Gọi số học sinh lúc đầu của nhóm đó là \(x\)(học sinh) \(x\inℕ^∗\).

Mỗi bạn lúc đầu trồng số cây là: \(\frac{120}{x}\)(cây) 

Số học sinh lúc sau là: \(x+3\)(học sinh) 

Mỗi bạn trồng số cây là: \(\frac{120}{x}-2\)(cây).

Ta có phương trình: \(\left(x+3\right)\left(\frac{120}{x}-2\right)=120\)

\(\Rightarrow120x+360-2x^2-6x=120x\)

\(\Leftrightarrow-2x^2-6x+360=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=12\left(tm\right)\\x=-15\left(l\right)\end{cases}}\)

xét bđt sau \(\left(\sqrt{a}+\sqrt{b}\right)^2=a+b+2\sqrt{ab}\ge\left(\sqrt{a+b}\right)^2\)

\(2\sqrt{ab}\ge0\)( luôn đúng)

\(\sqrt{a}+\sqrt{b}\ge\sqrt{a+b}\)

\(a,A=\sqrt{x-2}+\sqrt{6-x}\)

\(A=\sqrt{x-2}+\sqrt{6-x}\ge\sqrt{x-2+6-x}=\sqrt{4}=2\)

dấu "=" xảy ra khi và chỉ khi \(\orbr{\begin{cases}x=2\\x=6\end{cases}}\)

\(< =>MIN:A=2\)

áp dụng bđt bunhia cốp - xki với bộ số \(\left(\sqrt{x-2};\sqrt{6-x}\right);\left(1;1\right)\)

\(A^2=\left(\sqrt{x-2}+\sqrt{6-x}\right)^2\le\left(1^2+1^2\right)\left(x-2+6-x\right)\)

\(A^2\le2.4\)

\(A\le2\sqrt{2}\)

dấu "=" xảy ra khi và chỉ khi \(\frac{\sqrt{x-2}}{1}=\frac{\sqrt{6-x}}{1}\)

\(\sqrt{x-2}=\sqrt{6-x}\)

\(x=4\)

\(MAX:A=2\sqrt{2}\)

Đk: x \(\ge\)0 và x \(\ne\)1

Ta có: \(P=\frac{\sqrt{x}}{\sqrt{x}-1}+\frac{3}{\sqrt{x}+1}-\frac{6\sqrt{x}-4}{x-1}\)

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

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

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

b) Với x \(\ge\)0 và x \(\ne\)1 (2)

Ta có: \(P< \frac{1}{2}\) <=> \(\frac{\sqrt{x}-1}{\sqrt{x}+1}< \frac{1}{2}\) <=> \(2\sqrt{x}-2< \sqrt{x}+1\)

<=> \(\sqrt{x}< 3\) <=> x < 9 (1)

Từ (1) và (2) => \(0\le x< 9\) và x \(\ne\)1

Đk: x \(\ge\)0 và x \(\ne\)1

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

\(P=\frac{3x+3\sqrt{x}-9+\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)-\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)

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

\(P=\frac{3x+5\sqrt{x}-8}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}=\frac{3x+8\sqrt{x}-3\sqrt{x}-8}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)

\(P=\frac{\left(3\sqrt{x}+8\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}=\frac{3\sqrt{x}+8}{\sqrt{x}+2}\)

b) Với x \(\ge\)0 và x \(\ne\)1 (1), ta có:

\(P< \frac{15}{4}\) <=> \(\frac{3\sqrt{x}+8}{\sqrt{x}+2}< \frac{15}{4}\) <=> \(12\sqrt{x}+32< 15\sqrt{x}+30\)

<=> \(3\sqrt{x}>2\) <=> \(\sqrt{x}>\frac{2}{3}\) <=> \(x>\frac{4}{9}\)(2)

Từ (1) và (2) => \(x>\frac{4}{9}\) và x \(\ne\)1

\(A=\frac{\sqrt{x}}{\sqrt{x}+3}+\frac{2\sqrt{x}}{\sqrt{x}-3}-\frac{3x+9}{x-9}\)

\(A=\frac{\sqrt{x}\left(\sqrt{x}-3\right)+2\sqrt{x}\left(\sqrt{x}+3\right)-3x-9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)

\(A=\frac{x-3\sqrt{x}+2x+6\sqrt{x}-3x-9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)

