mn ơi giúp mình với tìm x
6x2-41x+48=0
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Những câu hỏi liên quan
16 tháng 5 2022
\(\Leftrightarrow6x^2-9x-32x+48=0\)
\(\Leftrightarrow3x\left(2x-3\right)-16\left(2x-3\right)=0\)
\(\Leftrightarrow\left(2x-3\right)\left(3x-16\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=\dfrac{16}{3}\end{matrix}\right.\)
2
10 tháng 1 2023
`a)2x^2+3(x-1)(x+1)=5x(x+1)`
`<=>2x^2+3x^2-3=5x^2+5x`
`<=>5x=-3`
`<=>x=-3/5`
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`b)(x-3)^3+3-x=0` nhỉ?
`<=>(x-3)^3-(x-3)=0`
`<=>(x-3)(x^2-1)=0`
`<=>[(x=3),(x^2=1<=>x=+-1):}`
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`c)5x(x-2000)-x+2000=0`
`<=>5x(x-2000)-(x-2000)=0`
`<=>(x-2000)(5x-1)=0`
`<=>[(x=2000),(x=1/5):}`
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`d)3(2x-3)+2(2-x)=-3`
`<=>6x-9+4-2x=-3`
`<=>4x=2`
`<=>x=1/2`
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`e)x+6x^2=0`
`<=>x(1+6x)=0`
`<=>[(x=0),(x=-1/6):}`
ND
Nguyễn Đức Trí
VIP
19 tháng 8 2023
\(\left(x-2\right)^5-\left(x-2\right)^3=0\)
\(\Rightarrow\left(x-2\right)^3\left(\left(x-2\right)^2-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-2=0\\\left(x-2\right)^2-1=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=2\\\left(x-2\right)^2=1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=2\\x-2=1\\x-2=-1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=2\\x=3\\x=1\end{matrix}\right.\)
Vậy \(x\in\left\{1;2;3\right\}\)
19 tháng 8 2023
⇒ ( x - 2)3 . (x - 2)2 - (x - 2)3 . 1 = 0 ⇒ ( x - 2)3 . [( x - 2)2 - 1] = 0
9 tháng 10 2021
1)
a) \(=15x^3-20x^2+10x\)
b) \(=3x^4-x^3+4x^2-9x^3+3x-12x=3x^4-10x^3+4x^2-9x\)
2)
a) \(\Rightarrow x\left(x^2-6x+12\right)=0\)
\(\Rightarrow x=0\)(do \(x^2-6x+12=\left(x^2-6x+\dfrac{36}{4}\right)+3=\left(x-\dfrac{6}{2}\right)^2+3\ge3>0\))
b) \(\Rightarrow\left(x+3\right)^3=0\Rightarrow x=-3\)
23 tháng 3 2022
(3x²-5x+2)+(3x²+5x)= bao nhiêu ạ
Giúp em vs ạ . Em cảm ơn
H9
HT.Phong (9A5)
CTVHS
10 tháng 8 2023
1) \(\left(x-3\right)^2-4=0\)
\(\Leftrightarrow\left(x-3-2\right)\left(x-3+2\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=1\end{matrix}\right.\)
2) \(x^2-2x=24\)
\(\Leftrightarrow x^2-2x-24=0\)
\(\Leftrightarrow x^2+4x-6x-24=0\)
\(\Leftrightarrow x\left(x+4\right)-6\left(x+4\right)=0\)
\(\Leftrightarrow\left(x+4\right)\left(x-6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-4\end{matrix}\right.\)
A
1
27 tháng 6 2017
\(A=\frac{3}{1-x}+\frac{4}{x}\ge\frac{\left(\sqrt{3}+2\right)^2}{1-x+x}=7+4\sqrt{3}\)
Dấu = xảy ra khi: \(x=\frac{2}{\sqrt{3}+2}\)
TT
4
10 tháng 4 2022
\(\left(2x-4\right)\left(3x+1\right)< 0\)
=> TH1: \(\begin{matrix}2x-4< 0\\3x+1>0\end{matrix}\)\(\Leftrightarrow\left\{{}\begin{matrix}2x< 4\\3x>-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x< 2\\x>-\dfrac{1}{3}\end{matrix}\right.\) (tm)
TH2: \(\begin{matrix}2x-4>0\\3x+1< 0\end{matrix}\)\(\Leftrightarrow\left\{{}\begin{matrix}2x>4\\3x< -1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>2\\x< -\dfrac{1}{3}\end{matrix}\right.\) (vô lí)
=> \(2>x>-\dfrac{1}{3}\)
\(6x^2-41x+48=0\)
\(\Leftrightarrow3x\left(2x-3\right)-16\left(2x-3\right)=0\)
\(\Leftrightarrow\left(2x-3\right)\left(3x-16\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-3=0\\3x-16=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=\dfrac{16}{3}\end{matrix}\right.\)
\(6x^2-41x+48=0\)
\(\Leftrightarrow6x^2-9x-32x+48=0\)
\(\Leftrightarrow3x\left(2x-3\right)-16\left(2x-3\right)=0\)
\(\Leftrightarrow\left(3x-16\right)\left(2x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-16=0\\2x-3=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}3x=16\Rightarrow x=\dfrac{16}{3}\\2x=3\Rightarrow x=\dfrac{3}{2}\end{matrix}\right.\)
Vậy.......................................