Tính GTNN của biểu thức:
c) C= (x2 - 3x + 1) (x2 - 3x + 1)
d) D= (x2 - 4x + 1) (x2 - 4x +5)
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Bài 1:
a: \(\Leftrightarrow x^2-5x+6< =0\)
=>(x-2)(x-3)<=0
=>2<=x<=3
b: \(\Leftrightarrow\left(x-6\right)^2< =0\)
=>x=6
c: \(\Leftrightarrow x^2-2x+1>=0\)
\(\Leftrightarrow\left(x-1\right)^2>=0\)
hay \(x\in R\)
Bài 3:
a) Ta có: \(A=25x^2-20x+7\)
\(=\left(5x\right)^2-2\cdot5x\cdot2+4+3\)
\(=\left(5x-2\right)^2+3>0\forall x\)(đpcm)
d) Ta có: \(D=x^2-2x+2\)
\(=x^2-2x+1+1\)
\(=\left(x-1\right)^2+1>0\forall x\)(đpcm)
Bài 1:
a) Ta có: \(A=x^2-2x+5\)
\(=x^2-2x+1+4\)
\(=\left(x-1\right)^2+4\ge4\forall x\)
Dấu '=' xảy ra khi x=1
b) Ta có: \(B=x^2-x+1\)
\(=x^2-2\cdot x\cdot\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{3}{4}\)
\(=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\forall x\)
Dấu '=' xảy ra khi \(x=\dfrac{1}{2}\)
\(a,=\left(x+1\right)\left(x+3\right)\\ b,=-5x^2+15x+x-3=\left(x-3\right)\left(1-5x\right)\\ c,=2x^2+2x+5x+5=\left(2x+5\right)\left(x+1\right)\\ d,=2x^2-2x+5x-5=\left(x-1\right)\left(2x+5\right)\\ e,=x^3+x^2-4x^2-4x+x+1=\left(x+1\right)\left(x^2-4x+1\right)\\ f,=x^2+x-5x-5=\left(x+1\right)\left(x-5\right)\)
a) `3x+5 =0`
`3x=-5`
`x=-5/3`
`b) -4x+8=0`
`-4x =-8`
`x=2`
`c) 3x -6=0`
`3x=6`
`x=2`
`d)x^2 +x =0`
`x(x+1) =0`
`=>[(x=0),(x=-1):}`
`e) x^2 -4 =0`
`x^2 =4`
`=> x = +-2`
`f) x^3 -27 =0`
`x^3 =27`
`=> x=3`
`g) 3x^2 +4 =0`
`3x^2 =-4`
`x^2 =-4/3(vô-lí)`
=> Đa thức ko có nghiệm
h) `x^3 -4x =0`
`x(x^2 -4) =0`
`=>[(x=0),(x^2=4 => x=+-2):}`
i) `2x^3 -32x =0`
`2x(x^2 -16)=0`
`=>[(2x=0),(x^2=16):}`
`=>[(x=0),(x=+-4):}`
a)C=(x2-3x+1)2>=0
c ) \(C=\left(x^2-3x+1\right)\left(x^2-3x+1\right)=\left(x^2-3x+1\right)^2\ge0\forall x\)
Dấu " = " xảy ra \(\Leftrightarrow x^2-3x+1=0\)
\(\Leftrightarrow x^2-3x+\dfrac{9}{4}-\dfrac{5}{4}=0\)
\(\Leftrightarrow\left(x-\dfrac{3}{2}\right)^2=\dfrac{5}{4}\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{3}{2}=\dfrac{\sqrt{5}}{2}\\x-\dfrac{3}{2}=\dfrac{-\sqrt{5}}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\sqrt{5}+3}{2}\\x=\dfrac{3-\sqrt{5}}{2}\end{matrix}\right.\)
Vậy Min C là : \(0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\sqrt{5}+3}{2}\\x=\dfrac{3-\sqrt{5}}{2}\end{matrix}\right.\)
d ) \(D=\left(x^2-4x+1\right)\left(x^2-4x+5\right)\)
\(=\left(x^2-4x+3-2\right)\left(x^2-4x+3+2\right)\)
\(=\left(x^2-4x+3\right)^2-4\ge-4\forall x\)
Dấu " = " xảy ra \(\Leftrightarrow x^2-4x+3=0\)
\(\Leftrightarrow x^2-3x-x+3=0\)
\(\Leftrightarrow x\left(x-3\right)-\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=3\end{matrix}\right.\)
Vậy Min D là : \(-4\Leftrightarrow\left[{}\begin{matrix}x=1\\x=3\end{matrix}\right.\)