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Đặt \(\dfrac{a}{b^2}=\dfrac{b^2}{c^3}=\dfrac{c^3}{a^4}=k\)
\(\Rightarrow\left\{{}\begin{matrix}a=k.b^2\\b^2=k.c^3\\c^3=k.a^4\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a=k.k.c^3=k^2c^3\\c^3=k.a^4\end{matrix}\right.\)
\(\Rightarrow a=k^2.k.a^4\)
\(\Rightarrow a=k^3a^4\)
\(\Rightarrow\left(ka\right)^3=1\)
\(\Rightarrow ka=1\)
\(\Rightarrow a=\dfrac{1}{k}\) (1)
Thế vào \(c^3=k.a^4\Rightarrow c^3=k.\dfrac{1}{k^4}=\dfrac{1}{k^3}\)
\(\Rightarrow c=\dfrac{1}{k}\) (2)
Thế vào \(b^2=kc^3\Rightarrow b^2=k.\dfrac{1}{k^3}=\dfrac{1}{k^2}\)
\(\Rightarrow b=\dfrac{1}{k}\) hoặc \(b=-\dfrac{1}{k}\) (3)
(1);(2);(3) \(\Rightarrow\left[{}\begin{matrix}a=b=c\\a=c=-b\end{matrix}\right.\)
TH1: \(a=b=c\)
\(\Rightarrow P=\left(1+\dfrac{a}{a}\right)\left(1+\dfrac{a}{a}\right)\left(1+\dfrac{a}{a}\right)=2.2.2=8\)
Th2: \(a=c=-b\)
\(\Rightarrow P=\left(1+\dfrac{-b}{b}\right)\left(1+\dfrac{b}{-b}\right)\left(1+\dfrac{-b}{-b}\right)=0.0.2=0\)
Bạn ơi thiếu đề rồi, cái biểu thức này không tính được đâu , mình nghĩ thế
a) Ta có: \(\left|2x-\dfrac{1}{3}\right|\ge0\forall x\)
\(\Leftrightarrow\left|2x-\dfrac{1}{3}\right|-\dfrac{7}{4}\ge-\dfrac{7}{4}\forall x\)
Dấu '=' xảy ra khi \(2x=\dfrac{1}{3}\)
hay \(x=\dfrac{1}{6}\)
Vậy: \(A_{min}=-\dfrac{7}{4}\) khi \(x=\dfrac{1}{6}\)
b) Ta có: \(\dfrac{1}{3}\left|x-2\right|\ge0\forall x\)
\(\left|3-\dfrac{1}{2}y\right|\ge0\forall y\)
Do đó: \(\dfrac{1}{3}\left|x-2\right|+\left|3-\dfrac{1}{2}y\right|\ge0\forall x,y\)
\(\Leftrightarrow\dfrac{1}{3}\left|x-2\right|+\left|3-\dfrac{1}{2}y\right|+4\ge4\forall x,y\)
Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}x-2=0\\3-\dfrac{1}{2}y=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=6\end{matrix}\right.\)
Vậy: \(B_{min}=4\) khi x=2 và y=6
b.ta chia B thành 10 nhóm mỗi nhóm có 6 hạng tử \(B=\left(2+2^2+2^3+2^4+2^5+2^6\right)+....+\left(2^{55}+2^{56}+2^{57}+2^{58}+2^{59}+2^{60}\right)\)
\(B\text{=}2\left(1+2+2^2+2^3+2^4+2^5\right)+...+2^{55}\left(1+2+2^2+2^3+2^4+2^5\right)\)
\(B\text{=}2.63+...+2^{56}.63\)
\(\Rightarrow B⋮63\)
\(\Rightarrow B⋮21\)
a)\(\dfrac{3}{10}\)-x=\(\dfrac{25}{30}\)-\(\dfrac{4}{30}\)
\(\dfrac{3}{10}-x=\dfrac{7}{10}\)
x = \(\dfrac{3}{10}-\dfrac{7}{10}\)
x=\(\dfrac{-4}{10}\)
b)\(\dfrac{-5}{8}+x=\dfrac{4}{9}-\dfrac{63}{9}\)
\(\dfrac{-5}{9}+x=\dfrac{-59}{9}\)
\(x=\dfrac{-59}{9}-\dfrac{-5}{9}\)
\(x=\dfrac{-64}{9}\)
c)=>2.18=(x-3).(x-3)
=>36=(x-3)\(^2\)
=>6\(^2\)=(x-3)\(^2\)
6= x-3
x=6+3=9
1) \(\dfrac{2}{15}\cdot6\dfrac{5}{11}+\dfrac{5}{11}\cdot\dfrac{-2}{15}-\dfrac{2}{15}\cdot2015^0\)
\(=\dfrac{2}{15}\cdot\dfrac{71}{11}-\dfrac{1}{11}\cdot\dfrac{2}{3}-\dfrac{2}{15}\cdot1\)
\(=\dfrac{142}{165}-\dfrac{2}{33}-\dfrac{2}{15}\)
\(=\dfrac{2}{3}\)
2) \(\dfrac{5}{2\cdot7}+\dfrac{3}{14\cdot11}+\dfrac{4}{11\cdot7}+\dfrac{1}{14\cdot15}+\dfrac{13}{15\cdot16}\)
\(=\dfrac{5}{14}+\dfrac{3}{154}+\dfrac{4}{77}+\dfrac{1}{210}+\dfrac{13}{240}\)
\(=\dfrac{39}{80}\)