Tìm x :
\(\frac{x+3}{3}=\frac{27}{x-3}\)
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M
26 tháng 4 2016
x . \(\frac{1}{2}\)- x.\(\frac{2}{3}\) + x.\(\frac{3}{4}\)- x. \(\frac{5}{6}\) = \(\frac{5}{6}\) -\(\frac{3}{4}\) + \(\frac{2}{3}\) -\(\frac{1}{2}\)
x . \(\frac{1}{2}\)- x.\(\frac{2}{3}\) + x.\(\frac{3}{4}\)- x. \(\frac{5}{6}\) = \(\frac{10}{12}\)-\(\frac{9}{12}\)+\(\frac{8}{12}\)-\(\frac{6}{12}\)
x . \(\frac{1}{2}\)- x.\(\frac{2}{3}\) + x.\(\frac{3}{4}\)- x. \(\frac{5}{6}\)= \(\frac{1}{4}\)=> x. (\(\frac{1}{2}\)- \(\frac{2}{3}\) + \(\frac{3}{4}\)- \(\frac{5}{6}\)) = \(\frac{1}{4}\)=> x.( \(\frac{6}{12}\)- \(\frac{8}{12}\)+\(\frac{9}{12}\)-\(\frac{10}{12}\))= \(\frac{1}{4}\)=> x. \(\frac{-1}{4}\)=\(\frac{1}{4}\)=> x = \(\frac{1}{4}\): \(\frac{-1}{4}\)=> x = -126 tháng 4 2016
=>x.(1/2-2/3+3/4)=1/4
=>x.7/12=1/4
=>x=1/4:7/12
=>x=1/4.12/7
=>x=3/7
N
5
DT
8 tháng 12 2017
=> \(\left(x-3\right).\left(x-3\right)=5.\left(9.8\right)\)
=> \(\left(x-3\right)^2=49\)
=> \(\orbr{\begin{cases}\left(x-3\right)^2=7^2\\\left(x-3\right)^2=\left(-7\right)^2\end{cases}}\) => \(\orbr{\begin{cases}x-3=7\\x-3=-7\end{cases}}\)=> \(\orbr{\begin{cases}x=7+3\\x=-7+3\end{cases}}\)=> \(\orbr{\begin{cases}x=10\\x=-4\end{cases}}\)
Vậy : x \(\varepsilon\){ 10 ; -4 }
P/s : \(\orbr{\begin{cases}\\\end{cases}}\) nghĩa là hoặc
LT
8 tháng 12 2017
(x-3)2=5.9,8
(x-3)2=49
TH1: x-3=7
=>x=7+3=10
TH2:x-3=-3
=> x=-3+3=0
Vay x=10;0
21 tháng 2 2020
\(\Leftrightarrow\left(x+40\right)\left(x-50\right)=0\Rightarrow\left[{}\begin{matrix}x=-40\\x=50\end{matrix}\right.\)
Giải rõ hơn nha
21 tháng 2 2020
\(\Leftrightarrow\frac{x^2-10x-29}{1971}+1+\frac{x^2-10x-27}{1973}+1-\frac{x^2-10x-1971}{29}-1-\frac{x^2-10x-1973}{27}-1=0\)
sai dấu r
NP
2
S
6 tháng 4 2019
\(\frac{x}{3}=\frac{2}{3}+\frac{-1}{7}\)
\(\frac{x}{3}=\frac{11}{21}\)
\(x=\frac{11.3}{21}\)(Áp dụng công thức \(\frac{a}{b}=\frac{c}{d}\)khi \(a.d=b.c\))
\(\Rightarrow x=\frac{11}{7}\)
~Học tốt~
6 tháng 4 2019
\(\frac{x}{3}=\frac{2}{3}+\frac{-1}{7}\)
\(\frac{x}{3}=\frac{2.7+\left(-1\right).3}{21}\)
\(\frac{x}{3}=\frac{11}{21}\)
\(\Leftrightarrow21x=33\)
\(\Leftrightarrow x=\frac{33}{21}=\frac{11}{7}\)
IM
8 tháng 9 2016
\(\frac{x+4}{2007}+\frac{x+8}{2003}=\frac{x+1}{2010}=\frac{x+3}{2008}\)
\(\Leftrightarrow\frac{x+4}{2007}=\frac{x+1}{2010}\)
\(\Leftrightarrow\left(x+4\right)2010=\left(x+1\right)2007\)
\(\Leftrightarrow2010x+8040=2007x+2007\)
\(\Leftrightarrow2010x-2007x=2007-8040\)
\(\Leftrightarrow3x=-6033\)
\(\Leftrightarrow x=-2011\)
8 tháng 9 2016
\(\frac{x+4}{2007}+\frac{x+8}{2003}=\frac{x+1}{2010}+\frac{x+3}{2008}\)
=>\(\left(\frac{x\text{+4}}{2007}+1\right)+\left(\frac{x+8}{2003}+1\right)=\left(\frac{x+1}{2010}+1\right)+\left(\frac{x+3}{2008}+1\right)\)
