\(a,\left(\sqrt{3}-1\right)^2=3-2\sqrt{3}+1=4-2\sqrt{3}\\ b,\sqrt{4-2\sqrt{3}}-\sqrt{3}=\left(\sqrt{3}-1\right)-\sqrt{3}=-1\)

a) Ta có:

(5^2n+1) + (2^n+4) + (2^n+1) = (25^n).5 - 5.(2^n) + (2^n).( 5 + 2^4 +2) = 5.( 25^n - 2^n ) + 23.2^n chia hết cho 23.  

Lời giải:

a) 

\(5^{2n+1}+2^{n+4}+2^{n+1}=5.25^n+16.2^n+2.2^n\)

\(\equiv 5.2^n+16.2^n+2.2^n\pmod {23}\)

\(\equiv 23.2^n\equiv 0\pmod {23}\)

Ta có đpcm.

b) 

\(2^{2n+2}+24n+14\) hiển nhiên chia hết cho $2(1)$

Mặt khác:

Nếu $n=3k+1$:

$2^{2n+2}+24n+14=2^{6k+4}+72k+38$

$=16.2^{6k}+72k+38\equiv 16+72k+38=54+72k\equiv 0\pmod 9$

Nếu $n=3k$:

$2^{2n+2}+24n+14=2^{6k+2}+72k+14=4.2^{6k}+72k+14$

$\equiv 4+72k+14=18+72k\equiv 0\pmod 9$

Nếu $n=3k+2$:

$2^{2n+2}+24n+14=2^{6k+6}+72k+62\equiv 1+72k+62$

$\equiv 63+72k\equiv 0\pmod 9$

Vậy tóm lại $2^{2n+2}+24n+14$ chia hết cho $9$ (2)

Từ $(1);(2)\Rightarrow 2^{2n+2}+24n+14\vdots 18$ (đpcm)

Ta có: \(A=1+4+4^2+...+4^{99}\)

\(\Rightarrow4A=4+4^2+4^3+...+4^{100}\)

\(\Rightarrow4A-A=\left(4+4^2+4^3+...+4^{100}\right)-\left(1+4+4^2+...+4^{99}\right)\)

\(\Rightarrow3A=4^{100}-1\)

\(\Rightarrow A=\dfrac{4^{100}-1}{3}\)

Vì \(4^{100}-1< 4^{100}\) nên \(\dfrac{4^{100}-1}{3}< \dfrac{4^{100}}{3}\)

\(\Rightarrow A< \dfrac{B}{3}\left(đpcm\right)\)

Vậy...

a.m+2>n+2

Ta có: m >n

=>m+2 > n+2 (cộng hai vế với 2)

do đó m+2>n+2

b, -2m < -2n

Ta có: m > n

=> -2m < -2n (nhân hai vế với -2)

do đó -2m<-2n

c,2m-5>2n-5

Ta có: m>n

=>2m>2n (nhân hai vế với 2)

=>2m-5>2n-5 ( cộng hai vế với -5)

do đó 2m-5>2n-5

d,4-3m<4-3n

Ta có :m>n

=> -3m<-3n (nhân hai vế với -3)

=> 4-3m<4-3n (cộng 2 vế với 4)

\(\Rightarrow\left\{{}\begin{matrix}\left(a-b\right)^2=0\\\left(b-c\right)^2=0\\\left(c-a\right)^2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=b\\b=c\\c=a\end{matrix}\right.\)

\(\Rightarrow a=b=c\left(đpcm\right)\)

Ta có: \(\left(a+b+c\right)^2=a^2+b^2+c^2+2ab+2bc+2ac\)

\(3\left(a^2+b^2+c^2\right)=3a^2+3b^2+3c^2\)
mà \(\left(a+b+c\right)^2=3\left(a^2+b^2+c^2\right)\Rightarrow a^2+b^2+c^2+2ab+2bc+2ca=3a^2+3b^2+3c^2\)

\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2bc-2ca=0\)
\(\Leftrightarrow a^2-2ab+b^2+b^2-2bc+c^2+c^2-2ac+a^2=0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)

Vì \(\left\{{}\begin{matrix}\left(a-b\right)^2\ge0\forall a,b\\\left(b-c\right)^2\ge0\forall b,c\\\left(c-a\right)^2\ge0\forall a,c\end{matrix}\right.\)

Mà \(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)
\(\Rightarrow\left\{{}\begin{matrix}\left(a-b\right)^2=0\\\left(b-c\right)^2=0\\\left(c-a\right)^2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=b\\b=c\\c=a\end{matrix}\right.\Leftrightarrow}a=b=c\Rightarrowđpcm}\)