Tìm GTLN của bt A = \(\frac{3x^2}{x^3+1}-\frac{1}{x+1}-\frac{x-1}{x^2-x+1}\).
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12 tháng 1 2019
a) A xác định \(\Leftrightarrow\hept{\begin{cases}3x\ne0\\x+1\ne0\\2-4x\ne0\end{cases}\Leftrightarrow\hept{\begin{cases}x\ne0\\x\ne-1\\x\ne\frac{1}{2}\end{cases}}}\)
\(A=\left(\frac{x+2}{3x}+\frac{2}{x+1}-3\right):\frac{2-4x}{x+1}-\frac{3x+1-x^2}{3x}\)
\(A=\left[\frac{\left(x+2\right)\left(x+1\right)}{3x\left(x+1\right)}+\frac{2\cdot3x}{3x\left(x+1\right)}-\frac{3\cdot3x\left(x+1\right)}{3x\left(x+1\right)}\right]\cdot\frac{x+1}{2\left(1-2x\right)}-\frac{3x+1-x^2}{3x}\)
\(A=\frac{x^2+3x+2+6x-9x^2-9x}{3x\left(x+1\right)}\cdot\frac{x+1}{2\cdot\left(1-2x\right)}-\frac{3x+1-x^2}{3x}\)
\(A=\frac{\left(-8x^2+2\right)\left(x+1\right)}{3x\left(x+1\right)2\left(1-2x\right)}-\frac{3x+1-x^2}{3x}\)
\(A=\frac{2\left(1-4x^2\right)}{3x\cdot2\left(1-2x\right)}-\frac{3x+1-x^2}{3x}\)
\(A=\frac{2\left(1-2x\right)\left(1-2x\right)}{3x\cdot2\left(1-2x\right)}-\frac{3x+1-x^2}{3x}\)
\(A=\frac{1+2x}{3x}-\frac{3x+1-x^2}{3x}\)
\(A=\frac{2x+1-3x-1+x^2}{3x}\)
\(A=\frac{x^2-x}{3x}\)
\(A=\frac{x\left(x-1\right)}{3x}\)
\(A=\frac{x-1}{3}\)
b) Thay x = 4 ta có :
\(A=\frac{4-1}{3}=\frac{3}{3}=1\)
c) Để A thuộc Z thì \(x-1⋮3\)
\(\Rightarrow x-1\in B\left(3\right)=\left\{0;3;6;...\right\}\)
\(\Rightarrow x\in\left\{1;4;7;...\right\}\)
Vậy.....
9 tháng 9 2019
1/a/
\(A=\frac{2}{xy}+\frac{3}{x^2+y^2}=\left(\frac{1}{xy}+\frac{1}{xy}+\frac{4}{x^2+y^2}\right)-\frac{1}{x^2+y^2}\)
\(\ge\frac{\left(1+1+2\right)^2}{\left(x+y\right)^2}-\frac{1}{\frac{\left(x+y\right)^2}{2}}=16-2=14\)
Dấu = xảy ra khi \(x=y=\frac{1}{2}\)
9 tháng 9 2019
b/
\(4B=\frac{4}{x^2+y^2}+\frac{8}{xy}+16xy=\left(\frac{4}{x^2+y^2}+\frac{1}{xy}+\frac{1}{xy}\right)+\left(\frac{1}{xy}+16xy\right)+\frac{5}{xy}\)
\(\ge\frac{\left(1+1+2\right)^2}{\left(x+y\right)^2}+2\sqrt{\frac{1}{xy}.16xy}+\frac{5}{\frac{\left(x+y\right)^2}{4}}\)
\(=16+8+20=44\)
\(\Rightarrow B\ge11\)
Dấu = xảy ra khi \(x=y=\frac{1}{2}\)
12 tháng 1 2019
\(A=\frac{1-x^2}{x}.\left(\frac{x^2}{x+3}-1\right)+\frac{3x^2-14x+3}{x^2+3x}\)
\(A=\frac{\left(x^2-x-3\right)\left(-x^2+1\right)}{x\left(x+3\right)}+\frac{3x^2-14x+3}{x^2+3x}\)
\(A=\frac{\left(x^2-x-3\right)\left(1-x^3\right)}{\left(x+3\right)x}+\frac{3x^2-14x+3}{x\left(x+3\right)}\)
\(A=\frac{\left(x^2-x-3\right)\left(1-x^2\right)+3x^2-14x+3}{\left(x+3\right)x}\)
\(A=\frac{-x^4+x^3+7x^2-15x}{x\left(x+3\right)}\)
\(A=\frac{x\left(-x^3+x^2+7x-15\right)}{x\left(x+3\right)}\)
\(A=\frac{-x^3+x^2+7x-15}{x+3}\)
\(A=-\frac{\left(x+3\right)\left(x^2-4x+5\right)}{x+3}\)
\(A=-\left(x^2-4x+5\right)\)
\(A=-x^2+4x-5\)
Trình độ hơi thấp, có gì sai sót xin bỏ qua cho :)
1 tháng 6 2019
\(1,\)\(\frac{x+2}{x+3}+\frac{x-1}{x+1}=\frac{2}{x^2+4x+3}+1\)
\(\Rightarrow\frac{\left(x+2\right)\left(x+1\right)}{\left(x+1\right)\left(x+3\right)}+\frac{\left(x-1\right)\left(x+3\right)}{\left(x+1\right)\left(x+3\right)}=\frac{2}{\left(x+1\right)\left(x+3\right)}+\frac{\left(x+1\right)\left(x+3\right)}{\left(x+1\right)\left(x+3\right)}\)
\(\Rightarrow\)\(x^2+3x+2+x^2-2x-3=2+x^2+4x+3\)
\(\Rightarrow x^2-3x-6=0\)
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1 tháng 6 2019
\(\frac{x+1}{x-2}+\frac{2x-1}{x-1}=\frac{2}{x^2-3x+2}+\frac{11}{2}\)
\(\Rightarrow\frac{2\left(x+1\right)\left(x-1\right)}{2\left(x-2\right)\left(x-1\right)}+\frac{2\left(2x-1\right)\left(x-2\right)}{2\left(x-1\right)\left(x-2\right)}\)\(=\frac{4}{2\left(x-1\right)\left(x-2\right)}+\frac{22\left(x-1\right)\left(x-2\right)}{2\left(x-1\right)\left(x-2\right)}\)
\(\Rightarrow2x^2-2+4x^2-10x+4=4+22x^2-66x+44\)
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