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Câu 1 : \(0,36\times630+0,6\times36\times6+3,6\)
\(=0,36\times10\times63+0,6\times6\times36+3,6\)
\(=3,6\times63+3,6\times36+3,6\)
\(=3,6\times\left(63+36+1\right)\)
\(=3,6\times100\)
\(=360\)
Câu 2 :\(3,5\times20,2+6,5\times27,9+3,5\times2,2-6,5\times5,5\)
\(=3,5\times\left(20,2+2,2\right)+6,5\times\left(27,9-5,5\right)\)
\(=3,5\times22,4+6,5\times22,4\)
\(=22,4\times\left(3,5+6,5\right)\)
\(=22,4\times10\)
\(=224\)
\(A=1.\left(2+2\right)+2.\left(3+2\right)+3.\left(4+2\right)+....+99.\left(100+2\right)\)
\(A = (1.2 + 2.3 + 3.4 + ... + 99.100) + (1.2 + 2.2 + 3.2 + ... + 99.2)\)
\(Đặt B = 1.2 + 2.3 + 3.4 + ... + 99.100\)
\(3B = 1.2.(3-0) + 2.3.(4-1) + 3.4.(5-2) + ... + 99.100.(101-98)\)
\(3B = 1.2.3 - 0.1.2 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 + ... + 99.100.101 - 98.99.100\)
\(3B = 99.100.101\)
\(B = 33.100.101 = 333300\)
\(A = 333300 + 2.(1 + 2 + 3 + ... + 99)\)
\(A = 333300 + 2.(1 + 99).99:2\)
\(A = 333300 + 100.99\)
\(A = 333300 + 9900\)
\(A = 343200\)
a. A = 1.4 + 2.5 + 3.6 +...+ 99.102
= 1( 2 +2) + 2(3+2) +...+ 99 (100 +2)
= 1.2 + 1.2 +2.3 + 2.2 +...+ 99 .100 +99 . 2
= ( 1.2 +2.3 + 3.4 +...+99 . 100) + 2(1 + 2 + 3+...+99)
= 333300 + 9900 = 343 200
b. B = 1.3 + 2.4 + 3.5 +...+ 99.101
= 1(2 +1) + 2(3 +1) + 3(4 +1) +...+ 99(100 +1)
= 1.2 + 1 + 2.3 + 2 + 3.4 +...+ 99. 100 +99
= ( 1.2 + 2.3 + 3.4 +...+ 99.100) + (1+2+...+99)
= 333300 + 4950 = 338 250
c. C = 4 + 12 + 24 +...+ 19404 + 19800
1/2C = 1.2 + 2.3 + 3.4 +...+ 98.99 + 99.100
1/2 C = 333300
C = 333300 : 1/2 = 666600
a) Đặt \(A=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{99.101}\)
\(2A=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\)
\(2A=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\)
\(2A=1-\frac{1}{101}=\frac{100}{101}\)
\(A=\frac{100}{101}\div2=\frac{50}{101}\)
b) Đặt \(B=\frac{1}{3.6}+\frac{1}{6.9}+\frac{1}{9.12}+\frac{1}{12.15}\)
\(3B=\frac{3}{3.6}+\frac{3}{6.9}+\frac{3}{9.12}+\frac{3}{12.15}\)
\(3B=\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+\frac{1}{9}-\frac{1}{12}+\frac{1}{12}-\frac{1}{15}\)
\(3B=\frac{1}{3}-\frac{1}{15}=\frac{4}{15}\)
\(B=\frac{4}{15}\div3=\frac{4}{45}\)
Đặt \(A=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{99.101}\)
\(2A=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\)
\(2A=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\)
\(2A=1-\frac{1}{101}=\frac{100}{101}\)
\(A=\frac{100}{101}\div2=\frac{50}{101}\)
a) 3,6 x 7,2 - 3,6 x 7,1
= 3,6 x (7,2 - 7,1)
= 3,6 x 0,1
= 0,36
b) 0,15 x 600 = 0,15 x 100 x 6
= 15 x 6
= 90
a) 3,6 x 7,2 - 3,6 x 7,1
= 3,6 x (7,2 - 7,1)
= 3,6 x 0,1
= 0,36
b) 0,15 x 600
= 0,15 x 100 x 6
= 15 x 6
= 90
=> x.6,3 = 19,5 - 11,625
=> x.6,3 = 7,875
=> x = 7,875 : 6,3
=> x = 1,25
Vậy x = 1,25
\(51\cdot72+105\cdot12:3\cdot6=3672+2520=6192\)
\(4\cdot8\cdot0,75-6\cdot3:3\cdot5=4\cdot8\cdot\frac{3}{4}-6\cdot5=24-30=-6\)
xin t ck
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51,72+105,12:3,6
=51,72+29,2
=80,92
4,8x0,75-6,3:3,5
=3,6-1,8
=1,8