CMR: \(\left(a+b\right)^3-\left(a-b\right)^3-2b^3=6a^2b\)
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CMR : \(\left(a+b\right)^3-\left(a-b\right)^3-2b^3=6a^2b\)
\(\left(a+b\right)^3-\left(a-b\right)^3-2b^3\)
\(=\left(a+b-a+b\right)[\left(a+b\right)^2+\left(a+b\right)\left(a-b\right)+\left(a-b\right)^2]-2b^3\)
\(=2b\left(a^2+2ab+b^2+a^2-b^2+a^2-2ab+b^2\right)-2b^2\)
\(=2b\left(3a^2+b^2\right)-2b^3\)
\(=2b\left(3a^2+b^2-b^2\right)\)
\(=2b\times3a^2=6a^2b\left(đpcm\right)\)
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Cho a, b, c > 0 . CMR :
\(\dfrac{a^3}{\left(2a+b\right)\left(2b+c\right)}+\dfrac{b^3}{\left(2b+c\right)\left(2c+a\right)}+\dfrac{c^3}{\left(2c+a\right)\left(2a+b\right)}\le\dfrac{a+b+c}{9}\)
Dấu >= hay <= vậy bạn? Bạn xem lại đề.
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Cho a,b,c>0. Cmr: \(\dfrac{a^3}{b\left(a+2b\right)}+\dfrac{2b^3}{c\left(2c+b\right)}+\dfrac{128c^3}{a\left(4a+c\right)}>a+2b+4c\)
Cho a,b,c >0, cmr \(\frac{a^3}{b\left(2b+a\right)}+\frac{2b^3}{c\left(2c+b\right)}+\frac{128c^3}{a\left(a+4c\right)}\ge a+2b+4c\)
Tính \(M=a^3+b^3+3ab\left(a^2+b^2\right)+6a^2b^2\left(a+b\right)\) biết a+b=1
\(M=a^3+b^3+3ab\left(a^2+b^2\right)+6a^2b^2\left(a+b\right)\)
\(=\left(a+b\right)^3-3ab\left(a+b\right)+3ab\left[\left(a+b\right)^2-2ab\right]+6a^2b^2\)
\(=1-3ab+3ab\left[1-2ab\right]+6a^2b^2\)
\(=1-3ab+3ab-6a^2b^2+6a^2b^2\)
=1
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Cho a,b,c là 3 số thực đôi một phân biệt. CMR:
\(3+\frac{\left(2a+b\right)\left(2b+c\right)}{\left(a-b\right)\left(b-c\right)}+\frac{\left(2b+c\right)\left(2c+a\right)}{\left(b-c\right)\left(c-a\right)}+\frac{\left(2c+a\right)\left(2a+b\right)}{\left(c-a\right)\left(a-b\right)}=\frac{2a+b}{a-b}+\frac{2b+c}{b-c}+\frac{2c+a}{c-a}\)
\(a^3+b^3+3ab\left(a^2+b^2\right)+6a^2b^2\left(a+b\right)\)
Đề thiếu: Cho a + b = 1. Tính giá trị biểu thức
Ta có:
\(a^3+b^3+3ab\left(a^2+b^2\right)+6a^2b^2\left(a+b\right)\)
\(=\left(a+b\right)\left(a^2-ab+b^2\right)+3ab\left[\left(a+b\right)^2-2ab\right]+6a^2b^2\left(a+b\right)\)
\(=\left(a+b\right)\left[\left(a+b\right)^2-3ab\right]+3ab\left[\left(a+b\right)^2-2ab\right]+6a^2b^2\left(a+b\right)\)
Thay a + b = 1
\(=1\left(1-3ab\right)+3ab\left(1-2ab\right)+6a^2b^2.1\)
\(=1-3ab+3ab-6a^2b^2+6a^2b^2\)
\(=1\)
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Rút gọn các biểu thức:
\(A=\left(5a+5\right)^2+10\left(a-3\right)\left(1+a\right)+a^2-6a+9\)
B = \(\left(6a-2\right)^2+4\left(3a-1\right)\left(1-2b\right)\left(2b-1\right)^2\)
rut gon bieu thuc :
a,\(\left(a+b\right)^3-\left(a-b\right)^3-6a^2b\)
b,\(\left(a+b\right)^3+\left(a-b\right)^3-6ab^2\)
a) \(\left(a+b\right)^3-\left(a-b\right)^3-6a^2b\)
\(\Leftrightarrow a^3+3a^2b+3ab^2+b^3-\left(a^3-3a^2b+3ab^2-b^3\right)-6a^2b\)
\(\Leftrightarrow a^3+3a^2b+3ab^2+b^3-a^3+3a^2b-3ab^2+b^3-6a^2b\)
\(\Leftrightarrow2b^3\)
b) \(\left(a+b\right)^3-\left(a-b\right)^3-6ab^2\)
\(\Leftrightarrow a^3+3a^2b+3ab^2+b^3-\left(a^3-3a^2b+3ab^2-b^3\right)-6ab^2\)
\(\Leftrightarrow a^3+3a^2b+3ab^2+b^3-a^3+3a^2b-3ab^2+b^3-6ab^2\)
\(\Leftrightarrow2b^3+6a^2b-6ab^2\)
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