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a ) 11,3 + 6,9 + 8,7,+ 13,1
= ( 11,3 + 8,7) + (6,9 + 13,1)
= 20 + 20
= 40
b) ( 1 + 3 + 5 + ...+ 2007 + 2009 + 2011 ) ( 125125 x127 - 127127 x 125)
= ( 1 + 3 + 5 + ... + 2007 + 2009 + 2011 ) ( 15890875 - 15890875 )
= ( 1 + 3 + 5 + ... + 2007 + 2009 + 2011 ) x 0
= 0
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 \(x=\dfrac{52}{5}\).
\(Toru\)
43 x 53 + 47 x 196 - 47 x 52
= 43 x 53 + 47 x ( 196 - 52 )
= 43 x 53 + 47 x 144 -> Hết thuận tiện rồi bạn ơi ( kiểm tra lại bài thử nha ).
a, Bội (6) = {0; 6}
b, Số đối của: -4 = 4 ; 0 = 0
c, \(3^2+10:2=9+10:2=9+5=14\)
Câu 2:
\(\left(15-\left[3^{20}:3^{19}+2022^0\right]\right):11=\left(15-\left[3^{20-19}+1\right]\right):11=\left(15-\left[3^1+1\right]\right):11\)
\(=\left(15-4\right):11=11:11=1\)
Câu 3:
\(2x-7=39\)
\(2x=39+7\)
\(2x=46\)
\(x=46:2\)
\(x=23\)
a; 23.17 + 23.83 - 14
= 23.(17 + 83) - 14
= 8.100 - 14
= 800 - 14
= 786
b; -176 + 197 - (-76 + 97)
= -176 + 197 + 76 - 97
= -(176 - 76) + (197 - 97)
= - 100 + 100
= 0
c; 2022 - [45 - (6 - 1)2 ] + 20210
= 2022 - [45 - 52] + 1
= 2022 + 1 - [45 - 25]
= 2023 - 20
= 2003
\(50\%+\dfrac{7}{12}-\dfrac{1}{2}\)
\(=\dfrac{1}{2}+\dfrac{7}{12}-\dfrac{1}{2}\)
\(=\left(\dfrac{1}{2}-\dfrac{1}{2}\right)+\dfrac{7}{12}\)
\(=\dfrac{7}{12}\)
_______________
\(2022\cdot67+2022\cdot43-2022\cdot10\)
\(=2022\cdot\left(67+43-10\right)\)
\(=2022\cdot100\)
\(=202200\)
_____________________
\(10,3+6,9+8,7+13,1\)
\(=\left(13,1+6,9\right)+\left(10,3+8,7\right)\)
\(=20+19\)
\(=39\)
___________________
\(17,58\times43+57\times17,58\)
\(=17,58\times\left(43+57\right)\)
\(=17,58\times100\)
\(=1758\)
tớ ghi sai nên cập nhật lại câu hỏi, phần 3 các bạn trả lời sớm mà bị sai thì bỏ qua cho tớ nhé