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HN
1
29 tháng 8 2015
A = 7/7.17 + 7/17.27 + 7/27.37 + ............ +7/1997.2007
A=7/10 ( 10/7.17 + 10/17.27 + 10/27.37 + ................+10/1997.2007)
A= 7/10 ( 1/7 -1/17 + 1/17 - 1/27 + 1/27 - 1/37 +...............+ 1/1997 - 1/2007)
A= 7/10 (1/7 - 1/2007)
A= 7/10 . 2000/14049
A=200/2007
bây h mk có vc rùi tích đúng nha tối mk lm típ cho
3 tháng 8 2016
a) \(6.8^{x-1}+8^{x+1}=6.8^{19}+8^{21}\)
\(\Rightarrow x-1+x+1=19+21\)
\(=2x=40\)
\(\Rightarrow x=20\)
b) \(4.3^{x-1}+2.3^{x+2}=4.3^6+2.3^9\)
\(\Rightarrow x-1+x+2=6+9\)
\(\Rightarrow2x+1=15\)
\(\Rightarrow2x=14\)
\(\Rightarrow x=7\)
NH
6
12 tháng 9 2017
1)
Ta có: \(\left(x-\frac{1}{2}\right)^3=8\Rightarrow\left(x-\frac{1}{2}\right)^3=2^3\)
\(\Rightarrow x-\frac{1}{2}=2\Rightarrow x=2+\frac{1}{2}=\frac{5}{2}\)
12 tháng 9 2017
\(\left(x-1\right)^3=\frac{8}{27}\Rightarrow\left(x-1\right)^3=\left(\frac{2}{3}\right)^3\)
\(\Rightarrow x-1=\frac{2}{3}\Rightarrow x=\frac{2}{3}+1=\frac{5}{3}\)
NQ
3
NC
23 tháng 8 2020
Bài làm:
Ta có:
Pt <=> \(\left(-8+x^2\right)^5=1\)
\(\Rightarrow-8+x^2=1\)
\(\Leftrightarrow x^2=9\)
\(\Rightarrow\orbr{\begin{cases}x=3\\x=-3\end{cases}}\)
Vậy \(\orbr{\begin{cases}x=3\\x=-3\end{cases}}\)
23 tháng 8 2020
dời trả lời nhanh z
định giúp bạn mink kiếm điểm ai ngờ...:))
12 tháng 7 2019
a) \(\frac{x}{x+1}=\frac{1}{2}\)
=> 2x = x + 1
=> 2x - x = 1
=> x = 1
b) \(\frac{x}{2}=\frac{x}{3}\)
=> 3x = 2x
=> 3x - 2x = 0
=> x = 0
c) \(\frac{x+1}{2}=\frac{x+1}{2017}\)
=> \(2017\left(x+1\right)=2\left(x+1\right)\)
=> 2017x + 2017 = 2x + 2
=> 2017x - 2x = 2 - 2017
=> 2015x = -2015
=> x = -2015 : 2015
=> x = -1
i) \(\frac{3}{x}=\frac{x}{2017}\)
=> x2 = 2017.3
=> x2 = 6051
=> \(\orbr{\begin{cases}x=\sqrt{6051}\\x=-\sqrt{6051}\end{cases}}\)
còn lại tự lm
12 tháng 7 2019
\(a,\frac{x}{x+1}=\frac{1}{2}\)
\(\Rightarrow x=\frac{1}{2}.\left(x+1\right)\)
\(\Rightarrow x=\frac{1}{2}x+\frac{1}{2}\)
\(\Rightarrow x-\frac{1}{2}x=\frac{1}{2}\)
\(\Rightarrow\frac{1}{2}x=\frac{1}{2}\)
\(\Rightarrow x=1\)
\(b,\frac{x}{2}=\frac{x}{3}\)
\(\Rightarrow x=\frac{x}{3}.2\)
