a) \(x=6\times1000+3\times100+2\times10+8=6000+300+20+8=6328\)

b) \(x=5\times1000+6\times10+7=5000+60+7=5067\)

c) \(x=9\times1000+9\times100+9=9000+900+9=9909\)

a) 1080 x 9 > 1020 x 9 (vì hai phép tính có cùng thừa số 9, mà 1080 > 1020)

b) 2 x 1000 x 5 = 1000 x 2 x 5 (vì hai phép tính có thừa số giống nhau)

c) Ta có: 200 + 4000 = 4200, bài tập đưa về so sánh 2400 × 2 và 4200 × 2

2400 x 2 < (200 + 4000) x 2 (vì hai phép tính có thừa số 2 giống nhau và 2400 < 4200)

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

\(A=\frac{1000^9+2}{1000^9-1}=\frac{1000^9-1+3}{1000^9-1}=\frac{1000^9-1}{1000^9-1}+\frac{3}{1000^9-1}=1+\frac{3}{1000^9-1}\)

\(B=\frac{1000^9+1}{1000^9-2}=\frac{1000^9-2+3}{1000^9-2}=\frac{1000^9-2}{1000^9-2}+\frac{3}{1000^9-2}=1+\frac{3}{1000^9-2}\)

Vì \(1000^9-1>1000^9-2\Rightarrow\frac{3}{1000^9-1}< \frac{3}{1000^9-2}\Rightarrow1+\frac{3}{1000^9-1}< 1+\frac{3}{1000^9-2}\Rightarrow A< B\)

Vậy A < B

x-992 chia 9=1000(dư 8)

x-992=1000 nhân 9 (dư 8)

x-992=9000(dư 8)

     x=9000 nhân 992 +8

     x=8928000+8

     x=8928008

x-992 chia 9=1000(dư 8)

x-992=1000 nhân 9 (dư 8)

x-992=9000(dư 8)

     x=9000 nhân 992 +8

     x=8928000+8

     x=8928008

9000 nha bạn

\(1000\)x \(9=9000\)

Nhã Phương ơi tk mình nha!!!

1000 x 9 = 9000

= 1000 + 1000 + 1000 + 1000 + 1000 + 1000 + 1000 + 1000 + 1000

= 9000

Đáp án : B

c

B