(5/12+1&4/3-0,25):5/8+0,75
1.tính A.-5/8-7/-12 B.3/4+(-5)/6-11/-12 C.1/6-1/36 D.15/48-(-5)/12 E.-1/7-1/8 F.-5/15-(-3)/25 G.3/4-(-5)/12-7/-24
a: \(=\dfrac{-5}{8}+\dfrac{7}{12}=\dfrac{-15}{24}+\dfrac{14}{24}=\dfrac{-1}{24}\)
b: \(=\dfrac{3}{4}-\dfrac{5}{6}+\dfrac{11}{12}=\dfrac{9}{12}-\dfrac{10}{12}+\dfrac{11}{12}=\dfrac{10}{12}=\dfrac{5}{6}\)
c: \(=\dfrac{6}{36}-\dfrac{1}{36}=\dfrac{5}{36}\)
d: \(=\dfrac{5}{12}+\dfrac{5}{12}=\dfrac{10}{12}=\dfrac{5}{6}\)
e: \(=\dfrac{-8}{56}-\dfrac{7}{56}=\dfrac{-15}{56}\)
f: \(=\dfrac{-5}{15}+\dfrac{3}{25}=\dfrac{-25}{75}+\dfrac{9}{75}=\dfrac{-16}{75}\)
Tính (theo mẫu).
Mẫu: \(\dfrac{1}{2}-\dfrac{5}{12}=\dfrac{6}{12}-\dfrac{5}{12}=\dfrac{6-5}{12}=\dfrac{1}{12}\) |
a) \(\dfrac{3}{4}-\dfrac{1}{8}\) b) \(\dfrac{2}{6}-\dfrac{5}{18}\) c) \(\dfrac{2}{5}-\dfrac{3}{20}\)
a) \(\dfrac{3}{4}-\dfrac{1}{8}=\dfrac{6}{8}-\dfrac{1}{8}=\dfrac{6-1}{8}=\dfrac{5}{8}\)
b) \(\dfrac{2}{6}-\dfrac{5}{18}=\dfrac{6}{18}-\dfrac{5}{18}=\dfrac{6-5}{18}=\dfrac{1}{18}\)
c) \(\dfrac{2}{5}-\dfrac{3}{20}=\dfrac{8}{20}-\dfrac{3}{20}=\dfrac{8-3}{20}=\dfrac{5}{20}=\dfrac{1}{4}\)
tính:
1/3+5/12
9/12-1/3
9/12-5/12
5/12+1/3
làm đầy đủ tick liền
1/3 + 5/12 = 4/12 + 5/12 = 9/12 = 3/4
9/12 - 1/3 = 9/12 - 4/12 = 5/12
9/12 - 5/12 = 4/12 = 1/3
5/12 + 1/3 = 5/12 + 4/12 = 9/12 = 3/4
Làm đầy đủ rồi đó
1/3 + 5/12 = 3/4
9/12 - 1/3 = 5/12
9/12 - 5/ 12 = 1/3
5/12 + 1/3 = 3/4
1/3+5/12=4/12+5/12=3/4
9/12-1/3=9/12-4/12=5/12
8/9 - 2/6 + 5/12=
5/12 + 3/4 + 1/3=
1/12 + 5/6 + 3/8=
8/9 - 2/6 + 5/12 = 34/36 - 12/36 + 15/36= 35/36
5/12 + 3/4 + 1/3 = 5/12 + 9/12 + 4/12 = 3/2
1/12 + 5/6 + 3/8= 2/24 +20/24 +9/24 = 31/24
Mọi người check xem đúng không:
<=>x(3x+2)+(x+1)^2 - (2x - 5)(2x+5)= -12
<=>3x^2 + 2x + (x+1)^2 - 2x^2 - 5^2= -12
<=>(3x^2 - 2x^2) + [ (x +1)^2 -5^2] + 2x = -12
<=>(3x - 2x)(3x + 2x)+ (x+1-5) (x+1+5) + 2x = -12
=> 3x - 2x = -12 ; 3x+2x = -12 ; x+1+5 = -12 ; x+1-5 = -12 hoặc 2x = -12
=> x = -12 ; 5x =-12 ; x+ 6 = -12 ; x -4 = -12 hoặc x = -12: 2
=> x= -12 ; x = -12:5; x = -12 :6; x = -12 + 4 hoặc x= -6
=> x= -12 x = -12/5; x = -2 ; x = -8 hoặc x = -6
đúng rồi nha bạn
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Mọi người check xem đúng không:
<=>x(3x+2)+(x+1)^2 - (2x - 5)(2x+5)= -12
<=>3x^2 + 2x + (x+1)^2 - 2x^2 - 5^2= -12
<=>(3x^2 - 2x^2) + [ (x +1)^2 -5^2] + 2x = -12
<=>(3x - 2x)(3x + 2x)+ (x+1-5) (x+1+5) + 2x = -12
=> 3x - 2x = -12 ; 3x+2x = -12 ; x+1+5 = -12 ; x+1-5 = -12 hoặc 2x = -12
=> x = -12 ; 5x =-12 ; x+ 6 = -12 ; x -4 = -12 hoặc x = -12: 2
=> x= -12 ; x = -12:5; x = -12 :6; x = -12 + 4 hoặc x= -6
=> x= -12 x = -12/5; x = -2 ; x = -8 hoặc x = -6
bạn làm sai rồi !
\(\Leftrightarrow x\left(3x+2\right)+\left(x+1\right)^2-\left(2x-5\right)\left(2x+5\right)=-12\)
\(\Leftrightarrow3x^2+2x+x^2+2x+1-4x^2+25=-12\)
\(\Leftrightarrow4x+26=-12\)
\(\Leftrightarrow4x=-38\)
\(\Leftrightarrow x=-\frac{19}{2}\)
Vậy tập nghiệm của phương trình là \(S=\left\{-\frac{19}{2}\right\}\)
1/5*1/12-1/5*1/12
Ko viết lại đề bài nha:
= 1/5 * (1/12 - 1/12)
= 1/5 * 0
= 0
\(\frac{1}{5}\times\frac{1}{12}-\frac{1}{5}\times\frac{1}{12}=\frac{1}{5}\times\left(\frac{1}{12}-\frac{1}{12}\right)=0\)
b1 : tìm x
a] [ 1/2x5 + 1/5x8 + 1/8x11 + ......+ 1/65x68 ] x X = 11/68
b] X - [ 12/2x4 + 12/4x6 + 12/ 6x8+...+12/48x50 ] = 1/5
c] X + [ 5/5 + 5/45 + 5/117+ 5/221 + ...+5/1845 ] =2
Tính giá trị của biểu thức a) A = (-10) + ( -34) + ( -54) b) B = 12 + ( -24) + 35 c) C = |- 1 + 2| +| - 2 + 1| + |- 1 + (-2)| d) D = [(-5) + ( -12)] + [( -12) + 5] + [-5 + 12]
c= (5^12+1)/(5^13+1) và D = (5^11+1)/(5^12+1) so sánh