\(\frac{1}{5.8} + \frac{1}{8.11} + ..+ \frac{1}{x.(x+3)}=\frac{101}{1540} \)

Nhân với 3 vào cả hai vế ta có:

\(\frac{3}{5.8} + \frac{3}{8.11} + ..+ \frac{3}{x.(x+3)}=\frac{303}{1540} \)

\(\frac{8-5}{5.8} + \frac{11-8}{8.11} + ..+ \frac{(x+3)-x}{x.(x+3)}=\frac{303}{1540} \)

\(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{x}-\frac{1}{x+3}=\frac{303}{1540}\)

\(\frac{1}{5}-\frac{1}{x+3}=\frac{303}{1540}\)

\(\frac{1}{x+3}=\frac{1}{5}-\frac{303}{1540}\)

\(\frac{1}{x+3}=\frac{1}{308}\)

\(x+3=308\)

\(x=308-3\)

\(x=305\)

Em cũng làm đễn đấy mà quên nhân 3 với \(\frac{101}{1540}\) nên cô dạy đội bồi dưỡng cho sai ( mà cũng hên là vô được :)) )

Lần sau nhớ cẩn thận hơn nha! :))) Dù sao cũng vào được  rồi. Cố lên! Chúc em may mắn:))

\(x.0=7\)

\(\Leftrightarrow0=7\)(Vô lý)

Vậy không có giả trị x nào thoả mãn yêu cầu bài toán.

x.0=7

x=7:0

x=          (ko chia đc vô lý)

câu này nữa bạn ơi 

8 [ x - 3 ] = 0

\(445-52+155+452\\ =\left(445+155\right)+\left(452-52\right)\\ =600+4000\\ =1000\\ 102\cdot55-102\cdot73+51\cdot36\\ =102\cdot\left(55-73\right)+51\cdot36\\ =102\cdot\left(-18\right)+102\cdot18\\ =0\\ 4\cdot5\cdot2\cdot25\cdot5-1000\\ =5000-1000\\ =4000\\ 8782-291\cdot13\\ =4999\\ 254+\left\{38\cdot\left[\left(42-16\right):2\right]\right\}\\ =254+\left[38\cdot\left(26:2\right)\right]\\ =254+\left(38\cdot13\right)\\ =748\\ \left(2791\cdot34+7882:14\right)\cdot0+1510-510:2\\ =1510-255\\ =1255\)

a. \(445-52+155+452=\left(445+155\right)+\left(452-52\right)=600+400=1000\)

b. \(102\cdot55-102\cdot73+51\cdot36=102\left(55-73+18\right)=102\cdot0=0\)

c.\(4\cdot5\cdot2\cdot25\cdot5-1000=4\cdot25\cdot5\cdot5\cdot2-1000=100\cdot50-1000=5000-1000=4000\)

d. \(8782-291\cdot13=8782-3783=4999\)

e. \(254+\left\{38\left[\left(42-16\right)\div2\right]\right\}=254+\left\{30\cdot13\right\}=254+494=748\)

f. \(\left(2791\cdot34+7882\div14\right)\cdot0+1510-510\div2=0+1510-255=1255\)\

\(\frac{1}{2}.x^2+x=0\)

\(\Leftrightarrow x.\left(\frac{1}{2}x+1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\\frac{1}{2}x+1=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\\frac{1}{2}x=-1\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=-2\end{cases}}\)

\(x^4\cdot x^7\cdot...\cdot x^{100}\)

\(=x^{4+7+...+100}\)

\(=x^{52\cdot33}=x^{1716}\)

\(x^1\cdot x^2\cdot x^3\cdot...\cdot x^{2006}\)

Ta có : \(x^1\cdot x^2=x^{1+2}=x^3\)

Tương tự : \(x^1\cdot x^2\cdot x^3=x^{1+2+3}=x^6\)

Áp dụng vào bài toán :

\(x^1\cdot x^2\cdot x^3\cdot...\cdot x^{2006}=x^{1+2+3+...+2006}\)

\(\Rightarrow x^{1+2+3+...+2006}=x^{2013021}\)

Sửa câu a : \(x^{1+2+3+...+2006}=x^{2033136}\)

 a, 7x . (x-10 ) = 0

=> \(\hept{\begin{cases}7x=0\\x-10=0\end{cases}}\) => \(\hept{\begin{cases}x=0\\x=10\end{cases}}\)

b,17 . (3x-6).(2x-8) = 0

          (3x-6).(2x-8) = 0 : 17

          (3x-6).(2x-8)  = 0

=>\(\hept{\begin{cases}3x-6=0\\2x-8=0\end{cases}}\)=>\(\hept{\begin{cases}x=2\\x=4\end{cases}}\)

Nhớ nha

\(5\cdot x+10\cdot9=990\)

\(\Rightarrow5\cdot x+90=990\)

\(\Rightarrow5\cdot x=900\)

\(\Rightarrow x=\dfrac{900}{5}\)

\(\Rightarrow x=180\)

_____________

\(1045:\left[215-\left(3\cdot x-24\right)\right]=5\)

\(\Rightarrow215-\left(3\cdot x-24\right)=1045:5\)

\(\Rightarrow215-\left(3\cdot x-24\right)=209\)

\(\Rightarrow3\cdot x-24=215-209\)

\(\Rightarrow3\cdot x-24=6\)

\(\Rightarrow3\cdot x=30\)

\(\Rightarrow x=10\)

`@` `\text {Ans}`

`\downarrow`

`5.x + 10.9 = 990`

`\Rightarrow 5x + 90 = 990`

`\Rightarrow 5x = 990 - 90`

`\Rightarrow 5x = 900`

`\Rightarrow x = 900 \div 5`

`\Rightarrow x = 180`

Vậy, `x = 180`

\(1045 \div [ 215 - (3 . x - 24 ) ] = 5\)

`\Rightarrow 1045 \div (215 - 3x + 24) = 5`

`\Rightarrow 191 + 3x = 1045 \div 5`

`\Rightarrow 191 + 3x = 209`

`\Rightarrow 3x = 209 - 191`

`\Rightarrow 3x =18`

`\Rightarrow x = 18 \div 3`

`\Rightarrow x = 6`

Vậy, `x = 6.`

5x+10.9=990

=>5x+90=990

=>5x=990-90

=>5x=900

=>x=900:5

=>x=180

______________

1045:[215-(3x-24)]=5

=>215-(3x-24)=1045:5

=>215-(3x-24)=209

=>3x-24=215-209

=>3x-24=6

=>3x=6+24

=>3x=30

=>x=30:3

=>x=10