Tìm max : a) \(P=\dfrac{2}{x^2+6x+12}\)
b) \(Q=3+\dfrac{4}{x^2+2x+2}\)
K
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23 tháng 7 2022
b: Đặt \(x^2-6x-2=a\)
Theo đề, ta có: \(a+\dfrac{14}{a+9}=0\)
=>(a+2)(a+7)=0
\(\Leftrightarrow\left(x^2-6x\right)\left(x^2-6x+5\right)=0\)
=>x(x-6)(x-1)(x-5)=0
hay \(x\in\left\{0;1;6;5\right\}\)
c: \(\Leftrightarrow\dfrac{-8x^2}{3\left(2x-1\right)\left(2x+1\right)}=\dfrac{2x}{3\left(2x-1\right)}-\dfrac{8x+1}{4\left(2x+1\right)}\)
\(\Leftrightarrow-32x^2=8x\left(2x+1\right)-3\left(8x+1\right)\left(2x-1\right)\)
\(\Leftrightarrow-32x^2=16x^2+8x-3\left(16x^2-8x+2x-1\right)\)
\(\Leftrightarrow-48x^2=8x-48x^2+18x+3\)
=>26x=-3
hay x=-3/26
KH
10 tháng 12 2018
1.
a) \(x\left(x+4\right)+x+4=0\)
\(\Leftrightarrow\left(x+1\right)\left(x+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+4=0\\x+1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-4\\x=-1\end{matrix}\right.\)
b) \(x\left(x-3\right)+2x-6=0\)
\(\Leftrightarrow\left(x+2\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=0\\x-3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-2\\x=3\end{matrix}\right.\)
10 tháng 12 2018
Bài 1:
a, \(x\left(x+4\right)+x+4=0\)
\(\Leftrightarrow x\left(x+4\right)+\left(x+4\right)=0\)
\(\Leftrightarrow\left(x+4\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+4=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-4\\x=-1\end{matrix}\right.\)
Vậy \(x=-4\) hoặc \(x=-1\)
b, \(x\left(x-3\right)+2x-6=0\)
\(\Leftrightarrow x\left(x-3\right)+2\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)
Vậy \(x=3\) hoặc \(x=-2\)
T
11 tháng 9 2023
\(a,-\dfrac{x}{2}+\dfrac{2x}{3}+\dfrac{x+1}{4}+\dfrac{2x+1}{6}=\dfrac{8}{3}\)
\(\Rightarrow-\dfrac{6x}{12}+\dfrac{8x}{12}+\dfrac{3\left(x+1\right)}{12}+\dfrac{2\left(2x+1\right)}{12}=\dfrac{8}{3}\)
\(\Rightarrow\dfrac{-6x+8x+3x+3+4x+2}{12}=\dfrac{8}{3}\)
\(\Rightarrow\dfrac{9x+5}{12}=\dfrac{8}{3}\)
\(\Rightarrow27x+15=96\)
\(\Rightarrow27x=81\)
\(\Rightarrow x=3\left(tm\right)\)
\(b,\dfrac{3}{2x+1}+\dfrac{10}{4x+2}-\dfrac{6}{6x+3}=\dfrac{12}{26}\)
\(\Rightarrow\dfrac{3}{2x+1}+\dfrac{10}{2\left(2x+1\right)}-\dfrac{6}{3\left(2x+1\right)}=\dfrac{6}{13}\)
\(\Rightarrow\dfrac{3}{2x+1}+\dfrac{5}{2x+1}-\dfrac{2}{2x+1}=\dfrac{6}{13}\)
\(\Rightarrow\dfrac{3+5-2}{2x+1}=\dfrac{6}{13}\)
\(\Rightarrow\dfrac{6}{2x+1}=\dfrac{6}{13}\)
\(\Rightarrow2x+1=13\)
\(\Rightarrow2x=12\)
\(\Rightarrow x=6\left(tm\right)\)
#Toru
H9
HT.Phong (9A5)
CTVHS
11 tháng 9 2023
a) \(-\dfrac{x}{2}+\dfrac{2x}{3}+\dfrac{x+1}{4}+\dfrac{2x+2}{6}=\dfrac{8}{3}\)
\(\Rightarrow\dfrac{-6x}{12}+\dfrac{8x}{12}+\dfrac{3\left(x+1\right)}{12}+\dfrac{2\left(2x+1\right)}{12}=\dfrac{4\cdot8}{12}\)
\(\Rightarrow-6x+8x+3x+3+4x+2=32\)
\(\Rightarrow9x+5=32\)
\(\Rightarrow9x=32-5\)
\(\Rightarrow9x=27\)
\(\Rightarrow x=\dfrac{27}{9}\)
\(\Rightarrow x=3\)
