Tính nhanh:
[(-13)*10+2*65](73*8394*4,22)
Tính nhanh:
[(-13)*10+2*65](73*8394*4,22)
\(=\left(-130+130\right)^{73\cdot8394\cdot4.22}=0^{73\cdot8394\cdot4.22}=0\)
Tìm x biết
a. \(|\)x-1\(|\)+ \(|\)x+3 \(|\) = 4
b. |2x+4| -3|4-x|= -5
c. |x-2010|+|x-2012|+ |x-2014|=2
a. |x-1|+ |x+3 | = 4
x-1+x+3=4
(x+x)-(1-3)=4
2x+2=4 ( trừ của trừ là cộng -(-2)=+2)
2x=4-2
x=2:2=1
b. |2x+4| -3|4-x|= -5
2x+4-3.4-x=-5
(2x-x)+4-(3.4)=-5
x+4-12=-5
x+4=-5+12
x+4=7
x=7-4=3
c. |x-2010|+|x-2012|+ |x-2014|=2
x-2010+x-2012+x-2014=2
(x+x+x)-(2010+2012+2014)=2
3x-6036=2
3x=2+6036
3x=6038
x=6038:3=3012
tick giùm đi nhé có j lần sau tui giải cho ^D^
Tính nhanh:
[(-13)*10+2*65](73*8394*4,22)
Tính : \(S_n=\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+...+\dfrac{1}{n\left(n+1\right)\left(n+2\right)}\)
\(S_n=\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+....+\dfrac{1}{n\left(n+1\right)}-\dfrac{1}{\left(n+1\right)\left(n+2\right)}\right)\)
\(S_n=\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{\left(n+1\right)\left(n+2\right)}\right)\)
\(S_n=\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{n\left(n+2\right)+1\left(n+2\right)}\right)\)
\(S_n=\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{n^2+2n+n+2}\right)\)
\(S_n=\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{n^2+3n+2}\right)\)
\(S_n=\dfrac{1}{4}-\dfrac{1}{2\left(n^2+3n+2\right)}\)
\(S_n=\dfrac{1}{4}-\dfrac{1}{2n^2+6n+4}\)
\(S_n=\dfrac{2n^2+6n+4}{4\left(2n^2+6n+4\right)}-\dfrac{4}{4\left(2n^2+6n+4\right)}\)
\(S_n=\dfrac{2n^2+6n+4}{8n^2+48n+16}-\dfrac{4}{8n^2+48n+16}\)
\(S_n=\dfrac{2n^2+6n}{8n^2+48n+16}\)
\(S_n=\dfrac{2\left(n^2+3n\right)}{2\left(4n^2+24n+8\right)}=\dfrac{n^2+3n}{4n^2+24n+8}\)
\(S_n=\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+...+\dfrac{1}{n\left(n+1\right)\left(n+2\right)}\\ 2S_n=\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+...+\dfrac{2}{n\left(n+1\right)\left(n+2\right)}\\ 2S_n=\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+...+\dfrac{1}{n\left(n+1\right)}-\dfrac{1}{\left(n+1\right)\left(n+2\right)}\\ =\dfrac{1}{1.2}-\dfrac{1}{\left(n+1\right)\left(n+2\right)}\\ =\dfrac{\left(n+1\right)\left(n+2\right)-2}{2\left(n+1\right)\left(n+2\right)}\\ =>S_n=\dfrac{\left(n+1\right)\left(n+2\right)-2}{4\left(n+1\right)\left(n+2\right)}\)
Giải sai r nhéLinh Nguyễn
\(S_n=\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+..........+\dfrac{1}{n\left(n+1\right)\left(n+2\right)}\)
\(2S_n=\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+....+\dfrac{2}{n\left(n+1\right)\left(n+2\right)}\)
\(2S_n=\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+....+\dfrac{1}{n\left(n+1\right)}-\dfrac{1}{\left(n+1\right)\left(n+2\right)}\)
\(2S_n=\dfrac{1}{1.2}-\dfrac{1}{\left(n+1\right)\left(n+2\right)}\)
\(2S_n=\dfrac{1}{2}-\dfrac{1}{\left(n+1\right)\left(n+2\right)}\)
\(\Rightarrow S_n=\dfrac{\dfrac{1}{2}-\dfrac{1}{\left(n+1\right)\left(n+2\right)}}{2}\)
Tính : S = 1.4 + 2.5 + 3.6 + 4.7 + ... + n ( n + 3 ) với n thuộc N*
Ta thấy:
1.4 = 1.(1 + 3) = 1.(1 + 1 + 2) = 1.(1 + 1)+ 2.1
2.5 = 2.(2 + 3) = 2.(2 + 1 + 2) = 2.(2 + 1)+ 2.2
3.6 = 3.(3 + 3) = 3.(3 + 1 + 2) = 3.(3 + 1)+ 2.3
4.7 = 4.(4 + 3) = 4.(4 + 1 + 2) = 4.(4 + 1)+ 2.4
. . . . . . . . . . .
