2^3 nhân 5+3^19:3^17-2022^0
28/25-17/19-3/25+2022/2023-2/19
28/25 - 17/19 - 3/25 + 2022/2023 - 2/19
= 28/25 - 3/25 - 17/19 + 2/19 - 2022/2023
= 1 - 1 - 2022/2023
= -2022/2023
a: =35/17-18/17-9/5+4/5
=1-1=0
b: =-7/19(3/17+8/11-1)
=7/19*18/187=126/3553
c: =26/15-11/15-17/3-6/13
=1-6/13-17/3
=7/13-17/3=-200/39
Bài1:Thực hiện phép tính:
a,(2^4.3.5^2):{450:[450-(4.5^3-2^3.5^2)]}
b,3^3.5^2-20{90-[164-2.(7^8:7^6+7^0)]}
c,[(18^7:18^6-17).2022-1986].5.1^2022-13^2.2020^0
Bài2:Tìm x:
a,(2^x+1)^2+3.(2^2+1)=2^2.10
b,3.(x-7)+2.(x+5)=41
(GIÚP MIK VỚI Ạ)
Bài 1.
\(a,\left(2^4\cdot3\cdot5^2\right):\left\{450:\left[450-\left(4\cdot5^3-2^3\cdot5^2\right)\right]\right\}\)
\(=\left(16\cdot3\cdot25\right):\left\{450:\left[450- \left(4\cdot125-8\cdot25\right)\right]\right\}\)
\(=\left(48\cdot25\right):\left\{450:\left[450-\left(500-200\right)\right]\right\}\)
\(=1200:\left[450:\left(450-300\right)\right]\)
\(=1200:\left(450:150\right)\)
\(=1200:3\)
\(=400\)
\(---\)
\(b,3^3\cdot5^2-20\left\{90-\left[164-2\cdot\left(7^8:7^6+7^0\right)\right]\right\}\)
\(=27\cdot25-20\left\{90-\left[164-2\cdot\left(7^2+1\right)\right]\right\}\)
\(=675-20\left\{90-\left[164-2\cdot\left(49+1\right)\right]\right\}\)
\(=675-20\left[90-\left(164-2\cdot50\right)\right]\)
\(=675-20\left[90-\left(164-100\right)\right]\)
\(=675-20\left(90-64\right)\)
\(=675-20\cdot26\)
\(=675-520\)
\(=155\)
\(---\)
\(c,\left[\left(18^7:18^6-17\right)\cdot2022-1986\right]\cdot5\cdot1^{2022}-13^2\cdot2020^0\)
\(=\left[\left(18-17\right)\cdot2022-1986\right]\cdot5\cdot1-169\cdot1\)
\(=\left(1\cdot2022-1986\right)\cdot5-169\)
\(=\left(2022-1986\right)\cdot5-169\)
\(=36\cdot5-169\)
\(=180-169\)
\(=11\)
Bài 2.
\(a) (2^x+1)^2+3\cdot(2^2+1)=2^2\cdot10\\\Rightarrow (2^x+1)^2+3\cdot(4+1)=4\cdot10\\\Rightarrow (2^x+1)^2+3\cdot5=40\\\Rightarrow (2^x+1)^2+15=40\\\Rightarrow (2^x+1)^2=40-15\\\Rightarrow (2^x+1)^2=25\\\Rightarrow (2^x+1)^2= (\pm 5)^2\\\Rightarrow \left[\begin{array}{} 2^x+1=5\\ 2^x+1=-5 \end{array} \right.\\ \Rightarrow \left[\begin{array}{} 2^x=4\\ 2^x=-6 (vô.lí) \end{array} \right. \\ \Rightarrow 2^x=2^2\\\Rightarrow x=2\)
Vậy \(x=2\).
