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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
5 x [(85 - 35 : 7) : 8 + 90] - 50
= 5 x [(85 - 5) : 8 + 90] - 50
= 5 x [80 : 8 + 90] - 50
= 5 x [10 + 90] - 50
= 5 x 100 - 50
= 500 - 50
= 450
\(5\times\left[\left(85-35:7\right):8+90\right]-50\\ =5\times\left[\left(85-5\right):8+90\right]-50\\ =5\times\left[80:8+90\right]-50\\ =5\times\left[10+90\right]-50\\ =5\times100-50\\ =500-50=450\)
5[(85 - 35 : 7) : 8 + 90] - 50
= 5[(85 - 5) : 8 + 90] - 50
= 5.[ 80 : 8 + 90] - 50
= 5.[ 10 + 90 ] - 50
= 5. 100 - 50
= 500 - 50
= 450
5[(85 - 35 : 7) : 8 + 90] - 50
= 5[(85 - 5) : 8 + 90] - 50
= 5.[ 80 : 8 + 90] - 50
= 5.[ 10 + 90 ] - 50
= 5. 100 - 50
= 500 - 50
= 450
5 . [(85 - 35 : 7) : 80 + 90] - 50
= 5 . [(85 - 5) : 80 + 90] - 50
= 5 . [80 : 80 +90] - 50
= 5 . [1 + 90] - 50
= 5 . 91 - 50
= 455 - 50 = 405
5 [ ( 85 - 35 : 7 ) : 80 + 90 ] - 50
= 5 [ ( 85 - 5 ) : 80 + 90 ] - 50
= 5 [ 80 : 80 + 90 ] - 50
= 5 [ 1 + 90 ] - 50
= 5 . 91 - 50
= 455 - 50
= 405
=))
\(5.\left[\left(85-35:7\right):8+90\right]\)
\(=5.\left[\left(85-5\right):8+90\right]\)
\(=5.\left(80:8+90\right)\)
\(=5.\left(10+90\right)\)
\(=5.100\)
\(=500\)
\(5\left[\left(85-35:7\right):8+90\right]\)
\(=5\left[\left(85-5\right):8=90\right]-50\)
\(=5\left[80:8+90\right]-50\)
\(=5\left[10+90\right]-50\)
\(=5.100-50\)
\(=450\)
\(a.83-\left(-756-17\right)+\left(50-756\right)=83+756+17+50-756=\left(83+17\right)+50+\left(756-756\right)=100+50=150\)
\(b.-105+\left(-34-95\right)-166=-105+\left(-34\right)-95-166=\left(-105-95\right)-\left(34+166\right)=\left(-200\right)-200=-400\)
\(c,-\left(39+228-407\right)+\left(118-161\right)=-39-228+407+118-161=\left(-39-161\right)-\left(228-118\right)+407=\left(-200\right)-110=-310\)
\(d,5^{10}:5^8+60:12+\left(-10\right)=5^2+5+\left(-10\right)=5\left(5+1-2\right)=5.4=20\)
\(e,-342-\left(161-342\right)-39=-342-161+342-39=\left(-342+342\right)-\left(161+39\right)=-200\)\(g,7^5:7^3+\left(-187-149\right)-213=7^2+\left(-187\right)-149-213=\left(49-149\right)-\left(187+213\right)=\left(-100\right)-400=-500\)
Câu 3:
a) \(\dfrac{12}{36}=\dfrac{12:12}{36:12}=\dfrac{1}{3}\)
\(\dfrac{-16}{20}=\dfrac{-16:4}{20:4}=\dfrac{-4}{5}\)
b) \(\dfrac{21}{105}=\dfrac{21:21}{105:21}=\dfrac{1}{5}\)
\(\dfrac{35}{150}=\dfrac{35:5}{150:5}=\dfrac{7}{30}\)
Câu 4:
a) \(\dfrac{3}{10}+\dfrac{5}{10}=\dfrac{3+5}{10}=\dfrac{8}{10}=\dfrac{4}{5}\)
b) Ta có: \(\left(-27\right)\cdot36+64\cdot\left(-27\right)+23\cdot\left(-100\right)\)
\(=\left(-27\right)\cdot\left(64+36\right)+23\cdot\left(-100\right)\)
\(=-27\cdot100-23\cdot100\)
\(=100\left(-27-23\right)\)
\(=-50\cdot100=-5000\)
c) \(\dfrac{5}{8}+\dfrac{3}{12}=\dfrac{15}{24}+\dfrac{6}{24}=\dfrac{21}{24}=\dfrac{7}{8}\)
d) Ta có: \(\dfrac{-2}{17}+\dfrac{3}{19}+\dfrac{-15}{17}+\dfrac{16}{19}+\dfrac{5}{6}\)
\(=\left(-\dfrac{2}{17}+\dfrac{-15}{17}\right)+\left(\dfrac{3}{19}+\dfrac{16}{19}\right)+\dfrac{5}{6}\)
\(=-1+1+\dfrac{5}{6}\)
\(=\dfrac{5}{6}\)
1)2763 + 152 = 2915
2)(-7) + (-14) = -21
3)(-35) + (-9) = -44
Tính:
\(5\times\left(\dfrac{105-35:5}{10}+90\right)-50=5\times\left(\dfrac{105-7}{10}+90\right)-50\)
\(=5\times\dfrac{98}{10}+5\times90-50=\dfrac{98}{2}+450-50\)
\(=49+400=449\)
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