\(x-\dfrac{1}{8}=\dfrac{4}{x-2}\) hay \(x-\dfrac{1}{8}=\dfrac{4}{x}-2\) vậy bạn?

\(\dfrac{x-1}{8}=\dfrac{4}{x-2}\)

=>\(\dfrac{\left(x-1\right)\left(x-2\right)}{8.\left(x-2\right)}=\dfrac{4.8}{\left(x-2\right).8}\)

=>khử mẫu

=>(x-1)(x-2)=32

.......

Là \(\dfrac{x+1}{8}=\dfrac{4}{x-2}\) nhoa

\(x^3-3x^2+3x-1=-8\)

\(\Leftrightarrow x-1=-2\)

hay x=-1

c) Ta có:
2x=5y=>x/5=y/2
Áp dụng tính chất dãy tỉ số bằng nhau, ta có:
 x/5=y/2=x-y/5-2=15/3=5
=> x=5.5=25; y=5.2=10
d)Đặt x/2=y/5=k
=> x=2k; y=5k=> xy=2k.5k=10k^2=10=> k^2=1=>k=\(\pm\)1
Với k=1=>x=2; y=5
Với k=-1=>x=-2; y=-5
 

\(\left(x-3\right)=\left(3-x\right)^2\)

\(\Leftrightarrow x-3=\left(x-3\right)^2\)

\(\Leftrightarrow\left(x-3\right)-\left(x-3\right)^2=0\)

\(\Leftrightarrow\left(x-3\right)\left[1-\left(x-3\right)\right]=0\)

\(\Leftrightarrow\left(x-3\right)\left(4-x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\4-x=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=4\end{matrix}\right.\)

___________

\(x^3+\dfrac{3}{2}x^2+\dfrac{3}{4}x+\dfrac{1}{8}=\dfrac{1}{64}\)

\(\Leftrightarrow x^3+3\cdot\dfrac{1}{2}\cdot x^2+3\cdot\left(\dfrac{1}{2}\right)^2\cdot x+\left(\dfrac{1}{2}\right)^3=\dfrac{1}{64}\)

\(\Leftrightarrow\left(x+\dfrac{1}{2}\right)^3=\left(\dfrac{1}{4}\right)^3\)

\(\Leftrightarrow x+\dfrac{1}{2}=\dfrac{1}{4}\)

\(\Leftrightarrow x=\dfrac{1}{4}-\dfrac{1}{2}\)

\(\Leftrightarrow x=-\dfrac{1}{4}\)

240

a)240

b)12; 24; 36; 48

` x : 1 / 2 + x : 1 / 3 + x : 1 / 4 + x = 2010`

`x . 2 + x . 3 + x . 4 + x = 2010`

`x ( 2 + 3 + 4 + 1 ) = 2010`

`x . 10 = 2010`

`x = 2010 : 10`

`x = 201`

Vậy ` x= 201`

\(=>x=2010:\left(2+3+4\right)=2010:9=\dfrac{670}{3}\)

Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\dfrac{x-1}{2}=\dfrac{y+3}{4}=\dfrac{z-5}{6}=\dfrac{-3x-4y+5z+3-12-25}{-3\cdot2-4\cdot4+5\cdot6}=\dfrac{16}{8}=2\)

Do đó: x=5; y=5; z=17

Giải:

\(\left(x-1\right)+\left(x-2\right)+...+\left(x-100\right)=-50\) 

\(\Rightarrow100x+\left(-1+-2+...+-100\right)=-50\) 

\(\Rightarrow100x-\left(1+2+...+100\right)=-50\) 

Số số hạng \(\left(1+2+...+100\right)\) là: \(\left(100-1\right):1+1=100\) 

Tổng dãy \(\left(1+2+...+100\right)\) là: \(\left(1+100\right).100:2=5050\) 

\(\Rightarrow100x-5050=-50\) 

\(\Rightarrow100x=-50+5050\) 

\(\Rightarrow100x=5000\) 

\(\Rightarrow x=5000:100\) 

\(\Rightarrow x=50\) 

Chúc bạn học tốt!

Ta có: \(\left(x-1\right)+\left(x-2\right)+...+\left(x-100\right)=-50\)

\(\Leftrightarrow100x-5050=-50\)

\(\Leftrightarrow100x=5000\)

hay x=50

\(a)6/4=9/4\) =)??

thiếu dữ liệu

Câu hỏi là gìv

x=35/19

y=-25/19

z=127/19

3.(x-1)/6=4.(y+3)/16=5.(z-5)/20

áp dụng tính cất dãy tỉ số = nhau ta có:

5z-25-3x+3-4y-12/6-16-20=16/-30=-8/15

x=-8/15 .2 +1=-1/15

y=-8/15 .4 -3=-77/15

z=-8/15 .4 +5=43/15