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Câu 14)
\(a,\\ =-\dfrac{3}{8}+\dfrac{8}{17}+\dfrac{-5}{8}-\dfrac{3}{5}+\dfrac{9}{17}\\ =\left(\dfrac{-3}{8}+\dfrac{-5}{8}\right)+\left(\dfrac{8}{17}+\dfrac{9}{17}\right)-\dfrac{3}{5}\\ =\left(-1\right)+1-\dfrac{3}{5}=0-\dfrac{3}{5}=\dfrac{-3}{5}\\ b,\\ =\dfrac{7}{15}.\dfrac{-15}{14}+\left(\dfrac{27}{16}-\dfrac{1}{8}\right):\dfrac{5}{8}\)
\(=\dfrac{-1}{2}+\dfrac{25}{16}.\dfrac{8}{5}=\dfrac{-1}{2}+\dfrac{5}{2}=2\\ c,\\ =\dfrac{2}{2}-\dfrac{2}{3}+\dfrac{2}{3}-\dfrac{2}{4}+.....+\dfrac{2}{99}-\dfrac{2}{100}\\ =1-\dfrac{1}{50}=\dfrac{49}{50}\)
Câu 15
\(a,2x+\dfrac{-1}{4}=\dfrac{3}{2}\\ 2x=\dfrac{3}{2}-\dfrac{-1}{4}=\dfrac{7}{4}\\ x=\dfrac{7}{4}:2=\dfrac{7}{8}\\ b,\dfrac{15}{x}=\dfrac{-3}{4}\\ x=\dfrac{15.4}{-3}=-20\)
Nếu là z+x thì mik biết làm nè:
Đặt x-y=2011(1)
y-z=-2012(2)
z+x=2013(3)
Cộng (1);(2);(3) lại với nhau ta được :
2x=2012=>x=1006
Từ (1) => y=-1005
Từ (3) => z=1007
Tuy có vẻ hơi muộn nhưng thôi
Nếu A là số tự nhiên ⇒ \(\dfrac{1}{10}\left(7^{2004}-3^{92^{94}}\right)\in N\)
\(\Rightarrow7^{2004}-3^{92^{94}}⋮10\)
Thật vậy, ta có :
72004 với lũy thừa là 2004 ⋮ 4
⇒ 72004 = ( .......... 9 )
392^94 với lũy thừa là 9294 mà 92 ⋮ 4 ⇒ 9294 ⋮ 4
⇒ 392^94 = ( .......... 9 )
⇒ 72004 - 392^94 = ( .......... 9 ) - ( ............ 9) = ( ........... 0 ) ⋮ 10
⇒ \(\dfrac{1}{10}\left(7^{2004}-3^{92^{94}}\right)\in N\)
A=1/10.(72004-392^94) là số tự nhiên.
Từ đề bài ta có:
\(T=\dfrac{1+2}{2}.\dfrac{1+3}{3}.\dfrac{1+4}{4}...\dfrac{1+98}{98}.\dfrac{1+99}{99}\)
\(=\dfrac{3}{2}.\dfrac{4}{3}.\dfrac{5}{4}...\dfrac{99}{98}.\dfrac{100}{99}\)
\(=\dfrac{100}{2}\)
\(=50\).
\(T=\left(\dfrac{1}{2}+1\right)\left(\dfrac{1}{3}+1\right)\left(\dfrac{1}{4}+1\right)...\left(\dfrac{1}{98}+1\right)\left(\dfrac{1}{99}+1\right)\)
\(T=\dfrac{3}{2}.\dfrac{4}{3}.\dfrac{5}{4}....\dfrac{99}{98}.\dfrac{100}{99}\)
\(T=\dfrac{3.4.5......99}{3.4.5......99}.\dfrac{100}{2}\)
\(T=50\)
\(=>9x+2=60:3\)
\(=>9x+2=20\)
\(=>9x=20-2\)
\(=>9x=18\)
\(=>x=18:2=2\)
Vậy số cần tìm là 2
CHÚC BẠN HỌC TỐT............
( 9x + 2 ) . 3 = 60
( 9x + 2 ) = 60 : 3
9x + 2 = 20
9x = 20 - 2
9x =18
x = 18 : 9
x = 2
2a/3b = 3b/4c = 4c/5d = 5d/2a (1)
ta có: 2a/3b=3b/4c=> 8ac=9b^2
4c/5d=5d/2a=> 8ac=25d^2
=> 9b^2=25d^2
=> b=5d/3
=> 3b=5d(*)
lại có: 3b/4c=4c/5d => 3b/4c=4c/3b (theo *)
=> 9b^2=16c^2
=> b=4c/3
=> 3b/4c=1
BT= 4*3b/4c (Vì các phân số = nhau)
=> BT=3b/c
Mà: 3b=4c ( Vì 3b/4c=1)
=> BT=4c/c=4
Vậy biểu thức trên = 4
\(1,\dfrac{8}{13}+\dfrac{5}{13}=\dfrac{8+5}{13}=\dfrac{13}{13}=1\)
\(2,\dfrac{-8}{7}+\dfrac{-6}{7}=\dfrac{\left(-8\right)+\left(-6\right)}{7}=\dfrac{-14}{7}=-2\)
\(5,\dfrac{7}{-9}-\dfrac{13}{9}=\dfrac{\left(-7\right)-13}{9}=\dfrac{-20}{9}\)
\(6,\dfrac{5}{7}-\dfrac{-3}{21}=\dfrac{15}{21}-\dfrac{-3}{21}=\dfrac{15-\left(-3\right)}{21}=\dfrac{18}{21}=\dfrac{6}{7}\)
\(9,\dfrac{7}{-10}-\dfrac{2}{-15}=\dfrac{-105}{150}-\dfrac{-20}{150}=\dfrac{-\left(105-20\right)}{150}=\dfrac{-85}{150}\)
\(10,\dfrac{6}{13}-\dfrac{-15}{39}=\dfrac{18}{39}-\dfrac{-15}{39}=\dfrac{18-\left(-15\right)}{39}=\dfrac{33}{39}=\dfrac{11}{13}\)
1) \(\dfrac{8}{13}+\dfrac{5}{13}=\dfrac{13}{13}=1\)
2 ) \(\dfrac{-8}{7}+\dfrac{-6}{7}=\dfrac{-14}{7}=-2\)
3 ) \(\dfrac{-4}{3}-\dfrac{-7}{3}\)\(=\dfrac{3}{3}=1\)
4 ) \(\dfrac{-8}{5}+\dfrac{6}{-5}=\dfrac{-8}{5}+\dfrac{-6}{5}\)\(=\dfrac{-14}{5}\)
5 ) \(\dfrac{7}{-9}-\dfrac{13}{9}=\dfrac{-7}{9}-\dfrac{13}{9}=\dfrac{-20}{9}\)
6 ) \(\dfrac{5}{7}-\dfrac{-3}{21}=\dfrac{5\times3}{7\times3}=\dfrac{15}{21}-\dfrac{-3}{21}=\dfrac{18}{21}\)\(=\dfrac{6}{7}\)
Đợi