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a, \(3^4\div3^2-\left[120-\left(2^6.2+5^2.2\right)\right]\)
\(=3^2-\left\{120-\text{[}2.\left(2^6+5^2\right)\text{]}\right\}\)
\(=3^2-\left(120-2\cdot89\right)\)
\(=9--58=9+58=67\)
1. \(a,3^4:3^2-\left[120-(2^6\cdot2+5^2\cdot2)\right]\)
\(=3^2-\left[120-\left\{(2^6+5^2)\cdot2\right\}\right]\)
\(=3^2-\left[120-\left\{(64+25)\cdot2\right\}\right]\)
\(=9-\left[120-89\cdot2\right]\)
\(=9-\left[120-178\right]=9-(-58)=67\)
b, Tương tự như bài a
2.a,\(4^x\cdot5+4^2\cdot2=2^3\cdot7+56\)
\(\Leftrightarrow4^x\cdot5+16\cdot2=8\cdot7+56\)
\(\Leftrightarrow4^x\cdot5+32=56+56\)
\(\Leftrightarrow4^x\cdot5+32=112\)
\(\Leftrightarrow4^x\cdot5=80\)
\(\Leftrightarrow4^x=16\Leftrightarrow4^x=4^2\Leftrightarrow x=2\)
\(b,24:(2x-1)^3-2=1\)
\(\Leftrightarrow24:(2x-1)^3=3\)
\(\Leftrightarrow(2x-1)^3=8\)
\(\Leftrightarrow(2x-1)^3=2^3\)
\(\Leftrightarrow2x-1=2\)
Làm nốt là xong thôi
1.
1+2+3+...+99+100
=[(100-1):1+1]x[(100+1):2]
=100x50,5
=5050
2.
a, x2017=x
=> x=1 hoặc x=-1
b, 2x+2=250:8
=> 2x+2=250:23
=> 2x+2=247
=> x+2=47
=> x= 45
c, 3x+3x+2=810
=> 3x+3x+2=34+36
=> x=4
chúc bạn học tốt k mình nha .
A) Ta có: 28 x ( 231 + 69) + 72 x ( 231 + 69)
= ( 28 + 72 ) x ( 231 + 69)
= 100 x 300
= 30 000
b) 2017 - { 52 x 22 - 11 [ 72 - 5 x 23 + 8 x (112 - 121) ] }
= 2017 - { 100 - 11 [ 49 - 40 + 8 x 0 ] }
= 2017 - { 100 - 1 [ 9 + 0 ]
= 2017 - { 100 - 1 x 9 }
= 2017 - { 99 x 9 }
= 2017 - 891
= 1126
c) 10 - { [ ( x : 3 + 17) : 10 + 2 x 24 ] : 10 } = 5
=> { [( x : 3 + 17 ) : 10 + 2 x 16 ] : 10 } = 10 - 5 = 5
=> [ ( x : 3 + 17) : 19 + 32 ] = 5 x 10 = 50
=> ( x : 3 + 17 ) : 19 = 50 - 32 = 18
=> x : 3 + 17 = 18 x 19 = 342
=> x : 3 = 342 - 17 = 325
=> x = 325 x 3
=> x = 975
Vậy x = 975
câu 1: 10+9(2^5+2^3):5-2018= 10+9(32+8):5-1= 10+9.40:5-1= 10+360:5-1= 10+72-1=92-1=91
câu 2:mình chưa học số âm bạn nhé
B,2(x+17):8=5
2(x+17)=5.8
2(x+17)=40
x+17=40:2
x+17=20
x=20-17
x=3
\(\left(3^{2016}\cdot11+3^{1018}\cdot50\right):\left(3^{2017}\cdot4^2\right)\)
\(=\left(3^{2016}\cdot11+3^{2016}\cdot3^2\cdot50\right):\left(3^{2017}\cdot4^2\right)\)
\(=\left(3^{2016}\cdot11+3^{2016}\cdot450\right):\left(3^{2017}\cdot16\right)\)
\(=\left[3^{2016}\cdot\left(11+450\right)\right]:\left(3^{2017}\cdot16\right)\)
\(=\left[3^{2016}\cdot461\right]:\left(3^{2017}\cdot16\right)\)
\(=\frac{3^{2016}\cdot461}{3^{2017}\cdot16}\)
\(=\frac{3^{2016}\cdot461}{3\cdot3^{2016}\cdot16}\)
\(=\frac{461}{3\cdot16}=\frac{461}{48}\)