\(---\)
\(b)3\cdot(x-7)+2\cdot(x+5)=41\\\Rightarrow 3\cdot x+3\cdot(-7)+2\cdot x+2\cdot5=41\\\Rightarrow 3x-21+2x+10=41\\\Rightarrow (3x+2x)+(-21+10)=41\\\Rightarrow 5x-11=41\\\Rightarrow 5x=41+11\\\Rightarrow 5x=52\\\Rightarrow x=\dfrac{52}{5}\)
Vậy \(x=\dfrac{52}{5}\).
\(Toru\)
Bài 13: Dấu <, =, >
10 … 10 + 3
11 + 2…. 2 + 11
9 … 10 + 9
10 … 10 + 0
17 – 4 … 14 - 3
18 – 4 … 12
15 … 15 – 1
17 + 1… 17 + 2
12+ 5 … 16
16 … 19 - 3
15 – 4 … 10 + 1
19 – 3 … 11
10 < 10 + 3
11 + 2=2 + 11
9 < 10 + 9
10 = 10 + 0
17 – 4 > 14 - 3
18 – 4 >12
15 > 15 – 1
17 + 1<17 + 2
12+ 5 > 16
16 =19 - 3
15 – 4 =10 + 1
19 – 3 >11
Giúp mình với mình đang cần gấp. 2^2.2^3-(2022^0+19):2^2
`2^2 .2^3-(2022^0+19):2^2`
`=4.8-(1+19):4`
`=4.8-20:4`
`=32-5`
`=27`
\(2^2\cdot2^3-\left(2022^0+19\right):2^2=4\cdot8-\left(1+19\right):4=32-20:4=32-5=27\)
2^2.2^3-(2022^0+19):2^2
=4.8-(1+19):4
=32-5=27
Tính thuận tiện
20080 x 17 + 1999 x 17 - ( 53 : 52 + 23 x 2)
22 x 18 + 23 21 + 23 x 61 - 23
2001 x 19 + 20010 x 19 : ( 33 : 33 + 17 x 150)
Tính bằng cách nhanh nhất:
a/ 2010-2012+2014-2016+2018-2020+2021-2022
b/ 1-3+5-7+9-11+13-15+17-19+21
TÍNH NHANH :
A . 2/3 . 15/17 + 2/3 . 2/17
B . 20/19 : 3/7 - 1/19 : 3/7
CHÚ Ý : DẤU CHẤM LÀ NHÂN
a, \(\frac{2}{3}\).\(\frac{15}{17}\)+\(\frac{2}{3}\).\(\frac{2}{17}\)
= \(\frac{2}{3}\).( \(\frac{15}{17}\)+ \(\frac{2}{17}\))
= \(\frac{2}{3}\). 1
=\(\frac{2}{3}\)
b,\(\frac{20}{19}\):\(\frac{3}{7}\)- \(\frac{1}{19}\): \(\frac{3}{7}\)
= ,\(\frac{20}{19}\).\(\frac{7}{3}\)- \(\frac{1}{19}\).\(\frac{7}{3}\)
= \(\frac{7}{3}\).(\(\frac{20}{19}\)-\(\frac{1}{19}\))
=\(\frac{7}{3}\).1
=\(\frac{7}{3}\)
TÍNH NHANH :
A . 2/3 . 15/17 + 2/3 . 2/17
B . 20/19 : 3/7 - 1/19 : 3/7
CHÚ Ý : DẤU CHẤM LÀ NHÂN
A. \(\frac{2}{3}.\frac{15}{17}+\frac{2}{3}.\frac{2}{17}\)
=\(\frac{2}{3}.\left(\frac{15}{17}+\frac{2}{17}\right)\)
=\(\frac{2}{3}.1\)
=\(\frac{2}{3}\)
B. \(\frac{20}{19}:\frac{3}{7}-\frac{1}{19}:\frac{3}{7}\)
=\(\left(\frac{20}{19}-\frac{1}{19}\right):\frac{3}{7}\)
=1:\(\frac{3}{7}\)
=\(\frac{7}{3}\)