\(\dfrac{148-x}{25}\) + \(\dfrac{169-x}{23}\) + \(\dfrac{186-x}{21}\) +\(\dfrac{199-x}{19}\) =10
Tìm x: \(\dfrac{148-x}{25}+\dfrac{169-x}{23}+\dfrac{186-x}{21}+\dfrac{199-x}{19}=10\)
148-x/25-1 + 169-x/23-2 + 186-x/21-3 + 199-x/19-4
123-x/25 + 123-x/23 + 123-x/21 + 123-x/19 =0
123-x=0 => x=123
\(\frac{148-x}{25}+\frac{169-x}{23}+\frac{186-x}{21}+\frac{199-x}{19}=10\)
\(\left(\frac{148-x}{25}-1\right)+\left(\frac{169-x}{23}-2\right)+\left(\frac{186-x}{21}-3\right)+\left(\frac{199-x}{19}-4\right)=0\)
=> \(\frac{123-x}{25}+\frac{123-x}{23}+\frac{123-x}{21}+\frac{123-x}{19}=0\)
=> \(\left(123-x\right)\left(\frac{1}{25}+\frac{1}{23}+\frac{1}{21}+\frac{1}{19}\right)=0\)
=> 123 - x = 0
=> x = 123
duongtiendung vế bên trái có thêm -1,-2,-3,-3 thì bên vế phải ,phải là 0+(-1)+(-2)+(-3)+(-4)
=-10 chứ = 0 sao đc
Tìm x biết
\(\dfrac{148-x}{25}+\dfrac{169-x}{23}+\dfrac{186-x}{21}+\dfrac{199-x}{19}\)=0
I : Giải các phương trình
a) \(\dfrac{x-2}{2000}+\dfrac{x-3}{1999}=\dfrac{x-4}{1998}+\dfrac{x-5}{1997}\)
b) \(\dfrac{148-x}{25}+\dfrac{169-x}{23}+\dfrac{186-x}{21}+\dfrac{199-x}{19}=10\)
c) \(\dfrac{2-x}{2017}-1=\dfrac{1-x}{2018}-\dfrac{x}{2019}\)
help me
\(a.\dfrac{x-2}{2000}+\dfrac{x-3}{1999}=\dfrac{x-4}{1998}+\dfrac{x-5}{1997}\\ \Leftrightarrow\dfrac{x-2}{2000}-1+\dfrac{x-3}{1999}-1=\dfrac{x-4}{1998}-1+\dfrac{x-5}{1997}-1\\ \Leftrightarrow\dfrac{x-2}{2000}-\dfrac{2000}{2000}+\dfrac{x-3}{1999}-\dfrac{1999}{1999}=\dfrac{x-4}{1998}-\dfrac{1998}{1998}+\dfrac{x-5}{1997}-\dfrac{1997}{1997}\\ \Leftrightarrow\dfrac{x-2002}{2000}+\dfrac{x-2002}{1999}=\dfrac{x-2002}{1998}+\dfrac{x-2002}{1997}\\ \Leftrightarrow\dfrac{x-2002}{2000}+\dfrac{x-2002}{1999}-\dfrac{x-2002}{1998}-\dfrac{x-2002}{1997}=0\\ \Leftrightarrow\left(x-2002\right)\left(\dfrac{1}{2000}+\dfrac{1}{1999}-\dfrac{1}{1998}-\dfrac{1}{1997}\right)=0\\ \)
\(Do:\dfrac{1}{2000}+\dfrac{1}{1999}-\dfrac{1}{1998}-\dfrac{1}{1997}\ne0\\ \Rightarrow x-2002=0\\ \Leftrightarrow x=2002\\ Vậy:S=\left\{2002\right\}\)
Mấy câu khác tương tự :v
b: \(\Leftrightarrow\left(\dfrac{148-x}{25}-1\right)+\left(\dfrac{169-x}{23}-2\right)+\left(\dfrac{186-x}{21}-3\right)+\left(\dfrac{199-x}{19}-4\right)=0\)
=>123-x=0
=>x=123
c: \(\Leftrightarrow\dfrac{x-2}{2017}+1=\dfrac{x-1}{2018}+\dfrac{x}{2019}\)
\(\Leftrightarrow\left(\dfrac{x-2}{2017}-1\right)=\left(\dfrac{x-1}{2018}-1\right)+\left(\dfrac{x}{2019}-1\right)\)
=>x-2019=0
=>x=2019
Giải phương trình:
a) \(\dfrac{15x}{4x^2+3x-4}\)\(-\)1=12(\(\dfrac{1}{x+4}\)+\(\dfrac{1}{3x-3}\))
b) \(\dfrac{148-x}{25}\)+\(\dfrac{169-x}{23}\)+\(\dfrac{186-x}{21}\)+\(\dfrac{199-x}{19}\)=0
c) ||x\(-\)2|+3|=5
c: =>|x-2|+3=-5 hoặc |x-2|+3=5
=>|x-2|=2
=>x-2=2 hoặc x-2=-2
=>x=4 hoặc x=0
148-x/29+169-x/23+186-x/21+199-x/19=10
\(\frac{148-x}{25}+\frac{169-x}{23}+\frac{186-x}{21}+\frac{199-x}{19}=10\)
Mk chi p bang 123 vi bam may tinh, con ck jai thi hk p!
148-x/25-169-x/23+186-x/4+199-x/19=10.tìm x
GIÚP MK VS Ạ!!!
\(\frac{148-x}{25}+\frac{169-x}{23}+\frac{186-x}{21}-\frac{199-x}{19}=10\)
Tìm x:
(148-x)/25+(169-x/23+(186-x)/21+(199-x)/10=10