\(A=\frac{3\sqrt{x}-9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\frac{3\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\frac{3}{\sqrt{x}+3}\)

b) Với x \(\ge\)0; x khác 9

Ta có: \(A=\frac{1}{3}\) <=> \(\frac{3}{\sqrt{x}+3}=\frac{1}{3}\) <=> \(\sqrt{x}+3=9\) <=> \(\sqrt{x}=6\) <=> x=  36 (tm)

c) Ta có: \(\sqrt{x}+3\ge3\) => \(\frac{3}{\sqrt{x}+3}\le\frac{3}{3}=1\) => A \(\le\)1

Dấu "=" xảy ra<=> x = 0

Vậy MaxA = 1 <=> x = 0

Ta có: \(\frac{1}{\left(n+1\right)\sqrt{n}+n.\sqrt{n+1}}=\frac{1}{\sqrt{n.\left(n+1\right)}.\left(\sqrt{n+1}+\sqrt{n}\right)}\)

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

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

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

\(T=\frac{1}{2\sqrt{1}+1\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+...+\frac{1}{100\sqrt{99}+99\sqrt{100}}\)

\(=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{99}}-\frac{1}{\sqrt{100}}\)

\(=1-\frac{1}{\sqrt{100}}=\frac{9}{10}\)

b) \(\Leftrightarrow1-\frac{1}{\sqrt{n+1}}=\frac{4}{5}\)( làm tắt nhé )

\(\Leftrightarrow\frac{1}{\sqrt{n+1}}=\frac{1}{5}\)

\(\Leftrightarrow\sqrt{n+1}=5\)

\(\Leftrightarrow n=24\)

Vậy ...

\(A=\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{3}+\sqrt{4}}+\frac{1}{\sqrt{5}+\sqrt{6}}+...+\frac{1}{\sqrt{79}+\sqrt{80}}\)

\(2A=\frac{2}{\sqrt{1}+\sqrt{2}}+\frac{2}{\sqrt{3}+\sqrt{4}}+\frac{2}{\sqrt{5}+\sqrt{6}}+...+\frac{2}{\sqrt{79}+\sqrt{80}}\)

\(>\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+\frac{1}{\sqrt{4}+\sqrt{5}}+...+\frac{1}{\sqrt{79}+\sqrt{80}}+\frac{1}{\sqrt{80}+\sqrt{81}}\)

\(=\frac{\sqrt{2}-1}{2-1}+\frac{\sqrt{3}-\sqrt{2}}{3-2}+\frac{\sqrt{4}-\sqrt{3}}{4-3}+...+\frac{\sqrt{81}-\sqrt{80}}{81-80}\)

\(=\sqrt{2}-1+\sqrt{3}-\sqrt{2}+\sqrt{4}-\sqrt{3}+...+\sqrt{81}-\sqrt{80}\)

\(=\sqrt{81}-1=8\)

suy ra đpcm. 

\(A=\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+...+\frac{1}{\sqrt{80}+\sqrt{81}}\)

\(=\frac{\sqrt{2}-1}{2-1}+\frac{\sqrt{3}-\sqrt{2}}{3-2}+\frac{\sqrt{4}-\sqrt{3}}{4-3}+...+\frac{\sqrt{81}-\sqrt{80}}{81-80}\)

\(=\sqrt{2}-1+\sqrt{3}-\sqrt{2}+\sqrt{4}-\sqrt{3}+...+\sqrt{81}-\sqrt{80}\)

\(=\sqrt{81}-1=8\)

\(B=\frac{1}{\sqrt{1}-\sqrt{2}}-\frac{1}{\sqrt{2}-\sqrt{3}}+\frac{1}{\sqrt{3}-\sqrt{4}}-...-\frac{1}{\sqrt{24}-\sqrt{25}}\)

\(=\frac{\sqrt{1}+\sqrt{2}}{1-2}-\frac{\sqrt{2}+\sqrt{3}}{2-3}+\frac{\sqrt{3}+\sqrt{4}}{3-4}-...-\frac{\sqrt{24}+\sqrt{25}}{24-25}\)

\(=-\sqrt{1}-\sqrt{2}+\sqrt{2}+\sqrt{3}-\sqrt{3}-\sqrt{4}+...+\sqrt{25}+\sqrt{24}\)

\(=\sqrt{25}-\sqrt{1}=4\)