Giải pt sau
a)x\(^2\)-10x=-25
b)\(\dfrac{x+4}{2000}+\dfrac{x+8}{1996}\dfrac{ }{ }\)=\(\dfrac{x+12}{1992}+\dfrac{x+16}{1988}\)
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LL
1 tháng 5 2018
a) \(\dfrac{\left(x+1\right)^2}{x^2-1}-\dfrac{\left(x-1\right)^2}{x^2-1}=\dfrac{16}{x^2-1}\)
=>\(\left(x+1\right)^2-\left(x-1\right)^2=16\)
=>\(x^2+2x+1-x^2+2x-1=16\)
=>4x=16=>x=4
b)\(\dfrac{12}{x^2-4}-\dfrac{x+1}{x-2}+\dfrac{x+7}{x+2}=0\)
=>\(\dfrac{12}{x^2-4}-\dfrac{\left(x+1\right)\left(x+2\right)}{x^2-4}+\dfrac{\left(x+7\right)\left(x-2\right)}{x^2-4}=0\)
=>\(12-\left(x+1\right)\left(x+2\right)+\left(x+7\right)\left(x-2\right)=0\)
=>\(12-x^2-3x-2+x^2+5x-14=0\)
=>2x-4=0=>2x=4=>x=2
c)\(\dfrac{12}{8+x^3}=1+\dfrac{1}{x+2}\)
=>\(\dfrac{12}{8+x^3}=\dfrac{x^3+8}{x^3+8}+\dfrac{x^2-2x+4}{x^3+8}\)
=>\(12=x^3+8+x^2-2x+4\)
=>\(x^3+x^2-2x=0\)
=>\(x^3-x+x^2-x=0\)
TL
3
TC
11 tháng 3 2023
`(x-2003)/16 +(x-1997)/11 +(x-1992)/9 +(x-1991)/7=10`
`<=>((x-2003)/16-1)+((x-1997)/11-2)+((x-1992)/9-3)+((x-1991)/7-4)=0`
`<=>(x-2019)/16+ (x-2019)/11 +(x-2019)/9+(x-2019)/7 =0`
`<=> (x-2019)(1/16+1/11+1/9+1/7)=0`
<=> x-2019=0`
`<=> x=2019`
4 tháng 1 2023
a: \(\Rightarrow\left(\dfrac{x+1}{35}+1\right)+\left(\dfrac{x+3}{33}+1\right)=\left(\dfrac{x+5}{31}+1\right)+\left(\dfrac{x+7}{29}+1\right)\)
=>x+36=0
=>x=-36
b: \(\Leftrightarrow\left(\dfrac{x-10}{1994}-1\right)+\left(\dfrac{x-8}{1996}-1\right)+\left(\dfrac{x-6}{1998}-1\right)+\left(\dfrac{x-4}{2000}-1\right)+\left(\dfrac{x-2}{2002}-1\right)=\left(\dfrac{x-2002}{2}-1\right)+\left(\dfrac{x-2000}{4}-1\right)+\left(\dfrac{x-1998}{6}-1\right)+\left(\dfrac{x-1996}{8}-1\right)+\left(\dfrac{x-1994}{10}-1\right)\)
=>x-2004=0
=>x=2004
G
3
D
8 tháng 2 2023
`[x+35]/1984-[x+30]/1989+[x+19]/2000+[x+23]/[1996=-2`
`<=>[x+35]/1984+1-[x+30]/1989-1+[x+19]/2000+1+[x+23]/1996+1=0`
`<=>[x+2019]/1984-[x+2019]/1989+[x+2019]/2000+[x+2019]/1996=0`
`<=>(x+2019)(1/1984-1/1989+1/2000+1/1996)=0`
`=>x+2019=0`
`<=>x=-2019`
8 tháng 2 2023
\(\dfrac{x+35}{1984}-\dfrac{x+30}{1989}+\dfrac{x+19}{2000}+\dfrac{x+23}{1996}\text{=}-2\)
\(\Leftrightarrow\dfrac{x+35}{1984}-\dfrac{x+30}{1989}+\dfrac{x+19}{2000}+\dfrac{x+23}{1996}+3-1\text{=}0\)
\(\Leftrightarrow\left(\dfrac{x+35}{1984}+1\right)-\left(\dfrac{x+30}{1989}+1\right)+\left(\dfrac{x+19}{2000}+1\right)+\left(\dfrac{x+23}{1996}+1\right)\text{=}0\)
\(\Leftrightarrow\dfrac{x+2019}{1984}-\dfrac{x+2019}{1989}+\dfrac{x+2019}{2000}+\dfrac{x+2019}{1996}\text{=}0\)
\(\Leftrightarrow\left(x+2019\right)\left(\dfrac{1}{1984}-\dfrac{1}{1989}+\dfrac{1}{2000}+\dfrac{1}{1996}\right)\text{=}0\)
\(\Leftrightarrow\left(x+2019\right)\text{=}0\)
\(\Leftrightarrow x\text{=}-2019\)
13 tháng 1 2018
bài 1:
\(\dfrac{x-10}{1994}+\dfrac{x-8}{1996}+\dfrac{x-6}{1998}=\dfrac{x-2002}{2}+\dfrac{x-2000}{4}+\dfrac{x-1998}{6}\)
<=>\(\left(\dfrac{x-10}{1994}-1\right)+\left(\dfrac{x-8}{1996}+-1\right)+\left(\dfrac{x-6}{1998}-1\right)=\left(\dfrac{x-2002}{2}-1\right)+\left(\dfrac{x-2000}{4}-1\right)+\left(\dfrac{x-1998}{6}-1\right)\)
<=>\(\dfrac{x-2004}{1994}+\dfrac{x-2004}{1996}+\dfrac{x-2004}{1998}=\dfrac{x-2004}{2}+\dfrac{x-2004}{4}+\dfrac{x-2004}{6}\)
<=>\(\dfrac{x-2004}{1994}+\dfrac{x-2004}{1996}+\dfrac{x-2004}{1998}-\dfrac{x-2004}{2}-\dfrac{x-2004}{4}-\dfrac{x-2004}{6}=0\)
<=>(x-2004)\(\left(\dfrac{1}{1994}+\dfrac{1}{1996}+\dfrac{1}{1998}-\dfrac{1}{2}-\dfrac{1}{4}-\dfrac{1}{6}\right)\)
vì 1/1994+1/1996+1/1998-1/2-1/4-1/6 khác 0
nên x-2004=0=>x=2004
vyaj.......
