Với \(k\in N;k\ne0\) ta có :

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

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

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

Áp dụng ta có :

\(M=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+....+\frac{1}{\sqrt{120}}-\frac{1}{\sqrt{121}}=1-\frac{1}{11}=\frac{10}{11}\)

\(\frac{3}{4}.\frac{8}{9}.\frac{15}{16}.\frac{24}{25}...\frac{63}{64}\)

\(=\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}.\frac{4.6}{5.5}...\frac{7.9}{8.8}\)

\(=\frac{1.3.2.4.3.5.4.6...7.9}{2.2.3.3.4.4.5.5...8.8}\)

\(=\frac{1.9}{2.8}=\frac{9}{16}\)

Hê lô

a/

\(A=\frac{3}{x+2}-\frac{2}{2-x}-\frac{8}{x^2-4}\)

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

\(=\frac{3x-6+2x+4-8}{\left(x+2\right)\left(x-2\right)}\)

\(=\frac{5x-10}{\left(x+2\right)\left(x-2\right)}=\frac{5}{x+2}\)

b/ Thay x = 3 thì ta được

\(\frac{5}{3+2}=1\)

B) biểu thức đó sẽ bằng 1

mình cần cách làm nhá bạn chứ lấy máy tính bấm cũng ra

bài1   A=\(\left(\frac{3-x}{x+3}\cdot\frac{x^2+6x+9}{x^2-9}+\frac{x}{x+3}\right):\frac{3x^2}{x+3}\)

=\(\left(-\frac{x-3\cdot\left(x+3\right)^2}{\left(x+3\right)^2\cdot\left(x-3\right)}+\frac{x}{x+3}\right):\frac{3x^2}{x+3}\)

=\(-\frac{x}{x+3}\cdot\frac{x+3}{3x^2}=\frac{-1}{3x}\)

b)  thế \(x=-\frac{1}{2}\)vào biểu thức A

 \(-\frac{1}{3\cdot\left(-\frac{1}{2}\right)}=\frac{2}{3}\)

c)  A=\(-\frac{1}{3x}< 0\)

VÌ (-1) <0  nên  3x>0

                        x >0

=2/10+3/10+4/10+......+13/10

=\(\frac{2+3+4+......+13}{10}\)

=90/10=9

k cho mình nha

=9 nhe cac ban

dấu sao kia là dấu nhân nhé

1. \(x=\frac{1}{9}\) thỏa mãn đk: \(x\ge0;x\ne9\)

Thay \(x=\frac{1}{9}\) vào A ta có:

\(A=\frac{\sqrt{\frac{1}{9}}+1}{\sqrt{\frac{1}{9}}-3}=-\frac{1}{2}\)

2. \(B=...\)

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

    \(B=\frac{3x-9\sqrt{x}+x+3\sqrt{x}-4x-6}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)

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

3. \(P=A:B=\frac{\sqrt{x}+1}{\sqrt{x}-3}:\frac{-6\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)

\(P=\frac{\sqrt{x}+3}{-6}\)

Vì \(\sqrt{x}+3\ge3\forall x\)\(\Rightarrow\frac{\sqrt{x}+3}{-6}\le\frac{3}{-6}=-\frac{1}{2}\)

hay \(P\le-\frac{1}{2}\)

Dấu "=" xảy ra <=> x=0