5^2+3^2-(5^16:25^7+3^4:3)
4/5 : ( 4/5 . -5/4) : ( 16/25 - 1/5 )
11/12 : ( 7/9 + -1/3 ) - ( 2/3 - 5/15 )
-2/5 . ( 9/8 + 16/32 -3/4 ) - 9/10
( 4/3 + 8/3 ) . ( 7/4 - 6/4 ) : ( 6/5 + 12/5 + 1/5 )
cách giải bài này như nào: 5/20+1/4+6/24+8/32=bao nhiêu
\(\dfrac{4}{5}\) : (\(\dfrac{4}{5}\) .- \(\dfrac{5}{4}\)) : (\(\dfrac{16}{25}\) - \(\dfrac{1}{5}\))
= \(\dfrac{4}{5}\) : (-1) : (\(\dfrac{16}{25}\) - \(\dfrac{5}{25}\))
= -\(\dfrac{4}{5}\) : \(\dfrac{11}{25}\)
= - \(\dfrac{4}{5}\) x \(\dfrac{25}{11}\)
= - \(\dfrac{20}{11}\)
\(\dfrac{11}{12}\) : (\(\dfrac{7}{9}\) + \(\dfrac{-1}{3}\)) - (\(\dfrac{2}{3}\) - \(\dfrac{5}{15}\))
= \(\dfrac{11}{12}\) : (\(\dfrac{7}{9}\) - \(\dfrac{3}{9}\)) - (\(\dfrac{2}{3}\) - \(\dfrac{1}{3}\))
= \(\dfrac{11}{12}\) : \(\dfrac{4}{9}\) - \(\dfrac{1}{3}\)
= \(\dfrac{11}{12}\) x \(\dfrac{9}{4}\) - \(\dfrac{1}{3}\)
= \(\dfrac{99}{48}\) - \(\dfrac{1}{3}\)
= \(\dfrac{99}{48}\) - \(\dfrac{16}{48}\)
= \(\dfrac{83}{48}\)
4/5 : ( 4/5 . -5/4) : ( 16/25 - 1/5 )
11/12 : ( 7/9 + -1/3 ) - ( 2/3 - 5/15 )
-2/5 . ( 9/8 + 16/32 -3/4 ) - 9/10
( 4/3 + 8/3 ) . ( 7/4 - 6/4 ) : ( 6/5 + 12/5 + 1/5 )
\(\dfrac{4}{5}\): (\(\dfrac{4}{5}\).-\(\dfrac{5}{4}\)) : (\(\dfrac{16}{25}\) - \(\dfrac{1}{5}\))
=\(\dfrac{4}{5}\) x - 1: (\(\dfrac{16}{25}\) - \(\dfrac{5}{25}\))
= - \(\dfrac{4}{5}\) : \(\dfrac{11}{25}\)
= - \(\dfrac{4}{5}\) x \(\dfrac{25}{11}\)
= - \(\dfrac{20}{11}\)
\(\dfrac{11}{12}\): (\(\dfrac{7}{9}\) + - \(\dfrac{1}{3}\)) - (\(\dfrac{2}{3}\) - \(\dfrac{5}{15}\))
= \(\dfrac{11}{12}\) : (\(\dfrac{7}{9}\) - \(\dfrac{3}{9}\)) - (\(\dfrac{2}{3}\) - \(\dfrac{1}{3}\))
= \(\dfrac{11}{12}\) : \(\dfrac{4}{9}\) - \(\dfrac{1}{3}\)
= \(\dfrac{11}{12}\) x \(\dfrac{9}{4}\) - \(\dfrac{1}{3}\)
= \(\dfrac{99}{48}\) - \(\dfrac{16}{48}\)
= \(\dfrac{83}{48}\)
- \(\dfrac{2}{5}\).(\(\dfrac{9}{8}\) + \(\dfrac{16}{32}\) - \(\dfrac{3}{4}\)) - \(\dfrac{9}{10}\)
= - \(\dfrac{2}{5}\).(\(\dfrac{9}{8}\) + \(\dfrac{4}{8}\) - \(\dfrac{6}{8}\)) - \(\dfrac{9}{10}\)
= - \(\dfrac{2}{5}\).(\(\dfrac{13}{8}\) - \(\dfrac{6}{8}\)) - \(\dfrac{9}{10}\)
= - \(\dfrac{2}{5}\).\(\dfrac{7}{8}\) - \(\dfrac{9}{10}\)
