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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
a)\(123-5:\left(x+4\right)=38\)
\(5:\left(x+4\right)=123-38\)
\(5:\left(x+4\right)=85\)
\(x+4=5:85\)
\(x=\dfrac{1}{17}-4\)
\(x=-\dfrac{67}{17}\)
b)\(70-5.\left(x-3\right)=45\)
\(5.\left(x-3\right)=70-45\)
\(5.\left(x-3\right)=35\)
\(x-3=35:5\)
\(x-3=7\)
\(x=7+3\)
\(x=10\)
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.
\(\dfrac{\dfrac{2}{5}+\dfrac{2}{7}-\dfrac{2}{9}-\dfrac{2}{11}}{\dfrac{4}{5}+\dfrac{4}{7}-\dfrac{4}{9}-\dfrac{4}{11}}=\dfrac{2.\left[\dfrac{1}{5}+\dfrac{1}{7}-\dfrac{1}{9}-\dfrac{1}{11}\right]}{4.\left[\dfrac{1}{5}+\dfrac{1}{7}-\dfrac{1}{9}-\dfrac{1}{11}\right]}\)\(=\dfrac{2}{4}=\dfrac{1}{2}\)
\(B=\dfrac{\dfrac{2}{5}+\dfrac{2}{7}-\dfrac{2}{9}-\dfrac{2}{11}}{\dfrac{4}{5}+\dfrac{4}{7}-\dfrac{4}{9}-\dfrac{4}{11}}=\dfrac{2.\left(\dfrac{1}{5}+\dfrac{1}{7}-\dfrac{1}{9}-\dfrac{1}{11}\right)}{4.\left(\dfrac{1}{5}+\dfrac{1}{7}-\dfrac{1}{9}-\dfrac{1}{11}\right)}=\dfrac{1}{2}\)
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



Bài 1:
a: Để A là phân số thì n-3<>0
=>n<>3
b: Thay n=0 vào A, ta được:
\(A=\frac{4}{0-3}=\frac{4}{-3}=-\frac43\)
Thay n=10 vào A, ta được:
\(A=\frac{4}{10-3}=\frac47\)
Thay n=-2 vào A, ta được:
\(A=\frac{4}{-2-3}=\frac{4}{-5}=-\frac45\)
Bài 2:
a: \(\frac{-4}{9}\cdot\frac{7}{15}+\frac{4}{-9}\cdot\frac{8}{15}\)
\(=\frac{-4}{9}\left(\frac{7}{15}+\frac{8}{15}\right)\)
\(=\frac{-4}{9}\cdot\frac{15}{15}=-\frac49\)
b: \(\frac{5}{-4}\cdot\frac{16}{25}+\frac{-5}{4}\cdot\frac{9}{25}\)
\(=\frac{-5}{4}\left(\frac{16}{25}+\frac{9}{25}\right)\)
\(=-\frac54\cdot\frac{25}{25}=-\frac54\)
c: \(4\frac{11}{23}-\frac{9}{14}+2\frac{12}{23}-\frac54\)
\(=\left(4+\frac{11}{23}+2+\frac{12}{23}\right)-\frac{9}{14}-\frac54\)
\(=7-\frac{9}{14}-\frac54=\frac{196}{28}-\frac{18}{28}-\frac{35}{28}=\frac{143}{28}\)
d: \(2\frac{13}{27}-\frac{7}{15}+3\frac{14}{27}-\frac{8}{15}\)
\(=\left(2+\frac{13}{27}+3+\frac{14}{27}\right)-\left(\frac{7}{15}+\frac{8}{15}\right)\)
=5+1-1
=5
e: \(11\frac14-\left(2\frac57+5\frac14\right)\)
\(=11+\frac14-2-\frac57-5-\frac14\)
\(=4-\frac57=\frac{23}{7}\)
g: \(\frac{7}{19}\cdot\frac{8}{11}+\frac{7}{19}\cdot\frac{3}{11}+\frac{12}{19}\)
\(=\frac{7}{19}\left(\frac{8}{11}+\frac{3}{11}\right)+\frac{12}{19}\)
\(=\frac{7}{19}+\frac{12}{19}=\frac{19}{19}=1\)
