\(\dfrac{x-7}{2010}\)+\(\dfrac{x-604}{471}\)+\(\dfrac{x-2}{403}\)=9
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19 tháng 2 2022
\(\Leftrightarrow\dfrac{x+1}{2010}+1+\dfrac{x+2}{2009}+1+...+\dfrac{x+2009}{2}+1+\dfrac{x+2010}{1}+1=0\)
=>x+2011=0
hay x=-2011
11 tháng 3 2018
đkxđ với mọi x
đặt a=x2+x+1
\(\dfrac{a}{a+1}+\dfrac{a+1}{a+2}=\dfrac{7}{6}\)
<=> \(\dfrac{6a\left(a+2\right)}{6\left(a+1\right)\left(a+2\right)}+\dfrac{6\left(a+1\right)^2}{6\left(a+1\right)\left(a+2\right)}=\dfrac{7\left(a+1\right)\left(a+2\right)}{6\left(a+1\right)\left(a+2\right)}\)
=> 6a(a+2) +6(a+1)2 =7(a+1)(a+2)
<=> 6a2+12a +6a2 +12a+6 =a2 +21a+14
<=> 12a2 -a2+24a-21a+6-14=0
<=> 11a2+3a-8=0
<=> 11a2 +11a-8a-8=0
<=> (11a2 +11a)-(8a+8)=0
<=> 11a(a+1)-8(a+1)=0
<=> (a+1)(11a-8)=0
=> a=-1 và a=\(\dfrac{8}{11}\)
thay a=x2+x+1 ta đc
x2+x+1=-1
<=> x2+x+2 =0 (vô nghiệm)
và x2+x+\(\dfrac{3}{11}\) =0(vô nghiệm )
vậy pt trên vô nghiệm
12 tháng 3 2018
c) \(8\left(x+\dfrac{1}{x}\right)^2+4\left(x^2+\dfrac{1}{x^2}\right)^2-4\left(x^2+\dfrac{1}{x^2}\right)\left(x+\dfrac{1}{x}\right)^2=\left(x+4\right)^2\left(2\right)\)ĐKXĐ : x # 0
( 2) <=> \(8\left(x+\dfrac{1}{x}\right)^2+4\left(x^2+\dfrac{1}{x^2}\right)\left[\left(x^2+\dfrac{1}{x^2}\right)-\left(x+\dfrac{1}{x}\right)^2\right]=\left(x+4\right)^2\)
\(< =>8\left(x+\dfrac{1}{x}\right)^2+4\left(x^2+\dfrac{1}{x^2}\right).\left(-2\right)=\left(x+4\right)^2\)
\(< =>8.\left[\left(x+\dfrac{1}{x}\right)^2-x^2-\dfrac{1}{x^2}\right]=\left(x+4\right)^2\)
\(< =>16=\left(x+4\right)^2\)
<=> x2 + 8x = 0
<=> x( x + 8) = 0
<=> x = 0 ( KTM ) hoặc x = - 8 ( TM )
Vậy,....
24 tháng 6 2021
`a)1/7xx2/7+1/7xx5/7+6/7`
`=1/7xx(2/7+5/7)+6/7`
`=1/7xx1+6/7`
`=1/7+6/7=1`
`b)6/11xx4/9+6/11xx7/9-6/11xx2/9`
`=6/11xx(4/9+7/9-2/9)`
`=6/11xx9/9`
`=6/11`
24 tháng 6 2021
Sorry nãy ghi thiếu.
`c)4/25xx5/8xx25/4xx24`
`=(4xx5xx25xx24)/(25xx8xx4)`
`=(4xx5xx24)/(4xx8)`
`=(5xx24)/8`
`=5xx3=15`
12 tháng 11 2021
a) \(\Leftrightarrow\dfrac{3}{2}:x=\dfrac{1}{2}\\ \Leftrightarrow x=\dfrac{3}{2}:\dfrac{1}{2}\\ \Leftrightarrow x=3\)
b) \(\Leftrightarrow x=\dfrac{7}{9}-\dfrac{2}{3}\\ \Leftrightarrow x=\dfrac{1}{9}\)
c) \(\Leftrightarrow x=\dfrac{8}{7}:\dfrac{6}{7}\\ \Leftrightarrow x=\dfrac{4}{3}\)
d) \(\Leftrightarrow x=\dfrac{9}{5}-\dfrac{3}{7}\\ \Leftrightarrow x=\dfrac{48}{35}\)
15 tháng 11 2021
a) x = 3
b) x = \(\dfrac{1}{9}\)
c) x = \(\dfrac{4}{3}\)
d)\(\dfrac{48}{35}\)
KN
2 tháng 3 2020
\(M=\left(x-1\right)\left(x+2\right)\left(x+3\right)\left(x+6\right)\)
\(=\left(x-1\right)\left(x+6\right)\left(x+2\right)\left(x+3\right)\)
\(=\left(x^2+5x-6\right)\left(x^2+5x+6\right)\)
\(=\left(x^2+5x\right)^2-36\ge-36\)
