=>x:x²-5x<0

=>x:x.x-5x<0

=>1.x-5x<0

=>x(1-5)<0

=>x.(-4)<0

=>x là một số nguyên bất kì

Dấu “=” xảy ra khi x=0 (Không chắc sai thì kệ :P)

\(x^2-5x\le0\)

\(\Leftrightarrow x\left(x-5\right)\le0\)

\(TH:x\left(x-5\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x-5=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=5\end{cases}}\)

\(TH:x\left(x-5\right)< 0\)

\(\Leftrightarrow\hept{\begin{cases}x\\x-5\end{cases}}\) trái dấu

Mà x > x - 5 nên \(\hept{\begin{cases}x>0\\x-5< 0\end{cases}}\Leftrightarrow\hept{\begin{cases}x>0\\x< 5\end{cases}}\Leftrightarrow x\in\left\{1;2;3;4\right\}\)

nguyên dương nhe :P nãy nhầm

\(A=\left(5x-1\right)+2\left(1-5x\right)\left(4+5x\right)\left(5x+4\right)^2\)

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

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

\(=\left(5x-1\right)\left(1-2\left(125x^3+300x^2+240x+64\right)\right)\)

\(=\left(5x-1\right)\left(1-250x^3-600x^2-480x-128\right)\)

\(=5x-1250x^4-3000x^3-2400x^2-640x-1+250x^3+600x^2+480x+128\)

\(=-1250x^4-2750x^3-1800x^2-110x+127\)

(Số hơi to)

\(B=\left(x-y\right)^3+\left(y+x\right)^3+\left(y-x\right)^3-3xy\left(x+y\right)\)

\(B=\left(x-y\right)^3+\left(y+x\right)^3-\left(x-y\right)^3-3xy\left(x+y\right)\)

\(B=\left(y+x\right)^3-3xy\left(x+y\right)\)

\(B=\left(x+y\right)\left[\left(x+y\right)^2-3xy\right]\)

\(B=\left(x+y\right)\left[x^2+2xy+y^2-3xy\right]\)

\(B=\left(x+y\right)\left(x^2-xy+y^2\right)\)

\(=x^3+y^3\)

a, (3x - 2)(4x + 3) = (2 - 3x)(x - 1)

\(\Leftrightarrow\) (3x - 2)(4x + 3) - (2 - 3x)(x - 1) = 0

\(\Leftrightarrow\) (3x - 2)(4x + 3) + (3x - 2)(x - 1) = 0

\(\Leftrightarrow\) (3x - 2)(4x + 3 + x - 1) = 0

\(\Leftrightarrow\) (3x - 2)(5x + 2) = 0

\(\Leftrightarrow\left[{}\begin{matrix}3x-2=0\\5x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{2}{3}\\x=\frac{-2}{5}\end{matrix}\right.\)

Vậy S = {\(\frac{2}{3}\); \(\frac{-2}{5}\)}

b, x² + (x + 3)(5x - 7) = 9

\(\Leftrightarrow\) x² - 9 + (x + 3)(5x - 7) = 0

\(\Leftrightarrow\) (x - 3)(x + 3) + (x + 3)(5x - 7) = 0

\(\Leftrightarrow\) (x + 3)(x - 3 + 5x - 7) = 0

\(\Leftrightarrow\) (x + 3)(6x - 10) = 0

\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\6x-10=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=\frac{5}{3}\end{matrix}\right.\)

Vậy S = {-3; \(\frac{5}{3}\)}

c, 2x² + 5x + 3 = 0

\(\Leftrightarrow\) 2x² + 2x + 3x + 3 = 0

\(\Leftrightarrow\) 2x(x + 1) + 3(x + 1) = 0

\(\Leftrightarrow\) (x + 1)(2x + 3) = 0

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

Vậy S = {-1; \(\frac{3}{2}\)}

d, \(\frac{3-2x}{2006}+\frac{3-2x}{2007}+\frac{3-2x}{2008}=\frac{3-2x}{2009}+\frac{3-2x}{2010}\)

\(\Leftrightarrow\) \(\frac{3-2x}{2006}+\frac{3-2x}{2007}+\frac{3-2x}{2008}-\frac{3-2x}{2009}-\frac{3-2x}{2010}=0\)

\(\Leftrightarrow\) (3 - 2x)\(\left(\frac{1}{2006}+\frac{1}{2007}+\frac{1}{2008}-\frac{1}{2009}-\frac{1}{2010}\right)\) = 0

\(\Leftrightarrow\) 3 - 2x = 0

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

Vậy S = {\(\frac{3}{2}\)}

Chúc bn học tốt!!

a)A(x) = 3x^3 - 4x^4 - 2x^3 + 4x^4 - 5x + 3 

=x^3-5x+3

bậc:3

hệ số tự do:3

hệ số cao nhất :3

B(x) = 5x^3 - 4x^2 - 5x^3 - 4x^2 - 5x - 3

=-8x^2-5x+3

bậc:2

hệ số tự do:3

hệ số cao nhất:3

b)A(x)+B(x)=x^3-8^2+10x+6

câu b mik ko đặt tính theo hàng dọc đc thông cảm nha

mk 0 bt nhưng ai chat nhìu thì kt bn với mk nha