CÂU B là sai các bạn đừng giải, mình xin lỗi

a)\(81^7-27^9-9^{13}=\left(3^4\right)^7-\left(3^3\right)^9-\left(3^2\right)^{13}\) 

 \(=3^{28}-3^{27}-3^{26}=3^{24}\left(3^4-3^3-3^2\right)\) 

 \(=3^{24}.45⋮45\) 

\(\Rightarrow81^7-27^9-9^{13}⋮45\left(đpcm\right)\)

#)Giải :

\(\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)\left(1-\frac{1}{4^2}\right)...\left(1-\frac{1}{n^2}\right)\)

\(=\frac{2^2-1}{2^1}.\frac{3^2-1}{3^2}.\frac{4^2-1}{4^2}.....\frac{n^2-1}{n^2}\)

\(=\frac{1.3}{2^2}.\frac{2.4}{3^2}.\frac{3.5}{4^2}.....\frac{\left(n-1\right)\left(n+1\right)}{n^2}\)

\(=\frac{\left[1.2.3.....\left(n-1\right)\right].\left[3.4.5.....\left(n+1\right)\right]}{\left(2.3.4.5.....n\right)\left(2.3.4.5.....n\right)}\)

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

๖²⁴ʱŤ.Ƥεɳɠʉїɳş༉ phải là \(2^2\)chứ không phải là \(2^1\)nhé bạn!        

Ta có: \(\frac{a}{2}=\frac{b}{3}=\frac{c}{4}\Leftrightarrow\frac{a^2}{4}=\frac{b^2}{9}=\frac{2c^2}{32}\)

Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:

\(\frac{a^2}{4}=\frac{b^2}{9}=\frac{2c^2}{32}=\frac{a^2-b^2+2c^2}{4-9+32}=\frac{108}{27}=4\)

\(\Rightarrow\hept{\begin{cases}\frac{a}{2}=4\\\frac{b}{3}=4\\\frac{c}{4}=4\end{cases}\Rightarrow}\hept{\begin{cases}a=8\\b=12\\c=16\end{cases}}\)

Vậy...

TA CÓ:\(\frac{a}{2}=\frac{b}{3}=\frac{c}{4}\)

\(\Leftrightarrow\frac{a^2}{4}=\frac{b^2}{9}=\frac{c^2}{16}\)

\(\Leftrightarrow\frac{a^2}{4}=\frac{b^2}{9}=\frac{2c^2}{32}\)

Áp dụng t/c dãy tỉ số bằng nhau, ta có:

\(\frac{a^2}{4}=\frac{b^2}{9}=\frac{2c^2}{32}=\frac{a^2-b^2+2c^2}{4-9+32}=\frac{108}{27}=4\)

\(\cdot\frac{a^2}{4}=4\Leftrightarrow a^2=16\Leftrightarrow a=\pm4\)

\(\cdot\frac{b^2}{9}=4\Leftrightarrow b^2=36\Leftrightarrow b=\pm6\)

\(\cdot\frac{2c^2}{32}=4\Leftrightarrow2c^2=128\Leftrightarrow c^2=64\Leftrightarrow c=\pm8\)

Vậy các cặp \(\left(a,b,c\right)\)thõa mãn là : \(\left(4;6;8\right)\)và \(\left(-4;-6;-8\right)\)

\(\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right)\left(1+\frac{1}{4}\right)........\left(1+\frac{1}{n}\right)\)

\(=\frac{3}{2}.\frac{4}{3}.\frac{5}{4}.......\frac{n+1}{n}\)

\(=\frac{3.4.5........\left(n+1\right)}{2.3.4........n}\)

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

a) \(\left(1+\frac{1}{2}\right).\left(1+\frac{1}{3}\right).\left(1+\frac{1}{4}\right)...\left(1+\frac{1}{n}\right)\)

= \(\frac{3}{2}.\frac{4}{3}.\frac{5}{4}....\frac{n+1}{n}\)

= \(\frac{3.4.5....\left(n+1\right)}{2.3.4..n}\)

= \(\frac{n+1}{2}\)

Đề là j thế bn ?

TÌM GIÁ TRỊ LỚN NHẤT CỦA BIỂU THỨC

a) \(\frac{3}{2}\). \(\frac{1}{3}\)+\(\frac{1}{2}\)

= 1/2 + 1/2

= 1

Mấy bài khác tương tự :)))

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

b) \(\left(\frac{1}{2}-\frac{1}{3}\right):\frac{1}{12}=\frac{1}{6}.12=2\)

c) \(\frac{5}{9}.\frac{7}{13}+\frac{5}{9}.\frac{9}{13}-\frac{5}{9}.\frac{3}{13}=\frac{5}{9}.\left(\frac{7}{13}+\frac{9}{13}-\frac{3}{13}\right)=\frac{5}{9}.1=\frac{5}{9}\)

\(\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{n}\right)\)

= \(\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{n-1}{n}\)

= \(\frac{1.2.3.4...\left(n-1\right)}{2.3.4...n}\)

= \(\frac{1}{n}\)

#)Giải :

\(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{n}\right)\)

\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.....\frac{n-1}{n}\)

\(=\frac{1.2.3.....n-1}{2.3.4.....n-1.n}=\frac{1}{n}\)