Ta có: \(a^3-3a^2+8a=9\)

\(\Leftrightarrow\left(a^3-3a^2+3a-1\right)+5a-8=0\)

\(\Leftrightarrow\left(a-1\right)^3+5a-8=0\)

Lại có: \(b^3-6b^2+17b=15\)

\(\Leftrightarrow\left(b^3-6b^2+12b-8\right)+5b-7=0\)

\(\Leftrightarrow\left(b-2\right)^3+5b-7=0\)

Cộng 2 vế trên lại ta được: \(\left(a-1\right)^3+\left(b-2\right)^3+5a+5b-15=0\)

\(\Leftrightarrow\left(a-1+b-2\right)\left[\left(a-1\right)^2-\left(a-1\right)\left(b-2\right)+\left(b-2\right)^2\right]+5\left(a+b-3\right)=0\)

\(\Leftrightarrow\left(a+b-3\right)\left[\left(a-1\right)^2-\left(a-1\right)\left(b-2\right)+\left(b-2\right)^2+5\right]=0\)

Mà \(\left(a-1\right)^2-\left(a-1\right)\left(b-2\right)+\left(b-2\right)^2+5\)

 \(=\left[\left(a-1\right)^2-\left(a-1\right)\left(b-2\right)+\frac{1}{4}\left(b-2\right)^2\right]+\frac{3}{4}\left(b-2\right)^2+5\)

\(=\left[a-1-\frac{1}{2}\left(b-2\right)\right]^2+\frac{3}{4}\left(b-2\right)^2+5>0\left(\forall a,b\right)\)

\(\Rightarrow a+b-3=0\Leftrightarrow a+b=3\)

Vậy a + b = 3

Tổng quát ta có: Với \(n\inℕ\)ta có:

\(\frac{1}{\sqrt{n}+\sqrt{n+1}}=\frac{\left(n+1\right)-n}{\sqrt{n}+\sqrt{n+1}}\)

\(=\frac{\left(\sqrt{n+1}-\sqrt{n}\right)\left(\sqrt{n+1}+\sqrt{n}\right)}{\sqrt{n}+\sqrt{n+1}}=\sqrt{n+1}-\sqrt{n}\)

Với \(n=2\)\(\Rightarrow\frac{1}{\sqrt{2}+\sqrt{3}}=\sqrt{3}-\sqrt{2}\)

Với \(n=3\)\(\Rightarrow\frac{1}{\sqrt{3}+\sqrt{4}}=\sqrt{4}-\sqrt{3}\)

...........................

Với \(n=79\)\(\Rightarrow\frac{1}{\sqrt{79}+\sqrt{80}}=\sqrt{80}-\sqrt{79}\)

\(\Rightarrow\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+.....+\frac{1}{\sqrt{79}+\sqrt{80}}\)

\(=\sqrt{3}-\sqrt{2}+\sqrt{4}-\sqrt{3}+......+\sqrt{80}-\sqrt{79}\)

\(=\sqrt{80}-\sqrt{2}=\sqrt{40.2}-\sqrt{2}=\sqrt{40}.\sqrt{2}-\sqrt{2}\)

\(=\sqrt{2}.\left(\sqrt{40}-1\right)>\sqrt{2}.\left(\sqrt{36}-1\right)\)

\(=\sqrt{2}.\left(6-1\right)=5\sqrt{2}>4\)( đpcm )

ĐKXĐ : \(\hept{\begin{cases}x\ge0\\x\ne1\\x\ne9\end{cases}}\)

a) \(=\left(\frac{x+2\sqrt{x}-7}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}+\frac{1-\sqrt{x}}{\sqrt{x}-3}\right)\div\left(\frac{\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}-\frac{\sqrt{x}+3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\right)\)

\(=\left(\frac{x+2\sqrt{x}-7}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}+\frac{\left(1-\sqrt{x}\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\right)\div\left(\frac{\sqrt{x}-1-\sqrt{x}-3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\right)\)

\(=\left(\frac{x+2\sqrt{x}-7}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}+\frac{3-2\sqrt{x}-x}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\right)\div\left(\frac{-4}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\right)\)

\(=\left(\frac{x+2\sqrt{x}-7+3-2\sqrt{x}-x}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\right)\div\left(\frac{-4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\right)\)

\(=\frac{-4}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\times\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}{-4}\)

\(=\frac{\sqrt{x}-1}{\sqrt{x}-3}\)

b) Để \(P\left(\sqrt{x}-3\right)=\left|x-3\right|\)

=> \(\frac{\sqrt{x}-1}{\sqrt{x}-3}\cdot\left(\sqrt{x}-3\right)=\left|x-3\right|\)(\(\hept{\begin{cases}x\ge0\\x\ne1\\x\ne9\end{cases}}\))

=> \(\sqrt{x}-1=\left|x-3\right|\)

=> \(\orbr{\begin{cases}\sqrt{x}-1=x-3\left(x\ge3\right)\\\sqrt{x}-1=3-x\left(1\le x< 3\right)\end{cases}}\)

=> \(\orbr{\begin{cases}x=4\\x=\frac{9-\sqrt{17}}{2}\end{cases}}\)

c) Em chịu T.T

Đặt \(A=\sqrt{9-2\sqrt{18}}+\sqrt{9+2\sqrt{18}}\)

=> \(A^2=9-2\sqrt{18}+2\sqrt{9-2\sqrt{18}}\cdot\sqrt{9+2\sqrt{18}}+9+2\sqrt{18}\)

\(=18+2\sqrt{9-2\sqrt{18}}\cdot\sqrt{9+2\sqrt{18}}\)

\(=18+2\sqrt{\left(9-2\sqrt{18}\right)\left(9+2\sqrt{18}\right)}\)

\(=18+2\sqrt{9^2-\left(2\sqrt{18}\right)^2}\)

\(=18+2\sqrt{81-72}\)

\(=18+6=24\)

=> \(A=\sqrt{24}\)

tính diện tích cần để trồng dừa, diện tích khu vườn. Trừ 2 cái cho nhau rồi chia cho 13