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Lời giải:

a)
Đặt $2^{10}=a; 3^{10}=b; 4^{10}=c$ trong đó $a,b,c>0$ và $a\neq b\neq c$

Khi đó:

Xét hiệu \(2^{30}+3^{30}+4^{30}-3.24^{10}=a^3+b^3+c^3-3abc\)

\(=(a+b+c)(a^2+b^2+c^2-ab-bc-ac)\)

\(=\frac{a+b+c}{2}[(a-b)^2+(b-c)^2+(c-a)^2]\)

Vì $a,b,c>0\Rightarrow a+b+c>0$

$a\neq b\neq c\Rightarrow (a-b)^2>0; (b-c)^2>0; (c-a)^2>0$

$\Rightarrow (a-b)^2+(b-c)^2+(c-a)^2>0$

Do đó:

$2^{30}+3^{30}+4^{30}-3.24^{10}=\frac{a+b+c}{2}[(a-b)^2+(b-c)^2+(c-a)^2]>0$

$\Rightarrow 2^{30}+3^{30}+4^{30}>3.24^{10}$

b)

Có: $4=\sqrt{16}>\sqrt{14}$

$\sqrt{33}>\sqrt{29}$

Cộng theo vế:

$4+\sqrt{33}>\sqrt{14}+\sqrt{29}$

Lời giải:

a)
Đặt $2^{10}=a; 3^{10}=b; 4^{10}=c$ trong đó $a,b,c>0$ và $a\neq b\neq c$

Khi đó:

Xét hiệu \(2^{30}+3^{30}+4^{30}-3.24^{10}=a^3+b^3+c^3-3abc\)

\(=(a+b+c)(a^2+b^2+c^2-ab-bc-ac)\)

\(=\frac{a+b+c}{2}[(a-b)^2+(b-c)^2+(c-a)^2]\)

Vì $a,b,c>0\Rightarrow a+b+c>0$

$a\neq b\neq c\Rightarrow (a-b)^2>0; (b-c)^2>0; (c-a)^2>0$

$\Rightarrow (a-b)^2+(b-c)^2+(c-a)^2>0$

Do đó:

$2^{30}+3^{30}+4^{30}-3.24^{10}=\frac{a+b+c}{2}[(a-b)^2+(b-c)^2+(c-a)^2]>0$

$\Rightarrow 2^{30}+3^{30}+4^{30}>3.24^{10}$

b)

Có: $4=\sqrt{16}>\sqrt{14}$

$\sqrt{33}>\sqrt{29}$

Cộng theo vế:

$4+\sqrt{33}>\sqrt{14}+\sqrt{29}$

\(A=\sqrt{6}+\sqrt{12}+\sqrt{30}+\sqrt{56}\)

\(B^2=\left(\sqrt{6}+\sqrt{30}\right)^2=36+2\sqrt{180}>36+26=62\)

B>7;\(\sqrt{30}>5;\sqrt{56}>7\)

A>7+5+7=19

A>19

ghi de sai ban oi

\(A=\sqrt{2}+\sqrt{6}+\sqrt{12}+\sqrt{20}+\sqrt{30}+\sqrt{42}\)

\(A< \sqrt{2,25}+\sqrt{6,25}+\sqrt{12,25}+\sqrt{20,25}+\sqrt{30,25}+\sqrt{42,25}=24=B\)

Vậy \(A< B\)

Chúc bạn học tốt ~ 

1.

 Ta có: \(A=\sqrt{31-2\sqrt{30}}=\sqrt{\left(\sqrt{30}-1\right)^2}=\left|\sqrt{30}-1\right|=\sqrt{30}-1\)

\(B=\sqrt{11-2\sqrt{30}}=\sqrt{\left(\sqrt{6}-\sqrt{5}\right)^2}=\left|\sqrt{6}-\sqrt{5}\right|=\sqrt{6}-\sqrt{5}\)

\(C=\sqrt{13-2\sqrt{30}}=\sqrt{\left(\sqrt{10}-\sqrt{3}\right)^2}=\left|\sqrt{10}-\sqrt{3}\right|=\sqrt{10}-\sqrt{3}\)

\(D=\sqrt{39-6\sqrt{30}}=\sqrt{\left(\sqrt{30}-3\right)^2}=\left|\sqrt{30}-3\right|=\sqrt{30}-3\)

\(A=\sqrt{31-2\sqrt{30}}=\sqrt{30}-1\)

\(B=\sqrt{11-2\sqrt{30}}=\sqrt{6}-\sqrt{5}\)

\(C=\sqrt{13-2\sqrt{30}}=\sqrt{10}-\sqrt{3}\)

\(D=\sqrt{39-6\sqrt{30}}=\sqrt{30}-3\)

1.

Ta có: \(E=\sqrt{37-6\sqrt{30}}=\sqrt{\left(3\sqrt{3}-\sqrt{10}\right)^2}=\left|3\sqrt{3}-\sqrt{10}\right|=3\sqrt{3}-\sqrt{10}\)

\(F=\sqrt{51-6\sqrt{30}}=\sqrt{\left(3\sqrt{5}-\sqrt{6}\right)^2}=\left|3\sqrt{5}-\sqrt{6}\right|=3\sqrt{5}-\sqrt{6}\)

\(G=\sqrt{59-6\sqrt{30}}=\sqrt{\left(3\sqrt{6}-\sqrt{5}\right)^2}=\left|3\sqrt{6}-\sqrt{5}\right|=3\sqrt{6}-\sqrt{5}\)

\(H=\sqrt{17-2\sqrt{30}}=\sqrt{\left(\sqrt{15}-\sqrt{2}\right)^2}=\left|\sqrt{15}-\sqrt{2}\right|=\sqrt{15}-\sqrt{2}\)

\(E=\sqrt{37-6\sqrt{30}}\\ =\sqrt{\left(3\sqrt{3}-\sqrt{10}\right)^2}\\ =\left|3\sqrt{3}-\sqrt{10}\right|\\ =3\sqrt{3}-\sqrt{10}\)

\(F=\sqrt{51-6\sqrt{30}}\\ =\sqrt{\left(3\sqrt{5}-\sqrt{6}\right)^2}\\ =\left|3\sqrt{5}-\sqrt{6}\right|\\ =3\sqrt{5}-\sqrt{6}\)

\(G=\sqrt{59-6\sqrt{30}}\\ =\sqrt{\left(3\sqrt{6}-\sqrt{5}\right)^2}\\ =\left|3\sqrt{6}-\sqrt{5}\right|\\ =3\sqrt{6}-\sqrt{5}\)

\(H=\sqrt{17-2\sqrt{30}}\\ =\sqrt{\left(\sqrt{15}-\sqrt{2}\right)^2}\\ =\left|\sqrt{15}-\sqrt{2}\right|=\sqrt{15}-\sqrt{2}\)

\(\sqrt{2}+\sqrt{6}+\sqrt{12}+...+\sqrt{110}\)\(=\sqrt{1.2}+\sqrt{2.3}+\sqrt{3.4}+...+\sqrt{10.11}\)

\(< \frac{1+2}{2}+\frac{2+3}{2}+\frac{3+4}{2}+...+\frac{10+11}{2}\)\(=\frac{1}{2}\left[\left(1+2+3+...+10\right)+\left(2+3+4+...+11\right)\right]\)\(=\frac{1}{2}\left(\frac{11.10}{2}+\frac{13.10}{2}\right)=\frac{1}{2}\left(55+65\right)=60\)

Vậy \(\sqrt{2}+\sqrt{6}+\sqrt{12}+...+\sqrt{110}< 60.\)

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