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Xét \(A=2^{30}+3^{30}+4^{30}=\left(2^3\right)^{10}+\left(3^3\right)^{10}+\left(2^2\right)^{30}=8^{10}+27^{10}+2^{60}\)

\(B=3^{20}+6^{20}+8^{20}=\left(3^2\right)^{10}+\left(6^2\right)^{10}+\left(2^3\right)^{20}=9^{10}+36^{10}+2^{60}\)

Vì \(8^{10}< 9^{10},27^{10}< 36^{10}\)nên A<B

2³⁰ = 2^3.10= 8¹⁰

3³⁰=3^3.10=27¹⁰

4³⁰=4^3.10=64¹⁰

Vế trái = 8¹⁰ + 27¹⁰ + 64¹⁰

3²⁰=3^2.10=9¹⁰

6²⁰=6^2.10=36¹⁰

8²⁰=8^2.10=64¹⁰

vế phải = 9¹⁰ + 36¹⁰ + 64¹⁰

Vì 64¹⁰=64¹⁰ ; 36¹⁰ > 27¹⁰ ; 9¹⁰ > 8¹⁰

=> vế phải > vế trái

ĐKXĐ: \(x\ne5\)

a) \(C=\frac{x+3}{5-x}=\frac{-2}{3}\Leftrightarrow3\left(x+3\right)=2\left(x-5\right)\Leftrightarrow x=-19\)

b) \(C=\frac{x+3}{5-x}< 1\)

Xét \(x>5\Rightarrow5-x< 0\Rightarrow x+3>5-x\Leftrightarrow x>1\)Vậy nhận \(x>5\)

Xét \(x< 5\Rightarrow5-x>0\Rightarrow x+3< 5-x\Leftrightarrow x< 1\)Vậy nhận \(x< 1\)

c) \(C=\frac{x+3}{5-x}=\frac{8}{5-x}-1\inℤ\)

Vì \(x\inℤ\)nên \(C\inℤ\)khi và chỉ khi \(8⋮\left(5-x\right)\Rightarrow\left(5-x\right)\inƯ\left(8\right)=\left\{\pm1;\pm2;\pm3;\pm4\right\}\)

\(\Rightarrow x\in\left\{1;2;3;4;6;7;8;9\right\}\)

a) Có \(\frac{a}{b}< \frac{c}{d}\)Nhân 2 vế cho b.d>0\(\Rightarrow\frac{abd}{b}< \frac{cbd}{d}\Leftrightarrow ad< bc\)(1)

+) Cộng 2 của (1) vế cho ab: \(ab+ad< ab+bc\Leftrightarrow a\left(b+d\right)< b\left(a+c\right)\)

Chia 2 vế cho b(b+d)>0: \(\frac{a\left(b+d\right)}{b\left(b+d\right)}< \frac{b\left(a+c\right)}{b\left(b+d\right)}\Leftrightarrow\frac{a}{b}< \frac{a+c}{b+d}\)

+) Cộng 2 vế của (1) cho cd: \(cd+ad< bc+cd\Leftrightarrow d\left(a+c\right)< c\left(b+d\right)\)

Chia 2 vế cho d(b+d)>0: \(\frac{d\left(a+c\right)}{d\left(b+d\right)}< \frac{c\left(b+d\right)}{d\left(b+d\right)}\Leftrightarrow\frac{a+c}{b+d}< \frac{c}{d}\)

Vậy ...............

b) Xét \(\frac{-1}{3}=\frac{-4}{12}\)và \(\frac{-1}{4}=\frac{-4}{16}\)

----> 3 số hữu tỉ ở giữa là \(\frac{-4}{13},\frac{-4}{14},\frac{-4}{15}\)

\(\frac{a}{b}< \frac{c}{d}\Rightarrow\frac{ad}{b}< c\Rightarrow ad< bc\)

