Tìm GTNN
J= x^2-8x+6/x^2-2x+1
K
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Những câu hỏi liên quan
LH
1
20 tháng 10 2021
a: \(x\left(x-3\right)+2x-6=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)
b: \(\left(x+1\right)^2-4\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=3\end{matrix}\right.\)
23 tháng 12 2023
Câu 5: B
Câu 6:
a: ĐKXĐ: \(x-2\ne0\)
=>\(x\ne2\)
b: ĐKXĐ: \(x+1\ne0\)
=>\(x\ne-1\)
8:
\(A=\dfrac{x^2+4}{3x^2-6x}+\dfrac{5x+2}{3x}-\dfrac{4x}{3x^2-6x}\)
\(=\dfrac{x^2+4-4x}{3x\left(x-2\right)}+\dfrac{5x+2}{3x}\)
\(=\dfrac{\left(x-2\right)^2}{3x\left(x-2\right)}+\dfrac{5x+2}{3x}\)
\(=\dfrac{x-2+5x+2}{3x}=\dfrac{6x}{3x}=2\)
7:
\(\dfrac{8x^3yz}{24xy^2}\)
\(=\dfrac{8xy\cdot x^2z}{8xy\cdot3y}\)
\(=\dfrac{x^2z}{3y}\)
NV
Nguyễn Việt Lâm
Giáo viên
26 tháng 12 2022
\(S=\dfrac{3x^2+8x+6}{x^2+2x+1}=\dfrac{-2\left(x^2+2x+1\right)+x^2+4x+4}{x^2+2x+1}=-2+\left(\dfrac{x+2}{x+1}\right)^2\ge-2\)
\(S_{min}=-2\) khi \(x=-2\)
23 tháng 10 2021
e: ta có: \(4x^2+4x-6=2\)
\(\Leftrightarrow4x^2+4x-8=0\)
\(\Leftrightarrow\left(x+2\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=1\end{matrix}\right.\)
f: Ta có: \(2x^2+7x+3=0\)
\(\Leftrightarrow\left(x+3\right)\left(2x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-\dfrac{1}{2}\end{matrix}\right.\)
9 tháng 10 2021
\(a,\Leftrightarrow\left[{}\begin{matrix}x+5=0\\2x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=-\dfrac{1}{2}\end{matrix}\right.\\ b,\Leftrightarrow\left(x+2\right)\left(x-3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=3\end{matrix}\right.\\ c,\Leftrightarrow2x^2-10x-3x-2x^2=26\\ \Leftrightarrow-13x=26\Leftrightarrow x=-2\\ d,\Leftrightarrow x^2-18x+16=0\\ \Leftrightarrow\left(x^2-18x+81\right)-65=0\\ \Leftrightarrow\left(x-9\right)^2-65=0\\ \Leftrightarrow\left(x-9+\sqrt{65}\right)\left(x-9-\sqrt{65}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=9-\sqrt{65}\\9+\sqrt{65}\end{matrix}\right.\)
\(e,\Leftrightarrow x^2-10x-25=0\\ \Leftrightarrow\left(x-5\right)^2-50=0\\ \Leftrightarrow\left(x-5-5\sqrt{2}\right)\left(x-5+5\sqrt{2}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=5+5\sqrt{2}\\x=5-5\sqrt{2}\end{matrix}\right.\\ f,\Leftrightarrow5x\left(x-1\right)-\left(x-1\right)=0\\ \Leftrightarrow\left(x-1\right)\left(5x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{5}\end{matrix}\right.\\ g,\Leftrightarrow2\left(x+5\right)-x\left(x+5\right)=0\\ \Leftrightarrow\left(2-x\right)\left(x+5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=-5\end{matrix}\right.\\ h,\Leftrightarrow x^2+2x+3x+6=0\\ \Leftrightarrow\left(x+3\right)\left(x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-2\end{matrix}\right.\\ i,\Leftrightarrow4x^2-12x+9-4x^2+4=49\\ \Leftrightarrow-12x=36\Leftrightarrow x=-3\)
