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Bài làm :
Bài 1 :
\(a,-2x^3y.\left(2x^2-3y+5y^2\right)\)
\(=-4x^5y+6x^3y^2-10x^3y^3\)
\(b,\left(x+1\right)\left(x^2-x+1\right)\)
\(=x^3-x^2+x+x^2-x+1\)
\(=x^3+1\)
\(c,\left(2x-1\right).\left(3x+2\right).\left(3-x\right)\)
\(=\left[\left(2x-1\right)\left(3x+2\right)\right]\left(3-x\right)\)
\(=\left(6x^2+4x-3x-2\right)\left(3-x\right)\)
\(=18x^2-6x^3+12x-4x^2-9x+3x^2-6+2x\)
\(=-6x^3+\left(18x^2-4x^2+3x^2\right)+\left(12x-9x+2x\right)-6\)
\(=-6x^3+17x^2+5x-6\)
Bài 2 :
\(\left(a+b\right).\left(a^3-a^2b+ab^2-b^3\right)\)
\(=a^4-a^3b+a^2b^2-ab^3+ba^3-a^2b^2+ab^3-b^4\)
\(=a^4+\left(-a^3b+ba^3\right)+\left(a^2b^2-a^2b^2\right)+\left(-ab^3+ab^3\right)-b^4\)
\(=a^4-b^4\)
=> đpcm
Học tốt nha
Bài 2:
Ta có: \(f\left(a\right)=6a^5-10a^4-5a^3+23a^2-29a+2005\)
\(=\left(6a^5-10a^4-2a^3\right)-\left(3a^3-5a^2-a\right)+\left(18a^2-30a-6\right)+2011\)
\(=2a^3\left(3a^2-5a-1\right)-a\left(3a^2-5a-1\right)+6\left(3a^2-5a-1\right)+2011\)
\(=\left(2a^3-a+6\right)\left(3a^2-5a-1\right)+2011\)
Mà \(3a^2-5a-1=0\)
\(\Rightarrow f\left(a\right)=2011\)
Vậy...
Ta có:
\(\left(a^3+3ab^2\right)^2=a^6+6a^4b^2+9a^2b^4=196\)
\(\left(b^3+3a^2b\right)^2=b^6+6a^2b^4+9a^4b^2=169\)
Lại có:
\(\left(a^3+3ab^2\right)^2-\left(b^3+3a^2b\right)^2=27\)
\(\Leftrightarrow a^6+6a^4b^2+9ab^4-b^6-6a^2b^4-9a^4b^2=27\)
\(\Leftrightarrow a^6-3a^4b^2+3a^2b^4-b^6=27\)
\(\Leftrightarrow\left(a^2-b^2\right)^3=27\)
\(\Leftrightarrow a^2-b^2=\sqrt[3]{27}=3\)
Áp dụng bất đẳng thức bu nhi a ta có
\(\left(a+2b\right)^2\le\left(1+2\right)\left(a^2+2b^2\right)=3.\left(a^2+2b^2\right)\le3.3c^2=9c^2\)
=> \(a+2b\le3c\)
Mà \(\frac{1}{a}+\frac{2}{b}=\frac{1}{a}+\frac{1}{b}+\frac{1}{b}\ge\frac{9}{a+2b}\ge\frac{9}{3c}=\frac{3}{c}\)
=> \(\frac{1}{a}+\frac{2}{b}\ge\frac{3}{c}\left(ĐPCM\right)\)
Lời giải:
Từ \(a+b+c+ab+bc+ac=0\)
\(\Rightarrow a+b+c+ab+bc+ac+abc+1=1\)
\(\Leftrightarrow (a+1)(b+1)(c+1)=1\)
Đặt \(\left\{\begin{matrix} a+1=x\\ b+1=y\\ c+1=z\end{matrix}\right.\Rightarrow xyz=1\)
Biểu thức trở thành:
\(A=\frac{1}{(a+2)+a+b+ab+1}+\frac{1}{(b+2)+b+c+bc+1}+\frac{1}{(c+2)+c+a+ac+1}\)
\(A=\frac{1}{(a+2)+(a+1)(b+1)}+\frac{1}{(b+2)+(b+1)(c+1)}+\frac{1}{(c+2)+(c+1)(a+1)}\)
\(A=\frac{1}{x+1+xy}+\frac{1}{y+1+yz}+\frac{1}{z+1+zx}\)
\(A=\frac{z}{xz+z+xyz}+\frac{zx}{yxz+xz+yz.xz}+\frac{1}{z+1+xz}\)
hay \(A=\frac{z}{xz+z+1}+\frac{xz}{1+xz+z}+\frac{1}{z+1+xz}\) (thay \(xyz=1\))
\(\Leftrightarrow A=\frac{z+xz+1}{xz+z+1}=1\)
Vậy \(A=1\)
Do abc=1nên ta được \(\frac{1}{ab+b+1}+\frac{1}{bc+c+1}+\frac{1}{ac+c+1}=\frac{abc}{ab+b+abc}+\frac{a}{abc+ac+a}+\frac{1}{ca+a+1}\)\(=\frac{ac}{1+a+ac}+\frac{a}{1+ac+a}+\frac{1}{ca+a+1}=1\)
Dấu "=" xảy ra khi a=b=c=1
Hình như shi thiếu bước đầu =)))
\(\frac{1}{a^2+2b^2+3}=\frac{1}{a^2+b^2+b^2+1+2}\le\frac{1}{2ab+2b+2}=\frac{1}{2}\cdot\frac{1}{ab+b+1}\)
Tương tự:\(\frac{1}{b^2+2c^2+3}\le\frac{1}{2}\cdot\frac{1}{bc+c+1};\frac{1}{c^2+2a^2+3}\le\frac{1}{2}\cdot\frac{1}{ca+a+1}\)
\(\Rightarrow LHS\le\frac{1}{2}\left(\frac{1}{ab+b+1}+\frac{1}{bc+c+1}+\frac{1}{ca+a+1}\right)=\frac{1}{2}\) Vì abc=1
\(1=x+y\ge2\sqrt{xy}\Rightarrow xy\le\frac{1}{4}\Rightarrow\left\{{}\begin{matrix}\frac{1}{xy}\ge4\\-xy\ge-\frac{1}{4}\end{matrix}\right.\)
\(A=1+\frac{9}{x^2y^2}-3\left(\frac{1}{x^2}+\frac{1}{y^2}\right)+x^3+y^3\)
\(A=1+\frac{9}{x^2y^2}-3\left(\frac{x^2+y^2}{x^2y^2}\right)+x^3+y^3\)
\(A=1+\frac{9}{x^2y^2}-3\left[\frac{\left(x+y\right)^2-2xy}{x^2y^2}\right]+\left(x+y\right)^3-3xy\left(x+y\right)\)
\(A=2+\frac{9}{x^2y^2}-3\left(\frac{1}{x^2y^2}-\frac{2}{xy}\right)-3xy\)
\(A=2+\frac{6}{\left(xy\right)^2}+\frac{6}{xy}-3xy\)
\(A\ge2+6.4^2+6.4-\frac{3.1}{4}=\frac{485}{4}\)
Dấu "=" xảy ra khi \(x=y=\frac{1}{2}\)