Tìm x:
|x-4|+|7-4|=1
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
HN
1
DN
9 tháng 8 2015
a,9/8-(2x+1)=7/4
<=>2x+1=-5/8
<=>2x=-13/8
<=>x=-13/16
b,3/4+1/4.x=7/4
<=>1/4.x=1
<=>x=4
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
19 tháng 8 2023
bài2 \(x\times\dfrac{15}{16}-x\times\dfrac{4}{16}=2\)
\(x\times\dfrac{11}{16}=2\)
\(x=2:\dfrac{11}{16}\)
\(x=\dfrac{32}{11}\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
19 tháng 8 2023
Bài 1 :
\(\dfrac{x}{16}\times\left(2017-1\right)=2\)
\(\dfrac{x}{16}\times2016=2\)
\(\dfrac{x}{16}=\dfrac{2}{2016}\)
\(x=\dfrac{2}{2016}\times16\)
\(x=\dfrac{1}{63}\)
VL
1
1 tháng 2 2017
(x-1/2).5/3=7/4-1/2
(x-1/2).5/3=5/4
x-1/2=5/4:5/3
x=3/4+1/2
x=5/4
(x-4/3).7/4=5-7/6
(x+4/3).7/4=23/6
x+4/3=23/6:7/4
x+4/3=46/21
x=46/21-4/3
x=6/7
NT
1
N
28 tháng 2 2022
a)
\(\Rightarrow\dfrac{47}{10}.x=\dfrac{1}{5}+\dfrac{2}{3}\)
\(\Rightarrow\dfrac{47}{10}.x=\dfrac{3}{15}+\dfrac{10}{15}\)
\(\Rightarrow\dfrac{47}{10}.x=\dfrac{13}{15}\)
\(\Rightarrow x=\dfrac{13}{15}:\dfrac{47}{10}\)
\(\Rightarrow x=\dfrac{13}{10}.\dfrac{10}{47}\)
\(\Rightarrow x=\dfrac{13}{47}\)
b)
\(\Rightarrow\dfrac{5}{7}:x=\dfrac{1}{6}-\dfrac{4}{5}\)
\(\Rightarrow\dfrac{5}{7}:x=\dfrac{5}{30}-\dfrac{24}{30}\)
\(\Rightarrow\dfrac{5}{7}:x=-\dfrac{19}{30}\)
\(\Rightarrow x=\dfrac{5}{7}:\left(-\dfrac{19}{30}\right)\)
\(\Rightarrow x=\dfrac{5}{7}.\left(-\dfrac{30}{19}\right)\)
\(\Rightarrow x=-\dfrac{150}{133}\)
17 tháng 11 2021
\(1,\Leftrightarrow x\left(x-9\right)=0\Leftrightarrow\left[{}\begin{matrix}x=9\\x=0\end{matrix}\right.\\ 2,\Leftrightarrow x^2-4x-x^2=7\Leftrightarrow-4x=7\Leftrightarrow x=-\dfrac{7}{4}\\ 3,\Leftrightarrow3x+2x-10=5\Leftrightarrow5x=15\Leftrightarrow x=3\\ 4,\Leftrightarrow\left(5x-1\right)\left(5x+1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{5}\\x=-\dfrac{1}{5}\end{matrix}\right.\\ 5,\Leftrightarrow\left(x-2\right)\left(3x-5\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{5}{3}\end{matrix}\right.\\ 6,\Leftrightarrow\left(x-7\right)\left(3x+4\right)=0\Leftrightarrow\left[{}\begin{matrix}x=7\\x=-\dfrac{4}{3}\end{matrix}\right.\)
\(7,\Leftrightarrow\left(2x-3\right)\left(2x+3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\\ 8,\Leftrightarrow\left(x-4\right)\left(10x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{5}\\x=4\end{matrix}\right.\\ 9,\Leftrightarrow2x^2-5x-2x^2=0\Leftrightarrow x=0\\ 10,\Leftrightarrow2x\left(x-2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\\ 11,\Leftrightarrow\left(4x-3\right)\left(3-2x\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=\dfrac{3}{2}\end{matrix}\right.\\ 12,\Leftrightarrow2x^2-10x-2x^2=3\Leftrightarrow-10x=3\Leftrightarrow x=-\dfrac{3}{10}\)
17 tháng 11 2021
