bo dau ngoac roi tinh :
A=(37,1-4,5)-(-4,5+37,1)
B=-(315,4+275)+4,315-(10-275)
C= - (\(\dfrac{3}{7}\)+\(\dfrac{3}{8}\))-(-\(\dfrac{3}{8}\)+\(\dfrac{4}{7}\))
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\(A=\left(37,1-4,5\right)-\left(-4,5+37,1\right)\)
\(=37,1-4,5+4,5-37,1\)
\(=37,1-37,1-4,5+4,5\)
\(=0.\)
\(B=-\left(315.4+275\right)+4.315-\left(10-275\right)\)
\(=-315.4-275+4.315-10+275\)
\(=-315.5+4.315-275+275-10\)
\(=0+0-10=-10.\)
\(C=-\left(\frac{3}{7}+\frac{3}{8}\right)-\left(-\frac{3}{8}+\frac{4}{7}\right)\)
\(=-\frac{3}{7}-\frac{3}{8}+\frac{3}{8}-\frac{4}{7}\)
\(=-\frac{3}{7}-\frac{4}{7}-\frac{3}{8}+\frac{3}{8}\)
\(=-1+0=-1.\)
A=37,1 - 4,5 + 4,5 - 37,1
A=(37,1 - 37,10) + ( 4,5 - 4,5 )
A = 0 + 0 = 0
B= -315.4 - 275 + 4.315 - 10 + 275
B=(-315.4 + 4.315) + ( 275-275) - 10
B= 0 + 0 - 10 = -10
C= -3/7 - 3/8 + 3/8 - 4/7
C = ( -3/7-4/7) + ( 3/8 - 3/8)
C=-7/7 + 0 = -7/7 = -1
a) [(-11,7) + 5,5] + [11,7 + (-2,5)]
= [11,7 + (-5,5)] + [11,7 + (-2,5)]
= 11,7 + (-5,5) + 11,7 + (-2,5)
= 11,7 + [(-5,5) + (-2,5)]
= 11,7 + (-8)
= 3,7
Mấy câu sau bn lm tương tự nha! ^^
a,[(-11.7)+5,5]+[11,7+(-2,5)] b,[(-6.8)+(-56,9)]+(2,8+56,9) c,(37,1-4,5)-(-4,5+37,1) d,-(315.4+275)+4.315-(10-275)
=[(-11,7)+11,7]+[5,5+(-2,5)] =[(-6,8)+2.8]+[(-56,9)+56,9] =37,1-4.5+4.5-37.1 =-315.4+4.315-(10-275+275)
=0+3 =(-4)+0 =(37,1-37,1)+(-4,5+4,5) =4.(-315+315)-10
=3 =(-4) =0+0 =4.0-10
=0 =0-10
=-10
e,(-2.5.0,375.0,4)-[0,125.3,25.(-8)] f,[(-30,27).0,5+(-9,73).0,5]:[3,116.0,8-(-1884).0,8]
=(-10.0,375)-(-1.3,25) ={0,5.[-30,27+(-9,73)]}:[0,8.(3,116+1,884)]
=-3,75-(-3,25) =[0,5.(-40)]:(0,8.5)
=-3,75+3,25 =-20:4
=0,5 =-5
đề của câu d, và 1 phần bị thiếu bên câu a, chỗ không có dấu =
a. = (50,9 - 50,8) * 49,1
= 1 * 49,1
= 49,1
mấy câu dưới tương tự vs cả lp 7 bài này quá dễ r mà
f, \(\dfrac{2^9.4^{10}}{8^8}=\dfrac{2^9.\left(2^2\right)^{10}}{\left(2^3\right)^8}=\dfrac{2^9.2^{20}}{2^{24}}=\dfrac{2^{29}}{2^{24}}=2^5=32\)
a: \(=\left(\dfrac{1}{3}-\dfrac{4}{3}\right)+\dfrac{14}{25}+\dfrac{11}{25}+\dfrac{2}{7}=\dfrac{2}{7}\)
b: \(=\dfrac{3}{7}-\dfrac{5}{2}-\dfrac{3}{5}+\dfrac{4}{7}+\dfrac{3}{2}-\dfrac{2}{5}=1-1-1=-1\)
c: \(=\dfrac{4}{25}+\dfrac{7}{5}\cdot\dfrac{5}{2}-2=\dfrac{4}{25}+\dfrac{7}{2}-2=\dfrac{83}{50}\)
a/\(-\dfrac{1}{6}+\left(-\dfrac{5}{8}+\dfrac{7}{6}\right)\)
\(=\left(-\dfrac{1}{6}+\dfrac{7}{6}\right)-\dfrac{5}{8}\)
\(=1-\dfrac{5}{8}\)