Bài 3:
a: \(\frac34\cdot\frac{16}{9}-\frac75:\frac{-21}{20}\)
\(=\frac{48}{36}+\frac75\cdot\frac{20}{21}\)
\(=\frac43+\frac{140}{105}=\frac43+\frac43=\frac83\)
b: \(2\frac13-\frac13\cdot\left\lbrack-\frac32+\left(\frac23+0,4\cdot5\right)\right\rbrack\)
\(=\frac73-\frac13\cdot\left\lbrack-\frac32+\frac23+2\right\rbrack\)
\(=\frac73-\frac13\cdot\left\lbrack\frac12+\frac23\right\rbrack=\frac73-\frac13\cdot\frac76=\frac73\left(1-\frac16\right)=\frac73\cdot\frac56=\frac{35}{18}\)
c: \(\left(20+9\frac14\right):2\frac14\)
\(=\left(20+9+\frac14\right):\frac94\)
=29,25:2,25
=13
d: \(\left(6-2\frac45\right)\cdot3\frac18-1\frac35:\frac14\)
\(=\left(6-2-\frac45\right)\cdot\frac{25}{8}-\frac85\cdot4\)
\(=\left(4-\frac45\right)\cdot\frac{25}{8}-\frac{32}{5}=\frac{16}{5}\cdot\frac{25}{8}-\frac{32}{5}=10-\frac{32}{5}=\frac{18}{5}\)
e: \(\frac{32}{15}:\left(-1\frac15+1\frac13\right)\)
\(=\frac{32}{15}:\left(-1-\frac15+1+\frac13\right)\)
\(=\frac{32}{15}:\left(\frac13-\frac15\right)=\frac{32}{15}:\frac{2}{15}=\frac{32}{2}=16\)
g: \(0,2\cdot\frac{15}{36}-\left(\frac25+\frac23\right):1\frac15\)
\(=\frac{3}{36}-\frac{6+10}{15}:\frac65\)
\(=\frac{1}{12}-\frac{16}{15}\cdot\frac56=\frac{1}{12}-\frac83\cdot\frac13=\frac{1}{12}-\frac89=\frac{3}{36}-\frac{32}{36}=-\frac{29}{36}\)
h: \(1\frac{13}{15}\cdot0,75-\left(\frac{8}{15}+0,25\right)\cdot\frac{24}{47}\)
\(=\frac{28}{15}\cdot\frac34-\left(\frac{8}{15}+\frac14\right)\cdot\frac{24}{47}\)
\(=\frac{21}{15}-\frac{32+15}{60}\cdot\frac{24}{47}=\frac{21}{15}-\frac{24}{60}=\frac75-\frac25=\frac55=1\)
g: \(5:\left(4\frac34-1\frac{25}{28}\right)-1\frac38:\left(\frac38+\frac{9}{20}\right)\)
\(=5:\left(\frac{19}{4}-\frac{53}{28}\right)-\frac{11}{8}:\frac{15+18}{40}\)
\(=5:\frac{80}{28}-\frac{11}{8}\cdot\frac{40}{33}=5\cdot\frac{28}{80}-\frac53=5\cdot\frac{7}{20}-\frac53=\frac74-\frac53=\frac{21-20}{12}=\frac{1}{12}\)
Bài 4:
a: \(x:3\frac{1}{15}=1\frac12\)
=>\(x:\frac{46}{15}=\frac32\)
=>\(x=\frac32\cdot\frac{46}{15}=\frac{23}{5}\)
b: \(5\frac47:x=13\)
=>\(\frac{39}{7}:x=13\)
=>\(x=\frac{39}{7}:13=\frac37\)
c: \(7x-3x=3,2\)
=>4x=3,2
=>x=0,8
h: \(\frac23x-\frac12x=\frac{5}{12}\)
=>\(x\left(\frac23-\frac12\right)=\frac{5}{12}\)
=>\(x\cdot\frac16=\frac{5}{12}\)
=>\(x=\frac{5}{12}:\frac16=\frac{5}{12}\cdot6=\frac52\)
i: \(2\frac14\left(x-7\frac13\right...