2^3 nhân 5+3^19:3^17-2022^0
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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
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 \(x=\dfrac{52}{5}\).
\(Toru\)
`2^2 .2^3-(2022^0+19):2^2`
`=4.8-(1+19):4`
`=4.8-20:4`
`=32-5`
`=27`
\(2^2\cdot2^3-\left(2022^0+19\right):2^2=4\cdot8-\left(1+19\right):4=32-20:4=32-5=27\)
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
=\(\frac{7}{3}\)
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:\(\frac{3}{7}\)
=\(\frac{7}{3}\)
2³.5 + 3¹⁹ : 3¹⁷ - 2022⁰
= 8.5 + 3² - 1
= 40 + 9 - 1
= 48
2^3 * 5+3^19:3^17-2022^0
= 8*5+3^19:3^17-1
=8*5+3^2-1
=8*5+9-1
=40+9-1
=48