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Đặt A = 12 + 32 + 52 + ... + 972 + 992
Đặt B = 22 + 42 + 62 + ... + 982
Khi đó A + B = 12 + 22 + 32 + ... + 982 + 992
= 1.1 + 2.2 + 3.3 + ... + 98.98 + 99.99
= 1.(2 - 1) + 2(3 - 1) + 3(4 - 1) + ... + 98(99 - 1) + 99(100 - 1)
= 1.2 + 2.3 + 3.4 + .... + 98.99 + 99.100 - (1 + 2 + 3 + ... + 99)
= 1.2 + 2.3 + 3.4 + .... + 98.99 + 99.100 - 99.(99 + 1):2
= 1.2 + 2.3 + 3.4 + .... + 98.99 + 99.100 - 5050
Đặt C = 1.2 + 2.3 + 3.4 + .... + 98.99 + 99.100
=> 3C = 1.2.3 + 2.3.3 + 3.4.3 + ... + 98.99.3 + 99.100.3
3C = 1.2.3 + 2.3.(4 - 1) + 3.4.(5 - 2) + ... + 98.99.(100 - 97) + 99.100.(101 - 98)
3C = 1.2.3 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 + .... + 98.99.100 - 97.98.99 + 99.100.101 - 98.99.100
3C = 99.100.101
C = 99.100.101 : 3 = 333 300
Khi đó A+ B = C - 5050 = 333 300 - 5050 = 328 250
Lại có B = 22 + 42 + 62 + ... + 982
= 22(12 + 22 + 32 + ... + 492)
= 4(12 + 22 + 32 + ... + 492)
Đặt D = 12 + 22 + 32 + ... + 492
= 1.1 + 2.2 + 3.3 + ... + 49.49
= 1.(2 - 1) + 2.(3 - 1) + 3.(4 - 1) + ... + 49(50 - 1)
= 1.2. + 2.3 + 3.4 + ... + 49.50 - (1 + 2 + 3 + 4 + ... + 49)
= 1.2. + 2.3 + 3.4 + ... + 49.50 - 49.(49 + 1) : 2
= 1.2 + 2.3 + 3.4 + ... + 49.50 - 1225
Khi đó : 1.2 + 2.3 + 3.4 + ... + 49.50
= (1.2.3 + 2.3.3 + ... + 49.50.3) : 3
= [1.2.3 + 2.3.(4 - 1) + ... + 49.50(51 - 48)] : 3
= (1.2.3 + 2.3.4 - 1.2.3 + ... + 49.50.51 - 48.49.50) : 3
= 49.50.51 : 3
= 41650
Khi đó D = 41650 - 1225 = 40425
Khi đó B = 40425 x 4 = 161700
Lại có : A + B = 328250
=> A + 161700 = 328250
=> A = 166550
Vậy 12 + 32 + 52 + ... + 972 + 992 = 166550
\(\frac{4}{5}-|x-\frac{1}{6}|=\frac{2}{3}\)
\(\Rightarrow|x-\frac{1}{6}|=\frac{2}{15}\)
\(\Rightarrow\orbr{\begin{cases}x-\frac{1}{6}=\frac{2}{15}\\x-\frac{1}{6}=-\frac{2}{15}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{3}{10}\\x=\frac{1}{30}\end{cases}}\)
Vậy.....
27 . 93 / 65 . 82
= 27 . (32)3 / (2.3)5 . (23)2
= 27 . 36 / 25 . 35 . 26
= 27 . 36 / 211 . 35
= 3/24
= 3/16
\(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{1990^2}\)
\(A< \frac{1}{2^2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{1989.1990}\)
\(A< \frac{1}{4}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{1989}-\frac{1}{1990}\)
\(A< \frac{1}{4}+\frac{1}{2}-\frac{1}{1990}< \frac{1}{4}+\frac{1}{2}\)
\(A< \frac{1}{4}+\frac{2}{4}=\frac{3}{4}\left(đpcm\right)\)
a) \(\frac{x+1}{99}+\frac{x+2}{98}=\frac{x+3}{97}+\frac{x+4}{96}\)
\(\Rightarrow\frac{x+1}{99}+1+\frac{x+2}{98}+1=\frac{x+3}{97}+1+\frac{x+4}{96}+1\)
\(\Rightarrow\frac{x+100}{99}+\frac{x+100}{98}-\frac{x+100}{97}-\frac{x+100}{96}=0\)
\(\Rightarrow\left(x+100\right)\left(\frac{1}{99}+\frac{1}{98}-\frac{1}{97}-\frac{1}{96}\right)=0\)
Vì 1/99 + 1/98 - 1/97 - 1/96 khác 0
=> x + 100 = 0 => x = -100
b) \(\frac{x-3}{47}+\frac{x-2}{48}=\frac{x-1}{49}+1\)
\(\Rightarrow\frac{x-3}{47}-1+\frac{x-2}{48}-1=\frac{x-1}{49}+1-2\)
\(\Rightarrow\frac{x-50}{47}+\frac{x-50}{48}-\frac{x-50}{49}=0\)
\(\Rightarrow\left(x-50\right)\left(\frac{1}{47}+\frac{1}{48}-\frac{1}{49}\right)=0\)
Vì 1/47 + 1/48 - 1/49 khác 0
Nên x -50 = 0 => x = 50
\(từ\frac{x}{y}=\frac{4}{7}\) \(\Rightarrow\frac{x}{4}=\frac{y}{7}\)
\(\frac{x}{4}=\frac{y}{7}=\frac{3x^2}{48}=\frac{4y^2}{196}=\frac{3x^2-4y^2}{48-196}=\frac{100}{-148}=\frac{25}{-37}\)
\(\Rightarrow x^2=\frac{25}{37}\cdot4=\frac{100}{37}\)
còn lại bn tự lm
Từ x/y=4/7 => x/7 = 4y
Đặt x/4 = y/7 =k
=> x=4k; y=7k
mà 3x^3 - 4y^2 = 100
hay: 3. ( 4k)^2 - 4. ( 7k)^2
nhân vào là ra rồi xét 2 trường hợp
nếu muốn giải cụ thể thì kb rồi mk trả lời đầy đủ hơn cho
- \(\frac{4^6.3^4.9^5}{6^{12}}=\frac{\left(2^2\right)^6.3^4.\left(3^2\right)^5}{\left(2.3\right)^{12}}=\frac{2^{12}.3^4.3^{10}}{2^{12}.3^{12}}=\frac{2^{12}.3^{14}}{2^{12}.3^{12}}=3^2=9\)
- \(\frac{3^{10}.11+9^5.5}{3^9.2^4}=\frac{3^{10}.11+\left(3^2\right)^5.5}{3^9.16}=\frac{3^{10}.11+3^{10}.5}{3^9.16}=\frac{3^{10}.\left(11+5\right)}{3^9.16}=\frac{3^{10}.16}{3^9.16}=3\)
- 2100 - 299 - 298 - ... - 22 - 2
= 2100 - (299 + 298 + ... + 22 + 2)
Đặt A = 299 + 298 + ... + 22 + 2
2A = 2100 + 299 + ... + 23 + 22
2A - A = (2100 + 299 + ... + 23 + 22) - (299 + 298 + ... + 22 + 2)
A = 2100 - 2
Ta có:
2100 - 299 - 298 - ... - 22 - 2
= 2100 - (2100 - 2)
= 2100 - 2100 + 2
= 0 + 2
= 2
- 38 : 36 + (22)4 : 29
= 32 + 28 : 29
\(=9+\frac{1}{2}\)
\(=\frac{18}{2}+\frac{1}{2}=\frac{19}{2}\)