tìm x:
\(\left\{^{10^{10}}\left[^2\left(^{10^9}x^{10^9}\right)^2\right]^{10^{10}}\right\}!=1\)
K
Khách
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HT
7 tháng 3 2021
Mình xin trình bày 2 cách, một là phân tích bình thường, 2 là xài L'Hospital. Bởi c3 ko ai cho xài L'Hospital để hack tự luận cả
C1: Normal
\(\left(2-x\right)+\left(2-x\right)^2+...+\left(2-x\right)^9-9\)
\(=\left[\left(2-x\right)-1\right]+\left[\left(2-x\right)^2-1\right]+...+\left[\left(2-x\right)^9-1\right]\)
\(=\left(2-x-1\right)+\left(2-x-1\right)\left(2-x+1\right)+\left(2-x-1\right)\left[\left(2-x\right)^2+\left(2-x\right)+1\right]+...+\left(2-x-1\right)\left[\left(2-x\right)^8+\left(2-x\right)^7+...+1\right]\)
\(=-\left(x-1\right)\left(1+2-x+1+\left(2-x\right)^2+\left(2-x\right)+1+....+\left(2-x\right)^8+\left(2-x\right)^7+...+1\right)\)
Lai co:
\(x+x^2+...+x^{10}-10=\left(x-1\right)+\left(x^2-1\right)+...+\left(x^{10}-1\right)\)
\(=\left(x-1\right)+\left(x-1\right)\left(x+1\right)+....+\left(x-1\right)\left(x^9+x^8+...+1\right)\)
\(=\left(x-1\right)\left[1+x+1+x^2+x+1+....+x^9+x^8+...+1\right]\)
\(\Rightarrow\lim\limits_{x\rightarrow1}....=\lim\limits_{x\rightarrow1}\dfrac{-[1+2-x+1+\left(2-x\right)^2+\left(2-x\right)+1+...+\left(2-x\right)^8+\left(2-x\right)^7+...+1]}{1+x+1+x^2+x+1+...+x^9+x^8+...+1}\)
\(=\lim\limits_{x\rightarrow1}\dfrac{-[9.1+8.\left(2-x\right)+7\left(2-x\right)^2+6\left(2-x\right)^3+5\left(2-x\right)^4+4\left(2-x\right)^5+3\left(2-x\right)^6+2\left(2-x\right)^7+\left(2-x\right)^8]}{10.1+9x^2+8x^3+7x^4+6x^5+5x^6+4x^7+3x^8+2x^9+x^{10}}\)
\(=\dfrac{-[1+2+3+...+9]}{1+2+3+...+10}=\dfrac{-45}{55}\)
HT
7 tháng 3 2021
C2: L'Hospital
\(=\lim\limits_{x\rightarrow1}\dfrac{-1-2\left(2-x\right)-3\left(2-x\right)^2-...-9\left(2-x\right)^8}{1+2x+3x^2+...+10x^9}\)
\(=\lim\limits_{x\rightarrow1}\dfrac{-1-2-3-...-9}{1+2+3+...+10}=-\dfrac{45}{55}\)
1 tháng 3 2018
Đặt \(A=x^{20}+x^{10}+1\)
\(x^{50}+x^{10}+1\)
\(=x^{50}-x^{20}+A\)
\(=x^{20}\left(x^{30}-1\right)+A\)
\(=x^{20}\left(x^{10}-1\right)A+A\)
\(=\left(x^{30}-x^{20}+1\right)A\)
mà \(\left(x^{30}-x^{20}+1\right)A⋮A\)
\(\Rightarrow\left(x^{50}+x^{10}+1\right)⋮\left(x^{20}+x^{10}+1\right)\)
Y
11 tháng 1 2023
\(8,1-\left(x-6\right)=4\left(2-2x\right)\)
\(\Leftrightarrow1-x+6=8-8x\)
\(\Leftrightarrow-x+8x=8-1-6\)
\(\Leftrightarrow7x=1\)
\(\Leftrightarrow x=\dfrac{1}{7}\)
\(9,\left(3x-2\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-2=0\\x+5=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-5\end{matrix}\right.\)
\(10,\left(x+3\right)\left(x^2+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x^2+2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=\varnothing\end{matrix}\right.\)
