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1)
\(y=x^2+5x-4\)
\(=x^2+2.x.\frac{5}{2}+\frac{25}{4}-\left(\frac{25}{4}+4\right)\)
\(=\left(x+\frac{5}{2}\right)^2-10,25\)
\(Min_y=-10,25\Leftrightarrow x=-\frac{5}{2}\)
2) \(y=2x^2-6x+5\)
\(=2\left(x^2-2.x.\frac{3}{2}+\frac{9}{4}\right)+\frac{1}{2}\)
\(=2\left(x-\frac{3}{2}\right)^2+\frac{1}{2}\)
\(Min_y=\frac{1}{2}\Leftrightarrow x=\frac{3}{2}\)
a) \(x^3-5x^2+8x-4\)
\(=x^3-2x^2-3x^2+6x+2x-4\)
\(=x^2\left(x-2\right)-3x\left(x-2\right)+2\left(x-2\right)\)
\(=\left(x-2\right)\left(x^2-3x+2\right)\)
\(=\left(x-2\right)\left(x^2-x-2x+2\right)\)
\(=\left(x-2\right)\left[x\left(x-1\right)-2\left(x-1\right)\right]\)
\(=\left(x-2\right)\left(x-1\right)\left(x-2\right)\)
b) \(A=10x^2-15x+8x-12+7\)
\(A=5x\left(2x-3\right)+4\left(2x-3\right)+7\)
\(A=\left(2x-3\right)\left(5x+4\right)+7\)
Dễ thấy \(\left(2x-3\right)\left(5x+4\right)⋮\left(2x-3\right)=B\)
Vậy để \(A⋮B\)thì \(7⋮\left(2x-3\right)\)
\(\Rightarrow2x-3\inƯ\left(7\right)=\left\{\pm1;\pm7\right\}\)
\(\Rightarrow x\in\left\{2;1;5;-2\right\}\)
Vậy.......
6) Ta có
\(A=\frac{x^3}{y+2z}+\frac{y^3}{z+2x}+\frac{z^3}{x+2y}\)
\(=\frac{x^4}{xy+2xz}+\frac{y^4}{yz+2xy}+\frac{z^4}{zx+2yz}\)
\(\ge\frac{\left(x^2+y^2+z^2\right)^2}{xy+2xz+yz+2xy+zx+2yz}\)
\(\Leftrightarrow A\ge\frac{1}{3\left(xy+yz+zx\right)}\ge\frac{1}{3\left(x^2+y^2+z^2\right)}=\frac{1}{3}\)
1: \(y=x^2+2\cdot x\cdot\dfrac{5}{2}+\dfrac{25}{4}-\dfrac{41}{4}\)
\(=\left(x+\dfrac{5}{2}\right)^2-\dfrac{41}{4}\ge-\dfrac{41}{4}\forall x\)
Dấu '=' xảy ra khi x=-5/2
2: \(y=2\left(x^2-2x+\dfrac{5}{2}\right)\)
\(=2\left(x^2-2x+1+\dfrac{3}{2}\right)\)
\(=2\left(x-1\right)^2+3\ge3\forall x\)
Dấu '=' xảy ra khi x=1
3: \(y=x^2-4x+4-3=\left(x-2\right)^2-3\ge-3\forall x\)
Dấu '=' xảy ra khi x=2
4: \(2x^2-8x+3\)
\(=2\left(x^2-4x+\dfrac{3}{2}\right)\)
\(=2\left(x^2-4x+4-\dfrac{5}{2}\right)\)
\(=2\left(x-2\right)^2-5\ge-5\forall x\)
Dấu '=' xảy ra khi x=2
Ta có: \(\frac{x^2y+2xy^2+y^3}{2x^2+xy-y^2}\)
\(=\frac{x^2y+xy^2+xy^2+y^3}{2x^2+2xy-xy-y^2}\)
\(=\frac{xy\left(x+y\right)+y^2\left(x+y\right)}{2x\left(x+y\right)-y\left(x+y\right)}\)
\(=\frac{\left(x+y\right)\left(xy+y^2\right)}{\left(2x-y\right)\left(x+y\right)}=\frac{xy+y^2}{2x-y}\left(đpcm\right)\)
Ta có: \(\frac{x^2+3xy+2y^2}{x^3+2x^2y-xy^2-2y^3}\)
\(=\frac{x^2+xy+2xy+2y^2}{x^2\left(x+2y\right)-y^2\left(x+2y\right)}\)
\(=\frac{x\left(x+y\right)+2y\left(x+y\right)}{\left(x^2-y^2\right)\left(x+2y\right)}\)
\(=\frac{\left(x+2y\right)\left(x+y\right)}{\left(x+y\right)\left(x-y\right)\left(x+2y\right)}=\frac{1}{x-y}\left(đpcm\right)\)
a) /x-5/=/2x+1/
=>x-5=2x+1 hoặc x-5=-(2x+1)
Th1 x-5=2x+1
-5-1=2x-x
x=-6
Thử lại thấy đúng
Th2: x-5=-(2x+1)
x-5=-2x-1
x+2x=-1+5
3x=4
x=4/3
Thử lại thấy đúng
Vậy x=-6 hoặc x=4/3
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