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\(1-x=\frac{29}{12}+\frac{32}{8}\)
\(\Rightarrow1-x=\frac{77}{12}\)
\(\Rightarrow x=1-\frac{77}{12}=\frac{-65}{12}\)
\(\frac{31}{8}.x-\frac{11}{4}=\frac{42}{12}.\frac{10}{8}-\frac{1}{3}\)
\(\Rightarrow\frac{31}{8}.x-\frac{11}{4}=\frac{35}{8}-\frac{1}{3}\)
\(\Rightarrow\frac{31}{8}.x-\frac{11}{4}=\frac{97}{24}\)
\(\Rightarrow\frac{31}{8}.x=\frac{97}{24}+\frac{11}{4}=\frac{163}{24}\)
\(\Rightarrow x=\frac{163}{24}:\frac{31}{8}=\frac{163}{93}\)
\(\dfrac{9}{10}+\dfrac{3}{5}\times\dfrac{1}{6}=\dfrac{9}{10}+\dfrac{3}{30}=\dfrac{27}{30}+\dfrac{3}{30}=1\)
\(\dfrac{31}{12}-\dfrac{7}{5}:\dfrac{14}{15}=\dfrac{31}{12}-\dfrac{7}{5}\times\dfrac{15}{14}=\dfrac{31}{12}-\dfrac{3}{2}=\dfrac{31}{12}-\dfrac{18}{12}=\dfrac{13}{12}\)
\(\dfrac{4}{7}\times\dfrac{2}{5}:\dfrac{10}{9}=\dfrac{14}{35}\times\dfrac{9}{10}=\dfrac{9}{25}\)
a: Ta có: \(\sqrt{x^2-4x+4}=\sqrt{4x^2-12x+9}\)
\(\Leftrightarrow\left|x-2\right|=\left|2x-3\right|\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-3=x-2\\2x-3=2-x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{5}{3}\end{matrix}\right.\)
c: Ta có: \(\sqrt{4x^2-4x+1}=\sqrt{x^2-6x+9}\)
\(\Leftrightarrow\left|2x-1\right|=\left|x-3\right|\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=x-3\\2x-1=3-x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{4}{3}\end{matrix}\right.\)
Lời giải:
\(\text{VT}=\sum \frac{a^2}{a+2b^3}=\sum (a-\frac{2ab^3}{a+2b^3})=3-2\sum \frac{ab^3}{a+2b^3}\)
Áp dụng BĐT AM-GM:
\(\sum \frac{ab^3}{a+2b^3}\leq \sum \frac{ab^3}{3\sqrt[3]{ab^6}}=\frac{1}{3}\sum \sqrt[3]{a^2}\leq \frac{1}{3}\sum \frac{a+a+1}{3}=\frac{1}{9}[2(a+b+c)+3]=1\)
$\Rightarrow \text{VT}\geq 3-2.1=1$. Ta có đpcm.
Dấu "=" xảy ra khi $a=b=c=1$
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98-78=?