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Ta có: \(n^5-5n^3+4n^2\)
\(=n^2\left(n^3-5n+4\right)\)
\(=n^2\left(n^3-n-4n+4\right)\)
\(=n^2\cdot\left[n\left(n-1\right)\left(n+1\right)-4\left(n-1\right)\right]\)
\(=n^2\left(n-1\right)\left(n^2+n-4\right)⋮120\)
\(\dfrac{1}{2.2}.\dfrac{2}{2.3}.....\dfrac{31}{64}=2^x\\ =>\dfrac{1}{2.2.2.....2.64}=2^x\\ \dfrac{1}{2^{30}.26}=2^x\\ =>\dfrac{1}{2^{36}}=2^x\\ =>2^{-36}=2^x\\ =>x=-36\)
b, Ta có : \(\dfrac{x}{3}=\dfrac{y}{4};\dfrac{y}{5}=\dfrac{z}{6}\Rightarrow\dfrac{x}{15}=\dfrac{y}{20}=\dfrac{z}{24}\)
Đặt \(x=15k;y=20k;z=24k\)
Thay vào A ta được : \(A=\dfrac{30k+60k+96k}{45k+80k+120k}=\dfrac{186k}{245k}=\dfrac{186}{245}\)
\(a,=\dfrac{\sqrt{x}-8+5}{\sqrt{x}-8}=1+\dfrac{5}{\sqrt{x}-8}\in Z\\ \Leftrightarrow\sqrt{x}-8\inƯ\left(5\right)=\left\{-5;-1;1;5\right\}\\ \Leftrightarrow\sqrt{x}\in\left\{3;7;9;13\right\}\\ \Leftrightarrow x\in\left\{9;49;81;169\right\}\left(tm\right)\\ b,=\dfrac{\sqrt{x}-2+7}{\sqrt{x}-2}=1+\dfrac{7}{\sqrt{x}-2}\in Z\\ \Leftrightarrow\sqrt{x}-2\inƯ\left(7\right)=\left\{-1;1;7\right\}\left(\sqrt{x}-2>-2\right)\\ \Leftrightarrow\sqrt{x}\in\left\{1;3;9\right\}\\ \Leftrightarrow x\in\left\{1;9;81\right\}\\ c,=\dfrac{2\left(\sqrt{x}+3\right)+2}{\sqrt{x}+3}=2+\dfrac{2}{\sqrt{x}+3}\in Z\\ \Leftrightarrow\sqrt{x}+3\inƯ\left(2\right)=\varnothing\left(\sqrt{x}+3>3\right)\\ \Leftrightarrow x\in\varnothing\)
đáp án là đpcm