Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
mk nghĩ đây là toán 8.
\(Pt\Leftrightarrow\left(x-2010\right)\left(\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+....+\frac{1}{72}\right)=\frac{16}{9}\Leftrightarrow\left(x-2010\right)\left(\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+....+\frac{1}{8.9}\right)=\frac{16}{9}\Leftrightarrow\left(x-2010\right)\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+....-\frac{1}{9}\right)=\frac{16}{9}\Leftrightarrow\left(x-2010\right).\frac{2}{9}=\frac{16}{9}\Leftrightarrow x-2010=8\Leftrightarrow x=2018.\text{ Vậy: x=2018}\)
a)x \(\in\){ -5; -4; -3; -2; -1;..;2005;2006;2007;2008;2009;2010}
b)x\(\in\){2004;2005;2006;2007;2008;2009;2010}
c)IxI\(\in\){0;1;2;3;4;5;6}
\(\Rightarrow\)x\(\in\){-6;-5;-4;-3;-2;-1;0;1;2;3;4;5;6}
-5 \(\le\)x\(\le\)2010 nên x \(\in\){-5; -4; -3; -2;...........; 2007; 2008; 2009; 2010}
2004\(\le\)x\(\le\)2010 nên x \(\in\){2004; 2005; 2006; 2007; 2008; 2009; 2010}
IxI \(\le\)6 nên x\(\in\){-6; -5; -4; -3; -2; -1; 0; 1; 2; 3; 4; 5; 6}
\(B=\left(1-\frac{1}{2010}\right)x\left(1-\frac{2}{2010}\right)x\left(1-\frac{3}{2010}\right)x...x\left(1-\frac{2011}{2010}\right)\)
\(B=\left(1-\frac{1}{2010}\right)x\left(1-\frac{2}{2010}\right)x\left(1-\frac{3}{2010}\right)x....x\left(1-\frac{2010}{2010}\right)x\left(1-\frac{2011}{2010}\right)\)
\(B=\left(1-\frac{1}{2010}\right)x\left(1-\frac{2}{2010}\right)x\left(1-\frac{3}{2010}\right)x...x\left(0\right)x\left(1-\frac{2011}{2010}\right)\)
\(B=0\)
Ta có : 95\(⋮\)5 ; 2010\(⋮\)5
\(\Rightarrow\)x+6\(⋮\)5
\(\Rightarrow\)12<x+6\(\le\)27
\(\Rightarrow\)x+6\(\in\){15;20;25}
\(\Rightarrow\)x\(\in\){9;14;19}
Vậy x\(\in\){9;14;19}
(Sai thì xin lỗi nha)
\(x+6+95+2010=x+1+5+95+2010=x+1+100+2010=x+1+2110\)
Vì \(2110⋮5\)\(\Rightarrow\)Để \(\left(x+6+95+2010\right)⋮5\)thì \(\left(x+1\right)⋮5\)(1)
Vì \(12< x\le21\)\(\Rightarrow12< x+1\le22\)(2)
Từ (1), (2) \(\Rightarrow x+1\in\left\{15;20\right\}\)\(\Rightarrow x\in\left\{14;19\right\}\)
Vậy \(x\in\left\{14;19\right\}\)
*TH1: *TH2:
x+10=2010. x+10=--2010
x=2010-10 x=(-2010) -10
x= 2000. x=-2020
Vậy x thuộc {2000;-2020}
Tk cho mk nha
x | -10 | ||
x+10 | - | 0 | + |
+) Nếu \(x< -10\) thì \(|x+10|=-x-10\)
\(pt\Leftrightarrow-x-10=2010\)
\(\Leftrightarrow-x=2020\)
\(\Leftrightarrow x=-2020\left(tm\right)\)
+) Nếu \(x\ge-10\) thì \(|x+10|=x+10\)
\(pt\Leftrightarrow x+10=2010\)
\(\Leftrightarrow x=2000\left(tm\right)\)
Vậy phương trình có tập nghiệm \(S=\left\{2000;-2020\right\}\)
Bài 1 :
72x+3 . 75-2x : 7x + 7x = 1
- > 7(2x+3)+(5-2x)-7 + 7x = 1
- > 71 + 7x = 1
- > 7x = 1 - 7 = -6 - > x thuộc rỗng
\(2010^5\left(x-60\right)=2010^6\)
=> \(x-60=\frac{2010^6}{2010^5}=2010\)
=> x = 2010 + 60 = 2070
Vậy x = 2070
20105.( x - 60 ) = 20106
<=> x - 60 = 20106 : 20105
<=> x - 60 = 2010
<=> x = 2070