1.Tìm x biết: |3 - x| + 3x - 1 = 0
2. Tìm x thỏa mãn \(\dfrac{x-10}{30}+\dfrac{x-14}{43}+\dfrac{x-5}{95}+\dfrac{x-148}{8}=0\)
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
NT
24 tháng 2 2018
\(\dfrac{x-10}{30}+\dfrac{x-14}{43}+\dfrac{x-5}{95}+\dfrac{x-148}{8}=0\\ \Rightarrow\left(\dfrac{x-10}{30}-3\right)+\left(\dfrac{x-14}{43}-2\right)+\left(\dfrac{x-5}{95}-1\right)+\left(\dfrac{x-148}{8}+6\right)=0\\ \Rightarrow\dfrac{x-100}{30}+\dfrac{x-100}{43}+\dfrac{x-100}{95}+\dfrac{x-100}{8}=0\\ \Rightarrow x-100=0\\ \Rightarrow x=100\)
NT
2
YN
7 tháng 1 2022
Answer:
Có:
\(\frac{x-10}{30}+\frac{x-14}{43}+\frac{x-5}{95}+\frac{x-148}{8}=0\)
\(\Rightarrow\frac{x-10}{30}+\frac{x-14}{43}+\frac{x-5}{95}+\frac{x+100}{8}-6=0\)
\(\Rightarrow\left(\frac{x-10}{30}-3\right)+\left(\frac{x-14}{43}-2\right)+\left(\frac{x-5}{95}-1\right)+\frac{x-100}{8}=0\)
\(\Rightarrow\frac{x-100}{30}+\frac{x-100}{43}+\frac{x-100}{95}+\frac{x-100}{8}=0\)
\(\Rightarrow\left(x-100\right)\left(\frac{1}{30}+\frac{1}{43}+\frac{1}{95}+\frac{1}{9}\right)=0\)
Mà \(\frac{1}{30}+\frac{1}{43}+\frac{1}{95}+\frac{1}{8}\ne0\Leftrightarrow\left(x-100\right)\left(\frac{1}{30}+\frac{1}{43}+\frac{1}{95}+\frac{1}{8}\right)=0\)
\(\Leftrightarrow x-100=0\Leftrightarrow x=100\)
23 tháng 2 2018
a, => (x-10/30 - 3) + (x-14/43 - 2) + (x-5/95 - 1) + x-100/8 = 0 ( vì x-148/8 = x-100/8 + 48/8 = x-100/8 + 6 )
=> x-100/30 + x-100/43 + x-100/95 + x-100/8 = 0
=> (x-100).(1/30 + 1/43 + 1/95 + 1/8) = 0
=> x-100 = 0 ( vì 1/30+1/43+1/95+1/8 > 0 )
=> x = 100
Vậy x = 100
Tk mk nha
14 tháng 7 2016
ta co( x-10 / 30 ) -3 + ( x -14 / 43 ) -2 + ( x-5 / 95 ) -1 + ( x- 148 / 8) + 6 =0
=> (x- 10 -90 / 30 ) + ( x- 14-86 / 43 ) + ( x- 5- 95 / 95) + ( x- 148 +48 ) =0
=> x-100 /30 + x-100 / 43 + x- 100 / 95 + x- 100 / 8 = 0
=> x-100 .( 1/30 + 1/43 +1/ 95 +1/ 8) =0
ma 1/30 + 1/43 + 1/ 95 + 1/8 khac 0 => x-100 =0 => x=100
O
18 tháng 4 2022
a) \(\left(x-\dfrac{1}{2}\right)\left(-3-\dfrac{x}{2}\right)=0\)
Th1 : \(x-\dfrac{1}{2}=0\)
\(x=0+\dfrac{1}{2}\)
\(x=\dfrac{1}{2}\)
Th2 : \(-3-\dfrac{x}{2}=0\)
\(\dfrac{x}{2}=-3\)
\(x=\left(-3\right)\cdot2\)
\(x=-6\)
Vậy \(x\) = \(\left(\dfrac{1}{2};-6\right)\)
b) \(x-\dfrac{1}{8}=\dfrac{5}{8}\)
\(x=\dfrac{5}{8}+\dfrac{1}{8}\)
\(x=\dfrac{3}{4}\)
c) \(-\dfrac{1}{2}-\left(\dfrac{3}{2}+x\right)=-2\)
\(\dfrac{3}{2}+x=-\dfrac{1}{2}-\left(-2\right)\)
\(\dfrac{3}{2}+x=\dfrac{3}{2}\)
\(x=\dfrac{3}{2}-\dfrac{3}{2}\)
\(x=0\)
d) \(x+\dfrac{1}{3}=\dfrac{-12}{5}\cdot\dfrac{10}{6}\)
\(x+\dfrac{1}{3}=-4\)
\(x=-4-\dfrac{1}{3}\)
\(x=-\dfrac{13}{3}\)
23 tháng 3 2021
Bài 1:
b) ĐKXĐ: \(x\ne3\)
Ta có: \(\dfrac{3-x}{20}=\dfrac{-5}{x-3}\)
\(\Leftrightarrow\dfrac{x-3}{-20}=\dfrac{-5}{x-3}\)
\(\Leftrightarrow\left(x-3\right)^2=100\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=10\\x-3=-10\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=13\left(nhận\right)\\x=-7\left(nhận\right)\end{matrix}\right.\)
Vậy: \(x\in\left\{13;-7\right\}\)
25 tháng 10 2021
\(b,P=\left[\dfrac{x\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}-1\right]:\dfrac{9-x^2+\left(x-3\right)\left(x+3\right)-\left(x-2\right)^2}{\left(x-2\right)\left(x+3\right)}\left(x\ne\pm3;x\ne2\right)\\ P=\left(\dfrac{x}{x+3}-1\right)\cdot\dfrac{\left(x-2\right)\left(x+3\right)}{9-x^2+x^2-9-\left(x-2\right)^2}\\ P=\dfrac{x-x-3}{x+3}\cdot\dfrac{\left(x-2\right)\left(x+3\right)}{-\left(x-2\right)^2}\\ P=\dfrac{-3}{-\left(x-2\right)}=\dfrac{3}{x-2}\)
Với \(x^3-4x=0\Leftrightarrow x\left(x-2\right)\left(x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\left(ktm\right)\\x=-2\end{matrix}\right.\)
Với \(x=0\Leftrightarrow P=\dfrac{3}{0-2}=-\dfrac{3}{2}\)
Với \(x=-2\Leftrightarrow P=\dfrac{3}{-2-2}=-\dfrac{3}{4}\)
Câu 2:
\(\dfrac{x-10}{30}+\dfrac{x-14}{43}+\dfrac{x-5}{95}+\dfrac{x-148}{8}=0\)
\(\Leftrightarrow\left(\dfrac{x-10}{30}-3\right)+\left(\dfrac{x-14}{43}-2\right)+\left(\dfrac{x-5}{95}-1\right)+\left(\dfrac{x-148}{8}+6\right)=0\)
=>x-100=0
hay x=100