\(\frac72\) -\(x\) = \(\frac23\) +\(\frac14\)
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NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
22 tháng 8
\(\frac{2}{x-3}\) ≤ \(\frac23\)
\(\frac{1}{x-3}\) ≤ \(\frac13\)
\(\frac{1}{x-3}-\frac13\) ≤ 0
\(\frac{3-x+3}{3\left(x-3\right)}\) ≤ 0
\(\frac{\left(3+3\right)-x}{3\left(x-3\right)}\) ≤ 0
\(\frac{6-x}{3\left(x-3\right)}\) ≤ 0
6 - \(x\) = 0 ⇒ \(x=6\); \(x-3=0\) ⇒ \(x=3\)
Lập bảng xét dấu ta có:
\(x\) | 3 6 |
-\(x+6\) | + + 0 - |
3\(x\) - 9 | - 0 + + |
3(\(\)\(x-3).\left(-x+6\right)\) | - || + 0 - |
Theo bảng trên ta có: \(x\) ≥ 6 hoặc \(x\) < 3
2N
3
LG
LMV gamer
CTVHS
23 tháng 8
\(\frac14-2x=5\)
\(2x=\frac14-5\)
\(2x=\frac{-19}{4}\)
\(x=-\frac{19}{4}:2\)
\(x=\frac{-19}{8}\)
VH
1
6 tháng 7 2022
\(P=\dfrac{\dfrac{8}{12}-\dfrac{3}{12}+\dfrac{5}{11}}{\dfrac{5}{12}+\dfrac{12}{12}-\dfrac{7}{11}}=\dfrac{\dfrac{5}{12}+\dfrac{5}{11}}{\dfrac{17}{12}-\dfrac{7}{11}}=\dfrac{115}{132}:\dfrac{103}{132}=\dfrac{115}{103}\)
23 tháng 7 2018
a) \(x+2x+3x+...+100x=-213\)
\(\Rightarrow x.\left(1+2+3+...+100\right)=-213\)
\(\Rightarrow x.5050=-213\Rightarrow x=\frac{-213}{5050}\)
b) \(\frac{1}{2}x-\frac{1}{3}=\frac{1}{4}-4\frac{1}{6}\)
\(\Rightarrow\frac{1}{2}x-\frac{1}{3}=\frac{1}{4}-\frac{25}{6}\)
\(\Rightarrow\frac{1}{2}x-\frac{1}{3}=\frac{-47}{12}\)
\(\Rightarrow\frac{1}{2}x=\frac{-43}{12}\Rightarrow x=\frac{-43}{6}\)
d) \(\frac{x+1}{3}=\frac{x-2}{4}\Rightarrow4\left(x+1\right)=3\left(x-2\right)\Rightarrow4x+4=3x-6\)
\(\Rightarrow4x-3x=-6-4\Rightarrow x=-10\)
c) \(3\left(x-2\right)+2\left(x-1\right)=10\)
\(\Rightarrow3x-6+2x-2=10\)
\(\Rightarrow5x=18\Rightarrow x=\frac{18}{5}\)
23 tháng 7 2018
a) \(x+2x+3x+4x+...+100x=-213\)
\(x.\left(1+2+3+4+...+100\right)=-213\)
\(x.5050=-213\)
\(x=-\frac{213}{5050}\)
b) \(\frac{1}{2}x-\frac{1}{3}=\frac{1}{4}-4\frac{1}{6}\)
\(\frac{1}{2}x-\frac{1}{3}=-\frac{47}{12}\)
\(\frac{1}{2}x=-\frac{43}{12}\)
\(x=\frac{-43}{6}\)
20 tháng 5 2018
a)P(x) = x^5 + 7x^4 - 9x^3 - 2x^2 - 1/4x
Q(x) = x^5 + 5x ^ 4 - 2x ^ 3 + 4x^2 - 1/4
b) P(x)+Q(x)
= (x^5 – 2x^2 + 7x^4 – 9x^3 – ¼ x ) + (5x^4 – x^5 + 4x^2 – 2x^3 – 1/4)
= x^5 – 2x^2 + 7x^4 – 9x^3 – ¼ x + 5x^4 – x^5 + 4x^2 – 2x^3 – 1/4
= (x^5 - x^5 ) + ( 7x^4 + 5x^4) + (-2x^3-9x^3) + ( -2x^2 +4x^2) + 1/4x+1/4
= 0 + 12x^4 + -11x^3 + 2x^2 + 1/4x + 1/4
= 12x^4 - 11x^3 + 2x^2 + 1/4x + 1/4
P(x) – Q(x)
= (x^5 – 2x^2 + 7x^4 – 9x^3 – ¼ x ) - (5x^4 – x^5 + 4x^2 – 2x^3 – 1/4)
= x^5 – 2x^2 + 7x^4 – 9x^3 – ¼ x - 5x^4 + x^5 - 4x^2 + 2x^3 + 1/4
=(x^5 + x^5 ) + ( 7x^4 - 5x^4) + (2x^3 - 9x^3) + ( -2x^2 - 4x^2) + 1/4x+1/4
= 2x^5 + 2x^4 + -7x^3 + -6x^2 + 1/4x + 1/4
=2x^5 + 2x^4 - 7x^3 - 6x^2 + 1/4x + 1/4
29 tháng 10 2018
\(Q=\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{4}\right)...\left(1-\dfrac{1}{2018}\right)\)
\(Q=\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}...\dfrac{2017}{2018}\)
\(Q=\dfrac{1.2.3...2017}{2.3.4...2018}\)
\(Q=\dfrac{1}{2018}\)
Vậy \(Q=\dfrac{1}{2018}\)
7/2 -x =11/12
x=7/2 - 11/12
x= 31/12
`7/2 -x =2/3 +1/4`
`7/2-x=8/12+3/12`
`7/2-x=11/12`
`x=7/2-11/12`
`x=42/12-11/12`
`x=31/12`
Vậy `x=31/12`