\(\dfrac{1}{a^3}+a\ge2\sqrt{\dfrac{a}{a^3}}=\dfrac{2}{a}\) ; \(\dfrac{1}{b^3}+b\ge\dfrac{2}{b}\) ; \(\dfrac{1}{c^3}+c\ge\dfrac{2}{c}\)

\(\Rightarrow\dfrac{1}{a^3}+\dfrac{1}{b^3}+\dfrac{1}{c^3}+a+b+c\ge2\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\) (1)

Lại có \(\dfrac{4a}{a^4+1}\le\dfrac{4a}{2\sqrt{a^4}}=\dfrac{4a}{2a^2}=\dfrac{2}{a}\)

Tương tự \(\dfrac{4b}{b^4+1}\le\dfrac{2}{b}\) ; \(\dfrac{4c}{c^4+1}\le\dfrac{2}{c}\)

\(\Rightarrow4\left(\dfrac{a}{a^4+1}+\dfrac{b}{b^4+1}+\dfrac{c}{c^4+1}\right)\le2\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\) (2)

Từ (1),(2)\(\Rightarrow\dfrac{1}{a^3}+\dfrac{1}{b^3}+\dfrac{1}{c^3}+a+b+c\ge4\left(\dfrac{a}{a^4+1}+\dfrac{b}{b^4+1}+\dfrac{c}{c^4+1}\right)\)

Dấu "=" xảy ra khi a=b=c=1

1) \(A=1+2+2^2+2^3+......+2^{2015}\)

\(\Leftrightarrow2A=2+2^2+2^3+......+2^{2016}\)

\(\Leftrightarrow2A-A=\left(2+2^2+2^3+......+2^{2016}\right)-\left(1+2+2^2+2^3+......+2^{2015}\right)\)

\(\Leftrightarrow A=2^{2016}-1\)

Vậy \(A=2^{2016}-1\)

6)Ta có: \(13+23+33+43+.......+103=3025\)

\(\Leftrightarrow2.13+2.23+2.33+2.43+.......+2.103=2.3025\)

\(\Leftrightarrow26+46+66+86+.......+206=6050\)

\(\Leftrightarrow\left(23+3\right)+\left(43+3\right)+\left(63+3\right)+\left(83+3\right)+.......+\left(203+3\right)=6050\)

\(\Leftrightarrow23+43+63+83+.......+203+3.10=6050\)

\(\Leftrightarrow23+43+63+83+.......+203+=6050-30\)

\(\Leftrightarrow23+43+63+83+.......+203+=6020\)

Vậy S=6020

b, B có 19 thừa số

=> \(-B=(1-\frac{1}{4})(1-\frac{1}{9})(1-\frac{1}{16})...(1-\frac{1}{400}) \)

<=>\(-B=\frac{(2-1)(2+1)(3-1)(3+1)(4-1)(4+1)...(20-1)(20+1)}{4.9.16...400} \)

<=>\(-B=\frac{(1.2.3.4...19)(3.4.5...21)}{(2.3.4.5.6...20)(2.3.4.5...20)} \)

<=>\(-B=\frac{21}{20.2} =\frac{21}{40} \)

<=>\(B=\frac{-21}{40} \)

1) \(\dfrac{x}{3}=\dfrac{y}{4}=t\Leftrightarrow\left\{{}\begin{matrix}x=3t\\y=4t\end{matrix}\right.\)

ta có \(x.y^2=324\Leftrightarrow3t.\left(4t\right)^2=324\)

\(\Leftrightarrow t^3=\dfrac{27}{4}\)

\(\Leftrightarrow t=\dfrac{3}{\sqrt[3]{4}}\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=3.\dfrac{3}{\sqrt[3]{4}}=\dfrac{9}{\sqrt[3]{4}}\\y=4.\dfrac{3}{\sqrt[3]{4}}=\dfrac{12}{\sqrt[3]{4}}\end{matrix}\right.\)

2) \(2^{x+1}.3^y=2^{2x}.3^x\)

\(\Leftrightarrow\left\{{}\begin{matrix}x+1=2x\\x=y\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=1\end{matrix}\right.\)

3) \(\dfrac{a}{b}=\dfrac{c}{d}\)

áp dụng dãy tỉ số = nhau ta có

\(\dfrac{a}{b}=\dfrac{c}{d}=\dfrac{a-c}{b-d}\)