\(1,\Leftrightarrow x\left(x-9\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=9\end{matrix}\right.\\ 2,\Leftrightarrow x^2-4x-x^2=7\\ \Leftrightarrow-4x=7\\ \Leftrightarrow x=\dfrac{-7}{4}\\ 3,\Leftrightarrow3x+2x-10=5\\ \Leftrightarrow5x=15\\ \Leftrightarrow x=3\\ 4,\Leftrightarrow\left(5x-1\right)\left(5x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{5}\\x=-\dfrac{1}{5}\end{matrix}\right.\)
\(5,\Leftrightarrow\left(x-2\right)\left(3x-5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{5}{3}\end{matrix}\right.\\ 6,\Leftrightarrow\left(3x+4\right)\left(x-7\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{4}{3}\\x=7\end{matrix}\right.\\ 7,\Leftrightarrow\left(2x-3\right)\left(2x+3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\)
\(8,\Leftrightarrow10x\left(x-4\right)+2\left(x-4\right)=0\\ \Leftrightarrow\left(x-4\right)\left(10x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=4\\x=-\dfrac{1}{5}\end{matrix}\right.\\ 9,\Leftrightarrow2x^2-5x-2x^2=0\\ \Leftrightarrow-5x=0\\ \Leftrightarrow x=0\\ 10,\Leftrightarrow2x\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)
\(11,\Leftrightarrow\left(2x-3\right)\left(4x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=\dfrac{3}{4}\end{matrix}\right.\\ 12,\Leftrightarrow2x^2-10x-2x^2=3\\ \Leftrightarrow-10x=3\\ \Leftrightarrow x=-\dfrac{3}{10}\)
30 tháng 8 2023
a: x*3/4=1/5
=>x=1/5:3/4=1/5*4/3=4/15
b: x*3/7=2/5
=>x=2/5:3/7=2/5*7/3=14/15
c: 1/3+2/9=2/12x
=>1/6x=3/9+2/9=5/9
=>x=5/9*6=30/9=10/3
d: 4/15*x-2/3=1/5
=>4/15*x=2/3+1/5=10/15+3/15=13/15
=>4x=13
=>x=13/4
e: x:1/7=2/3
=>x=2/3*1/7=2/21
f: 1/9:x=7/3
=>x=1/9:7/3=1/9*3/7=3/63=1/21
j: 1/4+5/12=8/3:x
=>8/3:x=3/12+5/12=8/12=2/3
=>x=4
h: =>7/4:x=1/5+1/2=7/10
=>x=7/4:7/10=10/4=5/2
HP
30 tháng 8 2023
thankkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkk
20 tháng 7 2023
Đk: `1 <=x <=7`.
Đặt `sqrt(7-x) = a, sqrt(x-1) = b`.
PT trở thành: `x + 4a = 4b + ab + 1`.
`<=> b^2 + 1 + 4a = 4b + ab + 1`.
`<=> b^2 - 4b = ab - 4a`
`<=> b(b-4) = a(b-4)`.
`<=> (b-a)(b-4) = 0`
`@ b = a <=> 7 -x = x - 1 <=> x = 4`.
`@ b = 4 <=> sqrt (x - 1) = 4 <=> x = 17 (ktm)`.
Vậy `x = 4`.
T
4
19 tháng 8 2017
a) 4/7 x X = 1/5 + 2/3
=> 4/7 x X = 3/15 + 10/15
=> 4/7 x X = 13/15
=> X = 13/15 : 4/7
=> X = 13/15 x 4/7 = 52/105
19 tháng 8 2017
a ) 4 / 7 x X = 1 / 5 + 2 / 3
= > 4 / 7 x X = 3 / 1 5 + 1 0 / 1 5
= > 4 / 7 x X = 1 3 / 1 5
= > X = 1 3 / 1 5 : 4 / 7
= > X = 1 3 / 1 5 x 4 / 7
= 5 2 / 1 0 5
\(\left|x-4\right|+\left|7-4\right|=1\)
\(\Leftrightarrow\left|x-4\right|+3=1\)
\(\Leftrightarrow\left|x-4\right|=-2\)
\(\Leftrightarrow\orbr{\begin{cases}x-4=2\\x-4=-2\end{cases}\Leftrightarrow\orbr{\begin{cases}x=6\\x=2\end{cases}}}\)
|x-4|+|7-4|=1
<=> |x-4|+3=1
<=>|x-4|=-2
Vì |x-4| \(\ge0\)
Mà -2<0
=> ko có giá trị x thỏa mãn
vậy........