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10 tháng 6 2018
\(x=7\Rightarrow\left\{{}\begin{matrix}4=x-3\\20=3x-1\end{matrix}\right.\)\(\Rightarrow P\left(7\right)=x^{100}-4x^{99}-20x^{98}-4x^{97}-...-20x^2-4x\\ =x^{100}-\left(x-3\right)x^{99}-\left(3x-1\right)x^{98}-\left(x-3\right)x^{97}-...-\left(3x-1\right)x^2-\left(x-3\right)x\\ =x^{100}-x^{100}+3x^{99}-3x^{99}+x^{98}-x^{98}+3x^{97}-...-3x^3+x^2-x^2+3x\\ =3x\\ =21\)
KB
30 tháng 9 2018
a ) \(\left(\dfrac{20x}{3y^2}\right):\left(\dfrac{4x^3}{5y}\right)=\dfrac{20x}{3y^2}.\dfrac{5y}{4x^3}=\dfrac{100xy}{12x^3y^2}=\dfrac{25}{3x^2y}\)
b ) Đ/k : \(x\ne-4\)
Ta có : \(\dfrac{4x+12}{\left(x+4\right)^2}:\dfrac{3\left(x+3\right)}{x+4}\)
\(=\dfrac{4\left(x+3\right)}{\left(x+4\right)^2}.\dfrac{x+4}{3\left(x+3\right)}\)
\(=\dfrac{4\left(x+3\right)\left(x+4\right)}{3\left(x+3\right)\left(x+4\right)^2}\)
\(=\dfrac{4}{3\left(x+4\right)}\)
\(=\dfrac{4}{3x+12}\)
H
26 tháng 12 2022
\(B1\\ a,2x+10y=2\left(x+5y\right)\\ b,x^2+4x+4=x^2+2.2x+2^2=\left(x+2\right)^2\\ c,x^2-y^2+10y-25\\ =\left(x^2-y^2\right)+5\left(2y-5\right)\\ =\left(x-y\right)\left(x+y\right)+5\left(2y-5\right)\\ B2\)
\(a,x^2-3x+x-3=0\\ =>x\left(x-3\right)+\left(x-3\right)=0\\ =>\left(x+1\right)\left(x-3\right)=0\\ =>\left[{}\begin{matrix}x+1=0\\x-3=0\end{matrix}\right.=>\left[{}\begin{matrix}x=-1\\x=3\end{matrix}\right.\\ b,2x\left(x-3\right)-\dfrac{1}{2}\left(4x^2-3\right)=0\\ =>2x^2-6x-2x^2+\dfrac{3}{2}=0\\ =>-6x=-\dfrac{3}{2}\\ =>x=\left(-\dfrac{3}{2}\right):\left(-6\right)\\ =>x=\dfrac{1}{4}\\ c,x^2-\left(x-3\right)\left(2x-5\right)=9\\ =>x^2-2x^2+6x+5x-15=9\\ =>-x^2+11-15-9=0\\ =>-x^2+11x-24=0\\ =>-x^2+8x+3x-24=0\\ =>-x\left(x-8\right)+3\left(x-8\right)=0\\ =>\left(3-x\right)\left(x-8\right)=0\\ =>\left[{}\begin{matrix}3-x=0\\x-8=0\end{matrix}\right.=>\left[{}\begin{matrix}x=3\\x=8\end{matrix}\right.\)
27 tháng 6 2017
a) ... \(\Leftrightarrow x^2\left(x-1\right)-4\left(x-1\right)^2=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-2\right)\left(x+2\right)=0\Leftrightarrow\hept{\begin{cases}x=1\\x=2\\x=-2\end{cases}}\)Vậy.....
b) ... \(\Leftrightarrow x^3\left(x-2\right)+10x\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x^3+10x\right)=0\)
\(\Leftrightarrow x\left(x-2\right)\left(x^2+10\right)=0\Leftrightarrow\hept{\begin{cases}x=0\\x=2\\x^2=-10\Rightarrow x\in\theta\end{cases}}\)(\(\theta\)là rỗng) Vậy.........
c) ... \(\Leftrightarrow2x-3=x+5\Leftrightarrow x=8\)Vậy.......
d) ... \(\Leftrightarrow x\left(x^2-16\right)=0\Leftrightarrow x\left(x-4\right)\left(x+4\right)=0\Leftrightarrow\hept{\begin{cases}x=0\\x=4\\x=-4\end{cases}}\)Vậy......
