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}

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

  -(-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

-(-2010) + 36.41 - 36.(-59) + (-2010)

= 2010 + 36.41 - 36.(-59) + (-2010)

= (-2010 + 2010) + 36.(-59 + 41)

= 0 + 36.(-18)

= 0 + (-648)

= -648

-75.(18 - 65) - 65.(75 - 18)

-75.18 - (-75).65 - 65.75 - 65.18

mk chỉ giải đc đến đây th, bn thông cảm nha

\(\dfrac{x+1}{65}+\dfrac{x+3}{63}\) = \(\dfrac{x+5}{61}\) + \(\dfrac{x+7}{59}\)

<=> \(\dfrac{x+1}{65}+1+\dfrac{x+3}{63}+1\) = \(\dfrac{x+5}{61}\) + 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

hình như bn sai đề đúng ko

\(\dfrac{x+1}{65}+\dfrac{x+3}{63}+\dfrac{x+5}{61}+\dfrac{x+7}{59}\)

\(\Leftrightarrow\dfrac{x+1}{65}+\dfrac{x+3}{63}-\dfrac{x+5}{61}-\dfrac{x+7}{59}=0\)

\(\left(\dfrac{x+1}{65}+1\right)+\left(\dfrac{x+3}{63}+1\right)-\left(\dfrac{x+5}{61}+1\right)-\left(\dfrac{x+7}{59}+1\right)\)

\(\Leftrightarrow\dfrac{x+66}{65}+\dfrac{x+66}{63}+\dfrac{x+66}{61}+\dfrac{x+66}{59}=0\)

\(\Leftrightarrow\left(x+66\right).\left[\left(\dfrac{1}{65}+\dfrac{1}{63}\right)-\left(\dfrac{1}{61}+\dfrac{1}{59}\right)\right]\)\(=0\)

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 \(x\in\left\{-66\right\}\)

11

9/10

11

9/10

 Cái sau là 9/10

\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+.....+\frac{1}{59\cdot60}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+.....+\frac{1}{59}-\frac{1}{60}\)

\(=1-\frac{1}{60}\)

\(=\frac{59}{60}\)

X=59/60 nha

\(B=\left(1-\frac{1}{2010}\right)x\left(1-\frac{2}{2010}\right)x\left(1-\frac{3}{2010}\right)x...x\left(1-\frac{2011}{2010}\right)\)

\(B=\left(1-\frac{1}{2010}\right)x\left(1-\frac{2}{2010}\right)x\left(1-\frac{3}{2010}\right)x....x\left(1-\frac{2010}{2010}\right)x\left(1-\frac{2011}{2010}\right)\)

\(B=\left(1-\frac{1}{2010}\right)x\left(1-\frac{2}{2010}\right)x\left(1-\frac{3}{2010}\right)x...x\left(0\right)x\left(1-\frac{2011}{2010}\right)\)

\(B=0\)

Phúc 6A phải k

xét các thừa số tích B có: \(1-\frac{2010}{2010}=0\)

Nên B = 0

`@` `\text {Ans}`

`\downarrow`

\(\dfrac{3}{5}\times\dfrac{96}{59}+\dfrac{3}{5}\times\dfrac{22}{59}\)

\(=\dfrac{3}{5}\times\left(\dfrac{96}{59}+\dfrac{22}{59}\right)\)

\(=\dfrac{3}{5}\times2\)

\(=\dfrac{6}{5}\)

\(\dfrac{3}{5}.\dfrac{96}{59}+\dfrac{3}{5}.\dfrac{22}{59}\)

\(=\dfrac{3}{5}\left(\dfrac{96}{59}+\dfrac{22}{59}\right)\)

\(=\dfrac{3}{5}.2=\dfrac{6}{5}\)

=\(\dfrac{3}{5}x\left(\dfrac{96}{59}+\dfrac{22}{59}\right)=\dfrac{3}{5}x2=\dfrac{6}{5}\)