14 h 15' - 3 h 45'= ?
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a,=\(\dfrac{8}{14}-\dfrac{1}{14}+\dfrac{5}{21}+\dfrac{3}{2}\)
=\(\dfrac{1}{2}+\dfrac{3}{2}+\dfrac{5}{21}\) =\(2+\dfrac{5}{21}\) =\(\dfrac{42}{21}+\dfrac{5}{21}\) =\(\dfrac{47}{21}\)
b,=\(\dfrac{11}{13}.\dfrac{12}{15}-\dfrac{7}{15}+\dfrac{14}{15}.\dfrac{11}{13}\)
=\(\dfrac{11}{13}.\left(\dfrac{12}{15}+\dfrac{14}{15}\right)-\dfrac{7}{15}\)
=\(\dfrac{11}{13}.\dfrac{26}{15}-\dfrac{17}{15}\) =\(\dfrac{22}{15}-\dfrac{17}{15}\) =\(\dfrac{5}{15}\) =\(\dfrac{1}{3}\)
c,=\(\left(\dfrac{3}{6}-\dfrac{2}{6}\right)^2\) =\(\left(\dfrac{1}{6}\right)^2\) =\(\dfrac{1}{36}\)
d,=câu này dễ mà
Ta có: \(\frac{15}{14}\left(\frac{14}{9}-\frac{3}{7}\right)-\frac{15}{14}\left(\frac{14}{6}-\frac{3}{7}\right)\)
\(=\frac{15}{14}\left[\left(\frac{14}{9}-\frac{3}{7}\right)-\left(\frac{14}{6}-\frac{3}{7}\right)\right]\)
\(=\frac{15}{14}\left(\frac{14}{9}-\frac{3}{7}-\frac{14}{6}+\frac{3}{7}\right)\)
\(=\frac{15}{14}\left(\frac{14}{9}-\frac{14}{6}\right)\)
\(=\frac{15}{14}\cdot\left(\frac{28}{18}-\frac{42}{18}\right)\)
\(=\frac{15}{14}\cdot\frac{-14}{18}\)
\(=\frac{-15}{18}=-\frac{5}{6}\)
\(A=\dfrac{14^{14}+1}{14^{15}+1}\)
\(\Rightarrow14.A=\dfrac{14^{15}+14}{14^{15}+1}\)
\(\Rightarrow14.A=\dfrac{14^{15}+1}{14^{15}+1}+\dfrac{13}{14^{15}+1}\)
\(\Rightarrow14.A=1+\dfrac{13}{14^{15}+1}\)
\(B=\dfrac{14^{15}+1}{14^{16}+1}\)
\(\Rightarrow14.B=\dfrac{14^{16}+14}{14^{16}+1}\)
\(\Rightarrow14.B=\dfrac{14^{16}+1}{14^{16}+1}+\dfrac{13}{14^{16}+1}\)
\(\Rightarrow14.B=1+\dfrac{13}{14^{16}+1}\)
Nhận xét: \(\dfrac{13}{14^{15}+1}>\dfrac{13}{14^{16}+1}\) (cùng tử, xét mẫu)
\(\Rightarrow A>B\)
Vậy \(A>B\)
a ) 1 3 b ) 1 4 c ) − 1 3
d ) 1 2 e ) 1 4 f ) 3 5
g ) 2 3 h ) − 3 2
Cho phản ứng xFeS2 + yH2SO4 (đặc, nóng) -> zFe(SO4)3 + tSO2 + aH2O. Các giá trị a ; x ; y ; t ; z lần lượt là:
A. 2 ; 14 ; 15 ; 1 ; 14
B. 1 ; 14 ; 14 ; 15 ; 2
C. 15 ; 2 ; 14 ; 14 ; 1
D. 14 ; 2 ; 14 ; 15 ; 1
Ta có:a5+b5=4
<=>(a5+b5)2=16
<=>a10+2a5b5+b10=16
<=>14+2a5b5=16(do a10+b10=14)
<=> 2a5b5=2
<=> a5b5=1
Lại có:
(a5+b5)(a10+b10)=4.14
<=>a15+a5b10+a10b5+b15=56
<=>a15+b15+a5b5(b5+a5)=56
<=>a15+b15+4=56( do a5+b5=4; a5b5=1)
<=> a15+b15=52
=>H=52
Vậy H=a15+b15=52
5 + ( x + 27 ) = 64
( x + 27 ) = 64 - 5 ( x + 27 ) = 59 x = 59 - 27 x = 32a) \(\Rightarrow x+27=59\Rightarrow x=32\)
b) \(\Rightarrow x-2=39\Rightarrow x=41\)
c) \(\Rightarrow x+5=-322\Rightarrow x=-327\)
d) \(\Rightarrow5x=35\Rightarrow x=7\)
e) \(\Rightarrow4\left(x-5\right)=56\Rightarrow x-5=14\Rightarrow x=19\)
f) \(\Rightarrow15+x=37\Rightarrow x=22\)
g) \(\Rightarrow7\left(13-x\right)=35\Rightarrow13-x=5\Rightarrow x=8\)
h) \(\Rightarrow10\left(x+1\right)=100\Rightarrow x+1=10\Rightarrow x=9\)
Đổi `14h15'= 13h75'`
Vậy `14h15' -3h45'= 13h75' -3h45'= 10h30'`
14h15' = 13h75'
13h75' - 3h45'= 10h30'