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\(\left(x+\frac{1}{2}\right)^4=16\)
\(\Rightarrow\left[\begin{array}{nghiempt}\left(x+\frac{1}{2}\right)^4=2^4\\\left(x+\frac{1}{2}\right)^4=-2^4\end{array}\right.\)
\(\Rightarrow\left[\begin{array}{nghiempt}x+\frac{1}{2}=2\\x+\frac{1}{2}=-2\end{array}\right.\)
\(\Rightarrow\left[\begin{array}{nghiempt}x=2-\frac{1}{2}\\x=-2-\frac{1}{2}\end{array}\right.\)
\(\Rightarrow\left[\begin{array}{nghiempt}x=\frac{3}{2}\\x=-\frac{5}{2}\end{array}\right.\)
( x + \(\frac{1}{2}\) )4 = 16
Vì 24 = 16 \(\Rightarrow\)x + \(\frac{1}{2}\) = 2
x = 2 - \(\frac{1}{2}\)
x = \(\frac{3}{2}\)
\(\frac{1}{2}x+\frac{3}{5}x=-\frac{2}{3}\)
\(=x\left(\frac{1}{2}+\frac{3}{5}\right)=-\frac{2}{3}\)
\(=x\cdot\frac{11}{10}=-\frac{2}{3}\)
\(\Rightarrow x=-\frac{2}{3}:\frac{11}{10}\)
\(\Rightarrow x=-\frac{20}{33}\)
Ta có: \(\frac{a}{b}< \frac{a+1}{b+1}\)
\(B=\frac{10^{2013}+1}{10^{2014}+1}< \frac{10^{2013}+1+9}{10^{2014}+1+9}=\frac{10^{2013}+10}{10^{2014}+10}=\frac{10\left(10^{2012}+1\right)}{10\left(10^{2013}+1\right)}=\frac{10^{2012}+1}{2^{2013}+1}=A\)
Vậy: \(A>B\)
Ta có:
\(10A=\frac{10\left(10^{2012}+1\right)}{10^{2013}+1}=\frac{10^{2013}+10}{10^{2013}+1}=\frac{10^{2013}+1+9}{10^{2013}+1}=\frac{10^{2013}+1}{10^{2013}+1}+\frac{9}{10^{2013}+1}=1+\frac{9}{10^{2013}+1}\)
\(10B=\frac{10\left(10^{2013}+1\right)}{10^{2014}+1}=\frac{10^{2014}+10}{10^{2014}+1}=\frac{10^{2014}+1+9}{10^{2014}+1}=\frac{10^{2014}+1}{10^{2014}+1}+\frac{9}{10^{2014}+1}=1+\frac{9}{10^{2014}+1}\)
Vì 102013+1<102014+1
\(\Rightarrow\frac{9}{10^{2013}+1}>\frac{9}{10^{2014}+1}\)
\(\Rightarrow1+\frac{9}{10^{2013}+1}>1+\frac{9}{10^{2014}+1}\)
\(\Rightarrow10A>10B\)
\(\Rightarrow A>B\)
Ta có
\(\frac{1}{1.6}+\frac{1}{6.11}+......+\frac{1}{\left(5n+1\right)\left(5n+6\right)}\)
\(=\frac{1}{5}\left(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}+.....+\frac{1}{5n+1}-\frac{1}{5n+6}\right)\)
\(=\frac{1}{5}\left(1-\frac{1}{5n+6}\right)\)
\(=\frac{1}{5}.\left[\frac{\left(5n+6\right)-1}{\left(5n+6\right)}\right]\)
\(=\frac{1}{5}.\frac{5n+5}{5n+6}\)
\(=\frac{n+1}{5n+6}\)
\(\Rightarrow\frac{1}{1.6}+\frac{1}{6.11}+......+\frac{1}{\left(5n+1\right)\left(5n+6\right)}=\frac{n+1}{5n+6}\) ( đpcm )
a) /2x + 1/ = 4
=> 2x + 1 = 4 hoặc 2x + 1 = -4
Nếu 2x + 1 = 4
=> x = \(\frac{3}{2}\)
Nếu 2x + 1 = -4
=> x = \(\frac{-5}{2}\)
Vậy x = \(\frac{3}{2}\) hoặc x = \(\frac{-5}{2}\)
b) 5x - 3 = 4x -7
=> ( 5x - 3 ) - ( 4x - 7 ) = 0
=> 5x - 3 - 4x + 7 = 0
=> ( 5x - 4x ) - ( 3 - 7 ) = 0
=> x + 4 = 0
=> x = -4
Vậy x = -4
a.
5 . (x : 3 - 4) = 15
x : 3 - 4 = 15 : 5
x : 3 - 4 = 3
x : 3 = 3 + 4
x : 3 = 7
x = 7 . 3
x = 21
b.
2x + 3 chia hết cho x + 1
<=> 2x + 2 + 1 chia hết cho x + 1
<=> 2(x + 1) + 1 chia hết cho x + 1
<=> 1 chia hết cho x + 1
<=> x + 1 thuộc Ư(1)
<=> x + 1 thuộc {-1 ; 1}
<=> x thuộc {-2 ; 0}
5( x : 3 - 4 ) = 15
x : 3 - 4 = 15: 5
x : 3 - 4 = 3
x : 3 = 3 + 4
x : 3 = 7
x = 7 . 3
x = 21
\(x^2\le4\)
\(\Leftrightarrow x^2\le2^2\)
\(\Leftrightarrow x=\left\{0;1;2;-1;-2\right\}\)
Thử lại : ta được kết quả đúng như trên
Vì :
\(x⋮12,25,30\) ( x \(\in\) N
=> x \(\in\) BCNN ( 12 ; 25 ; 30 )
Ta có : 12 = \(3.2^2\)
25 = 52
30 = 2.3.5
=> BCNN ( 12 ; 25 ; 30 ) = 3.22.52= 300
=> B ( 12 ; 25 ; 30 ) = { 0 ; 300 ; 600 ; 900 ....}
Vì : 0 < x < 800
=> x = 600
kết quả của mình cũng y chang thanks có gì giúp mình nha Phương An
a) /2x - \(\frac{1}{3}\) / =5
\(=>\left[\begin{array}{nghiempt}2x-\frac{1}{3}=5\\2x-\frac{1}{3}=-5\end{array}\right.\)
\(=>\left[\begin{array}{nghiempt}x=\frac{8}{3}\\x=-\frac{7}{3}\end{array}\right.\)
b)x3 - 4x = 0
<=> x(x2 - 4) = 0
\(=>\left[\begin{array}{nghiempt}x=0\\x^2-4=0\end{array}\right.\)
\(=>\left[\begin{array}{nghiempt}x=0\\x=\pm2\end{array}\right.\)
ko hỉu j hết. sorry