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a) \(\dfrac{2}{3}x-\dfrac{1}{2}=\dfrac{1}{10}\)
\(\dfrac{2}{3}x=\dfrac{1}{10}+\dfrac{1}{2}=\dfrac{3}{5}\)
\(x=\dfrac{3}{5}:\dfrac{2}{3}=\dfrac{9}{10}\)
b) \(\dfrac{39}{7}:x=13\)
\(x=\dfrac{\dfrac{39}{7}}{13}=\dfrac{3}{7}\)
c) \(\left(\dfrac{14}{5}x-50\right):\dfrac{2}{3}=51\)
\(\dfrac{14}{5}x-50=51\cdot\dfrac{2}{3}=34\)
\(\dfrac{14}{5}x=34+50=84\)
\(x=\dfrac{84}{\dfrac{14}{5}}=30\)
d) \(\left(x+\dfrac{1}{2}\right)\left(\dfrac{2}{3}-2x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=0\\\dfrac{2}{3}-2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=\dfrac{1}{3}\end{matrix}\right.\)
e) \(\dfrac{2}{3}x-\dfrac{1}{2}x=\dfrac{5}{12}\)
\(\dfrac{1}{6}x=\dfrac{5}{12}\)
\(x=\dfrac{5}{12}:\dfrac{1}{6}=\dfrac{5}{2}\)
g) \(\left(x\cdot\dfrac{44}{7}+\dfrac{3}{7}\right)\dfrac{11}{5}-\dfrac{3}{7}=-2\)
\(\left(x\cdot\dfrac{44}{7}+\dfrac{3}{7}\right)\cdot\dfrac{11}{5}=-2+\dfrac{3}{7}=-\dfrac{11}{7}\)
\(x\cdot\dfrac{44}{7}+\dfrac{3}{7}=-\dfrac{11}{7}:\dfrac{11}{5}=-\dfrac{5}{7}\)
\(\dfrac{44}{7}x=-\dfrac{5}{7}-\dfrac{3}{7}=-\dfrac{8}{7}\)
\(x=-\dfrac{8}{7}:\dfrac{44}{7}=-\dfrac{2}{11}\)
h) \(\dfrac{13}{4}x+\left(-\dfrac{7}{6}\right)x-\dfrac{5}{3}=\dfrac{5}{12}\)
\(\dfrac{25}{12}x-\dfrac{5}{3}=\dfrac{5}{12}\)
\(\dfrac{25}{12}x=\dfrac{5}{12}+\dfrac{5}{3}=\dfrac{25}{12}\)
\(x=1\)
Mỏi tay woa bn làm nốt nha!!
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\(A=1^2+2^2+3^2+....+10^2\\ A=1^{ }+\left(1+1\right)\cdot2+3\cdot\left(2+1\right)+.....+10\cdot\left(9+1\right)\\ A=1+2\cdot1+2+3\cdot2+3+....+10\cdot9+10\\ A=\left(1+2+3...+10\right)+\left(1\cdot2+3\cdot2+.....+10\cdot9\right)\)
Gọi 1+2+3+...+10 là P
Số số hạng là: (10 - 1) : 1 +1 = 10 (số)
P = (10+1) . 10 : 2 = 55
P = 55
Gọi \(1\cdot2+2\cdot3+....+9\cdot10\) là C
\(C=1\cdot2+2\cdot3+....+9\cdot10\\ 3\cdot C=1\cdot2\cdot3+2\cdot3\cdot3+....+9\cdot10\cdot3\\ 3\cdot C=1\cdot2\cdot3+2\cdot3\cdot\left(4-1\right)+....+9\cdot10\cdot\left(11-8\right)\\ 3\cdot C=1\cdot2\cdot3+2\cdot3\cdot4-1\cdot2\cdot3+.....+9\cdot10\cdot11-8\cdot9\cdot10\\ 3\cdot C=9\cdot10\cdot11\\ 3\cdot C=990\\ C=330\)
\(=>A=P+C\\ =>A=55+330\\ A=385\)
b)
\(B=5^2+10^2+15^2+...+50^2\\ B=5^2+\left(2\cdot5\right)^2+\left(3\cdot5\right)^2+....+\left(5\cdot10\right)^2\\ B=5^2+2^2\cdot5^2+3^2\cdot5^2+...+5^2\cdot10^2\\ B=5^2\cdot\left(1+2^2+3^2+....+10^2\right)\\ B=25\cdot\left(1+2^2+3^2+....+10^2\right)\)
\(\left(1+2^2+3^2+....+10^2\right)=A\)
\(=>B=25\cdot A\\ B=25\cdot385\\ B=9625\)
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a: \(20-\left[30-\left(5-1\right)^2\right]\)
\(=20-\left[30-4^2\right]\)
\(=20-14=6\)
b: \(71+\dfrac{50}{5+3\left(57-6\cdot7\right)}\)
\(=71+\dfrac{50}{5+3\cdot\left(57-42\right)}\)
