sai ở đâu : x = 0,999...
10x = 9.999...
x = 0.999...
--> 10x - x = 9
--> 9x = 9
--> x = 1
vậy 0.999... = 1
Những câu hỏi liên quan
Tính:
\(A=\sqrt{1+999...9^2+0,999...9^2}+0.999...9\)
\(\)( Có 100 số 9 )
Thách ai giải được:
x = 0,9999999999...
10x = 9,9999999999...
9x = 10x - 1x = 9,9999999999... - 0,9999999999...
9x = 9
x = 9 : 9
x = 1
1 = 0,99999999999...???
x = 0,9999999...
10x = 9,99999999...
9x = 10x - 1x = 9,9999999999... - 0,999999999...
9x = 9
x = 1
1 = 0,999999999999...???
Ai giải thích được không?
Theo mình thì là vì sự vô hạn của dãy số tự nhiên
Đúng 0
Bình luận (0)
\(\frac{3x}{5x+5y}-\frac{x}{10x-10y}\\ \left(\frac{3x}{1-3n}+\frac{2n}{3x+1}\right):\left(\frac{6x^2+10x}{1-6x+9x^2}\right)\\ \left(\frac{9}{x^3-9n}+\frac{1}{x+3}\right):\left(\frac{x}{3n+9}\right)\)
\(\frac{3x}{5x+5y}-\frac{x}{10x-10y}\)
= \(\frac{3x\left(x-y\right)}{5.2.\left(x+y\right)\left(x-y\right)}-\frac{x\left(x+y\right)}{10\left(x^2-y^2\right)}\)
= \(\frac{3x^2-3xy-x^2-xy}{10\left(x^2-y^2\right)}\)
= \(\frac{3x\left(x-y\right)}{10\left(x^2-y^2\right)}\)
= \(\frac{3x}{10\left(x+y\right)}\)
Đúng 0
Bình luận (0)
Tính \(\sqrt{1+999..9^2+0.999...9^2}\) (n chữ số 9)
1, x^4 +5x^3 +10x^2+ +15x+9=0
2. X^4 - 4x^3 - 9x^2 + 8x +4=0
2: Ta có: \(x^4-4x^3-9x^2+8x+4=0\)
\(\Leftrightarrow x^4-x^3-3x^3+3x^2-12x^2+12x-4x+4=0\)
\(\Leftrightarrow x^3\left(x-1\right)-3x^2\left(x-1\right)-12x\left(x-1\right)-4\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^3-3x^2-12x-4\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^3+2x^2-5x^2-10x-2x-4\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left[x^2\left(x+2\right)-5x\left(x+2\right)-2\left(x+2\right)\right]=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+2\right)\left(x^2-5x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+2=0\\x^2-5x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2\\x=\dfrac{5-\sqrt{33}}{2}\\x=\dfrac{5+\sqrt{33}}{2}\end{matrix}\right.\)
Vậy: \(S=\left\{1;-2;\dfrac{5-\sqrt{33}}{2};\dfrac{5+\sqrt{33}}{2}\right\}\)
Đúng 2
Bình luận (0)
1: Ta có: \(x^4+5x^3+10x^2+15x+9=0\)
\(\Leftrightarrow x^4+x^3+4x^3+4x^2+6x^2+6x+9x+9=0\)
\(\Leftrightarrow x^3\left(x+1\right)+4x^2\left(x+1\right)+6x\left(x+1\right)+9\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x^3+4x^2+6x+9\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left[x^3+3x^2+x^2+6x+9\right]=0\)
\(\Leftrightarrow\left(x+1\right)\left[x^2\left(x+3\right)+\left(x+3\right)^2\right]=0\)
\(\Leftrightarrow\left(x+1\right)\left(x+3\right)\left(x^2+x+3\right)=0\)
mà \(x^2+x+3>0\forall x\)
nên (x+1)(x+3)=0
\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-3\end{matrix}\right.\)
Vậy: S={-1;-3}
Đúng 0
Bình luận (0)
A= x^10-10x^9+10x^8-.....+10x^2-10x+1. Tính giá trị của A tại x=9
x=9
=>x+1=10
\(A=x^{10}-10x^9+10x^8-...+10x^2-10x+1\)
\(=x^{10}-x^9\left(x+1\right)+x^8\left(x+1\right)-...+x^2\left(x+1\right)-x\left(x+1\right)+1\)
\(=x^{10}-x^{10}-x^9+x^8+...+x^3+x^2-x^2-x+1\)
=-x+1
=-9+1=-8
Đúng 0
Bình luận (0)
Bài 1: Rút gọn
a)(x+9)(x-9)-x2
b)(10x-1)(10x+1)-(10x-1)2
c)(a+2b+3)(2a-2b-3)+(b-2c)2
d)(x-1)(x-2)-(x-2)(x+2)
a) (x+9)(x-9)-x2=x2-81-x2=-81
b) (10x-1)(10x+1)-(10x-1)2=100x2-1-100x2+20x-1=20x-2
d) (x-1)(x-2)-(x-2)(x+2)=x2-3x+2-x2+4=-3x+6
Đúng 0
Bình luận (0)
thực hiện phép tính
\(\dfrac{x^2-50}{3x^2-9x}:\dfrac{2x^2+10x}{x^2-9}\)
\(\dfrac{-3x^2}{2x+1}:\left(-\dfrac{9}{4x^2-1}\right)\)
\(\dfrac{x^2-50}{3x^2-9x}\div\dfrac{2x^2+10x}{x^2-9}\)
\(\Leftrightarrow\dfrac{x^2-50}{3x\left(x-3\right)}\div\dfrac{2x^2+10x}{\left(x-3\right)\left(x+3\right)}\)
MTC: 3x(x-3)(x+3)
\(\dfrac{(x^2-50)\left(x+3\right)}{3x\left(x-3\right)\left(x+3\right)}\div\dfrac{3x(2x^2+10x)}{3x\left(x-3\right)\left(x+3\right)}\)
\(\Rightarrow\)(x2-50)(x+3):3x(2x2+10x)
\(\Rightarrow\)(x3+3x2-50x-150):6x3+30x2
Đúng 1
Bình luận (0)