(1+x)+(3+x)+(5+x)+...+(59+x)=2010
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![](https://rs.olm.vn/images/avt/0.png?1311)
a) Ta có: \(7-\left(2x+4\right)=-\left(x+4\right)\)
\(\Leftrightarrow7-2x-4=-x-4\)
\(\Leftrightarrow-2x+3+x+4=0\)
\(\Leftrightarrow-x+7=0\)
\(\Leftrightarrow-x=-7\)
hay x=7
Vậy: S={7}
b) Ta có: \(\dfrac{2+x}{5}-0.5x=\dfrac{1-2x}{4}+0.25\)
\(\Leftrightarrow\dfrac{4\left(2+x\right)}{20}-\dfrac{0.5x\cdot20}{20}=\dfrac{5\left(1-2x\right)}{20}+\dfrac{20\cdot0.25}{20}\)
\(\Leftrightarrow4\left(2+x\right)-10x=5\left(1-2x\right)+5\)
\(\Leftrightarrow8+4x-10x=5-10x+5\)
\(\Leftrightarrow-6x+8=-10x+10\)
\(\Leftrightarrow-6x+8+10x-10=0\)
\(\Leftrightarrow4x-2=0\)
\(\Leftrightarrow4x=2\)
hay \(x=\dfrac{1}{2}\)
Vậy: \(S=\left\{\dfrac{1}{2}\right\}\)
d) Ta có: \(\dfrac{x-1}{59}+\dfrac{x-2}{58}+\dfrac{x-3}{57}=\dfrac{x-59}{1}+\dfrac{x-58}{2}+\dfrac{x-57}{3}\)
\(\Leftrightarrow\dfrac{x-1}{59}-1+\dfrac{x-2}{58}-1+\dfrac{x-3}{57}-1=\dfrac{x-59}{1}-1+\dfrac{x-58}{2}-1+\dfrac{x-57}{3}-1\)
\(\Leftrightarrow\dfrac{x-60}{59}+\dfrac{x-60}{58}+\dfrac{x-60}{57}=\dfrac{x-60}{1}+\dfrac{x-60}{2}+\dfrac{x-60}{3}\)
\(\Leftrightarrow\left(x-60\right)\left(\dfrac{1}{59}+\dfrac{1}{58}+\dfrac{1}{57}\right)-\left(x-60\right)\left(1+\dfrac{1}{2}+\dfrac{1}{3}\right)=0\)
\(\Leftrightarrow\left(x-60\right)\left(\dfrac{1}{59}+\dfrac{1}{58}+\dfrac{1}{57}-1-\dfrac{1}{2}-\dfrac{1}{3}\right)=0\)
mà \(\dfrac{1}{59}+\dfrac{1}{58}+\dfrac{1}{57}-1-\dfrac{1}{2}-\dfrac{1}{3}\ne0\)
nên x-60=0
hay x=60
Vậy: S={60}
![](https://rs.olm.vn/images/avt/0.png?1311)
Giải:
(x+1)+(x+3)+...+(x+59)=1710
=>(x+x+x+...+x)+(1+3+..+59)=1710
=>30\(\times\)x+900=1710 (30 số x)
=>30\(\times\)x=810
=>x=810:30=27
Vậy x=27
Chúc bạn hok tốt
Bỏ phép tính sang 1 bên, ta lập dãy số:
1; 3; 5;...; 59
Quy luật: Mỗi số hạng liên tiếp sẽ cách nhau 2 đơn vị
⇒ Ta có số số hạng của dãy số cũng như số chữ số x là:
(59 - 1) : 2 + 1 = 30 (số)
Tổng của dãy là: (1 + 59) x 30 : 2 = 900
⇒ Ta lập lại biểu thức trên như sau:
X x 30 + 900 = 1710
X x 30 = 1710 - 900
X x 30 = 810
X = 27
Vậy số cần tìm là: 27
Lần sau ghi cả câu hỏi nhe, HT
![](https://rs.olm.vn/images/avt/0.png?1311)
-(-2010) +36.41 - 36. (-59) + (-2010)
= 2010 + 1476 -2124 -2010
= (2010-2010) + (1476-2124)
= -648
-75.(18-65)-65.(75-18)
= -75. (-47) - 65. 57
= 3525 - 3705
= -180
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\dfrac{x+1}{65}+\dfrac{x+3}{63}\) = \(\dfrac{x+5}{61}\) + \(\dfrac{x+7}{59}\)
<=> \(\dfrac{x+1}{65}+1+\dfrac{x+3}{63}+1\) = \(\dfrac{x+5}{61}\) + 1 + \(\dfrac{x+7}{59}\) + 1
