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\(A=x^2+3x+7\)
\(=x^2+2.1,5x+2,25+4,75\)
\(=\left(x+1,5\right)^2+4,75\ge4,75\)
Vậy \(A_{min}=4,75\Leftrightarrow x=-1,5\)
\(B=2x^2-8x\)
\(=2\left(x^2-4x\right)\)
\(=2\left(x^2-4x+4-4\right)\)
\(=2\left[\left(x-2\right)^2-4\right]\)
\(=2\left(x-2\right)^2-8\ge-8\)
Vậy \(B_{min}=-8\Leftrightarrow x=2\)
\(A=3x^2-5x+3=3(x^2-\frac{5}{3}x)+3\)
\(=3(x^2-\frac{5}{3}x+\frac{5^2}{6^2})+\frac{11}{12}=3(x-\frac{5}{6})^2+\frac{11}{12}\)
Vì \((x-\frac{5}{6})^2\geq 0, \forall x\Rightarrow A\geq 3.0+\frac{11}{12}=\frac{11}{12}\)
Vậy A(min)$=\frac{11}{12}$ khi $x=\frac{5}{6}$
\(B=2x^2+2x+1=2(x^2+x+\frac{1}{4})+\frac{1}{2}\)
\(=2(x+\frac{1}{2})^2+\frac{1}{2}\geq 2.0+\frac{1}{2}=\frac{1}{2}\)
Vậy \(B_{\min}=\frac{1}{2}\) tại \((x+\frac{1}{2})^2=0\Leftrightarrow x=\frac{-1}{2}\)
C)
\(C=2x^2+y^2+10x-2xy+27\)
\(=(x^2+10x+25)+(x^2+y^2-2xy)+2\)
\(=(x+5)^2+(x-y)^2+2\)
Vì \((x+5)^2\ge 0, (x-y)^2\geq 0\Rightarrow C\geq 0+0+2=2\)
Vậy \(C_{\min}=2\) tại \(\left\{\begin{matrix} (x+5)^2=0\\ (x-y)^2=0\end{matrix}\right.\Leftrightarrow x=y=-5\)
1,
a,\(2x\left(3x^2-5x+3\right)\)
\(=6x^3-10x^2+6x\)
b,\(-2x\left(x^2+5x-3\right)\)
\(=-2x^3-10x^2+6x\)
c,\(-\dfrac{1}{2}x\left(2x^3-4x+3\right)\)
\(=-x^4+2x^2-\dfrac{3}{2}x\)
Bài 2:
a) \(\left(2x-1\right)\left(x^2-5-4\right)\)
\(=\left(2x-1\right)\left(x^2-9\right)\)
\(=2x^3-18x-x^2+9\)
b) \(-\left(5x-4\right)\left(2x+3\right)\)
\(=-\left(10x^2+15x-8x-12\right)\)
\(=-10x^2-7x+12\)
c) \(\left(2x-y\right)\left(4x^2-2xy+y^2\right)\)
\(=8x^3-y^3\)
4x(x-2005)-(x+2005)=0
4x(x-2005)+(x-2005)=0
(x-2005)(4x+1)=0
<=>x-2005=>x=2005
4x+1=0=>x=-1/4
b, (x+1)2-x-1=0
(x+1)2-(x+1)=0
(x+1)(x+1-1)=0
(x+1)x=0
<=>x+1=0=>x=-1
x =0
\(a,x^2\left(x-2x^3\right)\)
\(=x^3-2x^5\)
\(b,\left(x-2\right)\left(x-x^2+4\right)\)
\(=x^2-x^3+4x-2x+2x^2-8\)
\(=3x^2-x^3+2x-8\)
\(c,\left(x^2-1\right)\left(x^2+2x\right)\)
\(=x^4+2x^3-x^2-2x\)
\(d,\left(2x-1\right)\left(3x+2\right)\left(3-x\right)\)
\(=\left(6x^2+4x-3x-2\right)\left(3-x\right)\)
\(=\left(6x^2+x-2\right)\left(3-x\right)\)
\(=18x^2+3x-6-6x^3-x^2+2x\)
\(=17x^2+5x-6-6x^3-x^2\)
\(e,\left(x+3\right)\left(x^2+3x-5\right)\)
\(=x^3+3x^2-5x+3x^2+9x-15\)
\(=x^3+6x^2+4x-15\)
\(f,\left(xy-2\right)\left(x^3-2x-6\right)\)
\(=x^4y-2x^2y-6xy-2x^3+4x-12\)
\(g,\left(5x^3-x^2+2x-3\right)\left(4x^2-x+2\right)\)
\(=20x^5-4x^4+8x^3-12x^2-5x^4+x^3-2x^2+3x+10x^3-2x^2+4x-6\)
\(=20x^5-9x^4+19x^3-16x^2+7x-6\)
a. x2(x−2x3)= x3-2x5
b. (x−2)(x−x2+4)= x2-x3+4x-2x+2x2-8= -x3+3x2+2x-8
c. (x2−1)(x2+2x)= x4+2x3-x2-2x
d. (2x−1)(3x+2)(3−x) = (6x2+x-2)(3-x)=18x2-6x3+3x-x2-6+2x =-6x3+17x2+5x-6
e. (x+3)(x2+3x−5)= x3+3x2-5x+3x2+9x-15= x3+6x2+4x-15
f. (xy−2)(x3−2x−6)= x4y-2x2y-6xy-2x3+4x+12
g. (5x3−x2+2x−3)(4x2−x+2)= 20x5-9x4+19x3-12x2+7x-6
\(A=5x\left(4x^2-2x+1\right)-2x\left(10x^2-5x-2\right)\)
\(=20x^3-10x^2+5x-20x^3+10x^2+4x\)
\(=9x\)
Thay x=15 \(\Rightarrow A=9.15=135\)
\(B=6xy\left(xy-y^2\right)-8x^2\left(x-y^2\right)+5y^2\left(x^2-xy\right)\)
\(=6x^2y^2-6xy^3-8x^3+8x^2y^2+5x^2y^2-5xy^3\)
\(=19x^2y^2-11xy^3-8x^3\)
Thay x=1/2 ; y=2 vào B \(\Rightarrow19.\left(\frac{1}{2}\right)^2.2^2-11\cdot\frac{1}{2}\cdot2^3-8\cdot\left(\frac{1}{2}\right)^3\)
\(=19-44-1\)
\(=-26\)
\(B=\left(x-y-1\right)^2+3\left(y-2\right)^2+2005\text{ }\ge2005\)
\(C=\left(x^2+4x\right)^2-25\ge-25\)
\(2004.2006.\left(2005^2+1\right)=\left(2005-1\right)\left(2005+1\right)\left(2005^2+1\right)\)
\(=\left(2005^2-1\right)\left(2005^2+1\right)=2005^4-1< 2005^4\)
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