\(\Leftrightarrow\dfrac{a^4}{b^4}=\dfrac{c^4}{d^4}=\left(\dfrac{a-c}{b-d}\right)^4\left(1\right)\)

mà \(\dfrac{a^4}{b^4}=\dfrac{c^4}{d^4}=\dfrac{a^4+c^4}{b^4+c^4}\left(2\right)\)

từ (1)(2) suy ra đpcm

4) \(B=\dfrac{27^{15}.5^3.8^4}{25^2.81^{11}.2^{11}}=\dfrac{\left(3^3\right)^{15}.5^3.\left(2^3\right)^4}{\left(5^2\right)^2.\left(3^4\right)^{11}.2^{11}}=\dfrac{3^{45}.5^3.2^{12}}{5^4.3^{44}.2^{11}}=\dfrac{3.2}{5}=\dfrac{6}{5}\)

a) \(\dfrac{5+x}{4-x}=\dfrac{1}{2}\)

\(\Leftrightarrow2\left(5+x\right)=4-x\)

\(\Leftrightarrow2\left(5+x\right)-\left(4-x\right)=0\)

\(\Leftrightarrow10+2x-4+x=0\)

\(\Leftrightarrow6+3x=0\)

\(\Leftrightarrow3x=-6\)

\(\Leftrightarrow x=-2\)

Vậy x=-2

b) \(\dfrac{25}{14}=\dfrac{x+7}{x-4}\)

\(\Leftrightarrow25\left(x-4\right)=14\left(x+7\right)\)

\(\Leftrightarrow25\left(x-4\right)-14\left(x+7\right)=0\)

\(\Leftrightarrow25x-100-14x-98=0\)

\(\Leftrightarrow11x-198=0\)

\(\Leftrightarrow11x=198\)

\(\Leftrightarrow x=18\)

Vậy x=18

c) \(\dfrac{3x-5}{x+4}=\dfrac{5}{2}\)

\(\Leftrightarrow2\left(3x-5\right)=5\left(x+4\right)\)

\(\Leftrightarrow2\left(3x-5\right)-5\left(x+4\right)=0\)

\(\Leftrightarrow6x-10-5x-20=0\)

\(\Leftrightarrow x-30=0\)

\(\Leftrightarrow x=30\)

Vậy x=30

d) \(\dfrac{3x-1}{2x+1}=\dfrac{3}{7}\)

\(\Leftrightarrow7\left(3x-1\right)=3\left(2x+1\right)\)

\(\Leftrightarrow7\left(3x-1\right)-3\left(2x+1\right)=0\)

\(\Leftrightarrow21x-7-6x-3=0\)

\(\Leftrightarrow15x-10=0\)

\(\Leftrightarrow15x=10\)

\(\Leftrightarrow x=\dfrac{10}{15}=\dfrac{2}{3}\)

Vậy \(x=\dfrac{2}{3}\)

Áp dụng bất đẳng thức Bunyakovsky

\(\Rightarrow\sqrt{\left(\dfrac{8}{a^2}+\dfrac{9b^2}{2}+\dfrac{c^2a^2}{4}\right)\left[\left(\sqrt{2}\right)^2+\left(3\sqrt{2}\right)^2+2^2\right]}\ge\left(\sqrt{\dfrac{4}{a}+9b+ca}\right)^2\)

\(\Leftrightarrow2\sqrt{6}\sqrt{\dfrac{8}{a^2}+\dfrac{9b^2}{2}+\dfrac{c^2a^2}{4}}\ge\dfrac{4}{a}+9b+ac\)

Tương tự ta có \(\left\{{}\begin{matrix}2\sqrt{6}\sqrt{\left(\dfrac{8}{b^2}+\dfrac{9c^2}{2}+\dfrac{a^2b^2}{4}\right)}\ge\dfrac{4}{b}+9c+ab\\2\sqrt{6}\sqrt{\left(\dfrac{8}{c^2}+\dfrac{9a^2}{2}+\dfrac{b^2c^2}{4}\right)}\ge\dfrac{4}{c}+9a+bc\end{matrix}\right.\)

\(\Rightarrow2\sqrt{6}S\ge\dfrac{4}{a}+9a+\dfrac{4}{b}+9b+\dfrac{4}{c}+9c+ab+bc+ac\)

