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Bất đẳng thức cần chứng minh tương đương với:
\(a^3b^2-a^2b^3+b^3c^2-c^3b^2+c^3a^2-c^2a^3\ge0\)
\(\Leftrightarrow a^2b^2\left(a-b\right)+b^2c^2\left(b-c\right)+c^2a^2\left(c-a\right)\ge0\)
\(\Leftrightarrow a^2b^2\left(a-b\right)+b^2c^2\left(b-c\right)+c^2a^2\left(c-b+b-a\right)\ge0\)
\(\Leftrightarrow a^2b^2\left(a-b\right)+c^2a^2\left(b-a\right)+b^2c^2\left(b-c\right)+c^2a^2\left(c-b\right)\ge0\)
\(\Leftrightarrow\left(a^2b^2-c^2a^2\right)\left(a-b\right)+\left(b^2c^2-c^2a^2\right)\left(b-c\right)\ge0\)
\(\Leftrightarrow a^2\left(b^2-c^2\right)\left(a-b\right)+c^2\left(b^2-a^2\right)\left(b-c\right)\ge0\)
\(\Leftrightarrow\left[a^2\left(b+c\right)-c^2\left(a+b\right)\right]\left(a-b\right)\left(b-c\right)\ge0\)
\(\Leftrightarrow\left(a^2b+a^2c-c^2a-c^2b\right)\left(a-b\right)\left(b-c\right)\ge0\)
\(\Leftrightarrow\left[a\left(ab-c^2\right)+c\left(a^2-bc\right)\right]\left(a-b\right)\left(b-c\right)\ge0\) luôn đúng do \(a\ge b\ge c\ge0\)
cảm ơn bạn nhá, bạn trả lời giúp mình mấy câu hỏi về BĐT còn lại của mik đc ko? cảm ơn bn nhiều!
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a) \(\frac{a-1}{2}=\frac{b-2}{3}=\frac{c-3}{4}\Leftrightarrow\frac{2a-2}{4}=\frac{3b-6}{9}=\frac{c-3}{4}\)
Áp dụng t/c dãy tỉ số bằng nhau : \(\frac{2a-2}{4}=\frac{3b-6}{9}=\frac{c-3}{4}=\frac{2a+3b-c-2-6+3}{4+9-4}=\frac{45}{9}=5\)
Suy ra : \(\begin{cases}a=11\\b=17\\c=23\end{cases}\)
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Trả lời:
a, ( a + b )3 + ( a - b )3
= a3 + 3a2b + 3ab2 + b3 + a3 - 3a2b + 3ab2 - b3
= 2a3 + 6ab2
= 2a ( a2 + 3b2 ) (đpcm)
b, Sửa đề: ( a + b )3 - ( a - b )3
= a3 + 3a2b + 3ab2 + b3 - ( a3 - 3a2b + 3ab2 - b3 )
= a3 + 3a2b + 3ab2 + b3 - a3 + 3a2b - 3ab2 + b3
= 6a2b + 2b3
= 2b ( b2 + 2a2 )
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a) \(\left(a+b\right)^3+\left(a-b\right)^3=2a\left(a^2+3b^2\right)\)
\(\Leftrightarrow a^3+3a^2b+3ab^2+b^3+a^3-3a^2b+3ab^2-b^3-2a^3-6ab^2=0\)
\(\Leftrightarrow0=0\) ( đpcm) .
b) \(\left(a+b\right)^3-\left(a-b\right)^3=2b\left(b^2+3a^2\right)\)
\(\Leftrightarrow a^3+3a^2b+3ab^2+b^3-a^3+3a^2b-3ab^2+b^3-2a^3-6ab^2=0\)
\(\Leftrightarrow0=0\) ( luôn đúng )
Vậy đẳng thức được chứng minh.
Làm cách khác với "thị nở" :v.
a) \(\left(a+b\right)^3+\left(a-b\right)^3=2a\left(a^2+3b^2\right)\)
\(=\left[\left(a+b\right)+\left(a-b\right)\right]\left[\left(a+b\right)^2-\left(a+b\right)\left(a-b\right)+\left(a-b\right)^2\right]=2a\left(a^2+3b^2\right)\)
\(=\left(a+b+a-b\right)\left(a^2+2ab+b^2-a^2+b^2+a^2-2ab+b^2\right)=2a\left(a^2+3b^2\right)\)
\(=2a\left(a^2+3b^2\right)=2a\left(a^2+3b^2\right)\)
b) \(\left(a+b\right)^3-\left(a-b\right)^3=2b\left(b^2+3a^2\right)\)
\(=\left[\left(a+b\right)-\left(a-b\right)\right]\left[\left(a+b\right)^2+\left(a+b\right)\left(a-b\right)+\left(a-b\right)^2\right]=2b\left(b^2+3a^2\right)\)
\(=\left(a+b-a+b\right)\left(a^2+2ab+b^2+a^2-b^2+a^2-2ab+b^2\right)=2b\left(b^1+3a^2\right)\)\(=2b^2\left(b^2+3a^2\right)=2b^2\left(b^2+3a^2\right)\)
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a.\(\left(a+b\right)^3+\left(a-b\right)^3=2a\left(a^2+3b^2\right)\)
\(\Leftrightarrow a^3+3a^2b+3ab^2+b^3+a^3-3a^2b+3ab^2-b^3-2a^3-6ab^2=o\)
\(\Leftrightarrow0=0\)(đpcm)
b.\(\left(a+b\right)^3-\left(a-b\right)^3=2b\left(b^2+3a^2\right)\)
\(\Leftrightarrow a^3+3a^2b+3ab^2+b^3-a^3+3a^2b-3ab^2+b^3-2b^3-6a^2b=o\)
\(\Leftrightarrow0=0\)luôn đúng
Vậy đẳng thức được chứng minh
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a: \(=\left[\dfrac{3xy\left(x-2x^2y\right)}{3xy}+6x^2y-x\right]^2:\dfrac{1}{2}x^2\)
\(=\left[x-2x^2y+6x^2y-x\right]^2:\dfrac{1}{2}x^2\)
\(=\dfrac{16x^4y^2}{0.5x^2}=32x^2y^2\)
b: \(=\dfrac{7\left(a-b\right)^5+5\left(a-b\right)^3}{\left(a-b\right)^2}=7\left(a-b\right)^3+5\left(a-b\right)\)
c: \(=\dfrac{7\left(a-3b\right)^3+\left(a-3b\right)}{2\left(a-3b\right)}=\dfrac{7\left(a-3b\right)^2+1}{2}\)
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\(3y^2\left(a-3x\right)-a\left(a-3x\right)=\left(3y^2-a\right)\left(a-3x\right)\)