Dấu "=" xảy ra khi \(x\in\left\{0;-5\right\}\)
LT
2 tháng 3 2020
Giải PT \(\frac{x-6}{2010}+\frac{x-603}{471}+\frac{x-1}{403}=9\)
\(\Leftrightarrow\frac{x-6}{2010}+\frac{x-603}{471}+\frac{x-1}{403}-9=0\)
\(\Leftrightarrow\left(\frac{x-6}{2010}-1\right)+\left(\frac{x-603}{471}-3\right)+\left(\frac{x-1}{403}-5\right)=0\)
\(\Leftrightarrow\frac{x-2016}{2010}+\frac{x-2016}{471}+\frac{x-2016}{403}=0\)
\(\Leftrightarrow\left(x-2016\right)\left(\frac{1}{2010}+\frac{1}{471}+\frac{1}{403}\right)=0\)
Mà \(\left(\frac{1}{2010}+\frac{1}{471}+\frac{1}{403}\right)\ne0\)
\(\Leftrightarrow x-2016=0\Leftrightarrow x=2016\)
Vậy x=2016
b) \(M=\left(x-1\right)\left(x+2\right).\left(x+3\right)\left(x+6\right)\)
\(M=\left[\left(x-1\right)\left(x+6\right)\right].\left[\left(x+2\right).\left(x+3\right)\right]\)
\(M=\left(x^2+5x-6\right).\left(x^2+5x+6\right)=\left(x^2+5x\right)^2-36\)
Các bạn tự làm tiếp được rồi nhé
23 tháng 12 2018
\(a.\dfrac{3x-2}{5}+\dfrac{x-1}{9}=\dfrac{14x-3}{15}-\dfrac{2x+1}{9}\\ \Leftrightarrow\dfrac{27x-18}{45}+\dfrac{5x-5}{45}=\dfrac{42x-9}{45}-\dfrac{10x+5}{45}\\ \Rightarrow27x-18+5x-5=42x-9-10x-5\\ \Leftrightarrow32x-23=32x-14\\ \Leftrightarrow0x=9\\ \Rightarrow Phươngtrìnhvônghiệm\\ \Rightarrow S=\phi\)
\(b.\dfrac{x+3}{2}-\dfrac{2-x}{3}-1=\dfrac{x+5}{6}\\ \Leftrightarrow\dfrac{3x-9}{6}-\dfrac{4-2x}{6}-\dfrac{6}{6}=\dfrac{x+5}{6}\\ \Rightarrow3x-9-4+2x-6=x+5\\ \Leftrightarrow5x-19=x+5\\ \Leftrightarrow4x=24\\ \Rightarrow x=6\\ \Rightarrow S=\left\{6\right\}\)
4 tháng 1 2019
\(c.\dfrac{x+5}{2010}+\dfrac{x+4}{2011}+\dfrac{x+3}{2012}+\dfrac{x+2}{2013}=-4\\ \Leftrightarrow\dfrac{x+5}{2010}+1+\dfrac{x+4}{2011}+1+\dfrac{x+3}{2012}+1+\dfrac{x+2}{2013}+1=-4+4\\ \Rightarrow\dfrac{2015+x}{2010}+\dfrac{2015+x}{2011}+\dfrac{2015+x}{2012}+\dfrac{2015+x}{2013}=0\\ \Leftrightarrow\left(2015+x\right)\left(\dfrac{1}{2010}+\dfrac{1}{2011}+\dfrac{1}{2012}+\dfrac{1}{2013}\right)=0\)
Do \(\dfrac{1}{2010}+\dfrac{1}{2011}+\dfrac{1}{2012}+\dfrac{1}{2013}>0\)
nên \(2015+x=0\Rightarrow x=-2015\)
Câu d tương tự...thêm rồi chuyển vế sang :v
11 tháng 2 2022
1 x 7 x 6 x 9 1 x 7 x 3 x 2 x 3 x 3
----------------- = --------------------------- = \(\dfrac{1}{5}\)
7 x 9 x 6 x 5 7 x 3 x 3 x 3 x 2 x 5
\(\dfrac{x-7}{2010}+\dfrac{x-604}{471}+\dfrac{x-2}{403}=9\)
\(\Leftrightarrow\dfrac{x-7}{2010}-1+\dfrac{x-604}{417}-3+\dfrac{x-2}{403}-5=0\)
\(\Leftrightarrow\dfrac{x-2017}{2010}+\dfrac{x-2017}{471}+\dfrac{x-2017}{403}=0\)
\(\Leftrightarrow\left(x-2017\right)\left(\dfrac{1}{2010}+\dfrac{1}{471}+\dfrac{1}{403}\right)=0\)
Mà \(\dfrac{1}{2010}+\dfrac{1}{471}+\dfrac{1}{403}\ne0\)
\(\Rightarrow x-2017=0\Leftrightarrow x=2017\)