\(\Rightarrow x=\frac{2x}{3}\)
\(\Rightarrow3x=2x\)
\(\Rightarrow x=0\)
\(c,\frac{x+1}{2}=\frac{x+1}{2017}\)
\(\Rightarrow x+1=\frac{x+1}{2017}.2\)
\(\Rightarrow x+1=\frac{2x+2}{2017}\)
\(\Rightarrow2017x+2017=2x+2\)
\(\Rightarrow2017x-2x=2-2017\)
\(\Rightarrow2015x=-2015\)
\(\Rightarrow x=-1\)
\(i,\frac{3}{x}=\frac{x}{2017}\)
\(\Rightarrow x=3:\frac{x}{2017}\)
\(\Rightarrow x=\frac{6051}{x}\)
\(\Rightarrow x^2=6051\)
\(\Rightarrow x=\sqrt{6051}\)
\(o,\frac{x}{3}=\frac{x+1}{2}\)
\(\Rightarrow x=\frac{x+1}{2}.3\)
\(\Rightarrow x=\frac{3x+3}{2}\)
\(\Rightarrow2x=3x+3\)
\(\Rightarrow-x=3\)
\(\Rightarrow x=-3\)
\(m,\frac{x+1}{2}=\frac{x+2}{3}\)
\(\Rightarrow x+1=\frac{x+2}{3}.2\)
\(\Rightarrow x+1=\frac{2x+4}{3}\)
\(\Rightarrow3x+3=2x+4\)
\(\Rightarrow x=1\)
\(p,\frac{x+1}{2}=x\)
\(\Rightarrow2x=x+1\)
\(\Rightarrow x=1\)
\(m,\frac{2}{x}=\frac{x}{8}\)
\(\Rightarrow x=2:\frac{x}{8}\)
\(\Rightarrow x=\frac{16}{x}\)
\(\Rightarrow x^2=16\)
\(\Rightarrow x=4\)
\(Q,\frac{x^2}{2}=\frac{8}{x^2}\)
\(\Rightarrow x^2=\frac{8}{x^2}.2\)
\(\Rightarrow x^2=\frac{16}{x^2}\)
\(\Rightarrow x^4=16\)
\(\Rightarrow x=2\)
\(r,\frac{x^3}{2}=\frac{32}{x}\)
\(\Rightarrow x^3=\frac{32}{x}.2\)
\(\Rightarrow x^3=\frac{64}{x}\)
\(\Rightarrow x^4=64\)
\(\Rightarrow x=\sqrt[4]{64}\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
27 tháng 9 2024
Bài 1:
a; \(\dfrac{7}{8}\) + \(x\) = \(\dfrac{4}{7}\)
\(x\) = \(\dfrac{4}{7}\) - \(\dfrac{7}{8}\)
\(x\) = \(\dfrac{32}{56}\) - \(\dfrac{49}{56}\)
\(x=-\) \(\dfrac{49}{56}\)
Vậy \(x=-\dfrac{49}{56}\)
b; 6 - \(x\) = - \(\dfrac{3}{4}\)
\(x\) = 6 + \(\dfrac{3}{4}\)
\(x\) = \(\dfrac{24}{4}+\dfrac{3}{4}\)
\(x=\dfrac{27}{4}\)
Vậy \(x=\dfrac{27}{4}\)
c; \(\dfrac{1}{-5}\) + \(x\) = \(\dfrac{3}{4}\)
\(x\) = \(\dfrac{3}{4}\) + \(\dfrac{1}{5}\)
\(x=\dfrac{15}{20}\) + \(\dfrac{4}{20}\)
\(x=\dfrac{19}{20}\)
Vậy \(x=\dfrac{19}{20}\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
27 tháng 9 2024
Bài 1:
d; - 6 - \(x\) = - \(\dfrac{3}{5}\)
\(x\) = - 6 + \(\dfrac{3}{5}\)
\(x=-\dfrac{30}{5}\) + \(\dfrac{3}{5}\)
\(x=-\dfrac{27}{5}\)
Vậy \(x=-\dfrac{27}{5}\)
e; - \(\dfrac{2}{6}\) + \(x\) = \(\dfrac{5}{7}\)