\(\Leftrightarrow2\sqrt{6}S\ge\dfrac{4}{a}+a+8a+\dfrac{4}{b}+b+8b+\dfrac{4}{c}+c+8c+ab+bc+ca\)

Áp dụng bất đẳng thức Cauchy

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{4}{a}+a\ge2\sqrt{4}=4\\\dfrac{4}{b}+b\ge2\sqrt{4}=4\\\dfrac{4}{c}+c\ge2\sqrt{4}=4\end{matrix}\right.\)

\(\Rightarrow\dfrac{4}{a}+a+8a+\dfrac{4}{b}+b+8b+\dfrac{4}{c}+c+8c+ab+bc+ca\ge12+8a+8b+8c+ab+bc+ac\)

\(\Rightarrow2\sqrt{6}S\ge12+8a+8b+8c+ab+bc+ac\)

\(\Leftrightarrow2\sqrt{6}S\ge12+2a+bc+2b+ac+2c+ab+6\left(a+b+c\right)\)

Áp dụng bất đẳng thức Cauchy

\(\Rightarrow2a+bc\ge2\sqrt{2abc}\)

Tượng tự ta có \(2b+ac\ge2\sqrt{2abc}\) ; \(2c+ab\ge2\sqrt{2abc}\)

\(\Rightarrow12+2a+bc+2b+ac+2c+ab+6\left(a+b+c\right)\ge6\left(a+b+c+\sqrt{2abc}\right)+12\)

\(\Rightarrow2\sqrt{6}S\ge6\left(a+b+c+\sqrt{2abc}\right)+12\)

Theo đề bài ta có \(a+b+c+\sqrt{2abc}\ge10\)

\(\Rightarrow6\left(a+b+c+\sqrt{2abc}\right)+12\ge72\)

\(\Rightarrow S\ge\dfrac{72}{2\sqrt{6}}=6\sqrt{6}\) ( đpcm )

Dấu " = " xảy ra khi \(a=b=c=2\)

a: \(=-8\cdot\left(\dfrac{3}{4}-\dfrac{1}{4}\right):\left(\dfrac{9}{4}-\dfrac{7}{6}\right)\)

\(=-8\cdot\dfrac{1}{2}:\dfrac{27-14}{12}\)

\(=-4\cdot\dfrac{12}{13}=\dfrac{-48}{13}\)

b: \(=\left(\dfrac{10}{3}+\dfrac{5}{2}\right):\left(\dfrac{19}{6}-\dfrac{21}{5}\right)-\dfrac{11}{31}\)

\(=\dfrac{35}{6}:\dfrac{-31}{30}-\dfrac{11}{31}\)

\(=\dfrac{-35}{6}\cdot\dfrac{30}{31}-\dfrac{11}{31}=-6\)

-\(\dfrac{1}{4}\)- -5+\(\dfrac{1}{3}\)-\(\dfrac{3}{2}\)-3-\(\dfrac{7}{4}\)+\(\dfrac{4}{3}\)

=-\(\dfrac{1}{4}\)+5+\(\dfrac{1}{3}\)-\(\dfrac{3}{2}\)-3-\(\dfrac{7}{4}\)+\(\dfrac{4}{3}\)

=-(\(\dfrac{1}{4}\)+\(\dfrac{7}{4}\))+(5-3)+(\(\dfrac{1}{3}\)+\(\dfrac{4}{3}\))-\(\dfrac{3}{2}\)

=-2+2+\(\dfrac{5}{3}\)-\(\dfrac{3}{2}\)

=\(\dfrac{1}{6}\)

a: \(=\left(\dfrac{-48}{12}+\dfrac{-8}{12}+\dfrac{21}{12}\right)\cdot\dfrac{-12}{13}\)

\(=\dfrac{-35}{12}\cdot\dfrac{-12}{13}=\dfrac{35}{13}\)

b: \(=\dfrac{-3}{6}+\dfrac{5}{6}-\dfrac{312}{100}+\dfrac{51}{10}\)

\(=\dfrac{1}{3}-\dfrac{312}{100}+\dfrac{51}{10}=\dfrac{347}{150}\)

c: \(=\left(\dfrac{48}{300}+\dfrac{175}{300}-\dfrac{135}{100}\right)\cdot\dfrac{5}{2}+\dfrac{1}{4}\)

\(=\dfrac{88}{300}\cdot\dfrac{5}{2}+\dfrac{1}{4}=\dfrac{59}{60}\)