nếu
1+1=2
2+2=8
3+3=18
vậy 4+4=?
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`2023 \times 14 + 2023 \times 83+97`
`= 2023 \times (14+83) + 97`
`= 2023 \times 97 + 97`
`= 196231 +97`
`= 196328`
`78 \times 31+78 \times 24+78 \times 17+78 \times 22 \times 72`
`= 78 \times (31+24+17+22 \times 72)`
`= 78 \times (31+24+17+1584)`
`= 78 \times 1656`
`= 129168`
`371 \times 69+371 \times 14+629 \times 83`
`= 371 \times (69+14) + 629 \times 83`
`= 371 \times 83 + 629 \times 83`
`= 83 \times (371+629)`
`= 83 \times 1000`
`= 83000`
`91 \times 51+49 \times 163-49 \times 72`
`= 91 \times 51 + 49 \times (163-72)`
`= 91 \times 51+49 \times 91`
`= 91 \times (51+49)`
`= 91 \times 100`
`= 9100`
`3 \times 12 \times 17 + 4 \times 9 \times 81 + 2 \times 6 \times 6`
`= 9 \times 4 \times 17 + 4 \times 9 \times 81 + 4 \times 9 \times 2`
`= 9 \times 4 \times (17+81+2)`
`= 36 \times 100`
`= 3600`
`42 \times 17 + 58 \times 83 \times 42 \times 83 + 42 \times 83 + 17`
`= 17 \times (42+58) + 42 \times 83 \times (58 \times 83 +1)`
`= 17 \times 100 + 3486 \times 4815`
`= 1700 + 16785090`
`= 16786790`
1: =7/2+23/5=35/10+46/10=81/10
2: =6/8+53/8=59/8
3: =43/8+65/7=821/56
2:
a: \(2^2\cdot3-4=4\cdot3-4=12-4=8\)
b: \(16-2^3\cdot2=16-2^4=16-16=0\)
c: \(4^2-4\cdot2=16-8=8\)
d: \(3^3-2\cdot3^2=27-2\cdot9=27-18=9\)
e: \(7^2-9\cdot2^2=49-9\cdot4=49-36=13\)
f: \(2^2\cdot3+4^2=4\cdot3+16=12+16=28\)
Bài 1:
a: \(13+21\cdot5-\left(198:11-8\right)\)
\(=13+105-18+8\)
=21+87
=108
b: \(272:16-5+4\cdot\left(30-5-255:17\right)\)
\(=17-5+4\cdot\left(30-5-15\right)\)
\(=12+4\cdot10=12+40=52\)
c: \(15\cdot24-14\cdot5\cdot\left(145:5-27\right)\)
\(=360-70\left(29-27\right)\)
=360-140
=220
d: \(18\cdot3-18\cdot2+3\cdot\left(\dfrac{51}{17}\right)\)
\(=18\left(3-2\right)+3\cdot3\)
=18+9
=27
e: \(64+115+36-25\cdot8\cdot6\cdot2^2\cdot3+4^2\)
\(=100+115-200\cdot6\cdot4\cdot3+16\)
\(=231-14400=-14169\)
f: \(15\cdot8-\left(17-30+83\right)-144:6\)
\(=120-24-\left(100-30\right)\)
=96-70
=26
a,=\(\dfrac{\left(2-\dfrac{1}{3}+\dfrac{1}{4}\right).12}{\left(2+\dfrac{1}{6}-\dfrac{1}{4}\right).12}\)+\(\dfrac{\left(\dfrac{3}{5}-\dfrac{1}{4}+\dfrac{1}{2}\right).20}{\left(\dfrac{1}{2}+\dfrac{3}{4}-\dfrac{2}{5}\right).20}\)
=\(\dfrac{24-4+3}{24+2-3}\) +\(\dfrac{12-5+10}{10+15-8}\)(nhân từng số hạng với 12;20)
=\(\dfrac{23}{23}\)+\(\dfrac{17}{17}\) =1+1=2
b,=\(\dfrac{5.\left(\dfrac{1}{79}\right)+5.\left(\dfrac{1}{83}\right)+\dfrac{1}{17}}{17.\left(\dfrac{1}{79}\right)+17.\left(\dfrac{1}{83}\right)+\dfrac{1}{5}}\)=\(\dfrac{5.\left(\dfrac{1}{79}+\dfrac{1}{83}\right)+\dfrac{1}{17}}{17.\left(\dfrac{1}{79}+\dfrac{1}{83}\right)+\dfrac{1}{5}}\)
4+4=32
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