\(=71+\dfrac{50}{5+3\cdot15}=71+\dfrac{50}{50}=72\)
c: \(4\cdot\left\{270:\left[50-\left(2^5+45:5\right)\right]\right\}\)
\(=4\cdot\left\{270:\left[50-32-9\right]\right\}\)
\(=4\cdot\left\{\dfrac{270}{50-41}\right\}=4\cdot\dfrac{270}{9}=4\cdot30=120\)
d: \(411-\left[\dfrac{\left(107+3\right)}{5}-2^2\right]\)
\(=411-\left[\dfrac{110}{5}-4\right]\)
=410-22+4
=410-18
=392
e: \(450-5\left[3^2\left(7^5:7^3-41\right)-12\right]+18\)
\(=450-5\left[9\cdot\left(7^2-41\right)-12\right]+18\)
\(=450-5\cdot\left[9\cdot8-12\right]+18\)
=468-5*60
=468-300
=168
f:
\(102-150:\left[18-2\cdot\left(10-8\right)^2\right]+1018^0\)
\(=102-150:\left[18-2\cdot4\right]+1\)
\(=103-\dfrac{150}{18-8}=103-15=88\)
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a: \(61\cdot45+61\cdot23-68\cdot51\)
\(=61\left(45+23\right)-68\cdot51\)
\(=68\cdot61-68\cdot51\)
\(=68\left(61-51\right)=68\cdot10=680\)
b: \(3\cdot5^2-\left(75-4\cdot2^3\right)\)
\(=75-75+4\cdot8\)
\(=4\cdot8=32\)
c: \(36:\left\{2^2\cdot5-\left[30-\left(5-1\right)^2\right]\right\}\)
\(=\dfrac{36}{20-30+4^2}\)
\(=\dfrac{36}{-10+16}=\dfrac{36}{6}=6\)
d: \(\left(12\cdot49-3\cdot2^2\cdot7^2\right):\left(2020\cdot2021\right)\)
\(=\dfrac{\left(12\cdot49-12\cdot49\right)}{2020\cdot2021}=0\)
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Bài 1: Thực hiện các phép tính sau:
\(a)\)Chưa rỏ đề
\(b)\)\(5025\div5-25\div5\)
\(=\)\(1005-5\)
\(=\)\(1000\)
\(c)\)\(218-180\div2\div9\)
\(=\)\(218-10\)
\(=\)\(208\)
\(d)\)\(\left(328-8\right)\div32\)
\(=\)\(320\div32\)
\(=\)\(10\)
Bài 1:
a) ( Tôi không nhìn rõ đầu bài )
b) 5025 : 5 - 25 : 5
= ( 5025 - 25 ) : 5
= 5000 : 5
= 1000
c) 218 - 180 : 2 : 9
= 218 - 180 : ( 2 . 9 )
= 218 - 180 : 18
= 218 - 10
= 208
d) ( 328 - 8 ) : 32
= 320 : 32
= 10
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a) \(3.5^2+15.2^2-26\div2\)
= 3.25 + 15.4 - 13
= 75 + 60 - 13
= 135 - 13
= 122
b) \(5^3.2-100\div4+2^3.5\)
= 125.2 - 25 + 8.5
= 250 - 25 + 40
= 225 + 40
= 265
c)\(6^2\div9+50.2-3^3.33\)
= 36 : 9 + 100 - 9.33
= 4 + 100 - 297
= 104 - 297
= -193
d)\(3^2.5+2^3.10-81\div3\)
= 9.5 + 8.10 - 27
= 45 + 80 - 27
= 125 - 27
= 98
e) \(5^{13}\div5^{10}-25.2^2\)
= 53 - 25.4
= 125 - 100
= 25
f) \(20\div2^2+5^9\div5^8\)
= 20 : 4 + 5
= 5 + 5
= 10
a, Ta có : S = 1*2 + 2*3 +3*4 + .... + 50*51
3S=1*2*3+2*3*3+3*4*3+....+50*51*3
3S=1*2*3+2*3*(4-1)+3*4*(5-2)+....+50*51*(52-49)
3S=1*2*3+2*3*4-1*2*3+3*4*5-2*3*4+...+50*51*52-49*50*51
3S=50*51*52
S=(50*51*52)/3=442000
b,Ta có 12 + 22 + 32 + ....... + n2=\(\frac{n\cdot\left(n+1\right)\cdot\left(2n+1\right)}{6}\)
=> 12 + 22 + 32 + ....... + 502= \(\frac{50\cdot\left(50+1\right)\cdot\left(2\cdot50+1\right)}{6}\)
=\(\frac{50\cdot51\cdot101}{6}\)= 42925