<=> \(\dfrac{x+66}{65}+\dfrac{x+66}{63}\) = \(\dfrac{x+66}{61}\) + \(\dfrac{x+66}{59}\)
<=> \(\dfrac{x+66}{65}+\dfrac{x+66}{63}\) - \(\dfrac{x+66}{61}\) - \(\dfrac{x+66}{59}\) = 0
<=> (x + 66) . (\(\dfrac{1}{65}+\dfrac{1}{63}+\dfrac{1}{61}+\dfrac{1}{59}\)) = 0
<=> x + 66 = 0
<=> x = -66
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\dfrac{x+1}{65}+\dfrac{x+3}{63}+\dfrac{x+5}{61}+\dfrac{x+7}{59}\)
\(\Leftrightarrow\dfrac{x+1}{65}+\dfrac{x+3}{63}-\dfrac{x+5}{61}-\dfrac{x+7}{59}=0\)
\(\left(\dfrac{x+1}{65}+1\right)+\left(\dfrac{x+3}{63}+1\right)-\left(\dfrac{x+5}{61}+1\right)-\left(\dfrac{x+7}{59}+1\right)\)
\(\Leftrightarrow\dfrac{x+66}{65}+\dfrac{x+66}{63}+\dfrac{x+66}{61}+\dfrac{x+66}{59}=0\)
\(\Leftrightarrow\left(x+66\right).\left[\left(\dfrac{1}{65}+\dfrac{1}{63}\right)-\left(\dfrac{1}{61}+\dfrac{1}{59}\right)\right]\)\(=0\)
Do \(\dfrac{1}{65}< \dfrac{1}{63}< \dfrac{1}{61}< \dfrac{1}{59}\)
\(\Rightarrow\left(\dfrac{1}{65}+\dfrac{1}{63}\right)-\left(\dfrac{1}{61}+\dfrac{1}{59}\right)< 0\)
Vậy để \(\left(x+66\right).\left[\left(\dfrac{1}{65}+\dfrac{1}{63}\right)-\left(\dfrac{1}{61}+\dfrac{1}{59}\right)\right]=0\)
\(\Leftrightarrow x+66=0\)
\(\Leftrightarrow x=-66\)
Vậy \(x\in\left\{-66\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+.....+\frac{1}{59\cdot60}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+.....+\frac{1}{59}-\frac{1}{60}\)
\(=1-\frac{1}{60}\)
\(=\frac{59}{60}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(B=\left(1-\frac{1}{2010}\right)x\left(1-\frac{2}{2010}\right)x\left(1-\frac{3}{2010}\right)x...x\left(1-\frac{2011}{2010}\right)\)
\(B=\left(1-\frac{1}{2010}\right)x\left(1-\frac{2}{2010}\right)x\left(1-\frac{3}{2010}\right)x....x\left(1-\frac{2010}{2010}\right)x\left(1-\frac{2011}{2010}\right)\)
\(B=\left(1-\frac{1}{2010}\right)x\left(1-\frac{2}{2010}\right)x\left(1-\frac{3}{2010}\right)x...x\left(0\right)x\left(1-\frac{2011}{2010}\right)\)
\(B=0\)
![](https://rs.olm.vn/images/avt/0.png?1311)
`@` `\text {Ans}`
`\downarrow`
\(\dfrac{3}{5}\times\dfrac{96}{59}+\dfrac{3}{5}\times\dfrac{22}{59}\)
\(=\dfrac{3}{5}\times\left(\dfrac{96}{59}+\dfrac{22}{59}\right)\)
\(=\dfrac{3}{5}\times2\)
\(=\dfrac{6}{5}\)
( 1 + x ) + ( 3 + x ) + ( 5 + x ) + ... + ( 59 + x ) = 2010
( 1 + 3 + 5 + ... + 59 ) + ( x + x + x + ... + x ) = 2010
30x = 2010 - 900
30x = 1110
x = 1110 : 30
x = 37
<=>25x+1+3+5+7+...+59=2010
<=>25x+625 =2010
<=>25x =2010-625
<=>25x =1385
<=>x =1385:25
<=>x =55,4
Vậy x=55,4