Câu 2 :

b) \(\frac{x}{3}=\frac{-2}{9}\)

=> x = \(\frac{-2}{9}.3\) = \(\frac{-2}{3}\)

c) \(0,5x-\frac{2}{3}x=\frac{7}{12}\)

=> \(\frac{1}{2}x-\frac{2}{3}x=\frac{7}{12}\)

=> \(-\frac{1}{6}\)x = \(\frac{7}{12}\)

=> x = \(\frac{7}{12}:\frac{-1}{6}\)

=> x =\(\frac{-7}{2}\)

Đề 1 câu 5 :

\(3B=3^2+3^3+3^4+...+3^{201}\)

\(\Rightarrow2B=3B-B=3^{201}-3\)

\(\Rightarrow2B+3=\left(3^{201}-3\right)+3=3^{201}\)

Do đó n = 201

b: ĐKXĐ: x>=-1

\(\sqrt{x+1}=x+1\)

\(\Leftrightarrow\left\{{}\begin{matrix}x>=-1\\\left(x+1\right)^2=x+1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left(x+1\right)\cdot x=0\\x>=-1\end{matrix}\right.\Leftrightarrow x\in\left\{0;-1\right\}\)

c: \(\sqrt{x-1}=1-x\)

ĐKXĐ: \(\left\{{}\begin{matrix}x-1>=0\\1-x< =0\end{matrix}\right.\Leftrightarrow x=1\)

Do đó: x=1 là nghiệm của phương trình

d: \(2x+3+\dfrac{4}{x-1}=\dfrac{x^2+3}{x-1}\)(ĐKXĐ: x<>1)

\(\Leftrightarrow\left(2x+3\right)\left(x-1\right)+4=x^2+3\)

\(\Leftrightarrow2x^2-2x+3x-3+4-x^2-3=0\)

\(\Leftrightarrow x^2+x-2=0\)

=>(x+2)(x-1)=0

=>x=-2(nhận) hoặc x=1(loại)

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

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

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

ĐK: \(x^4-4x^3+14x-11\ge0\) (*)

\(PT\Leftrightarrow\left\{{}\begin{matrix}x\le1\\x^4-4x^3+14x-11=x^2-2x+1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\le1\\x^4-4x^3-x^2+16x-12=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\le1\\\left(x-1\right)\left(x-2\right)\left(x-3\right)\left(x+2\right)=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\)(tm)

e/ ĐKXĐ: \(x\ge1\)

\(\Leftrightarrow x+3-\sqrt{x-1}=4\)

\(\Leftrightarrow\sqrt{x-1}=x-1\)

\(\Leftrightarrow x-1=x^2-2x+1\)

\(\Leftrightarrow x^2-3x+2=0\Rightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)

f/ \(\Leftrightarrow\left\{{}\begin{matrix}x+5\ge0\\x^3+x^2+6x+28=\left(x+5\right)^2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge-5\\x^3-4x+3=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge-5\\\left(x-1\right)\left(x^2+x-3\right)=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=1\\x=\frac{-1\pm\sqrt{13}}{2}\\\end{matrix}\right.\)

1/\(4x^4+12x^3-47x^2+12x+4=0\)

\(\Leftrightarrow\left(x-2\right)\left(4x^3+20x^2-7x-2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(2x-1\right)\left(2x^2+11x+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\frac{1}{2}\\x=\frac{-11\pm\sqrt{105}}{4}\end{matrix}\right.\)

Vậy ....

1,

\(a^3+b^3\ge ab\left(a+b\right)\)

\(\Leftrightarrow\left(a+b\right)\left(a^2-ab+b^2\right)-ab\left(a+b\right)\ge0\)

\(\Leftrightarrow\left(a+b\right)\left(a-b\right)^2\ge0\) ( đúng )

Dấu "=" \(\Leftrightarrow a=b\)

a) Áp dụng BĐT trên ta có:

\(\Sigma\left(\frac{1}{a^3+b^3+abc}\right)\le\Sigma\left(\frac{1}{ab\left(a+b\right)+abc}\right)=\Sigma\left[\frac{1}{ab}\cdot\left(\frac{1}{a+b+c}\right)\right]=\frac{1}{a+b+c}\cdot\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)=\frac{a+b+c}{\left(a+b+c\right)\cdot abc}=\frac{1}{abc}\)

Dấu "=" khi \(a=b=c\)

b) \(\Sigma\left(\frac{1}{a^3+b^3+1}\right)\le\Sigma\left(\frac{1}{ab\left(a+b\right)+abc}\right)=\Sigma\left[\frac{1}{ab}\cdot\left(\frac{1}{a+b+c}\right)\right]=\frac{1}{abc}=1\)

Dấu "=" khi \(a=b=c=1\)

c) \(\Sigma\left(\frac{1}{a+b+1}\right)\le\Sigma\left(\frac{1}{\sqrt[3]{ab}\left(\sqrt[3]{a}+\sqrt[3]{b}\right)+\sqrt[3]{abc}}\right)=\Sigma\left[\frac{1}{\sqrt[3]{ab}\left(\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}\right)}\right]\)

\(=\frac{1}{\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}}\cdot\left(\frac{1}{\sqrt[3]{ab}}+\frac{1}{\sqrt[3]{bc}}+\frac{1}{\sqrt[3]{ca}}\right)=\frac{\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}}{\left(\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}\right)\cdot\sqrt[3]{abc}}=\frac{1}{\sqrt[3]{abc}}=1\)

Dấu "=" khi \(a=b=c=1\)