x+2x+3x+...+1000x=5005x\(10^4\)

ta có:x+2x+3x+...+1000x=x(1+2+3+...+1000)

                                     =x((1000+1):2x1000)

                                     =x500500

\(\Rightarrow\) x500500=5005x\(10^4\) 

x5005.\(100\)=5005x\(10^4\) 

x5005.\(10^2\)=5005x\(10^4\) 

\(\Rightarrow x.10^2=10^4\)

x=\(10^4:10^2\)

x=\(10^2\) 

x=100

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

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

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

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

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

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

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

___________

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

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

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

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

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

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

b: \(x^3+\dfrac{1}{27}=\left(x+\dfrac{1}{3}\right)\left(x^2-\dfrac{1}{3}x+\dfrac{1}{9}\right)\)

c: \(x^3-3x^2+3x-1=\left(x-1\right)^3\)

e: \(a^2y^2-2axby+b^2x^2\)

\(=\left(ay\right)^2-2\cdot ay\cdot bx+\left(bx\right)^2\)

\(=\left(ay-bx\right)^2\)

f: \(100-\left(3x-y\right)^2\)

\(=\left(10-3x+y\right)\left(10+3x-y\right)\)

g: \(64x^2-\left(8a+b\right)^2\)

\(=\left(8x\right)^2-\left(8a+b\right)^2\)

\(=\left(8x-8a-b\right)\left(8x+8a+b\right)\)

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

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

hay x=-1

Bài 2: 

a: =>x-45=-20

hay x=25

b: =>35+30-5x=-12+112

=>65-5x=100

=>5x=-35

hay x=-7

c: =>x-124=1000

hay x=1124

d: \(\Leftrightarrow46-\left(3x-2\right)^3=-18\)

\(\Leftrightarrow\left(3x-2\right)^3=64\)

=>3x-2=4

hay x=2

Bài 2: 

a: =>x-45=-20

hay x=25

b: =>35+30-5x=-12+112=100

=>65-5x=100

=>5x=-35

hay x=-7

c: =>x-124=1000

hay x=1124

d: \(\Leftrightarrow46-\left(3x-2\right)^3=-18\)

\(\Leftrightarrow\left(3x-2\right)^3=64\)

=>3x-2=4

hay x=2

a, \(\left(x+2\right)^3-x\left(x^2+6x-3\right)=0\Leftrightarrow x^3+4x^2+4x+2x^2+8x+8-x^3-6x^2+3x=0\)

\(\Leftrightarrow15x+8=0\Leftrightarrow x=-\frac{8}{15}\)

b, \(\left(x+4\right)^3-x\left(x+6\right)^2=7\Leftrightarrow12x+64=0\Leftrightarrow x=-\frac{19}{4}\)làm tắt:P 

Tự làm nốt nhé 

áp dụng tính chất dãy tỉ số bằng nhau ta có:

\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{2x}{4}=\dfrac{3y}{9}=\dfrac{2x-3y}{4-9}=-\dfrac{54}{5}\)

\(\dfrac{x}{2}=-\dfrac{54}{5}\Rightarrow x=-\dfrac{54}{5}.2=-\dfrac{108}{5}\)

\(\dfrac{y}{3}=-\dfrac{54}{5}\Rightarrow y=-\dfrac{54}{5}.3=-\dfrac{162}{5}\)

Vậy \(x=-\dfrac{108}{5};y=-\dfrac{162}{5}\)

Ta có: \(\dfrac{x}{2}=\dfrac{y}{3}\)

nên \(\dfrac{2x}{4}=\dfrac{3y}{9}\)

mà 2x-3y=54

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

\(\dfrac{2x}{4}=\dfrac{3y}{9}=\dfrac{2x-3y}{4-9}=\dfrac{-54}{5}\)

Do đó: \(x=-\dfrac{108}{5};y=-\dfrac{162}{5}\)