\(=\dfrac{8}{8}-\dfrac{5}{8}=\dfrac{3}{8}\)
b/\(\left(\dfrac{2}{5}\right)^2+5\dfrac{1}{2}\cdot\left(4,5-2\right)+\dfrac{2^3}{\left(-4\right)}\)
\(=\dfrac{4}{25}+\dfrac{11}{2}\cdot\dfrac{5}{2}+-\dfrac{2^3}{2^2}\)
\(=\dfrac{4}{25}+\dfrac{55}{4}+\left(-2\right)\)
\(=\dfrac{16}{100}+\dfrac{1375}{100}-\dfrac{200}{100}\)
\(=\dfrac{1191}{100}=11,91\)
\(a.\dfrac{5}{3}.8,1=\dfrac{40,5}{3}\)
\(b.\dfrac{8}{5}.\left(-4,5\right)=\dfrac{-36}{5}\)
\(c.\dfrac{14}{5}.75\%=\dfrac{10,5}{5}=2,1\)
\(d.\dfrac{11}{8}.\dfrac{37}{12}=\dfrac{407}{96}\)
`a)4/5+5 1/2 xx (4,5-2)+7/10`
`=4/5+11/2*2,5+7/10`
`=0,8+2,2+0,7`
`=3+0,7=3,7`
`b)125%xx 17/4:(1 5/16-0,5)+2008`
`=1,25xx4,25:13/16+2008`
`=85/13+2008`
`=2014 7/13`
`c)5/11+(16/11+1)`
`=5/11+1+5/11+1`
`=2+10/11=32/11`
`d)3/17+11/4+5/8+14/17+3/8`
`=3/17+14/17+5/8+3/8+11/4`
`=1+1+11/4`
`=19/4`
a)
\(\dfrac{4}{5}+5\dfrac{1}{2}x\left(4,5-2\right)=\dfrac{7}{10}\)
<=> \(\dfrac{11}{2}x\times2,5=\dfrac{7}{10}-\dfrac{4}{5}=\dfrac{-1}{10}\)
<=> \(\dfrac{55}{4}x=\dfrac{-1}{10}< =>x=\dfrac{-2}{275}\)
b) \(125\%\times\dfrac{17}{4}:\left(1\dfrac{5}{16}-0,5\right)+2008\)
= \(\dfrac{85}{16}:\left(\dfrac{21}{16}-\dfrac{1}{2}\right)+2008=\dfrac{85}{16}:\dfrac{13}{16}+2008=\dfrac{26189}{13}\)
c) \(\dfrac{5}{11}+\left(\dfrac{16}{11}+1\right)\)
= \(\dfrac{21}{11}+1=\dfrac{32}{11}\)
d) \(\left(\dfrac{3}{17}+\dfrac{14}{17}\right)+\left(\dfrac{5}{8}+\dfrac{3}{8}\right)+\dfrac{11}{4}\)
= 1 + 1 + \(\dfrac{11}{4}\) = \(\dfrac{19}{4}\)
b: \(A=\dfrac{10^7-8+13}{10^7-8}=1+\dfrac{13}{10^7-8}\)
\(B=\dfrac{10^8-7+13}{10^8-7}=1+\dfrac{13}{10^8-7}\)
mà \(10^7-8< 10^8-7\)
nên A>B
c: \(\dfrac{1}{10}A=\dfrac{10^{1992}+1}{10^{1992}+10}=1-\dfrac{9}{10^{1992}+10}\)
\(\dfrac{1}{10}B=\dfrac{10^{1993}+1}{10^{1993}+10}=1-\dfrac{9}{10^{1993}+10}\)
mà \(\dfrac{9}{10^{1992}+10}>\dfrac{9}{10^{1993}+10}\)
nên A<B
Áp dụng công thức bỏ dấu ngoặc:
+ có dấu trừ đằng trước-> đổi dấu tất cả các hạng tử trong ngoặc
+ có dấu cộng đằng trước-> để nguyên dấu các hạng tử trong ngoặc
\(A=\left(37,1-4,5\right)-\left(-4,5\right)+37,1\)
\(A=37,1-4,5+4,5+37,1\)
\(A=2.37,1=74,2\)
\(B=-\left(315,4+275\right)+4,315-\left(10-275\right)\)
\(B=-315,4-275+4,315-10+275\)
\(B=-315,4+4,315-10=-321,085\)
\(C=-\left(\dfrac{3}{7}+\dfrac{3}{8}\right)-\left(-\dfrac{3}{8}+\dfrac{4}{7}\right)\)
\(C=-\dfrac{3}{7}-\dfrac{3}{8}+\dfrac{3}{8}-\dfrac{4}{7}\)
\(C=-1\)
Chúc bạn học tốt!!!