\(j,\Leftrightarrow x^2\left(x+1\right)+\left(x+1\right)=0\Leftrightarrow\left(x^2+1\right)\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x^2=-1\left(vô.lí\right)\\x=-1\end{matrix}\right.\Leftrightarrow x=-1\\ k,\Leftrightarrow x^2\left(x-1\right)=4\left(x-1\right)^2\\ \Leftrightarrow x^2\left(x-1\right)-4\left(x-1\right)^2=0\\ \Leftrightarrow\left(x-1\right)\left(x^2-4x+4\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x-2\right)^2=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
XO
21 tháng 12 2021
\(S=\dfrac{3x^2-8x+6}{x^2-2x+1}=\dfrac{2x^2-4x+2+x^2-4x+4}{x^2-2x+1}\)
\(=\dfrac{2\left(x-1\right)^2+\left(x-2\right)^2}{\left(x-1\right)^2}=2+\dfrac{\left(x-2\right)^2}{\left(x-1\right)^2}\ge2\)
=> MIN S = 2
Dấu "=" xảy ra <=> x - 2 = 0
<=> x = 2
Vậy Min S = 2 khi x = 2
NL
2
H9
HT.Phong (9A5)
CTVHS
20 tháng 8 2023
a) \(\left(x+2\right)^3-x^2\left(x+6\right)=0\)
\(\Leftrightarrow x^3+6x^2+12x+8-x^3-6x^2=0\)
\(\Leftrightarrow12x+8=0\)
\(\Leftrightarrow12x=-8\)
\(\Leftrightarrow x=-\dfrac{8}{12}\)
\(\Leftrightarrow x=-\dfrac{2}{3}\)
b) \(\left(2x+3\right)^3-8x\left(x+1\right)\left(x-1\right)=9x\left(4x-3\right)\)
\(\Leftrightarrow8x^3+36x^2+54x+27-8x\left(x^2-1\right)=36x^2-27x\)
\(\Leftrightarrow8x^3+36x^2+54x+27-8x^3+8x=36x^2-27x\)
\(\Leftrightarrow8x^3-8x^3+36x^2-36x^2+54x+27x+8x+27=0\)
\(\Leftrightarrow89x+27=0\)
\(\Leftrightarrow x=-\dfrac{27}{89}\)
c) \(\left(2-x\right)^3+\left(2+x\right)^3-12x\left(x+1\right)=0\)
\(\Leftrightarrow8-12x+6x^2-x^3+8+12x+6x^2+x^3-12x^2-12x=0\)
\(\Leftrightarrow\left(x^3-x^3\right)+\left(6x^2+6x^2-12x^2\right)-\left(12x-12x\right)+12x+\left(8+8\right)=0\)
\(\Leftrightarrow12x+16=0\)
\(\Leftrightarrow x=-\dfrac{16}{12}\)
\(\Leftrightarrow x=-\dfrac{4}{3}\)
20 tháng 8 2023
`#040911`
`a)`
`(x + 2)^3 - x^2(x + 6) = 0`
`<=> x^3 + 6x^2 + 12x + 8 - x^3 - 6x^2 = 0`
`<=> (x^3 - x^3) + (6x^2 - 6x^2) + 12x = 0`
`<=> 12x = 0`
`<=> x = 0`
Vậy, `x = 0.`
`b)`
`(2x + 3)^3 - 8x(x - 1)(x + 1) = 9x(4x - 3)`
`<=> 8x^3 + 36x^2 + 54x + 27 - 8x(x^2 - 1) = 36x^2 - 27x`
`<=> 8x^3 + 36x^2 + 54x + 27 - 8x^3 + 8x - 36x^2 + 27x = 0`
`<=> (8x^3 - 8x^3) + (36x^2 - 36x^2) + (54x + 8x + 27x) + 27 = 0`
`<=> 89x + 27 = 0`
`<=> 89x = -27`
`<=> x = -27/89`
Vậy, `x = -27/89`
`c)`
`(2 - x)^3 + (2 + x)^3 - 12x(x + 1) = 0`
`<=> 8 - 12x + 6x^2 - x^3 + 8 + 12x + 6x^2 + x^3 - 12x^2 - 12x = 0`
`<=> (-x^3 + x^3) + (12x - 12x - 12x) + (6x^2 + 6x^2 - 12x^2) + (8 + 8)=0`
`<=> -12x + 16 = 0`
`<=> -12x = -16`
`<=> 12x = 16`
`<=> x=4/3`
Vậy, `x = 4/3.`