\(x\) = \(\dfrac{5}{7}\) + \(\dfrac{2}{6}\)
\(x\) = \(\dfrac{15}{21}\) + \(\dfrac{1}{3}\)
\(x=\dfrac{15}{21}\) + \(\dfrac{7}{21}\)
\(x=\dfrac{22}{21}\)
Vậy \(x=\dfrac{22}{21}\)
f; - 8 - \(x\) = - \(\dfrac{5}{3}\)
\(x\) = \(-\dfrac{5}{3}\) + 8
\(x\) = \(\dfrac{-5}{3}\) + \(\dfrac{24}{3}\)
\(x\) = \(\dfrac{-19}{3}\)
Vậy \(x=-\dfrac{19}{3}\)
NL
0
HT
3 tháng 8 2021
4, Q = |x+\(\frac{1}{5}\) | -x +\(\frac{4}{7}\)
xét x \(\ge\) \(-\frac{1}{5}\)
Ta Có Q = |x+\(\frac{1}{5}\) | -x + \(\frac{4}{7}\) = x+\(\frac{1}{5}\) - x +\(\frac{4}{7}\) = \(\frac{27}{35}\) (1)
xét x \(< -\frac{1}{5}\)
Ta có Q = | x +\(\frac{1}{5}\) | - x + \(\frac{4}{7}\) = -x - \(\frac{1}{5}\) - x + \(\frac{4}{7}\) = -2x + \(\frac{13}{35}\)
với x \(< -\frac{1}{5}\)
=> -2x \(>\) \(\frac{2}{5}\)
=> -2x + \(\frac{13}{35}\) \(>\frac{27}{35}\) (2)
Từ (1) và (2) => MinQ = \(\frac{27}{35}\) khi \(x\ge-\frac{1}{5}\)
5 , D = |x| + |8-x|
D = |x| + |8-x| \(\ge\) |x+8-x| = |8| = 8
Dấu ''='' xảy ra khi x(8-x) \(\ge\) 0 <=> 0\(\le\)x\(\le\) 8
Vậy MinD = 8 khi \(0\le x\le8\)
6,L= |x - 2012| + |2011 - x|
L = |x-2012| + |2011-x| \(\ge\) | x-2012 + 2011 - x | = |-1| = 1
Dấu ''= '' xảy ra khi ( x-2012)(2011-x) \(\ge\) 0
HT
3 tháng 8 2021
làm nốt câu 6 nãy ấn nhầm
<=> 2011\(\le\) x \(\le\) 2012
Vậy MinL = 1 khi \(2011\le x\le2012\)
7 , E = | x- \(\frac{2006}{2007}\) | + |x-1|
Ta có :
E = |x-\(\frac{2006}{2007}\) | + |1-x|
E = | x - \(\frac{2006}{2007}\) | + |1-x| \(\ge\) | x - \(\frac{2006}{2007}\) + 1 - x | = \(\frac{1}{2007}\)
Dấu ''='' xảy ra khi (x- \(\frac{2006}{2007}\) ) ( 1-x ) \(\ge0\) <=> \(\frac{2006}{2007}\le x\le1\)
Vậy MinE = \(\frac{1}{2007}\) khi \(\frac{2006}{2007}\le x\le1\)
8 ,F = | x -\(\frac{1}{4}\) | + | \(x-\frac{3}{4}\) |
Ta có :
F = | x - \(\frac{1}{4}\) | + | \(\frac{3}{4}\) - x |
F = | x - \(\frac{1}{4}\) | + | \(\frac{3}{4}\) -x | \(\ge\) | x - \(\frac{1}{4}\) + \(\frac{3}{4}\) -x | = \(\frac{1}{2}\)
Dấu ''='' xảy ra khi ( x-\(\frac{1}{4}\) ) ( \(\frac{3}{4}-x\) ) \(\ge\) 0 <=> \(\frac{1}{4}\le x\le\frac{3}{4}\)
Vậy MinF = \(\frac{1}{2}\) khi \(\frac{1}{4}\le x\le\frac{3}{4}\)