b) \(\dfrac{3}{2x+1}+\dfrac{10}{4x+2}-\dfrac{6}{6x+3}=\dfrac{12}{26}\) (ĐK: \(x\ne-\dfrac{1}{2}\))
\(\Rightarrow\dfrac{3}{2x+1}+\dfrac{10}{2\left(2x+1\right)}-\dfrac{6}{3\left(2x+1\right)}=\dfrac{6}{13}\)
\(\Rightarrow\dfrac{3}{2x+1}+\dfrac{5}{2x+1}-\dfrac{2}{2x+1}=\dfrac{6}{13}\)
\(\Rightarrow\dfrac{6}{2x+1}=\dfrac{6}{13}\)
\(\Rightarrow2x+1=13\)
\(\Rightarrow2x=12\)
\(\Rightarrow x=\dfrac{12}{2}\)
\(\Rightarrow x=6\left(tm\right)\)
15 tháng 4 2018
a)
\(A=\dfrac{2x^2-16x+41}{x^2-8x+22}=\dfrac{2\left(x^2-8x+22\right)-3}{x^2-8x+22}\)
\(A-2=-\dfrac{3}{x^2-8x+22}=-\dfrac{3}{\left(x-4\right)^2+6}\ge-\dfrac{3}{6}=-\dfrac{1}{2}\)
\(A\ge\dfrac{3}{2}\) khi x =4
31 tháng 3 2017
a) \(\dfrac{4x+3}{5}-\dfrac{6x-2}{7}=\dfrac{5x+4}{3}+3\)
\(\Leftrightarrow21\left(4x+3\right)-15\left(6x-2\right)=35\left(5x+4\right)+315\)
\(\Leftrightarrow84x+63-90x+30=175x+140+315\)
\(\Leftrightarrow84x-90x-175x=140+315-63-30\)
\(\Leftrightarrow-181x=362\)
\(\Leftrightarrow x=-\dfrac{362}{181}=-2\)
Vậy: Tập ngiệm của phương trình là: \(S=\left\{-2\right\}\)
b) \(\dfrac{x+4}{5}-x+4=\dfrac{x}{3}-\dfrac{x-2}{2}\)
\(\Leftrightarrow6\left(x+4\right)-30x+120=10x-15\left(x-2\right)\)
\(\Leftrightarrow6x+24-30x-120=10x-15x+30\)
\(\Leftrightarrow6x-30x-10x+15x=30-24+120\)
\(\Leftrightarrow-19x=126\)
\(\Leftrightarrow x=-\dfrac{126}{19}\)
Vậy: Tập ngiệm của phương trình là: \(S=\left\{-\dfrac{126}{19}\right\}\)
c) \(\dfrac{x+2}{3}+\dfrac{3\left(2x-1\right)}{4}-\dfrac{5x-3}{6}=x+\dfrac{5}{12}\)
\(\Leftrightarrow4\left(x+2\right)+9\left(2x-1\right)-2\left(5x-3\right)=12x+5\)
\(\Leftrightarrow4x+8+18x-9-10x+6=12x+5\)
\(\Leftrightarrow4x+18x-10x-12x=5-8+9-6\)
\(\Leftrightarrow0x=0\)
Vậy: Tập ngiệm của phương trình là: \(S=\left\{R\right\}\)
_Chúc bạn học tốt_
25 tháng 2 2021
`a,x^3-8 ne 0`
`=>x^3 ne 8`
`=>x ne 2`
`b,2x^2+5x+3 ne 0`
`=>2x^2+2x+3x+3 ne 0`
`=>2x(x+1)+3(x+1) ne 0`
`=>(x+1)(2x+3) ne 0`
`=>x ne -1,-3/2`
`c,x^2-4 ne 0`
`=>x^2 ne 4`
`=>x ne 2,-2`
25 tháng 2 2021
a) ĐK:
\(x^3-8\ne0\\ \Leftrightarrow x\ne2\)
b) ĐK:
\(2x^2+5x+3\ne0\\ \Leftrightarrow\left[{}\begin{matrix}x\ne-1\\x\ne-\dfrac{3}{2}\end{matrix}\right.\)
c) ĐK:
\(x^2-4\ne0\\ \Leftrightarrow x\ne\pm2\)
13 tháng 11 2018
a) \(x-1-\dfrac{x^2-4}{x+1}=\dfrac{\left(x-1\right)\left(x+1\right)-\left(x^2-4\right)}{x+1}=\dfrac{x^2-1-x^2+4}{x+1}=\dfrac{3}{x+1}\)
b) \(\dfrac{3x-1}{2\left(3x+1\right)}+\dfrac{3x+1}{2\left(3x-1\right)}-\dfrac{6x}{\left(3x-1\right)\left(3x+1\right)}\)
\(=\dfrac{\left(3x-1\right)^2+\left(3x+1\right)^2-12x}{2\left(3x-1\right)\left(3x+1\right)}\)
\(=\dfrac{9x^2-6x+1+9x^2+6x+1-12x}{2\left(3x-1\right)\left(3x+1\right)}\)
\(=\dfrac{18x^2-12x+2}{2\left(3x-1\right)\left(3x+1\right)}\)
\(=\dfrac{9x^2-6x+1}{\left(3x-1\right)\left(3x+1\right)}\)
\(=\dfrac{\left(3x-1\right)^2}{\left(3x-1\right)\left(3x+1\right)}=\dfrac{3x-1}{3x+1}\)
