Tìm x:
5x+4 + 5x+3 = 5x+2 + 5x+1
2006 2007 2008 2009
Giúp mik nhanh nka!!!
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=>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 \(\hept{\begin{cases}x>0\\x-5< 0\end{cases}}\Leftrightarrow\hept{\begin{cases}x>0\\x< 5\end{cases}}\Leftrightarrow x\in\left\{1;2;3;4\right\}\)
\(A=\left(5x-1\right)+2\left(1-5x\right)\left(4+5x\right)\left(5x+4\right)^2\)
\(=\left(5x-1\right)-2\left(5x-1\right)\left(5x+4\right)^3\)
\(=\left(5x-1\right)\left(1-2\left(5x+4\right)^3\right)\)
\(=\left(5x-1\right)\left(1-2\left(125x^3+300x^2+240x+64\right)\right)\)
\(=\left(5x-1\right)\left(1-250x^3-600x^2-480x-128\right)\)
\(=5x-1250x^4-3000x^3-2400x^2-640x-1+250x^3+600x^2+480x+128\)
\(=-1250x^4-2750x^3-1800x^2-110x+127\)
(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
\(\frac{5x+4}{2006}+\frac{5x+3}{2007}=\frac{5x+2}{2008}+\frac{5x+1}{2009}\)
\(\Leftrightarrow\frac{5x+4}{2006}+1+\frac{5x+3}{2007}+1=\frac{5x+2}{2008}+1+\frac{5x+1}{2009}+1\)
\(\Leftrightarrow\frac{5x+2010}{2006}+\frac{5x+2010}{2007}=\frac{5x+2010}{2008}+\frac{5x+2010}{2009}\)
\(\Leftrightarrow\left(5x+2010\right)\left(\frac{1}{2006}+\frac{1}{2007}\right)=\left(5x+2010\right)\left(\frac{1}{2008}+\frac{1}{2009}\right)\)
\(\Leftrightarrow5x+2010=0\)
\(\Leftrightarrow5x=-2010\)
\(\Leftrightarrow x=-402\)
2 4 6 và 8