Tìm GTNN của 7x^2-x+1
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![](https://rs.olm.vn/images/avt/0.png?1311)
\(=2x^2\left(2x+1\right)+2x\left(2x+1\right)+\left(2x+1\right)\)
\(=\left(2x^2+2x+1\right)\left(2x+1\right)\)
Ta có :
4x3+6x2+4x+1
= (4x3+2x2)+(4x2+2x)+(2x+1)
= 2x2(2x+1)+2x(2x+1)+(2x+1)
= (2x2+2x+1)(2x+1)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(A=\left(37^3+12^3\right):49-37\times12\)
\(=\left(37+12\right)\left(37^2+12^2-37\times12\right):49-37\times12\)
\(=37^2+12^2-2\times37\times12\)
\(=\left(37-12\right)^2=25^2=625\)
b) \(B=\left(52^3-48^3\right):4+52\times48\)
\(=\left(52-48\right)\left(52^2+48^2+52\times48\right):4+52\times48\)
\(=52^2+48^2+2\times52\times48\)
\(=\left(52+48\right)^2=100^2=10000\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(-x^3+3x^2-3x+1\) (tại x=6, ta có)
\(=x\left(-x^2+3x-3\right)+1\)
\(=6\left(-6^2+3.6-3\right)+1\)
\(=6\left(-36+18-3\right)+1\)
\(=6.\left(-21\right)+1=-125\)
Ta có :
-x3+3x2-3x+1
= -(x3-1)+(3x2-3x)
= -(x-1)(x2+x+1)+3x(x-1)
=(x-1)(3x-x2-x-1)
= (x-1)(-x2+2x-1)
= (x-1)[-(x2-x)+(x-1)]
= (x-1)[-x(x-1)+(x-1)]
= (x-1)2.(1-x) (1)
Thay x = 6 vào (1) ta được :
-x3+3x2-3x+1=(x-1)2.(1-x)
= (6-1)2.(1-6)
= 52.(-5) = -125
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có:
\(\left\{{}\begin{matrix}x^2-yz=a\\y^2-zx=b\\z^2-xy=c\end{matrix}\right.\) ⇒\(\left\{{}\begin{matrix}x^3-xyz=ax\\y^3-xyz=by\\z^3-xyz=cz\end{matrix}\right.\)
⇒\(ax+by+cz=x^3+y^3+z^3-3xyz\)
\(=\left(x+y\right)^3+z^3-3xy\left(x+y\right)-3xyz\)
\(=\left(x+y+z\right)^3-3\left(x+y\right)z\left(x+y+z\right)-3xy\left(x+y+z\right)\)
\(=\left(x+y+z\right)\left[\left(x+y+z\right)^2-3\left(x+y\right)z-3xy\right]\)
⇒\(ax+by+cz⋮x+y+z\)
em ghi sai đề hay sao á
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,=5xy\left(3x-y+4x^2\right)\)
\(b,=\left(x-4\right)^2-16=\left(x-4+4\right)\left(x-4-4\right)\)
\(=x\left(x-8\right)\)
\(c,=\left(x-5\right)^2-9^2=\left(x-14\right)\left(x+4\right)\)
\(d,=\left(x-y\right)^2-\left(x+5\right)^2=\left(2x-y+5\right)\left(-y-5\right)\)
a) \(15x^2y-5xy^2+20x^3y=5xy\left(3x-y+4x^2\right)\)
b) \(x^2+16-8x-16=x^2-8x=x\left(x-8\right)\)
c) \(x^2+25-10x-81=\left(x-5\right)^2-9^2=\left(x-14\right)\left(x+4\right)\)
d) \(x^2-2xy+y^2-x^2-10x-25=\left(x-y\right)^2-\left(x+5\right)^2=-\left(2x-y+5\right)\left(y+5\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(Với\) \(\left\{{}\begin{matrix}x=8\\y=9\end{matrix}\right.\) \(thay\) \(vào\), \(ta\) \(có\):
\(A=x\left(x-y\right)+y\left(x+y\right)\)
\(=8\left(8-9\right)+9\left(8+9\right)\)
\(=8.-1+9.17=-8+153=145\)
\(7x^2-x+1=7\left(x^2-\dfrac{x}{7}+\dfrac{1}{196}\right)+\dfrac{27}{28}\)
\(=7\left(x-\dfrac{1}{14}\right)^2+\dfrac{27}{28}\ge\dfrac{27}{28}\forall x\)
\(Min=\dfrac{27}{28}\Leftrightarrow x=\dfrac{1}{14}\)
A= 7x2 - x + 1
A= 7( x2 - 2.1/14x + 1/196) + 27/28
A= 7(x - 1/14)2 + 27/28
A = 7(x - 1/14)2 ≥ 0 ⇔ 7(x-1/14)2 +27/28 ≥ 27/28
A(min)= 27/28 ⇔ x = 1/14