2 | + | 2 | + … + | 2 | = | |
3 x 5 | 5 x 7 | 111 x 113 |
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Ta có: \(\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+...+\dfrac{2}{111\cdot113}\)
\(=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{111}-\dfrac{1}{113}\)
\(=\dfrac{1}{3}-\dfrac{1}{113}\)
\(=\dfrac{110}{339}\)
2/3x5 + 2/5x7 + ....+2/111x113
=1/3-1/5 + 1/5 - 1/7 +....+1/111-1/113
=1/3 - 1/113
=110/339
a: =(1-2-3+4)+(5-6-7+8)+...+(21-22-23+24)
=0+0+...+0
=0
b: =-7111+53+711-153=-6400-100=-6500
c: =-43(-1-296+296)=-43*(-1)=43
1.a,=(54+45+1).113
=100.113
=11300
b,=(3/7+8/14)+(4/9+10/18)
=1+1
=2
2.a,=13/10+1/3
=49/30
b,=12/9.(1/12+1/6)
=12/9.1/4
=1/3
c,=3/4.3/2
=9/8
d,=3/2-1/3
=7/6
1:tính bằng cách thuận tiện nhất:
a)54 x 113 + 45 x 113 + 113
= 54 x 113 + 45 x 113 + 113x1
=113 x(54+45+1)
= 113x100
=1300
b)3/7 + 4/9 + 8/14 + 10/18
=(3/7+8/14)+(4/9+10/18)
= 1 + 1
=2
k) 4 x 113 x 25 - 5 x 112 x 20
= (4 x 25) x 113 - (5 x 20) x 112
= 100 x 113 - 100 x 112
= 100 x (113 - 112)
= 100
p) 1235 x 6789 x (630 - 315 x 2) : 1996
= 1235 x 6789 x (630 - 630) : 1996
= 1235 x 6789 x 0 : 1996
= 0
l) 54 x 113 + 45 x 113 + 113
= 113 (54 + 45 + 1)
= 113 x 100
= 11300
u) (500 x 9 - 250 x 18) x (1 + 2 + 3 + ... + 9)
= (2 x 250 x 9 - 250 x 2 x 9) x (1 +2 + 3 + ... + 9)
= 0 x (1 + 2 + 3 + ... + 9)
= 0
2)\(x\left(x+1\right)\left(x+2\right)\left(x+3\right)=[x\left(x+3\right)].\left[\left(x+1\right)\left(x+2\right)\right]=\left(x^2+3x\right)\left(x^2+3x+2\right)\)
Đặt y=\(x^2+3x\) ta được:\(y\left(y+2\right)=y^2+2y=\left(y+1\right)^2-1\)
Vậy GTNN của x(x+1)(x+2)(x+3) là -1
bài 4
\(\left\{{}\begin{matrix}a+b=170\\\left[{}\begin{matrix}\dfrac{2}{5}a+\dfrac{1}{2}b=21\left(1\right)\\\dfrac{1}{2}a+\dfrac{2}{5}b=21\left(2\right)\end{matrix}\right.\end{matrix}\right.\)
\(\left[{}\begin{matrix}\dfrac{2}{5}a+\dfrac{1}{2}b=\dfrac{4\left(a+b\right)+b}{10}=21\Rightarrow b=210-4.170=-470\\\dfrac{1}{2}a+\dfrac{2}{5}b=\dfrac{4\left(a+b\right)+a}{10}=21\Rightarrow210-4.170=-470\end{matrix}\right.\)
\(470+170=640\)
Đáp số 640
d,155-10(x+1)=55
10(x+1) = 155 - 55
10(x+1) = 100
x+1 = 100 : 10
x+1 = 10
x = 10 - 1 =9
e,6(x+\(2^3\))+40=100
6(x+\(2^3\)) = 100 - 40
6(x+\(2^3\)) = 60
(x+\(2^3\)) = 60 : 6
(x+\(2^3\)) = 10
x = 10 - \(2^3\)
x = 10 - 8 = 2
f,\(2^2\)(x+\(3^2\))-5=55
\(2^2\)(x+\(3^2\)) = 55 + 5
\(2^2\)(x+\(3^2\)) = 60
(x+\(3^2\)) = 60 : \(2^2\)
(x+\(3^2\)) = 60 : 4
(x+\(3^2\)) = 15
x + 9 = 15
x = 15-9=6.
Hok tốt !
d) \(155-10\left(x+1\right)=55\)
\(\Rightarrow10\left(x+1\right)=155-55\)
\(\Rightarrow10\left(x+1\right)=100\)
\(\Rightarrow x+1=100:10\)
\(\Rightarrow x+1=10\)
\(\Rightarrow x=10-1\)
\(\Rightarrow x=9\)
Vậy x = 9
e) \(6\left(x+2^3\right)+40=100\)
\(\Rightarrow6\left(x+2^3\right)=100-40\)
\(\Rightarrow6\left(x+2^3\right)=60\)
\(\Rightarrow x+2^3=60:6\)
\(\Rightarrow x+2^3=10\)
\(\Rightarrow x+8=10\)
\(\Rightarrow x=10-8\)
\(\Rightarrow x=2\)
Vậy x = 2
f) \(2^2\left(x+3^2\right)-5=55\)
\(\Rightarrow2^2\left(x+3^2\right)=55+5\)
\(\Rightarrow2^2\left(x+3^2\right)=60\)
\(\Rightarrow4\left(x+9\right)=60\)
\(\Rightarrow x+9=60:4\)
\(\Rightarrow x+9=15\)
\(\Rightarrow x=15-9\)
\(\Rightarrow x=6\)
Vậy x = 6
\(\frac{2}{3\times5}+\frac{2}{5\times7}+...+\frac{2}{111\times113}\)
\(=\frac{5-3}{3\times5}+\frac{7-5}{5\times7}+...+\frac{113-111}{111\times113}\)
\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{111}-\frac{1}{113}\)
\(=\frac{1}{3}-\frac{1}{113}\)
\(=\frac{110}{339}\)