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a. Chu vi là \(\left(12+5\right).2=34\left(m\right)\)
Diện tích là \(12.5=60\left(m^2\right)=600000\left(cm^2\right)\)
b. Cần lát \(600000:\left(40.40\right)=375\) viên gạch
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3x . 2 + 15 = 33
3x . 2 = 33 - 15 = 18
3x = 18 : 2 = 9 = 32
=> x = 2
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\(\Leftrightarrow164-4\left(x-5\right)=80\\ \Leftrightarrow4\left(x-5\right)=84\\ \Leftrightarrow x-5=21\Leftrightarrow x=26\)
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\(\frac{2^{15}\cdot9^4}{6^3.8^3}=\frac{2^{15}.3^8}{2^3.3^3.2^9}=\frac{2^{15}.3^8}{2^{12}.3^3}=2^3.3^5=8.243=1944\)
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200 : [ 100 - ( 52. 20100 +50 ) ] + ( 72 - 6. 23 )2021
= 200 : [ 100 - ( 25 . 1 + 50 ) ] + ( 49 - 6 . 8 ) 2021
= 200 : [ 100 - ( 25 + 50 ) ] + ( 49 - 48 ) 2021
= 200 : [ 100 - 75 ] + 12021
= 200 : [ 100 - 75 ] + 1
= 200 : 25 + 1
= 8 + 1
= 9
\(200:[100-\left(5^2.2021^0+50\right)]+\left(7^2-6.2^3\right)^{2021}\)
\(200:[100-\left(25+50\right)]+\left(49-48\right)^{2021}\)
\(200:25+1\)
=8+1=9
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\(\left|x+\frac{4}{15}\right|-\left|-3,75\right|=-\left|-2,15\right|\)
=> \(\left|x+\frac{4}{15}\right|-3,75=-2,15\)
=> \(\left|x+\frac{4}{15}\right|=-2,15+3,75\)
=> \(\left|x+\frac{4}{15}\right|=1,6=\frac{8}{5}\)
=> \(\orbr{\begin{cases}x+\frac{4}{15}=\frac{8}{5}\\x+\frac{4}{15}=\frac{-8}{5}\end{cases}}\)=> \(\orbr{\begin{cases}x=\frac{4}{3}\\x=\frac{-28}{15}\end{cases}}\)
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1, \(\dfrac{2}{x}\) \(\in\) N; ( Đk \(x\) \(\ne\) 0)
\(\dfrac{2}{x}\) \(\in\) N ⇔ 2 ⋮ \(x\) ⇒ \(x\) \(\in\)Ư(2) ⇒ \(x\) \(\in\) {1; 2}
2, \(\dfrac{3}{x}\) \(\in\) N (đk \(x\) \(\ne\) 0)
\(\dfrac{3}{x}\) \(\in\) N ⇔ 3 ⋮ \(x\) ⇒ \(x\) \(\in\) Ư(3) = {1; 3}
3, \(\dfrac{4}{x}\) \(\in\) N (đk \(x\) \(\ne\) 0)
\(\dfrac{4}{x}\) \(\in\) N ⇔ 4 ⋮ \(x\) ⇒ \(x\) \(\in\) Ư(4) = { 1; 2; 4}
4, \(\dfrac{5}{x}\) \(\in\) N (đk \(x\) \(\ne\) 0)
\(\dfrac{5}{x}\) \(\in\) N ⇔5 ⋮ \(x\) ⇒ \(x\) \(\in\) Ư(5) = { 1; 5}
5, \(\dfrac{6}{x}\) \(\in\) N đk \(x\) \(\ne\) 0;
\(\dfrac{6}{x}\) \(\in\) N ⇔ 6 ⋮ \(x\) ⇒ \(x\) \(\in\) Ư(6) = {1; 2; 3; 6}
6, \(\dfrac{9}{x+1}\) \(\in\)N đk \(x\) \(\ne\) -1
\(\dfrac{9}{x+1}\) \(\in\) N ⇔ 9 ⋮ \(x\) + 1 ⇒ \(x+1\) \(\in\) Ư(9) = {1; 3; 9}
⇒ \(x\) \(\in\) {0; 2; 8}
7, \(\dfrac{8}{x+1}\) \(\in\) N (đk \(x\) \(\ne\) -1)
\(\dfrac{8}{x+1}\) \(\in\) N ⇔ 8 ⋮ \(x\) + 1 ⇒ \(x\) + 1 \(\in\) Ư*) = { 1; 2; 4; 8}
\(x\in\) {0; 1; 3; 7}