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Áp dụng bất đẳng thức Cauchy - Schwarz
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{ab}{c}+\dfrac{bc}{a}\ge2\sqrt{\dfrac{ab^2c}{ca}}=2\sqrt{b^2}=2b\\\dfrac{bc}{a}+\dfrac{ca}{b}\ge2\sqrt{\dfrac{abc^2}{ab}}=2\sqrt{c^2}=2c\\\dfrac{ab}{c}+\dfrac{ca}{b}\ge2\sqrt{\dfrac{a^2bc}{bc}}=2\sqrt{a^2}=2a\end{matrix}\right.\)
\(\Rightarrow2\left(\dfrac{ab}{c}+\dfrac{bc}{a}+\dfrac{ca}{b}\right)\ge2\left(a+b+c\right)\)
\(\Rightarrow\dfrac{ab}{c}+\dfrac{bc}{a}+\dfrac{ca}{b}\ge a+b+c\) ( đpcm )
Dấu " = " xảy ra khi \(a=b=c\)
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a) https://hoc24.vn/hoi-dap/question/398481.html
b)
a2 + b2 + c2 = ab + ac + bc
<=> 2a2 + 2b2 + 2c2 = 2ac + 2ab + 2bc
<=> (a2 - 2ac + c2) + (a2 - 2ab + b2) + (b2 - 2bc + c2) = 0
<=> (a - b)2 + (a - c)2 + (b - c)2 = 0
<=> a = b = c
1. Ta có:
\(\left(a^2+b^2\right)\left(x^2+y^2\right)=\left(ax+by\right)^2\)
=> \(a^2x^2+a^2y^2+b^2x^2+b^2y^2=a^2x^2+2axby+b^2y^2\)
=> \(a^2y^2+b^2x^2=2axby\)
=> \(a^2y^2+b^2x^2-2axby=0\)
=> \(a^2y^2+b^2x^2-2aybx=0\)
=> \(\left(ay-bx\right)^2=0\)
Mà \(\left(ay-bx\right)^2\ge0\)
Dấu '' = '' xảy ra \(\Leftrightarrow\) \(ay-bx=0\)
\(\Leftrightarrow\) \(ay=bx\)
\(\Leftrightarrow\) \(\dfrac{a}{x}=\dfrac{b}{y}\)
2. Ta có:
\(a^2+b^2+c^2=ab+bc+ac\)
=> \(2a^2+2b^2+2c^2=2ab+2bc+2ac\)
=> \(2a^2+2b^2+2c^2-2ab-2bc-2ac=0\)
=> \(\left(a^2-2ab+b^2\right)+\left(a^2-2ac+c^2\right)+\left(b^2-2bc+c^2\right)=0\)
=> \(\left(a-b\right)^2+\left(a-c\right)^2+\left(b-c\right)^2=0\)
Ta thấy:
\(\left(a-b\right)^2\ge0\); \(\left(a-c\right)^2\ge0\); \(\left(b-c\right)^2\ge0\)
=> \(\left(a-b\right)^2+\left(a-c\right)^2+\left(b-c\right)^2\ge0\)
Mà \(\left(a-b\right)^2+\left(a-c\right)^2+\left(b-c\right)^2=0\)
Dấu '' = '' xảy ra \(\Leftrightarrow\) \(\left\{{}\begin{matrix}a-b=0\\a-c=0\\b-c=0\end{matrix}\right.\)
\(\Leftrightarrow\)\(\left\{{}\begin{matrix}a=b\\a=c\\b=c\end{matrix}\right.\)
\(\Leftrightarrow\) a = b = c
\(\text{a) }\left(\dfrac{1}{2}a^2x^4+\dfrac{4}{3}\:ax^3-\dfrac{2}{3}ax^2\right):\left(-\dfrac{2}{3}\:ax^2\right)\\ =-3ax^2-2x+1\)
\(\text{b) }4\left(\dfrac{3}{4}x-1\right)+\left(12x^2-3x\right):\left(-3x\right)-\left(2x+1\right)\\ =3x-4-4x+1-2x-1\\ =-3x-4\)
kết quả cuối cùng là: a. -\(\dfrac{3}{4}ax^2-2x+1\)
b. \(\)-\(3x-4\)
sai đề à VT=0; VP#0
mk k bt, hình như là đúng mà