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a/ \(8x-x^2\)
\(=-\left(x^2-8x\right)\)
\(=-\left(x^2-2\cdot4x+16-16\right)\)
\(=-\left(x-4\right)^2+16\)
Có \(\left(x-4\right)^2\ge0\)
\(\Rightarrow-\left(x-4\right)^2\le0\)
\(\Rightarrow-\left(x-4\right)^2+16\le16\)
\(\Rightarrow GTLN\left(8x-x^2\right)=16\)
với \(\left(x-4\right)^2=0;x=4\)
b/ \(\frac{3}{x^2-4x+10}\)
Xét mẫu số ta có : \(x^2-4x+10\)
\(=x^2-2\cdot2x+4-4+10\)
\(=\left(x-2\right)^2-4+10\)
\(=\left(x-2\right)^2+6\)
Có \(\left(x-2\right)^2\ge0\)\(\Rightarrow\left(x-2\right)^2+6\ge6\)
\(\Rightarrow\frac{3}{\left(x-2\right)^2+6}\le\frac{3}{6}\)
\(\Rightarrow GTLN\frac{3}{x^2-4x+10}=\frac{3}{6}\)
với \(\left(x-2\right)^2=0;x=2\)
c/ cái này f GTNN chứ bạn, mik thấy kq ra dương , bạn ktra giúp mik nha.
\(x^2+y^2\)
Có \(x+y=2\Rightarrow x=2-y\)
\(x^2+y^2\)
\(=\left(2-y\right)^2+y^2\)
\(=4-4y+y^2+y^2\)
\(=4-4y+y^2\)
\(=2y^2-4y+4\)
\(=2\left(y^2-2y+2\right)\)
\(=2\left(y^2-2\cdot1y+1+1\right)\)
\(=2\left[\left(y-1\right)^2+1\right]\)
\(=2\left(y-1\right)^2+2\)
Có \(\left(y-1\right)^2\ge0\Rightarrow\left(y-1\right)^2+2\ge2\)
\(\Rightarrow GTNN2\left(y-1\right)^2+2\ge2\)
với \(\left(y-1\right)^2=0;y=1\)
\(\Rightarrow GTNN\left(x^2+y^2\right)\ge2\)với\(x=1;y=1\)
a,x(2x-1)-(x-1)^2-x^2=0
<=>x(2x-1-x)-(x-1)^2=0
<=>x(x-1)-(x-1)^2=0
<=>(x-x+1)(x-1)=0
<=>x-1=0
<=>x=1
b,(x+2)^3-x^3-6x^2=4
<=>x^3+6x^2+12x+8-x^3-6x^2=4
<=>12x+8=4
<=>x=-1/3
tick mik nha
`a)x(2x-1)-(x-1)^2-x^2=0`
`<=>2x^2-x-x^2+2x-1-x^2=0`
`<=>x-1=0`
`<=>x=1`
Vậy `x=1.`
`b)(x+2)^3-x^3-6x^2=4`
`<=>x^3+6x^2+12x+8-x^3-6x^2=4`
`<=>12x+8=4`
`<=>12x=-4`
`<=>x=-1/3`
Vậy `x=-1/3.`
1.
a) \(2x^4-4x^3+2x^2\)
\(=2x^2\left(x^2-2x+1\right)\)
\(=2x^2\left(x-1\right)^2\)
b) \(2x^2-2xy+5x-5y\)
\(=\left(2x^2-2xy\right)+\left(5x-5y\right)\)
\(=2x\left(x-y\right)+5\left(x-y\right)\)
\(=\left(x-y\right)\cdot\left(2x+5\right)\)
2 .
a,
\(4x\left(x-3\right)-x+3=0\)
⇒\(4x\left(x-3\right)-\left(x-3\right)=0\)
⇒\(\left(x-3\right)\left(4x-1\right)=0\)
⇒\(\left[{}\begin{matrix}x-3=0\\4x-1=0\end{matrix}\right.\)
⇔\(\left[{}\begin{matrix}x=3\\4x=1\end{matrix}\right.\)
⇔\(\left[{}\begin{matrix}x=3\\x=\dfrac{1}{4}\end{matrix}\right.\)
vậy \(x\in\left\{3;\dfrac{1}{4}\right\}\)
b,
\(\)\(\left(2x-3\right)^2-\left(x+1\right)^2=0\)
⇒\(\left(2x-3-x-1\right)\left(2x-3+x+1\right)\) = 0
⇒\(\left(x-4\right)\left(3x-2\right)=0\)
⇔\(\left[{}\begin{matrix}x-4=0\\3x-2=0\end{matrix}\right.\)
⇔\(\left[{}\begin{matrix}x=4\\3x=2\end{matrix}\right.\)
⇔\(\left[{}\begin{matrix}x=4\\x=\dfrac{2}{3}\end{matrix}\right.\)
vậy \(x\in\left\{4;\dfrac{2}{3}\right\}\)
2 x 2 + 4 x + 8 x 3 - 3 x 2 - x + 3 : P = x 3 - 8 x + 1 x - 3
Tìm GTLN:
\(B=21-8x-2x^2\)
\(=-2\left(x^2+4x+4\right)+8+21\)
\(=-2\left(x+2\right)^2+29\)
Với mọi giá trị của x ta có:
\(-2\left(x+2\right)^2\le0\Rightarrow-2\left(x+2\right)^2+29\le29\)
Dấu "=" xảy ra khi
\(x+2=0\Rightarrow x=-2\)
Vậy Max B = 29 khi x = -2
Tìm x :
\(A=\left(x+2\right)^3+x\left(x+3\right)\left(x-3\right)-6x^2=29\)
\(A=x^3+6x^2+12x+8+x^3-9x-6x^2=29\)
\(A=2x^3+3x+8=29\)