20 tháng 11 2022
c: \(=\dfrac{1}{x-2}-\dfrac{x+4}{\left(x-2\right)\left(x+2\right)}-\dfrac{2}{x\left(x+2\right)}\)
\(=\dfrac{x\left(x+2\right)-x^2-4x-2x+4}{x\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{x^2+2x-x^2-6x+4}{x\left(x-2\right)\left(x+2\right)}=\dfrac{-4x+4}{x\left(x-2\right)\left(x+2\right)}\)
d: Sửa đề: \(\dfrac{2x^2+1}{x^3+1}-\dfrac{x-1}{x^2-x+1}-\dfrac{1}{x+1}\)
\(=\dfrac{2x^2+1-x^2+1-x^2+x-1}{\left(x+1\right)\left(x^2-x+1\right)}\)
\(=\dfrac{x+1}{\left(x+1\right)\left(x^2-x+1\right)}=\dfrac{1}{x^2-x+1}\)
29 tháng 11 2022
a: \(=\dfrac{x^3-x^2+x-1}{\left(x-2\right)\left(x+2\right)}-\dfrac{x-2}{\left(x+2\right)\left(x+1\right)}-\dfrac{3x}{\left(x-2\right)\left(x+1\right)}+\dfrac{2x+5}{\left(x+1\right)\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{\left(x-1\right)\left(x^2+1\right)\left(x+1\right)-x^2+4x-4-3x^2-6x+2x+5}{\left(x+2\right)\left(x+1\right)\left(x-2\right)}\)
\(=\dfrac{x^4-1-4x^2+1}{\left(x+2\right)\left(x-2\right)\left(x+1\right)}=\dfrac{x^2\left(x-2\right)\left(x+2\right)}{\left(x+2\right)\left(x-2\right)\left(x+1\right)}\)
=x^2/x+1
b: Sửa đề: \(\dfrac{19x^2-30x+9}{2x^3+54}-\dfrac{x-3}{2x^2+6x}-\dfrac{3x^2}{2x^2-6x+18}\) \(=\dfrac{19x^2-30x+9}{2\left(x+3\right)\left(x^2-3x+9\right)}-\dfrac{x-3}{2x\left(x+3\right)}-\dfrac{3x^2}{2\left(x^2-3x+9\right)}\)
\(=\dfrac{19x^3-30x^2+9x-\left(x-3\right)\left(x^2-3x+9\right)-3x^3\left(x+3\right)}{2x\left(x+3\right)\left(x^2-3x+9\right)}\)
\(=\dfrac{19x^3-30x^2+9x-3x^4-9x^3-\left(x^3-3x^2+9x-3x^2+9x-27\right)}{2x\left(x+3\right)\left(x^2-3x+9\right)}\)
\(=\dfrac{-3x^4+10x^3-30x^2+9x-x^3+6x^2-18x+27}{2x\left(x+3\right)\left(x^2-3x+9\right)}\)
\(=\dfrac{-3x^4+10x^3-24x^2-9x+27}{2x\left(x+3\right)\left(x^2-3x+9\right)}\)
26 tháng 11 2022
a: \(=\dfrac{5x+30}{x\left(x-3\right)\left(x+2\right)}+\dfrac{4x+12}{x\left(x+2\right)}-\dfrac{x}{x-3}\)
\(=\dfrac{5x+30+4\left(x^2-9\right)-x^2\left(x+2\right)}{x\left(x+2\right)\left(x-3\right)}\)
\(=\dfrac{5x+30+4x^2-36-x^3-2x^2}{x\left(x+2\right)\left(x-3\right)}\)
\(=\dfrac{-x^3+2x^2+5x-6}{x\left(x+2\right)\left(x-3\right)}\)
\(=\dfrac{-x^3-2x^2+4x^2+8x-3x-6}{x\left(x+2\right)\left(x-3\right)}\)
\(=\dfrac{\left(x+2\right)\left(-x^2+4x-3\right)}{x\left(x+2\right)\left(x-3\right)}=\dfrac{-\left(x-1\right)}{x}\)
b: \(=\dfrac{x}{x+1}-\dfrac{1}{\left(x-1\right)\left(x+1\right)}+\dfrac{x^5+x+1}{\left(x+1\right)\left(x^3-1\right)}\)
\(=\dfrac{x^2-x-1}{\left(x+1\right)\left(x-1\right)}+\dfrac{x^3-x^2+1}{\left(x+1\right)\left(x-1\right)}\)
\(=\dfrac{x^2-x-1+x^3-x^2+1}{\left(x+1\right)\left(x-1\right)}=\dfrac{x^3-x}{\left(x+1\right)\left(x-1\right)}=x\)
a) \(P_{max}=\left(x^2+6x+12\right)_{min}=\left[\left(x^2+2.3x+9\right)+3\right]_{min}=\left[\left(x+3\right)^2+3\right]_{min}=3\)\(P_{max}=P\left(-3\right)=\dfrac{2}{3}\)
bn làm giúp mk câu b vs