\(a^3+1+1\ge3\sqrt[3]{a^3.1.1}=3a\)

Thiết lập tương tự hai BĐT còn lại và cộng theo vế,ta có:

\(a^3+b^3+c^3+6\ge3\left(a+b+c\right)\)

\(\Leftrightarrow a^3+b^3+c^3+6\ge39\)

\(\Leftrightarrow a^3+b^3+c^3\ge33\)

Mà theo đề bài \(a^3+b^3+c^3=27< 33\rightarrow\)vô lí.

Do đó đề sai!

\(M=\frac{1}{16x^2}+\frac{1}{4y^2}+\frac{1}{z^2}=\left(x^2+y^2+z^2\right)\left(\frac{1}{16x^2}+\frac{1}{4y^2}+\frac{1}{z^2}\right)\)

Áp dụng bđt bunhicopxki ta có:

\(\left(x^2+y^2+z^2\right)\left(\frac{1}{16x^2}+\frac{1}{4y^2}+\frac{1}{z^2}\right)\ge\left(x\cdot\frac{1}{4x}+y\cdot\frac{1}{2y}+z\cdot\frac{1}{z}\right)^2=\left(\frac{1}{4}+\frac{1}{2}+1\right)^2=\frac{49}{16}\)

Dấu "=" xảy ra <=> \(x=\sqrt{\frac{1}{7}};y=\sqrt{\frac{2}{7}};z=\sqrt{\frac{4}{7}}\)

  12x² - 32x² +25x -6

= x ( 12x - 32x + 25 - 6 ) 

= x ( -20x + 19 ) 

bị sai đề đúng không?????????

k cho mk mấy cái đi a...hi..hi...

\(a^2+b^2+c^2=ab+bc+ac\)

\(\Leftrightarrow2\left(a^2+b^2+c^2\right)=2\left(ab+bc+ac\right)\)

\(\Leftrightarrow2a^2+2b^2+2c^2=2ab+2bc+2ac\)

\(\Leftrightarrow a^2+a^2+b^2+b^2+c^2+c^2-2ab-2bc-2ac=0\)

\(\Leftrightarrow\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(a^2-2ac+c^2\right)=0\)

\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(a-c\right)^2=0\)

\(\Rightarrow\hept{\begin{cases}a-b=0\\b-c=0\\a-c=0\end{cases}\Rightarrow\hept{\begin{cases}a=b\\b=c\\a=c\end{cases}\Rightarrow}a=b=c\left(đpcm\right)}\)

Dạng nì mik làm rồi ak =_= nhưng sai thì bạn ib mik nha =))

Có :  \(a^2+b^2+c^2=ab+bc+ca.\)

\(\Rightarrow2a^2+2b^2+2c^2=2ab+2bc+2ac\)

\(\Rightarrow2a^2+2b^2+2c^2-2ab-2bc-2ac=0\)

\(\Rightarrow\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ac+a^2\right)=0\)

\(\Rightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\) (*)

Ta có  (**)\(\left(a-b\right)^2\ge0\)

         \(\left(b-c\right)^2\ge0\)

         \(\left(c-a\right)^2\ge0\)

Từ (*) và (**) => cả a,b,c đều lớn hơn hoặc bằng 0

\(\Rightarrow a-b=0;b-c=0;c-a=0\)

\(\Rightarrow a=b=c\left(đpcm\right)\)

hok tốt ạ

a) \(A=\frac{4x}{x+2}+\frac{2}{x-2}+\frac{5x-6}{4-x^2}\)

\(A=\frac{4x\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{2\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\frac{5x-6}{\left(x-2\right)\left(x+2\right)}\)

\(A=\frac{4x^2-8x+2x+4-5x+6}{\left(x-2\right)\left(x+2\right)}\)

\(A=\frac{4x^2-11x+10}{\left(x-2\right)\left(x+2\right)}\)

\(a,A=\frac{4x}{x+2}+\frac{2}{x-2}+\frac{5x-6}{4-x^2}\)

\(=\frac{4x}{x+2}+\frac{2}{x-2}+\frac{6-5x}{x^2-4}\)

\(=\frac{4x\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{2\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{6-5x}{\left(x-2\right)\left(x+2\right)}\)

\(=\frac{4x^2-8x+2x+4+6-5x}{\left(x-2\right)\left(x+2\right)}\)

\(=\frac{4x^2-11x+10}{\left(x-2\right)\left(x+2\right)}\)

giúp mik vs

ta có: (a^2-a+2012)(a^2-a+2014)-3

=(a^2-a+2013-1)(a^2-a+2013+1)-3

=(a^2-a+2013)^2-1-3

=(a^2-a+2013)^2-4

=(a^2-a+2013-2)(a^2-a+2013+2)

=(a^2-a+2011)(a^2-a+2015)

chúc bn học tốt