n(n + 3) = n(n + 1) + 2n
Vậy C = 1.2 + 2.1 + 2.3 + 2.2 + 3.4 + 2.3 + . . . + n(n + 1) + 2n
= 1.2 + 2 +2.3 + 4 + 3.4 + 6 + . . . + n(n + 1) + 2n
= [1.2 +2.3 +3.4 + . . . + n(n + 1)] + (2 + 4 + 6 + . . . + 2n)
Mà 1.2 + 2.3 + 3.4 + … + n.(n + 1) = \(\dfrac{n.\left(n+1\right).\left(n+2\right)}{3}\)
Và 2 + 4 + 6 + . . . + 2n = \(\dfrac{\left(2n+2\right).n}{2}\)
⇒C = \(\dfrac{n.\left(n+1\right).\left(n+2\right)}{3}+\dfrac{\left(2n+2\right).n}{2}-\dfrac{n.\left(n+1\right).\left(n+5\right)}{3}\)
Dựa theo công thức tự thiết kế do các anh em trong đoàn ( những con người ẩn danh ) là : { k . ( k + 3 ) = k . ( k + 1 ) + 2 . k }
Ta có :
S = 1 . 4 + 2 . 5 + 3 . 6 + . . . + n . ( n + 3 )
S = ( 1 . 2 + 2 . 1 ) + ( 2 . 3 + 2 . 2 ) + . . . + [ n . ( n + 1 ) + 2 . n ]
S = ( 1 . 2 + 2 . 3 + . . . + n . ( n + 1 ) ) + ( 2 . 1 + 2 . 2 + . . . + 2 . n )
Dựa theo công thức số 37 và 55 quyển 7 của các em trong đoàn .
Ta có :
S = [ n . ( n + 1 ) . ( n + 2 ) ] + ( \(n^2\)+ n ) ]
Tính tổng B = 3 - 32 + 33 - 34 + ... +32013 - 32014 + 32015 - 32016
Tìm x,y biết: \(|x-2009|+|x-2010|+|x-2012|=3\)
Cho tỉ lệ thức \(\dfrac{3x-y}{x+y}=\dfrac{3}{4}\). Tìm giá trị của tỉ số \(\dfrac{x}{y}\)
\(\dfrac{3x-y}{x+y}=\dfrac{3}{4}\)
\(\Rightarrow4\left(3x-y\right)=3\left(x+y\right)\)
\(\Rightarrow12x-4y=3x+3y\)
\(\Rightarrow12x-4y-3y=3x\)
\(\Rightarrow12x-7y=3x\)
\(\Rightarrow12x-3x=7y\)
\(\Rightarrow9x=7y\)
\(\Rightarrow\dfrac{x}{7}=\dfrac{y}{9}\)
\(\Rightarrow\dfrac{x}{y}=\dfrac{7}{9}\)
Tìm tỉ số \(\dfrac{x}{y}\), biết \(\dfrac{x-y}{x+2y}=\dfrac{3}{4}\)
Giải:
Theo đề ra, ta có:
\(\dfrac{x-y}{x+2y}=\dfrac{3}{4}\)
\(\Leftrightarrow3.\left(x+2y\right)=4.\left(x-y\right)\)
\(\Leftrightarrow3x+6y=4x-4y\)
\(\Leftrightarrow6y+4y=4x-3x\)
\(\Leftrightarrow10y=x\)
\(\Leftrightarrow\dfrac{x}{y}=10\)
Vậy ...
Chúc bạn học tốt!
Dễ thấy \(y\ne0\)
Ta có: \(\dfrac{x-y}{x+2y}=\dfrac{3}{4}\Leftrightarrow\dfrac{\dfrac{x}{y}-\dfrac{y}{y}}{\dfrac{x}{y}+\dfrac{2y}{y}}=\dfrac{3}{4}\)
\(\Leftrightarrow\dfrac{\dfrac{x}{y}-1}{\dfrac{x}{y}+2}=\dfrac{3}{4}\)
Đặt \(\dfrac{x}{y}=k\) với k là số hữu tỉ
Khi đó có \(\dfrac{k-1}{k+2}=\dfrac{3}{4}\Leftrightarrow\dfrac{k+2}{4}=\dfrac{k-1}{3}=\dfrac{k+2-\left(k-1\right)}{4-3}=3\)
\(\Rightarrow k=10\)
Vậy \(\dfrac{x}{y}=10\)
Cho a,b,c khác 0 và đôi một khác nhau thỏa mãn: \(\dfrac{b+c}{bc}=\dfrac{2}{a}\). Chứng minh:\(\dfrac{b}{c}=\dfrac{a-b}{c-a}\)
\(\dfrac{b+c}{bc}=\dfrac{2}{a}< =>\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{2}{a}\)
\(< =>\dfrac{1}{b}-\dfrac{1}{a}+\dfrac{1}{c}-\dfrac{1}{a}=0\)
\(< =>\dfrac{a-b}{ab}+\dfrac{a-c}{ac}=0\)
\(< =>\dfrac{a-b}{ab}=\dfrac{a-c}{ac}\)
\(< =>\dfrac{ab}{ac}=\dfrac{a-b}{c-a}< =>\dfrac{b}{c}=\dfrac{a-b}{c-a}\)