Giải pt:
\(x^2+\dfrac{4x^2}{(x+2)^2}=12\)
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BC
1
27 tháng 9 2021
Tham khảo:
Giải phương trình: \(\sqrt{12-\dfrac{3}{x^2}}+\sqrt{4x^2-\dfrac{3}{x^2}}=4x^2\) - Hoc24
MV
12 tháng 8 2017
2.
\(\dfrac{x+5}{2006}+\dfrac{x+4}{2007}+\dfrac{x+3}{2008}< \dfrac{x+9}{2002}+\dfrac{x+10}{2001}+\dfrac{x+11}{2000}\\ \Leftrightarrow\dfrac{x+5}{2006}+1+\dfrac{x+4}{2007}+1+\dfrac{x+3}{2008}+1< \dfrac{x+9}{2002}+1+\dfrac{x+10}{2001}+1+\dfrac{x+11}{2000}+1\\ \Leftrightarrow\dfrac{x+2011}{2006}+\dfrac{x+2011}{2007}+\dfrac{x+2011}{2008}< \dfrac{x+2011}{2002}+\dfrac{x+2011}{2001}+\dfrac{x+2011}{2000}\\ \Leftrightarrow\dfrac{x+2011}{2006}+\dfrac{x+2011}{2007}+\dfrac{x+2011}{2008}-\dfrac{x+2011}{2002}-\dfrac{x+2011}{2001}-\dfrac{x+2011}{2000}< 0\\ \Leftrightarrow\left(x+2011\right)\left(\dfrac{1}{2006}+\dfrac{1}{2007}+\dfrac{1}{2008}-\dfrac{1}{2002}-\dfrac{1}{2001}-\dfrac{1}{2000}\right)< 0\\ \Leftrightarrow\left(x+2011\right)\left(\dfrac{1}{2006}-\dfrac{1}{2002}+\dfrac{1}{2007}-\dfrac{1}{2001}+\dfrac{1}{2008}-\dfrac{1}{2000}\right)< 0\)
Vì \(\left\{{}\begin{matrix}\dfrac{1}{2006}< \dfrac{1}{2002}\\\dfrac{1}{2007}< \dfrac{1}{2001}\\\dfrac{1}{2008}< \dfrac{1}{2000}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{2006}-\dfrac{1}{2002}< 0\\\dfrac{1}{2007}-\dfrac{1}{2001}< 0\\\dfrac{1}{2008}-\dfrac{1}{2000}< 0\end{matrix}\right.\Rightarrow\left(\dfrac{1}{2006}-\dfrac{1}{2002}+\dfrac{1}{2007}-\dfrac{1}{2001}+\dfrac{1}{2008}-\dfrac{1}{2000}\right)< 0\)
\(\Rightarrow x>0\)
Vậy \(x>0\)
KN
24 tháng 2 2018
\(\dfrac{x^2+2x+2}{x+1}+\dfrac{x^2+8x+20}{x+4}=\dfrac{x^2+4x+6}{x+2}+\dfrac{x^2+6x+12}{x+3}\)\(\Leftrightarrow\)\(\dfrac{x^2+2x+1+1}{x+1}+\dfrac{x^2+8x+16+4}{x+4}=\dfrac{x^2+4x+4+2}{x+2}+\dfrac{x^2+6x+9+3}{x+3}\)
\(\Leftrightarrow\) \(\dfrac{\left(x+1\right)^2+1}{x+1}+\dfrac{\left(x+4\right)^2+4}{x+4}=\dfrac{\left(x+2\right)^2+2}{x+2}+\dfrac{\left(x+3\right)^2+3}{x+3}\)
\(\Leftrightarrow\) \(x+1+\dfrac{1}{x+1}+x+4+\dfrac{4}{x+4}=x+2+\dfrac{2}{x+2}+x+3+\dfrac{3}{x+3}\)
\(\Leftrightarrow\) \(\dfrac{1}{x+1}\) + \(\dfrac{4}{x+4}\) - \(\dfrac{2}{x+2}\) - \(\dfrac{3}{x+3}\) = x + 2 + x + 3 - x - 1 - x - 4
\(\Leftrightarrow\) \(\dfrac{1}{x+1}\) + \(\dfrac{4}{x+4}\) - \(\dfrac{2}{x+2}\) - \(\dfrac{3}{x+3}\) = 0
\(\Leftrightarrow\) \(\dfrac{1}{x+1}\) + \(\dfrac{4}{x+4}\) = \(\dfrac{2}{x+2}\) + \(\dfrac{3}{x+3}\)
\(\Leftrightarrow\) \(\dfrac{x+4}{\left(x+1\right)\left(x+4\right)}\) + \(\dfrac{4\left(x+1\right)}{\left(x+1\right)\left(x+4\right)}\) = \(\dfrac{2\left(x+3\right)}{\left(x+3\right)\left(x+2\right)}\) + \(\dfrac{3\left(x+2\right)}{\left(x+2\right)\left(x+3\right)}\)
\(\Leftrightarrow\) \(\dfrac{x+4+4x+4}{x^2+5x+4}\) = \(\dfrac{2x+6+3x+6}{x^2+5x+6}\)
\(\Leftrightarrow\) \(\dfrac{5x+8}{x^2+5x+4}\) = \(\dfrac{5x+12}{x^2+5x+6}\)
Đặt 5x + 8 = y; x2 + 5x + 4 = t, ta có:
\(\dfrac{y}{t}\) = \(\dfrac{y+4}{t+2}\)
\(\Leftrightarrow\) \(\dfrac{y\left(t+2\right)}{t\left(t+2\right)}\) = \(\dfrac{t\left(y+4\right)}{t\left(t+2\right)}\)
\(\Leftrightarrow\) yt + 2y = yt + 4t
\(\Leftrightarrow\) 2y = 4t
\(\Leftrightarrow\) 2(5x + 8) = 4(x2 + 5x + 4)
\(\Leftrightarrow\) 10x + 16 = 4x2 + 20x + 16
\(\Leftrightarrow\) 16 - 16 = 4x2 + 20x - 10x
\(\Leftrightarrow\) 0 = 4x2 + 10x
\(\Leftrightarrow\) 2x(2x + 5) = 0
\(\Leftrightarrow\)\(\left\{{}\begin{matrix}x=0\\2x+5=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\x=-\dfrac{5}{2}\end{matrix}\right.\)
CHÚC BN HOK TỐT...
