Số tiền lãi sau một tháng người đó thu được là : 

60 000 000 : 100 x 0,8 = 480 000 ( đồng )

                          Đs : 480 000 đồng

\(\dfrac{x+2}{x-5}-\dfrac{x-9}{5-x}-\dfrac{x-8}{5-x}\)

\(=\dfrac{x+2}{x-5}+\dfrac{x-9}{x-5}+\dfrac{x-8}{x-5}\)

\(=\dfrac{\left(x+2\right)+\left(x-9\right)+\left(x-8\right)}{x-5}\)

\(=\dfrac{x+2+x-9+x-8}{x-5}\)

\(=\dfrac{3x-15}{x-5}\)

\(=\dfrac{3\left(x-5\right)}{x-5}\)

\(=3\)

\(\dfrac{-9}{x}=\dfrac{\dfrac{-3}{4}}{49}\)

\(\Leftrightarrow\dfrac{-9}{x}=\dfrac{-3}{196}\)

\(\Leftrightarrow x.\left(-3\right)=\left(-9\right).196\)

\(\Leftrightarrow x.\left(-3\right)=-1764\)

\(\Leftrightarrow x=\left(-1764\right):\left(-3\right)\)

\(\Leftrightarrow x=588\)

Vậy \(x=588\)

Gọi số kg gạo xay xát từ 300 kg thóc là x.

Đổi 1 tạ thóc = 100 kg thóc

Vì số thóc và số gạo là 2 đại lượng tỉ lệ thuận nên ta có:

\(\dfrac{100}{60}=\dfrac{300}{x}\)

\(\Rightarrow x=\dfrac{60.300}{100}=\dfrac{18000}{100}=180\)

Vậy 300 kg thóc thì xay xát được 180 kg gạo

\(\dfrac{1}{x^2-4x-5}=\dfrac{1}{\left(x^2+x\right)-\left(5x+5\right)}=\dfrac{1}{x\left(x+1\right)-5\left(x+1\right)}=\dfrac{1}{\left(x+1\right)\left(x-5\right)}\)

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

MTC \(x\left(x-3\right)\left(x+1\right)\left(x-5\right)\)

\(\dfrac{1}{x^2-4x-5}=\dfrac{1}{\left(x^2+x\right)-\left(5x+5\right)}=\dfrac{1}{x\left(x+1\right)-5\left(x+1\right)}\\ =\dfrac{1}{\left(x+1\right)\left(x-5\right)}=\dfrac{x\left(x-3\right)}{x\left(x-3\right)\left(x+1\right)\left(x-5\right)}=\dfrac{x^2-3x}{x\left(x-3\right)\left(x+1\right)\left(x-5\right)}\)

\(\dfrac{2}{x^2-2x-x}=\dfrac{2}{x^2-3x}=\dfrac{2}{x\left(x-3\right)}=\dfrac{2\left(x+1\right)\left(x-5\right)}{x\left(x-3\right)\left(x+1\right)\left(x-5\right)}=\dfrac{\left(2x+2\right)\left(x-5\right)}{x\left(x-3\right)\left(x+1\right)\left(x-5\right)}\\ =\dfrac{2x^2-10x+2x-10}{x\left(x-3\right)\left(x+1\right)\left(x-5\right)}=\dfrac{2x^2-8x-10}{x\left(x-3\right)\left(x+1\right)\left(x-5\right)}\)

\(\dfrac{x^3-16x}{x^3-3x^2-4x}=1\)

\(\Leftrightarrow\dfrac{x^3-16x}{x^3-3x^2-4x}-1=0\)

\(\Leftrightarrow\dfrac{x^3-16x}{x^3-3x^2-4x}-\dfrac{x^3-3x^2-4x}{x^3-3x^2-4x}=0\)

\(\Leftrightarrow\dfrac{x^3-16x-\left(x^3-3x^2-4x\right)}{x^3-3x^2-4x}=0\)

\(\Leftrightarrow x^3-16x-\left(x^3-3x^2-4x\right)=0\)

\(\Leftrightarrow x^3-16x-x^3+3x^2+4x=0\)

\(\Leftrightarrow3x^2-12x=0\)

\(\Leftrightarrow3x\left(x-4\right)=0\)

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

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

Vậy x=0 hoặc x=4

(x-5)/3=(y+8)/5 

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

(x-5)/3=(y+8)/5=[(x-5)-(y+8)]/(3-5)=(x-5-y-8)/(-2)=[(x-y)-(5+8)]/(-2)=(8-13)/(-2)=(-5)/(-2)=5/2

nên (x-5)/3=5/2 => x-5=5/2*3 =>x=15/2+5=25/2

       (y+8)/5=5/2 =>y+8=5/2*5 =>y=25/2-8=9/2

2)

Theo đề ta có:

m chia 11, 13, 17 đều dư 7 \(\Rightarrow m-7⋮11;13;17\)

\(\Rightarrow m-7\in BC\left(11;13;17\right)\)

Vì 11; 13;17 là các số nguyên tố

\(\Rightarrow BCNN\left(11;13;17\right)=11.13.17=2431\)

\(\Rightarrow BC\left(11;13;17\right)\in\left\{0;2431;7293;9724;12155;......\right\}\)

Vì m-7 là số tự nhiên lớn nhất là có 4 chữ số

\(\Rightarrow m-7=9724\)

\(\Rightarrow m=9724+7\)

\(\Rightarrow m=9731\)

Theo t/c dãy tỉ số = nhau:

\(\frac{x-5}{3}=\frac{y+8}{5}=\frac{x-5-y-8}{3+5}=\frac{x-y-13}{8}=\frac{8-13}{8}=-\frac{5}{8}\)

=> \(\frac{x-5}{3}=-\frac{5}{8}\Rightarrow x-5=-\frac{5}{8}.3=-\frac{15}{8}\Rightarrow x=-\frac{15}{8}+5=\frac{25}{8}\)

=> \(\frac{y+8}{5}=-\frac{5}{8}\Rightarrow y+8=-\frac{5}{8}.5=-\frac{25}{8}\Rightarrow y=-\frac{25}{8}-8=-\frac{89}{8}\)

Vậy x=25/8; y=-89/8.

1) \(3x^3-48x=0\)

\(\Leftrightarrow3x\left(x^2-16\right)=0\)

\(\Leftrightarrow3x\left(x^2-4^2\right)=0\)

\(\Leftrightarrow3x\left(x-4\right)\left(x+4\right)=0\)

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

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

Vậy x=0 ; x=4 ; x=-4

b) \(x^3+x^2-4x=4\)

\(\Leftrightarrow x^3+x^2-4x-4=0\)

\(\Leftrightarrow\left(x^3+x^2\right)-\left(4x+4\right)=0\)

\(\Leftrightarrow x^2\left(x+1\right)-4\left(x+1\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(x^2-4\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(x^2-2^2\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(x-2\right)\left(x+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\x-2=0\\x+2=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=2\\x=-2\end{matrix}\right.\)

Vậy x=-1 ; x=2 ; x=-2