(3< \sqrt{10}\)\(\Rightarrow2.3< 2\sqrt{10}\)\(\Rightarrow6< 2\sqrt{10}\)\(\Rightarrow2\sqrt{10}-6>0\)(2)

Từ (1) và (2) \(\Rightarrow\sqrt{\left(3-\sqrt{10}\right)^2}+\sqrt{\left(2\sqrt{10}-6\right)^2}\)

\(=\left|3-\sqrt{10}\right|+\left|2\sqrt{10}-6\right|\)

\(=\sqrt{10}-3+2\sqrt{10}-6=3\sqrt{10}-9\)

4. Vì \(x>0\)\(\Rightarrow x.\sqrt{\frac{9}{x}}+5\sqrt{x}=\sqrt{x^2.\frac{9}{x}}+5\sqrt{x}=\sqrt{9x}+5\sqrt{x}\)

\(=3\sqrt{x}+5\sqrt{x}=8\sqrt{x}\)

Thiếu 1 phương trình :

\(4x^2-4\left(2n+1\right)x+4n^2+96mnp+1=0\)

Định lí PYTAGO cho tam giác ABC vuông tại A: \(BC^2=AB^2+AC^2=2AB^2\Rightarrow BC=AB\sqrt{2}\)

Xét tam giác ABC vuông tại A: \(sinB=\frac{AC}{BC}=\frac{AB}{AB\sqrt{2}}=\frac{\sqrt{2}}{2}\)\(cosB=\frac{AB}{BC}=\frac{AB}{AB\sqrt{2}}=\frac{\sqrt{2}}{2}\)

\(tanB=\frac{AC}{AB}=\frac{AB}{AB}=1\), \(cotB=\frac{AB}{AC}=\frac{AB}{AB}=1\)

Vì tam giác ABC vuông cân tại A-->B=45⁰

Vậy \(sin45^0=cos45^0\frac{\sqrt{2}}{2},tan45^0=cot45^0=1\)

,

Ta có: \(5\sqrt{x-1}-\sqrt{36x-36}+\sqrt{9x-9}=\sqrt{8x+12}\)   \(\left(ĐK:x\ge1\right)\)

    \(\Leftrightarrow5\sqrt{x-1}-6\sqrt{x-1}+3\sqrt{x-1}=\sqrt{8x+12}\)

    \(\Leftrightarrow2\sqrt{x-1}=\sqrt{8x+12}\)

    \(\Leftrightarrow\left(2\sqrt{x-1}\right)^2=\left(\sqrt{8x+12}\right)^2\)

    \(\Leftrightarrow4.\left(x-1\right)=8x+12\)

    \(\Leftrightarrow4x-4=8x+12\)

    \(\Leftrightarrow-4x=16\)

    \(\Leftrightarrow x=-4\left(L\right)\)

Vậy \(S=\varnothing\)

\(5\sqrt{x-1}-\sqrt{36\left(x-1\right)}+\sqrt{9\left(x-1\right)}=\sqrt{4\left(2x+3\right)}\) 

\(5\sqrt{x-1}-6\sqrt{x-1}+3\sqrt{x-1}=2\sqrt{2x+3}\) 

\(2\sqrt{x-1}=2\sqrt{2x+3}\) 

\(\sqrt{x-1}=\sqrt{2x+3}\) 

\(\hept{\begin{cases}2x+3\ge0\\x-1=2x-3\end{cases}}\) 

\(\hept{\begin{cases}2x\ge-3\\x-2x=-3+1\end{cases}}\) 

\(\hept{\begin{cases}x\ge-\frac{3}{2}\\-x=-2\end{cases}}\) 

\(\hept{\begin{cases}x\ge-\frac{3}{2}\\x=2\end{cases}}\) 

\(\Rightarrow x=2\)

Ta có: \(5a\sqrt{64ab^3}-\sqrt{3}.\sqrt{12a^3b^3}-5b\sqrt{81a^3b}\)

    \(=5a.8b.\sqrt{ab}-\sqrt{3}.2\sqrt{3}.ab.\sqrt{ab}-5b.9a\sqrt{ab}\)

    \(=40ab.\sqrt{ab}-6ab.\sqrt{ab}-45ab.\sqrt{ab}\)

    \(=40ab.\sqrt{ab}-6ab.\sqrt{ab}-45ab.\sqrt{ab}\)

    \(=-11ab\sqrt{ab}\)

\(5a\sqrt{64ab^3}\) \(-\sqrt{3}\)\(.\sqrt{12a^3b^3}\)\(-5b\sqrt{81a^3b}\)

\(=40ab\sqrt{ab}\)\(-6ab\sqrt{ab}-45ab\sqrt{ab}\)

\(=\left(40ab-6ab-45ab\right)\sqrt{ab}\)

\(=-11ab\sqrt{ab}\)

sao tôi lại thấy tên tôi nhỉ ?

Machi!Rồi bạn trong đội tuyển văn không?

a)\(\frac{3}{2}\sqrt{6}+2\sqrt{\frac{2}{3}}-4\sqrt{\frac{3}{2}}=\frac{3}{2}\sqrt{6}+2\frac{\sqrt{6}}{3}-4\frac{\sqrt{6}}{2}\)

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

b) \(\left(x\sqrt{\frac{6}{x}}+\sqrt{\frac{2x}{3}}+\sqrt{6x}\right):\sqrt{6x}=\left(x.\frac{\sqrt{6x}}{x}+\frac{\sqrt{6x}}{3}+\sqrt{6x}\right):\sqrt{6x}\)

\(=1+\frac{1}{3}+1=2\frac{1}{3}\)

a) Ta có: \(\sqrt{125}-4\sqrt{45}+3\sqrt{20}-\sqrt{80}\)

         \(=5\sqrt{5}-4.3\sqrt{5}+3.2\sqrt{5}-4\sqrt{5}\)

         \(=5\sqrt{5}-12\sqrt{5}+6\sqrt{5}-4\sqrt{5}\)

         \(=-5\sqrt{5}\)

         \(\approx-11,18033989\)

b, \(5\sqrt{a}-4b\sqrt{25a^2}+5a\sqrt{16ab^2}-2\sqrt{9a}\)

\(=5\sqrt{a}-20ab+5a.4\sqrt{a}b-6\sqrt{a}\)

\(=-\sqrt{a}-20ab+20a\sqrt{a}b\)