4 tháng 3 2018
a)\(x\left(x+1\right)\left(x+2\right)\left(x+3\right)=24\)
\(\Leftrightarrow x\left(x+3\right)\left(x+1\right)\left(x+2\right)-24=0\)
\(\Leftrightarrow\left(x^2+3x\right)\left(x^2+3x+2\right)-24=0\)
Đặt \(x^2+3x+1=t\)
\(\Leftrightarrow\left(t-1\right)\left(t+1\right)-24=0\)
\(\Leftrightarrow t^2-25=0\)
\(\Leftrightarrow\left(t-5\right)\left(t+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}t=5\\t=-5\end{matrix}\right.\)
TH1:t=5\(\Rightarrow x^2+3x+1=5\)
\(\Leftrightarrow x^2+3x-4=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-4\end{matrix}\right.\)
TH2:t=-5\(\Rightarrow x^2+3x+1=-5\)
\(\Leftrightarrow x^2+3x+6=0\)(vô nghiệm)
Vậy ...
4 tháng 3 2018
b)\(\Leftrightarrow2\left(x^4-10x^2+9\right)=0\)
\(\Leftrightarrow x^4-9x^2-x^2+9=0\)
\(\Leftrightarrow x^2\left(x^2-9\right)-\left(x^2-9\right)=0\)
\(\Leftrightarrow\left(x^2-9\right)\left(x^2-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-3\\x=1\\x=-1\end{matrix}\right.\)
NL
2
H
22 tháng 10 2023
`2(x-3)-4x=0`
`<=> 2x-6-4x=0`
`<=> -2x-6=0`
`<=>-2x=6`
`<=>x=-3`
__
`x^2-2x+1=25`
`<=>(x-1)^2=25`
`<=> (x-1)^2 = (+- 5)^2`
\(\Leftrightarrow\left[{}\begin{matrix}x-1=5\\x-1=-5\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=6\\x=-4\end{matrix}\right.\)
22 tháng 10 2023
a: 2(x-3)-4x=0
=>2x-6-4x=0
=>-2x-6=0
=>2x+6=0
=>2x=-6
=>x=-3
b: \(x^2-2x+1=25\)
=>\(\left(x-1\right)^2=25\)
=>\(\left[{}\begin{matrix}x-1=5\\x-1=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-4\end{matrix}\right.\)
21 tháng 1
\(\dfrac{20x^2+120x+180}{\left(3x+5\right)^2-4x^2}+\dfrac{5x^2-25}{9x^2-\left(2x+5\right)^2}-\dfrac{\left(2x+3\right)^2-x^2}{3\left(x^2+8x+15\right)}\)
\(=\dfrac{20\left(x^2+6x+9\right)}{\left(3x+5+2x\right)\left(3x+5-2x\right)}+\dfrac{5\left(x-5\right)\left(x+5\right)}{\left(3x-2x-5\right)\left(3x+2x+5\right)}-\dfrac{\left(2x+3-x\right)\left(2x+3+x\right)}{3\left(x+3\right)\left(x+5\right)}\)
\(=\dfrac{20\left(x+3\right)^2}{5\left(x+1\right)\cdot\left(x+5\right)}+\dfrac{5\left(x-5\right)\left(x+5\right)}{5\left(x+1\right)\left(x-5\right)}-\dfrac{\left(x+3\right)\cdot3\left(x+1\right)}{3\left(x+3\right)\left(x+5\right)}\)
\(=\dfrac{4\left(x+3\right)^2}{\left(x+1\right)\left(x+5\right)}+\dfrac{x+5}{x+1}-\dfrac{x+1}{x+5}\)
\(=\dfrac{4\left(x+3\right)^2+\left(x+5\right)^2-\left(x+1\right)^2}{\left(x+1\right)\left(x+5\right)}\)
\(=\dfrac{4x^2+24x+36+x^2+10x+25-x^2-2x-1}{\left(x+1\right)\cdot\left(x+5\right)}\)
\(=\dfrac{4x^2+32x+60}{\left(x+1\right)\left(x+5\right)}=\dfrac{4\left(x^2+8x+15\right)}{\left(x+1\right)\left(x+5\right)}\)
\(=\dfrac{4\left(x+3\right)\cdot\left(x+5\right)}{\left(x+1\right)\left(x+5\right)}=\dfrac{4x+12}{x+1}\)
\(\left(x+1\right)^2=4x^2+20x+25\)
<=> \(\left(x+1\right)^2=\left(2x+5\right)^2\)
<=> \(x+1=2x+5\)
=> \(x=-4\)
Vũ Minh Tuấn Băng Băng 2k6 Hưng Nguyễn Lê Việt Nguyễn Lê Phước Thịnh ........
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