= - \(\dfrac{7}{20}\) - \(\dfrac{18}{20}\)
= - \(\dfrac{5}{4}\)
4 [-2(8:4)+15(-3)-(-12)
3(25:5-14:2)-5(6:2)
[-15:(-3)]-3[2(5-9):3]
-16:(-4)[7-2(15:3)]
-7[8-3(14:7)-12:(-4)]-3(-2)
2[3-9:(-3)+2(5-7)]-18:(-9)
-16:(-8)+5[3-15:5+2] (-3+4)
4.[-2(8:4) + 15(-3) - (-12)]
= 4.[-2.2 - 45 + 12]
= 4.[-4 - 45 + 12]
= 4.[-49 + 12]
= 4.[-37]
= - 148
3.(25 : 5 - 14 : 2) - 5.(6:2)
= 3.(5 - 7) - 5.3
= 3.(-2) - 15
= - 6 - 15
= -21
[-15:(-3)] - 3.[2.(5 - 9):3]
= 5 - 3.[(2.-4):3]
= 5 - 3.[-8:3]
= 5 + 24:3
= 5 + 8
= 13
4[-2(8:4)+15:(-3)-(-12)
3(25:5-14:2)-5(6:2)
[-15:(-3)]-3[2(5-9):3)]
-16:(-4)[7-2(15:3)]
-7[8-3(14:7)-12:(-4)]-3(-2)
2[3-9:(-3)+2(5-7)]-18:(-9)
-16:(-8)+5[3-15:5+2] (-3+4)
1/ (\(\left(-\dfrac{2}{3}\right)\)\(^2\) x \(\dfrac{-9}{8}\) - 25% x \(\dfrac{-16}{5}\)
2/ -1\(\dfrac{2}{5}\) x 75% + \(\dfrac{-7}{5}\) x 25%
3/ -2\(\dfrac{3}{7}\) x (-125%) + \(\dfrac{-17}{7}\) x 25%
4/ (-2)\(^3\) x (\(\dfrac{3}{4}\) x 0.25) : (2\(\dfrac{1}{4}\) - 1\(\dfrac{1}{6}\))
1) Ta có: \(\left(-\dfrac{2}{3}\right)^2\cdot\dfrac{-9}{8}-25\%\cdot\dfrac{-16}{5}\)
\(=\dfrac{4}{9}\cdot\dfrac{-9}{8}-\dfrac{1}{4}\cdot\dfrac{-16}{5}\)
\(=\dfrac{-1}{2}+\dfrac{4}{5}\)
\(=\dfrac{-5}{10}+\dfrac{8}{10}=\dfrac{3}{10}\)
2) Ta có: \(-1\dfrac{2}{5}\cdot75\%+\dfrac{-7}{5}\cdot25\%\)
\(=\dfrac{-7}{5}\cdot\dfrac{3}{4}+\dfrac{-7}{5}\cdot\dfrac{1}{4}\)
\(=\dfrac{-7}{5}\left(\dfrac{3}{4}+\dfrac{1}{4}\right)=-\dfrac{7}{5}\)
3) Ta có: \(-2\dfrac{3}{7}\cdot\left(-125\%\right)+\dfrac{-17}{7}\cdot25\%\)
\(=\dfrac{-17}{7}\cdot\dfrac{-5}{4}+\dfrac{-17}{7}\cdot\dfrac{1}{4}\)
\(=\dfrac{-17}{7}\cdot\left(\dfrac{-5}{4}+\dfrac{1}{4}\right)\)
\(=\dfrac{17}{7}\)
4) Ta có: \(\left(-2\right)^3\cdot\left(\dfrac{3}{4}\cdot0.25\right):\left(2\dfrac{1}{4}-1\dfrac{1}{6}\right)\)
\(=\left(-8\right)\cdot\left(\dfrac{3}{4}\cdot\dfrac{1}{4}\right):\left(\dfrac{9}{4}-\dfrac{7}{6}\right)\)
\(=\left(-8\right)\cdot\dfrac{3}{16}:\dfrac{54-28}{24}\)
\(=\dfrac{-3}{2}\cdot\dfrac{24}{26}\)
\(=\dfrac{-72}{52}=\dfrac{-18}{13}\)
1/2×(2×x-3)+105/2=-137/2 A,-12×25-25×75-25×13 B,-50/7×49/10-35/2×-10/7+ -25/3× -9/5 C,-354+265-156+125 D,-74+40-50+16-35 E,-2/3×4/5+-4/5×4/3-0,125