bài 2:
\(\dfrac{x-85}{15}+\dfrac{x-74}{13}+\dfrac{x-67}{11}+\dfrac{x-64}{9}=10\)
<=>\(\left(\dfrac{x-85}{15}-1\right)+\left(\dfrac{x-74}{13}-2\right)+\left(\dfrac{x-67}{11}-3\right)+\left(\dfrac{x-64}{9}-4\right)=0\)
<=>\(\dfrac{x-100}{15}+\dfrac{x-100}{13}+\dfrac{x-100}{11}+\dfrac{x-100}{9}=0\)
<=>\(\left(x-100\right)\left(\dfrac{1}{15}+\dfrac{1}{13}+\dfrac{1}{11}+\dfrac{1}{9}\right)=0\)
vì 1/15+1/13+1/11+1/9 khác 0
=>x-100=0<=>x=100
22 tháng 4 2021
a) ĐKXĐ: \(x\notin\left\{2;-2\right\}\)
Ta có: \(\dfrac{x+1}{x-2}-\dfrac{5}{x+2}=\dfrac{12}{x^2-4}+1\)
\(\Leftrightarrow\dfrac{\left(x+1\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\dfrac{5\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}=\dfrac{12}{\left(x-2\right)\left(x+2\right)}+\dfrac{x^2-4}{\left(x-2\right)\left(x+2\right)}\)
Suy ra: \(x^2+3x+2-5x+10=12+x^2-4\)
\(\Leftrightarrow x^2-2x+12-8-x^2=0\)
\(\Leftrightarrow-2x+4=0\)
\(\Leftrightarrow-2x=-4\)
hay x=2(loại)
Vậy: \(S=\varnothing\)
22 tháng 4 2021
b) Ta có: \(\left|2x+6\right|-x=3\)
\(\Leftrightarrow\left|2x+6\right|=x+3\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+6=x+3\left(x\ge-3\right)\\-2x-6=x+3\left(x< -3\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-x=3-6\\-2x-x=3+6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\left(nhận\right)\\x=-3\left(loại\right)\end{matrix}\right.\)
Vậy: S={-3}
10 tháng 11 2021
\(a,ĐK:...\\ PT\Leftrightarrow x^2-6x=x^2-7x+10\\ \Leftrightarrow x=10\left(tm\right)\\ b,ĐK:...\\ PT\Leftrightarrow2x\left(4-x\right)-\left(2-2x\right)\left(8-x\right)=\left(8-x\right)\left(4-x\right)\\ \Leftrightarrow8x-2x^2+16+18x-2x^2=32-12x+x^2\\ \Leftrightarrow3x^2-38x+16=0\left(casio\right)\\ c,ĐK:...\\ PT\Leftrightarrow2x\left(x-4\right)-4x=0\\ \Leftrightarrow2x^2-12x=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\left(tm\right)\\x=6\left(tm\right)\end{matrix}\right.\)
TC
31 tháng 1 2023
`(x+19)/3 +(x+13)/5 = (x+7)/7 + (x+1)/9`
`<=> x/3 + 19/3 +x/5 +13/5 = x/7 +1 +x/9 +1/9`
`<=> x/3 +x/5 -x/7 -x/9 = 1+1/9 -19/3 -13/5`
`<=> x (1/3 +1/5 -1/7 -1/9) = -118/45`
`<=> x * 88/315 = -352/45`
`<=> x = -28`
Vậy `S={-28}`
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 \(x>0\)
a) \(x^2-10x=-25\)
\(\Leftrightarrow x^2-10x+25=0\)
\(\Leftrightarrow\left(x-5\right)^2=0\)
\(\Leftrightarrow x=5\)
b) \(\dfrac{x+4}{2000}+\dfrac{x+8}{1996}=\dfrac{x+12}{1992}+\dfrac{x+16}{1988}\)
\(\Leftrightarrow\dfrac{x+4}{2000}+1+\dfrac{x+8}{1996}+1=\dfrac{x+12}{1992}+1+\dfrac{x+16}{1988}+1\)
\(\Leftrightarrow\dfrac{x+2004}{2000}+\dfrac{x+2004}{1996}-\dfrac{x+2004}{1992}-\dfrac{x+2004}{1988}=0\)
\(\Leftrightarrow\left(x+2004\right)\left(\dfrac{1}{2000}+\dfrac{1}{1996}-\dfrac{1}{1992}-\dfrac{1}{1988}\right)=0\)
\(\Leftrightarrow x+2004=0\)(vì \(\dfrac{1}{2000}+\dfrac{1}{1996}-\dfrac{1}{1992}-\dfrac{1}{1988}\ne0\))
\(\Leftrightarrow x=-2004\)
c.ơn.mik lm đc r nha