Chào

1 a/ Trục căn thức ở mẫu

\(VT=\frac{-\sqrt{1}+\sqrt{2}}{2-1}+\frac{-\sqrt{2}+\sqrt{3}}{3-2}+...+\frac{-\sqrt{47}+\sqrt{48}}{48-47}\)\(=-\sqrt{1}+\sqrt{2}-\sqrt{2}+\sqrt{3}-....-\sqrt{47}+\sqrt{48}=\sqrt{48}-1>3=VP\)

b/

\(2\left(10+3\sqrt{11}\right)=11+2.\sqrt{11}.3+9=\left(\sqrt{11}+3\right)^2\)

\(VT=\left(\sqrt{11}-3\right)\sqrt{2}\sqrt{10+3\sqrt{11}}=\left(\sqrt{11}-3\right)\left(\sqrt{11}+3\right)=11-9=2=VP\)

2/

\(B=\left(5+\sqrt{21}\right)\left(\sqrt{7}-\sqrt{3}\right)\sqrt{2\left(5+\sqrt{3}.\sqrt{7}\right)}\)

\(2\left(5+\sqrt{21}\right)=7+2\sqrt{7}.\sqrt{3}+3=\left(\sqrt{7}+\sqrt{3}\right)^2\)

\(B=\left(5+\sqrt{21}\right)\left(\sqrt{7}-\sqrt{3}\right)\left(\sqrt{7}+\sqrt{3}\right)=\left(5+\sqrt{21}\right).4\)

\(=20+4\sqrt{21}\)

A chắc không rút gọn được.

Điều kiện: \(a\ge0;a\ne1\)

  \(A=\frac{\sqrt{a}+3}{\sqrt{a}-3}\)

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

\(=\frac{\sqrt{a}-3}{\sqrt{a}-3}+\frac{6}{\sqrt{a}-3}=1+\frac{6}{\sqrt{a}-3}\)

mà 1 \(\in\)Z

\(\Rightarrow\frac{6}{\sqrt{a}-3}\in Z\)

\(\Rightarrow\left(\sqrt{a}-3\right)\inƯ\left(2\right)\)

\(\Rightarrow\left(\sqrt{a}-3\right)\in\left\{1;-1;2;-2\right\}\)

\(\Rightarrow\sqrt{a}\in\left\{2;0;3;-1\right\}\)

\(\Rightarrow a\in\left\{4;0;9\right\}\)

Vậy là ta đã hoàn thành bài

Ta có:

(n2−8)2+36

=n4−16n2+64+36

=n4+20n2+100−36n2

=(n2+10)2−(6n)2

=(n2+10+6n)(n2+10−6n)

Mà để (n2+10+6n)(n2+10−6n) là số nguyên tố thì n2+10+6n=1 hoặc n2+10−6n=1

Mặt khác ta có n2+10−6n<n2+10+6n  n2+10−6n=1 (n thuộc N) 

 n2+9−6n=0 hay (n−3)2=0  n=3

Vậy với n=3 thì (n2−8)2+36 là số nguyên tố
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(a - 1)^3 + a^3 + (a + 1)^3=a^3 - 3a^2 + 3a - 1 + a^3 + a^3 + 3a^2 + 3a +1 = 3a^3 + 6a 
= 3a(a^2 + 2) = 3a(a^2 - 1) + 9a 
= 3(a - 1)a(a + 1) + 9a 
vì tích của 3 số tự nhiên liên tiếp chia hết cho 3 nên 3(a - 1)a(a + 1) chia hết cho 9 
Mặt khác 9a chia hết cho 9 nên 
==>3(a - 1)a(a + 1) + 9a (đpcm)

Cho 1 đúntg nha

\(\frac{1}{2\sqrt{2}}-\frac{3}{2}\sqrt{\frac{9}{2}}+\frac{2}{5}\sqrt{30}=\frac{1}{2\sqrt{2}}-\frac{3}{2}.\frac{\sqrt{9}}{\sqrt{2}}+\frac{2\sqrt{30}}{5}\)

\(=\frac{1}{2\sqrt{2}}-\frac{9}{2\sqrt{2}}+\frac{2\sqrt{30}}{5}=\frac{-8}{2\sqrt{2}}+\frac{2\sqrt{30}}{5}\)

\(=\frac{-4}{\sqrt{2}}+\frac{2\sqrt{30}}{5}=\frac{-4\sqrt{2}}{2}+\frac{2\sqrt{30}}{5}\)

\(=\frac{-20\sqrt{2}}{10}+\frac{4\sqrt{30}}{10}=\frac{-20\sqrt{2}+4\sqrt{30}}{10}=\frac{2.\left(2\sqrt{30}-10\sqrt{2}\right)}{10}\)

\(=\frac{2\sqrt{30}-10\sqrt{2}}{5}\)

\(\frac{1}{2\sqrt{2}}-\frac{3}{2}\sqrt{\frac{9}{2}}+\frac{2}{5}\sqrt{30}=\frac{1}{2\sqrt{2}}-\frac{3}{2}.\frac{\sqrt{9}}{\sqrt{2}}+\frac{2\sqrt{30}}{5}=\frac{\sqrt{2}}{4}-\frac{9\sqrt{2}}{4}+\frac{2\sqrt{30}}{5}=-2\sqrt{2}+\frac{2\sqrt{30}}{5}=\frac{2\sqrt{30}-10\sqrt{2}}{5}\)