2
11 tháng 1 2023
`8)1-(x-5)=4(2-2x)`
`<=>1-x+5=8-6x`
`<=>5x=2<=>x=2/5`
`9)(3x-2)(x+5)=0`
`<=>[(x=2/3),(x=-5):}`
`10)(x+3)(x^2+2)=0`
Mà `x^2+2 > 0 AA x`
`=>x+3=0`
`<=>x=-3`
`11)(5x-1)(x^2-9)=0`
`<=>(5x-1)(x-3)(x+3)=0`
`<=>[(x=1/5),(x=3),(x=-3):}`
`12)x(x-3)+3(x-3)=0`
`<=>(x-3)(x+3)=0`
`<=>[(x=3),(x=-3):}`
`13)x(x-5)-4x+20=0`
`<=>x(x-5)-4(x-5)=0`
`<=>(x-5)(x-4)=0`
`<=>[(x=5),(x=4):}`
`14)x^2+4x-5=0`
`<=>x^2+5x-x-5=0`
`<=>(x+5)(x-1)=0`
`<=>[(x=-5),(x=1):}`
BT
22 tháng 3 2017
<=> (1-1/10)(x-1)+x/10=x-9/10
<=> 9x/10-9/10+x/10=x-9/10
<=> x=x
Như vậy, phương trình thỏa mãn với mọi x
HN
2
3 tháng 11 2017
\(9\left(x+1\right)^2-\left(3x-2\right)\left(3x+2\right)=10\)
\(9\left(x^2+2x+1\right)-\left(9x^2-4\right)=10\)
\(9x^2+18x+9-9x^2+4-10=0\)
\(18x+3=0\)
\(18x=-3\)
\(x=\frac{-3}{18}\)
\(x=\frac{-1}{6}\)
vậy \(x=\frac{-1}{6}\)
LM
26 tháng 12 2017
Ta có:
\(9\left(x+1\right)^2-\left(3x-2\right)\left(3x+2\right)=10\)
\(\Rightarrow9\left(x^2+2x+1\right)-\left(9x^2-4\right)=10\)
\(\Rightarrow9x^2+18x+9-9x^2+4=10\)
\(\Rightarrow9x^2-9x^2+18x+13=10\)
\(\Rightarrow18x=10-13\)
\(\Rightarrow18x=-3\)
\(\Rightarrow18x=-\frac{1}{6}\)
5 tháng 6 2015
Vê trái:
\(=\frac{2}{\left(x-1\right)\left(x+1\right)}+\frac{4}{\left(x-2\right)\left(x+2\right)}+...+\frac{20}{\left(x-10\right)\left(x+10\right)}\)
\(=\frac{\left(x+1\right)-\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}+\frac{\left(x+2\right)-\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}+...+\frac{\left(x+10\right)-\left(x-10\right)}{\left(x+10\right)\left(x-10\right)}\)
\(=\frac{1}{x-1}-\frac{1}{x+1}+\frac{1}{x-2}-\frac{1}{x+2}+...+\frac{1}{x-10}-\frac{1}{x+10}\)
\(=\left(\frac{1}{x-1}+\frac{1}{x-2}+...+\frac{1}{x-10}\right)-\left(\frac{1}{x+1}+\frac{1}{x+2}+...+\frac{1}{x+10}\right)\)
Vế phải:
\(=\frac{\left(x+1\right)-\left(x-10\right)}{\left(x-10\right)\left(x+1\right)}+\frac{\left(x+2\right)-\left(x-9\right)}{\left(x-9\right)\left(x+2\right)}+...+\frac{\left(x+10\right)-\left(x-1\right)}{\left(x-1\right)\left(x+10\right)}\)
\(=\frac{1}{x-10}-\frac{1}{x+1}+\frac{1}{x-9}-\frac{1}{x+2}+...+\frac{1}{x-1}-\frac{1}{x+10}\)
\(=\left(\frac{1}{x-1}+\frac{1}{x-2}+...+\frac{1}{x-10}\right)-\left(\frac{1}{x+1}+\frac{1}{x+2}+...+\frac{1}{x+10}\right)\) = vế phải
=> đpcm
30 tháng 10 2021
1) (x−1):0,16=−9:(1−x)
\(\Rightarrow\)(x-1):0,16= 9:(-1):(x-1)
\(\Rightarrow\)(x-1):0,16=9:(x-1)
\(\Rightarrow\)(x-1).(x-1)= 9. 0,16
\(\Rightarrow\)(x-1)\(^2\)= 1,44=1,2\(^2\)=(-1,2)\(^2\)
\(\Rightarrow\)x-1=1,2\(\Rightarrow\)x=2,2
hoặc x-1= -1,2\(\Rightarrow\)x= -0,2
Vậy x =2,2 ; x=0,2
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