a/ \(n^3-n=\left(n-1\right)n\left(n+1\right)\) 

Ta có : \(n\in Z\Leftrightarrow n-1;n;n+1\in Z\) và là 3 số nguyên liên tiếp

\(\Leftrightarrow n^3-n⋮6\left(đpcm\right)\)

b/ \(A=n^5-n=n\left(n^4-1\right)=n\left(n^2-1\right)\left(n^2+1\right)=\left(n-1\right)n\left(n+1\right)\left(n^2+1\right)\)

Ta có : \(n\left(n-1\right)\left(n+1\right)⋮6\)

+) Nếu \(n=5k\Leftrightarrow n\left(n-1\right)\left(n+1\right)⋮5\Leftrightarrow A⋮30\)

+) Nếu \(n=5k+1\Leftrightarrow n\left(n-1\right)\left(n+1\right)⋮5\Leftrightarrow A⋮30\)

+) Nếu \(n=5k+2\Leftrightarrow n\left(n-1\right)\left(n+1\right)⋮5\Leftrightarrow A⋮30\)

+) Nếu \(n=5k+3\Leftrightarrow n\left(n-1\right)\left(n+1\right)⋮5\Leftrightarrow A⋮30\)

+) Nếu \(n=5k+4\Leftrightarrow n\left(n-1\right)\left(n+1\right)⋮30\Leftrightarrow A⋮30\)

Vậy...

a: \(n^3-n\)

\(=n\left(n^2-1\right)\)

\(=\left(n-1\right)\cdot n\cdot\left(n+1\right)\)

Vì n-1, n và n+1 là ba số tự nhiên liên tiếp nên \(\left(n-1\right)\cdot n\cdot\left(n+1\right)⋮3\)

b: Ta có: \(n^5-n\)

\(=n\left(n^4-1\right)\)

\(=n\left(n^2-1\right)\left(n^2+1\right)\)

\(=n\left(n-1\right)\left(n+1\right)\left(n^2+1\right)⋮30\)

\(A=\left(8.3^3\right)^5.49.7^{13}\)

\(\Rightarrow A=\left(2^3.3^3\right)^5.7^2.7^{13}\)

\(\Rightarrow A=\left[\left(2.3\right)^3\right]^5.7^{15}\)

\(\Rightarrow A=\left(6^3\right)^5.7^{15}\)

\(\Rightarrow A=6^{15}.7^{15}=\left(7.6\right)^{15}=42^{15}⋮42\)

Gọi O là tâm của hình vuông ABCD

=>O là trung điểm chung của AC và BD

a:S.ABCD là hình chóp tứ giác đều nên SO vuông góc (ABCD)

mà \(SO\subset\left(SAC\right)\)

nên \(\left(SAC\right)\perp\left(ABCD\right)\)

b: BD vuông góc SO

BD vuông góc AC

\(SO,AC\subset\left(SAC\right)\)

=>\(BD\perp\left(SAC\right)\)

=>\(\left(SAC\right)\perp\left(SBD\right)\)

a) Ta có:

\(\left. \begin{array}{l}\left. \begin{array}{l}SA \bot SB\\SA \bot SC\end{array} \right\} \Rightarrow SA \bot \left( {SBC} \right)\\SA \subset \left( {SAB} \right)\end{array} \right\} \Rightarrow \left( {SAB} \right) \bot \left( {SBC} \right)\)

b) Ta có:

\(\left. \begin{array}{l}\left. \begin{array}{l}SA \bot SB\\SA \bot SC\end{array} \right\} \Rightarrow SA \bot \left( {SBC} \right)\\SA \subset \left( {SCA} \right)\end{array} \right\} \Rightarrow \left( {SCA} \right) \bot \left( {SBC} \right)\)

c) Ta có:

\(\left. \begin{array}{l}\left. \begin{array}{l}SA \bot SB\\SB \bot SC\end{array} \right\} \Rightarrow SB \bot \left( {SCA} \right)\\SB \subset \left( {SAB} \right)\end{array} \right\} \Rightarrow \left( {SAB} \right) \bot \left( {SCA} \right)\)

\(B=\left(\dfrac{15\left(\sqrt{6}-1\right)}{\left(\sqrt{6}-1\right)\left(\sqrt{6}+1\right)}+\dfrac{4\left(\sqrt{6}+2\right)}{\left(\sqrt{6}-2\right)\left(\sqrt{6}+2\right)}-\dfrac{12\left(3+\sqrt{6}\right)}{\left(3-\sqrt{6}\right)\left(3+\sqrt{6}\right)}\right)\left(\sqrt{6}+11\right)\)