N
1
30 tháng 6 2022
\(\Leftrightarrow\dfrac{x^2+2x+1-1}{x+1}+\dfrac{x^2+8x+16+4}{x+4}=\dfrac{x^2+4x+4+2}{x+2}+\dfrac{x^2+6x+9+3}{x+3}\)
\(\Leftrightarrow x+1-\dfrac{1}{x+1}+x+4+\dfrac{4}{x+4}=x+2+\dfrac{2}{x+2}+x+3+\dfrac{3}{x+3}\)
\(\Leftrightarrow2x+5-\dfrac{1}{x+1}+\dfrac{4}{x+4}=2x+5+\dfrac{2}{x+2}+\dfrac{3}{x+3}\)
=>-x-4+4x+4=2x+6+3x+6
=>3x=5x+12
=>-2x=12
hay x=-6(nhận)
10 tháng 11 2021
\(a,ĐK:...\\ PT\Leftrightarrow x^2-6x=x^2-7x+10\\ \Leftrightarrow x=10\left(tm\right)\\ b,ĐK:...\\ PT\Leftrightarrow2x\left(4-x\right)-\left(2-2x\right)\left(8-x\right)=\left(8-x\right)\left(4-x\right)\\ \Leftrightarrow8x-2x^2+16+18x-2x^2=32-12x+x^2\\ \Leftrightarrow3x^2-38x+16=0\left(casio\right)\\ c,ĐK:...\\ PT\Leftrightarrow2x\left(x-4\right)-4x=0\\ \Leftrightarrow2x^2-12x=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\left(tm\right)\\x=6\left(tm\right)\end{matrix}\right.\)
10 tháng 5 2023
g: =>12x+1>=36x+12-24x-3
=>12x+1>=12x+9(loại)
h: =>6(x-1)+4(2-x)<=3(3x-3)
=>6x-6+8-4x<=9x-9
=>2x+2<=9x-9
=>-7x<=-11
=>x>=11/7
i: =>4x^2-12x+9>4x^2-3x
=>-12x+9>-3x
=>-9x>-9
=>x<1
15 tháng 2 2022
\(Đk:\) \(x\ne1,x\ne2,x\ne3\)
\(\Rightarrow\dfrac{x+4}{\left(x-2\right)\left(x-1\right)}+\dfrac{x+1}{\left(x-3\right)\left(x-1\right)}=\dfrac{2x+5}{\left(x-3\right)\left(x-1\right)}\)
\(\Rightarrow\dfrac{\left(x+4\right)\cdot\left(x-3\right)+\left(x+1\right)\left(x-2\right)}{\left(x-2\right)\left(x-1\right)\left(x-3\right)}=\dfrac{\left(2x+5\right)\left(x-2\right)}{\left(x-3\right)\left(x-1\right)\left(x-2\right)}\)
\(\Rightarrow x^2-3x+4x-12+x^2-2x+x-2=2x^2-4x+5x-10\)
\(\Rightarrow0x-14=x-10\)
\(\Rightarrow x=-4\left(tmđk\right)\)
Đáng lẽ không giúp đâu vì dài nhưng thôi đành cố gắng vậy :((
`x^2+(4x^2)/(x+2)^2=12(x ne -2)`
`<=>x^2-2.x.(2x)/(x+2)+(4x^2)/(x+2)^2=12-(4x^2)/(x+2)`
`<=>(x-(2x)/(x+2))^2=12-(4x^2)/(x+2)`
`<=>((x^2+2x-2x)/(x+2))^2=12-(4x^2)/(x+2)`
`<=>(x^2/(x+2))^2=12-(4x^2)/(x+2)`
`<=>(x^2/(x+2))^2+6((x^2)/(x+2))-2((x^2)/(x+2))-12=0`
`<=>((x^2)/(x+2))((x^2)/(x+2)+6)-2((x^2)/(x+2)+6)=0`
`<=>((x^2)/(x+2)+6)(x^2/(x+2)-2)=0`
`+)x^2/(x+2)+6=0`
`<=>x^2+6x+12=0`
`<=>(x+3)^2+3=0` vô lý
`+)x^2/(x+2)-2=0`
`<=>x^2-2x-4=0`
`<=>(x-1)^2-5=0`
`<=>x=+-\sqrt{5}+1`
Vậy `S={\sqrt{5}+1,-\sqrt{5}+1}`