Cho C={ 1/52 ;1/53 ;1/54 ;1/55 ;1/56 ; . . . . . . . . . . ;1/100 }
Hãy so sánh C với 1/2
cho a= 23 .3 ;b=32 . 52 ;c=2.5. khi đó UCLN=(a,b,c)là
A, 23 3.5 B,1 C.23 , 32, 52 D,30
viết là sao để ra kết quả luôn nha
cảm ơn nhiều ❤
Cho hệ phương trình sau:
{x+y√5=0x√5+3y=1−√5{x+y5=0x5+3y=1−5
Ta tìm được nghiệm của hệ là :
A x=√5−52;y=−√5−12x=5−52;y=−5−12
B x=√5−52;y=√5−12x=5−52;y=5−12
C x=1−√52,y=−√5+52x=1−52,y=−5+52
D x=−√5+52;y=√5−12x=−5+52;y=5−12
Tìm các số tự nhiên a,b,c sao cho 52/9=5+1/a+1/b+c
đề nek
Tìm các số tự nhiên a,b,c sao cho:
\(\frac{52}{9}=\frac{5+1}{a}+\frac{1}{b+c}\)
cho B=1/4+1/4^2+...+1/4^2014 và C=1/52.(35/1.3+35/3.5+.....+35/103.105).so sánh B và C
Ta có B = \(\frac{1}{4}+\frac{1}{4^2}+...+\frac{1}{4^{2014}}\)
=> 4B = \(1+\frac{1}{4}+...+\frac{1}{4^{2013}}\)
Lấy 4B trừ B theo vế ta có :
4B - B = \(\left(1+\frac{1}{4}+...+\frac{1}{4^{2013}}\right)-\left(\frac{1}{4}+\frac{1}{4^2}+...+\frac{1}{4^{2014}}\right)\)
=> 3B = \(1-\frac{1}{4^{2014}}\)
=> B = \(\left(1-\frac{1}{4^{2014}}\right):3=\frac{1}{3}-\frac{1}{3.4^{2014}}\)
Lại có C = \(\frac{1}{52}\left(\frac{35}{1.3}+\frac{35}{3.5}+...+\frac{35}{103.105}\right)=\frac{1}{52}.\frac{35}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{103.105}\right)\)
\(=\frac{35}{104}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{103}-\frac{1}{105}\right)\)
\(=\frac{35}{104}.\left(1-\frac{1}{105}\right)=\frac{35}{104}.\frac{104}{105}=\frac{1}{3}\)
Vì \(\frac{1}{3}-\frac{1}{3.4^{104}}< \frac{1}{3}\Rightarrow B< C\)
Vậy B < C
Bài 1: tìm x
c) \(|\) x \(|\) = 3,5
d) \(|\) x \(|\) = -2,7
e) \(|\) 1 - x \(|\) + 0,73 = 3
f) 52 . 73 . 11x + 52 . 72 .11 = 0
g) (3 . 5 + 5 . 7)x + ( 3 . 5 + 5 . 7) + (3 . 5 + 5 . 7) = 0
h) 52 . 72 . 112x - 52 . 72 . 114 = 0
l) \(|\) x + \(\dfrac{3}{4}\) \(|\) - 5 = -2
c) \(\left|x\right|=3,5\Rightarrow\left[{}\begin{matrix}x=3,5\\x=-3,5\end{matrix}\right.\)
d) \(\left|x\right|=-2,7\Rightarrow x\in\varnothing\)
l) \(\left|x+\dfrac{3}{4}\right|-5=-2\Rightarrow\left|x+\dfrac{3}{4}\right|=3\)
\(\Rightarrow\left[{}\begin{matrix}x+\dfrac{3}{4}=3\\x+\dfrac{3}{4}=-3\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=3-\dfrac{3}{4}\\x=-3-\dfrac{3}{4}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{9}{4}\\x=\dfrac{15}{4}\end{matrix}\right.\)
Đính chính câu l \(x=-\dfrac{15}{4}\) không phải \(x=\dfrac{15}{4}\)
c) x = 3,5 hoặc -3,5
d) x = -2,7
e) x = -1,27
f) x = 0
g) x = -2