\(ad< bc\Rightarrow\frac{ad}{b}< c\Rightarrow\frac{a}{b}< \frac{c}{d}\)

a) Vì b>0,d>0 nên khi nhân 2 vế của 1 BĐT cho b hoặc d thì dấu của BĐT không đổi

Có\(\frac{a}{b}< \frac{c}{d}\)nhân 2 vế BĐT cho b.d>0\(\Rightarrow\frac{a.b.d}{b}< \frac{c.b.d}{d}\Leftrightarrow ad< bc\)

b) Tương tự câu a ta chia 2 vế BĐT cho b.d

\(ad< bc\Rightarrow\frac{ad}{bd}< \frac{bc}{bd}\Leftrightarrow\frac{a}{b}< \frac{c}{d}\)

a) \(\frac{7^3.5^8}{49.25^4}=\frac{7^3.5^8}{7^2.\left(5^2\right)^4}=7.\frac{5^8}{5^8}=7\)

b) \(\frac{3^9.25.5^3}{15.625.3^8}=\frac{3.3^8.5^2.5^3}{3.5.5^4.3^8}=\frac{5^5}{5^5}=1\)

c) Đề hơi sai roi bạn oi

d) \(\left(\frac{2}{5}-\frac{1}{2}\right)^2+\left(\frac{1}{2}+\frac{3}{5}\right)^2=\left(\frac{-1}{10}\right)^2+\left(\frac{11}{10}\right)^2=\frac{1}{100}+\frac{121}{100}=\frac{61}{50}\)

Khanh Nguyễn Ngọc  :câu d ko sai bạn nha dấu "/" là trên nhó

Xét OC//AD=> \(\widehat{COA}=\widehat{DAB}\)(2 góc đồng vị)

Xét 2 tam giác AOC và BAD: OC=AD, \(\widehat{COA}=\widehat{DAB}\), OA=AB

=> \(\Delta AOC=\Delta BAD\left(c.g.c\right)\Rightarrow\widehat{CAO}=\widehat{DBA}\)ở vị trí đồng vị => AC//BD

giúp tớ với 

\(-0,35.\frac{3}{14}.3\frac{5}{7}.\left(-\frac{4}{21}\right)\)
\(=-\frac{3}{40}.\frac{26}{7}.\left(-\frac{4}{21}\right)\)
\(=-\frac{39}{140}.\left(-\frac{4}{21}\right)=\frac{13}{245}\)

\(\frac{315-x}{101}+\frac{313-x}{103}+\frac{311-x}{105}+\frac{309-x}{107}+4=0\)

\(\Leftrightarrow\left(1+\frac{315-x}{101}\right)+\left(1+\frac{313-x}{103}\right)+\left(1+\frac{311-x}{105}\right)+\left(1+\frac{309-x}{107}\right)=0\)

\(\Leftrightarrow\frac{315-x+101}{101}+\frac{313-x+103}{103}+\frac{311-x+105}{105}+\frac{309-x+107}{107}=0\)

\(\Leftrightarrow\frac{416-x}{101}+\frac{416-x}{103}+\frac{416-x}{105}+\frac{416-x}{107}=0\)

\(\Leftrightarrow\left(416-x\right)\left(\frac{1}{101}+\frac{1}{103}+\frac{1}{105}+\frac{1}{107}\right)=0\)

\(\Leftrightarrow416-x=0\)

\(\Leftrightarrow x=416\)

\(\frac{315-x}{101}+\frac{313-x}{103}+\frac{311-x}{105}+\frac{309-x}{107}+4=0\)

\(\Leftrightarrow\left(\frac{315-x}{101}+1\right)+\left(\frac{313-x}{103}+1\right)+\left(\frac{311-x}{105}+1\right)+\left(\frac{309-x}{107}+1\right)=0\)

\(\Leftrightarrow\frac{315-x+101}{101}+\frac{313-x+103}{103}+\frac{311-x+105}{105}+\frac{309-x+107}{107}=0\)

\(\Leftrightarrow\frac{416-x}{101}+\frac{416-x}{103}+\frac{416-x}{105}+\frac{416-x}{107}=0\)

\(\Leftrightarrow\left(416-x\right)\left(\frac{1}{101}+\frac{1}{103}+\frac{1}{105}+\frac{1}{107}\right)=0\)

Vì \(\left(\frac{1}{101}+\frac{1}{103}+\frac{1}{105}+\frac{1}{107}\right)\ne0\)

\(\Rightarrow416-x=0\Leftrightarrow x=416\)