=>\(\frac{x+2011}{2007}+\frac{x+2011}{2003}=\frac{x+2011}{2010}+\frac{x+2011}{2008}\)
=>\(\frac{x+2011}{2007}+\frac{x+2011}{2003}-\frac{x+2011}{2010}-\frac{x+2011}{2008}=0\)
=>\(x+2011\left(\frac{1}{2007}+\frac{1}{2003}-\frac{1}{2010}-\frac{1}{2008}\right)=0\)
Mà \(\frac{1}{2007}+\frac{1}{2003}-\frac{1}{2010}-\frac{1}{2008}\ne0\)
=> x+2011=0
=>x=-2011
Vậy x = -2011
16 tháng 11 2021
a, A=15√x−11x+2√x−3+3√x−21−√x−2√x+3√x+3A=15x−11x+2x−3+3x−21−x−2x+3x+3
=15√x−11x−√x+3√x−3−3√x−2√x−1−2√x+3√x+3=15x−11x−x+3x−3−3x−2x−1−2x+3x+3
=15√x−11√x(√x−1)+3(√x−1)−3√x−2√x−1−2√x+3√x+3=15x−11x(x−1)+3(x−1)−3x−2x−1−2x+3x+3
=15√x−11(√x−1)(√x+3)−3√x−2√x−1−2√x+3√x+3=15x−11(x−1)(x+3)−3x−2x−1−2x+3x+3
=15√x−11−(3√x−2)(√x+3)−(2√x+3)(√x−1)(√x−1)(√x+3)=15x−11−(3x−2)(x+3)−(2x+3)(x−1)(x−1)(x+3)
=15√x−11−(3x+9√x−2√x−6)−(2x−2√x+3√x−3)(√x−1)(√x+3)=15x−11−(3x+9x−2x−6)−(2x−2x+3x−3)(x−1)(x+3)
=15√x−11−3x−9√x+2√x+6−2x+2√x−3√x+3(√x−1)(√x+3)=15x−11−3x−9x+2x+6−2x+2x−3x+3(x−1)(x+3)
=7√x−5x−8(√x−1)(√x+3)
3 tháng 1 2018
\(\frac{x-4}{2021}+\frac{x-3}{2020}=\frac{x-2}{2019}+\frac{x-1}{2018}\)
\(\Leftrightarrow\left(\frac{x-4}{2021}+1\right)+\left(\frac{x-3}{2020}+1\right)=\left(\frac{x-2}{2019}+1\right)+\left(\frac{x-1}{2018}+1\right)\)
\(\Leftrightarrow\frac{x+2017}{2021}+\frac{x+2017}{2020}=\frac{x+2017}{2019}+\frac{x+2017}{2018}\)
\(\Leftrightarrow\frac{x+2017}{2021}+\frac{x+2017}{2020}-\frac{x+2017}{2019}-\frac{x+2017}{2018}=0\)
\(\Leftrightarrow\left(x+2017\right)\left(\frac{1}{2021}+\frac{1}{2020}-\frac{1}{2019}-\frac{1}{2018}\right)=0\)
Mà \(\left(\frac{1}{2021}+\frac{1}{2020}-\frac{1}{2019}-\frac{1}{2018}\right)\ne0\)
\(\Leftrightarrow x+2017=0\)
\(\Leftrightarrow x=-2017\)
Vậy ..
3 tháng 1 2018
=> (x-4/2021 +1) + (x-3/2020 +1) = (x-2/2019 +1)+ (x-1/2018 +1)
=> x+2017/2021 + x+2017/2020 = x+2017/2019 + x+2017/2018
=> x+2017/2018 + x+2017/2018 - x+2017/2020 - x+2017/2021 = 0
=> (x+2017).(1/2018+1/2019+1/2020+1/2021) = 0
=> x+2017 = 0 ( vì 1/2018+1/2019+1/2020+1/2021 > 0 )
=> x=-2017
Vậy x=-2017
k mk nha
17 tháng 12 2016
Ta có: \(\frac{1+x}{3}=\frac{3+x}{5}\)
=> 5.(1+x) = 3.(3+1)
=> 5 + 5x = 9 + 3x
=> 5x - 3x = 9 - 5
=> 2x = 4
=> x = 2
Thế x = 2 vào \(\frac{1+x}{3}=\frac{8+2x}{3y}\)
Ta được: \(\frac{1+2}{3}=\frac{8+2.2}{3.y}\)= 1 = \(\frac{12}{3y}\)
=> y = 4
Vậy x = 2; y = 4
17 tháng 12 2016
Nguyễn Huy TúTrương Hồng Hạnhsoyeon_Tiểubàng giảiHoàng Lê Bảo NgọcTrần Việt Linh
Ta có:
\(\frac{x+3}{3}=\frac{27}{x-3}\)
\(\Rightarrow\left(x+3\right)\left(x-3\right)=27.3\)
\(\Leftrightarrow x^2-9=27.3\)
\(\Leftrightarrow x^2-9=81\)
\(\Leftrightarrow x^2=81+9\)
\(\Leftrightarrow x^2=90\)
\(\Leftrightarrow x=\sqrt{90}=3\sqrt{10}\)
Cái đoạn(x+3)(x-3)=x2-9 là mình dùng hằng đẳng thức của lớp 8
x+3/3=27/x-3
=> (x+3).(x-3)=27.3
=> x.(x-3)+3.(x-3)=81
=> x2-3x+3x-9=81
=> x2-9=81
=> x2=72
=> x= căn 72