a) \(\left(2y-1\right)^{1000}-\left(3+y\right)^{1000}=0\)

\(\Rightarrow\left(2y-1\right)^{1000}=\left(3+y\right)^{1000}\)

\(\Rightarrow2y-1=3+y\)

\(2y-y=3+1\)

\(y=4\)

b) \(\left(x-\frac{2}{9}\right)^3=\left(\frac{2}{3}\right)^6\)

\(\left(x-\frac{2}{9}\right)^3=\left(\left(\frac{2}{3}\right)^2\right)^3\)

\(\Rightarrow x-\frac{2}{9}=\left(\frac{2}{3}\right)^2\)

\(x-\frac{2}{9}=\frac{4}{9}\)

\(x=\frac{2}{3}\)

c) \(\left(2x-1\right)^6=\left(2x-1\right)^8\)

\(\left(\left(2x-1\right)^3\right)^2=\left(\left(2x-1\right)^4\right)^2\)

\(\Rightarrow\left(2x-1\right)^3=\left(2x-1\right)^4\)

\(8x^3-1=16x^4-1\)

\(16x^4-8x^3=0\)

\(8x^3\left(2x-1\right)=0\)

Nếu \(8x^3=0\) thì \(x^3=0\Rightarrow x=0\)

Nếu \(2x-1=0\)thì \(2x=1\Rightarrow x=\frac{1}{2}\)

Vậy x=0 và x=1/2

f: Ta có: \(x\left(2x-9\right)-4x+18=0\)

\(\Leftrightarrow\left(2x-9\right)\left(x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{9}{2}\\x=2\end{matrix}\right.\)

g: Ta có: \(4x\left(x-1000\right)-x+1000=0\)

\(\Leftrightarrow\left(x-1000\right)\left(4x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1000\\x=\dfrac{1}{4}\end{matrix}\right.\)

f. x(2x - 9) - 4x + 18 = 0

<=> x(2x - 9) - 2(2x - 9) = 0

<=> (x - 2)(2x - 9) = 0

<=> \(\left[{}\begin{matrix}x-2=0\\2x-9=0\end{matrix}\right.\)

<=> \(\left[{}\begin{matrix}x=2\\x=\dfrac{9}{2}\end{matrix}\right.\)

g. 4x(x - 1000) - x + 1000 = 0

<=> 4x(x - 1000) - (x - 1000) = 0

<=> (4x - 1)(x - 1000) = 0

<=> \(\left[{}\begin{matrix}4x-1=0\\x-1000=0\end{matrix}\right.\)

<=> \(\left[{}\begin{matrix}x=\dfrac{1}{4}\\x=1000\end{matrix}\right.\)

h. 2x(x - 4) - 6x²(-x + 4) = 0

<=> 2x(x - 4) + 6x²(x - 4) = 0

<=> (2x + 6x²)(x - 4) = 0

<=> 2x(1 + 3x)(x - 4) = 0

<=> \(\left[{}\begin{matrix}2x=0\\1+3x=0\\x-4=0\end{matrix}\right.\)

<=> \(\left[{}\begin{matrix}x=0\\x=\dfrac{-1}{3}\\x=4\end{matrix}\right.\)

i. 2x(x - 3) + x² - 9 = 0

<=> 2x(x - 3) + (x - 3)(x + 3) = 0

<=> (2x + x + 3)(x - 3) = 0

<=> (3x + 3)(x + 3) = 0

<=> \(\left[{}\begin{matrix}3x+3=0\\x+3=0\end{matrix}\right.\)

<=> \(\left[{}\begin{matrix}x=-1\\x=-3\end{matrix}\right.\)

j. 9x - 6x² + x³ = 0

<=> x(9 - 6x + x²) = 0

<=> x(3 - x)² = 0

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

<=> \(\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\)

 4x(x – 1000) – x + 1000 = 0

(4x-1)(x-1000)                 =0

⇔x=1/4 hoặc  1000

Đáp án cần chọn là: A

Ta có (x−4).1000=0nên x–4=0(vì 1000≠0)

Suy ra

x=0+4

x=4.

Vậy x=4.

Ta có;

x+2x+3x+4x+...+9x=1000-100

=>x+2x+3x+4x+...+9x=900

=>(1+2+3+4+....+9)x=900

=>[(1+9).9:2]x=900

=>45x=900

x=900:45=20

cho tớ đúng nha

Ta có;

x+2x+3x+4x+...+9x=1000-100

=>x+2x+3x+4x+...+9x=900

=>(1+2+3+4+....+9)x=900

=>45x=900

x=900:45=20

khi tớ đăng câu hỏi thì bạn nào trả lời nhanh và đúng nhất sẽ được ****(giống robert lewandoski ấy)