h) x = 0
l) x = 9/4
x = -15/4
\(Cho:\frac{52}{9}=5+\frac{1}{a+\frac{1}{b+\frac{1}{c}}}\)
a,b,c thuộc N
tính a + b + c
Ta có \(\frac{52}{9}=5+\frac{7}{9}=5+\frac{1}{\frac{9}{7}}\)
=\(5+\frac{1}{1+\frac{2}{7}}=5+\frac{1}{1+\frac{1}{\frac{7}{2}}}\)
\(=5+\frac{1}{1+\frac{1}{3+\frac{1}{2}}}\)
Vậy a = 1
b=3
c=2
Tìm các stn a,b,c sao cho:
\(\frac{52}{9}=5+\frac{1}{a+\frac{1}{b+\frac{1}{c}}}\)
Ai nhanh mk k cho
Ta có :
\(\frac{52}{9}=5+\frac{7}{9}\)
\(\frac{7}{9}=\frac{1}{\frac{9}{7}}=\frac{1}{1+\frac{2}{7}}\)
\(\frac{2}{7}=\frac{1}{\frac{7}{2}}=\frac{1}{1+\frac{5}{2}}\)
\(\frac{5}{2}=\frac{1}{\frac{2}{5}}\)
\(\Rightarrow\frac{52}{9}=5+\frac{1}{1+\frac{1}{1+\frac{1}{\frac{2}{5}}}}\)
\(\Rightarrow\hept{\begin{cases}a=1\\b=1\\c=\frac{2}{5}\end{cases}}\)
Tính giá trị biểu thức:
a) A = -11/23.10/-13 +-11/13.3/-23-2/23
b) B = 1/1.2.3 + 1/2.3.4 + ... + 1/18.19.20
c) C = 52/1.6 + 52/6.11 + 52/11.16 +...+ 52/31.35
d) D = 1. 2 .3 + 2 .4. 6+ 4 .8 .12 + 7 .14 .21/ 1. 3. 5 + 2. 6. 10 + 4 .12 .20 + 7 .21 .35 + 3/5
GIÚP MIK VS AK!
\(B=\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+\frac{1}{3\cdot4\cdot5}+\frac{1}{18\cdot19\cdot20}\)
\(B=\frac{1}{2}\left(\frac{2}{1\cdot2\cdot3}+\frac{2}{2\cdot3\cdot4}+\frac{2}{3\cdot4\cdot5}+\frac{2}{18\cdot19\cdot20}\right)\)
\(B=\frac{1}{2}\left(\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+\frac{1}{3\cdot4}-\frac{1}{4\cdot5}+...+\frac{1}{18\cdot19}-\frac{1}{19\cdot20}\right)\)
\(B=\frac{1}{2}\left(\frac{1}{1\cdot2}-\frac{1}{19\cdot20}\right)\)
\(B=\frac{1}{2}\cdot\frac{189}{380}=\frac{189}{760}\)
\(C=\frac{52}{1\cdot6}+\frac{52}{6\cdot11}+\frac{52}{11\cdot16}+...+\frac{52}{31\cdot36}\)
\(C=\frac{52}{5}\left(\frac{5}{1\cdot6}+\frac{5}{6\cdot11}+\frac{5}{11\cdot16}+...+\frac{6}{31\cdot36}\right)\)
\(C=\frac{52}{5}\left(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}+...+\frac{1}{31}-\frac{1}{36}\right)\)
\(C=\frac{52}{5}\cdot\left(1-\frac{1}{36}\right)\)
\(C=\frac{91}{9}\)
a) Ta có: \(\left(x-1\right)^2=1\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=1\\x-1=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=0\end{matrix}\right.\)
Vậy: \(x\in\left\{0;2\right\}\)
b) Ta có: x(3x+9)=0
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\3x+9=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\3x=-9\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\end{matrix}\right.\)
Vậy: \(x\in\left\{0;-3\right\}\)