\(=\left(\dfrac{15\left(\sqrt{6}-1\right)}{5}+\dfrac{4\left(\sqrt{6}+2\right)}{2}-\dfrac{12\left(3+\sqrt{6}\right)}{3}\right)\left(\sqrt{6}+11\right)\)

\(=\left[3\left(\sqrt{6}-1\right)+2\left(\sqrt{6}+2\right)-4\left(3+\sqrt{6}\right)\right]\left(\sqrt{6}+11\right)\)

\(=\left(3\sqrt{6}-3+2\sqrt{6}+4-12-4\sqrt{6}\right)\left(\sqrt{6}+11\right)\)

\(=\left(\sqrt{6}-11\right)\left(\sqrt{6}+11\right)\)

\(=6-121=-115\) là số nguyên (đpcm)

b) Ta có: \(B=\left(\dfrac{15}{\sqrt{6}+1}+\dfrac{4}{\sqrt{6}-2}-\dfrac{12}{3-\sqrt{6}}\right)\left(\sqrt{6}+11\right)\)

\(=\left(\dfrac{15\left(\sqrt{6}-1\right)}{5}+\dfrac{4\left(\sqrt{6}+2\right)}{2}-\dfrac{12\left(3+\sqrt{6}\right)}{3}\right)\left(\sqrt{6}+11\right)\)

\(=\left(3\sqrt{6}-3+2\sqrt{6}+4-12-4\sqrt{6}\right)\left(\sqrt{6}+11\right)\)

\(=\left(\sqrt{6}-11\right)\left(\sqrt{6}+11\right)\)

=6-121=-115

a) Ta có:

      \(\sqrt 2 \sin \left( {x - \frac{\pi }{4}} \right) = \sqrt 2 \left( {\sin x\cos \frac{\pi }{4} + \cos x\sin \frac{\pi }{4}} \right) = \sqrt 2 \left( {\sin x.\frac{{\sqrt 2 }}{2} + \cos x.\frac{{\sqrt 2 }}{2}} \right) = \sin x + \cos x\)

b) Ta có:

\(\tan \left( {\frac{\pi }{4} - x} \right) = \frac{{\tan \frac{\pi }{4} - \tan x}}{{1 + \tan \frac{\pi }{4}\tan x}} = \frac{{1 - \tan x}}{{1 + \tan x}}\;\)

câu b thôi nhé bạn

2¹⁷+2¹⁴=\(2^{14}x\left(2^3+1\right)\)=2¹⁴ x 9 \(⋮\)9

=> \(\left(2^{17}+2^{14}\right)⋮9\)

a ) Ta có : \(15^3-25^2=\left(5.3\right)^3-\left(5^2\right)^2=5^3.3^3-5^{2.2}\)

\(=5^3.27-5^4=5^3.27-5^3.5=5^3.\left(27-5\right)=5^3.22\)

Vì \(22⋮11\Rightarrow5^3.22⋮11\Rightarrow\left(15^3-25^2\right)⋮11\)

Vậy \(\left(15^3-25^2\right)⋮11\)

b ) Ta có : \(2^{17}+2^{14}=2^{14}.2^3+2^{14}=2^{14}.8+2^{14}=2^{14}.\left(8+1\right)=2^{14}.9^{14}\)

nhầm

số 9^14 là số 9 nha bạn

nó là : \(2^{14}.\left(8+1\right)=2^{14}.9\)

Vì \(9⋮9\Rightarrow2^{14}.9⋮9\)\(\Rightarrow\)\(\left(2^{17}+2^{14}\right)⋮9\)

Vậy \(\left(2^{17}+2^{14}\right)⋮9\)

cả câu a nữa

15³-25²=\(5^3x3^3-5^4\)=\(5^3x\left(3^3-1\right)\)=\(5^3\) x 22 \(⋮\)11

tham khảo:

a) Tam giác ABD có HK là đường trung bình nên HK//BD

Vì ABCD là hình vuông nên AC⊥BD. Suy ra AC⊥HK

Vì SH⊥(ABCD) nên SH⊥AC

Ta có: AC⊥SH,AC⊥HK nên AC⊥(SHK)

b) Ta có tam giác AHD và tam giác DKC bằng nhau nên DH⊥CK

Mà SH⊥(ABCD) nên SH⊥CK

Suy ra CK⊥(SDH)