1/2×(2×x-3)+105/2=-137/2
1/2×(2×x-3)=-137/2-105/2
1/2×(2×x-3)=-242/2
2×x-3=-242/2:1/2
2×x-3=-242
2.x=(-242)+3
2.x=239
x=239:2
x=239/2
A,-12×25-25×75-25×13
=[(-12)-75-13].25
=(-100).25
-2500
B,-50/7×49/10-35/2×-10/7+ -25/3× -9/5
=(-35)-(-25)+(-13)
=-23
C,-354+265-156+125
=[(-354)-156]+(265+125)
=(-510)+277
=-233
D,-74+40-50+16-35
=(-74)+40+(-50)+16+(-35)
=[(-74)+(-35)]+[40+(-50)+16]
=(-109)+26
=-83
E,-2/3×4/5+-4/5×4/3-0,125
=-2/3.4/5+4/5.(-4/3)-0,125
=4/5.[-2/3+(-4/3)]-0,125
=4/5.(-2)-0,125
=-8/5-0,125
=(-1,6)+(-0,125)
=-1,725
1) 15 - 98 + 91 - 75 + 98 - 91 - 13
2) 5 - ( 4 - 7 + 12 ) + ( 4 - 7 + 12 ) - 10
3) 20 - ( 25 - 98 + 12 ) + ( 25 - 98 - 72 )
4) - ( 17 - 4 + 13 + | -17 | + 13 )
5) | -16 | - ( 5 - 8 + 7) - ( - 5 ) - 8
6) -20 + ( 3 - 16 + 7 ) - ( 3 - 16 + 7 )
7) 32 - ( 5 - 8 + 32 ) + ( 5 - 8 + 6 )
8) -20 - ( 25 - 11 + 8 ) + ( 25 - 8 + 20 )
1)-73
2)-5
3)-64
4)-56
5)9
6)-20
7)6
8)-2
Tick nha
đăng tùng câu một thui bạn
bạn làm thế chắc tụi mk xỉu gần hết
A = 9/8 - 8/9 + 3/25 + 1/4 - 5/16 + 19/25 - 1/9 + 2/25 - 1/81
B = -1/3- 8/35 - 2/9 - 1/35 + 4//5 + -4/9 + 3/7
\(A=\dfrac{9}{8}-\dfrac{8}{9}+\dfrac{3}{24}+\dfrac{1}{4}-\dfrac{5}{16}+\dfrac{19}{25}-\dfrac{1}{9}+\dfrac{2}{25}-\dfrac{1}{81}\)
\(=\dfrac{9}{8}+\dfrac{1}{4}-\dfrac{5}{16}+\dfrac{1}{8}-\dfrac{8}{9}-\dfrac{1}{9}-\dfrac{1}{81}+\dfrac{19}{25}+\dfrac{2}{25}\)
\(=\dfrac{10}{8}+\dfrac{1}{4}-\dfrac{5}{16}-1-\dfrac{1}{81}+\dfrac{21}{25}\)
\(=\dfrac{20+4-5}{16}-\dfrac{82}{81}+\dfrac{21}{25}\)
\(=\dfrac{19}{16}-\dfrac{82}{81}+\dfrac{21}{25}\)
\(=\dfrac{32891}{16\cdot81\cdot25}\)
b: \(B=-\dfrac{1}{3}-\dfrac{8}{35}-\dfrac{2}{9}-\dfrac{1}{35}+\dfrac{4}{5}-\dfrac{4}{9}+\dfrac{3}{7}\)
\(=\dfrac{-1}{3}-\dfrac{2}{9}-\dfrac{4}{9}-\dfrac{8}{35}-\dfrac{1}{35}+\dfrac{4}{5}+\dfrac{3}{7}\)
\(=\dfrac{-3-2-4}{9}+\dfrac{-9}{35}+\dfrac{28+15}{35}\)
\(=-1+\dfrac{-9+43}{35}=-1+\dfrac{34}{35}=-\dfrac{1}{35}\)
Tính P=16^7.5^4-16^7 .5^3 / 16^7. 25^2- 16^7.5^3
\(P=\dfrac{16^7\cdot5^3\left(5-1\right)}{16^7\cdot\left(25^2-5^3\right)}=\dfrac{5^3\cdot2^2}{5^4-5^3}=\dfrac{5^3\cdot2^2}{5^3\cdot\left(5-1\right)}=1\)