(x+1/5)2 + 17/25 = 26/25
x = ????
(x + 1/5)^2 + 17/25=26/25
(x+1/5)^2 + 17/5 = 26/25
B= x20+25x19+25x18+25x17+...+25x2+25x+25 với x=-24
(x+1/5)^2+17/25=26/25
(x+1/5)^2=26/25-17/25=9/25
x+1/5=\(\sqrt{\dfrac{9}{25}}\)
=3/5 =>x=3/5-1/5=2/5
\(\left(x+\dfrac{1}{5}\right)^2+\dfrac{17}{25}=\dfrac{26}{25}\)
\(\left(x+\dfrac{1}{5}\right)^2=\dfrac{9}{25}\)
\(\left[{}\begin{matrix}x+\dfrac{1}{5}=\dfrac{3}{5}\\x+\dfrac{1}{5}=-\dfrac{3}{5}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{5}\\x=-\dfrac{4}{5}\end{matrix}\right.\)
\(\left(x+\dfrac{1}{5}\right)^2=\dfrac{26}{25}+\dfrac{17}{25}\\ \left(x+\dfrac{1}{5}\right)^2=\dfrac{43}{25}\\ x+\dfrac{1}{5}=\dfrac{\sqrt{45}}{5}\\ x=\dfrac{\sqrt{45}}{5}-\dfrac{1}{5}\\ x=\dfrac{-1+\sqrt{43}}{5}\)
Tính \(E=x^{20}+25x^{19}+25x^{18}+25x^{17}+...+25x^3+25x^2+25x+25\)
Với x=-24
x=-24
=>-x=24
=>-x+1=25
thay -x+1=25 vào E ta được:
E=x20+(-x+1)x19+(-x+1)x18+(-x+1)x17+...+(-x+1)x3+(-x+1)x2+(-x+1)x+(-x+1)
=x20-x20+x19-x19+x18-x18+x17-...-x4+x3-x3+x2-x2+x-x+1
=1
Vậy với x=-24 thì E=1
x = ‐24
=> ‐ X = 24
=> ‐ X + 1 = 25
thay ‐x+1=25 vào E ta được:
E = x 20 + ﴾‐ x + 1﴿ x 19 + ﴾‐ x + 1﴿ x 18 + ﴾‐ x + 1﴿ x 17 + ... + ﴾‐ x + 1﴿ x 3 + ﴾‐ x + 1 ﴿ x 2 + ﴾‐ x + 1﴿ x + ﴾‐ x + 1﴿
= x 20 ‐x 20 + x 19 ‐x 19 + x 1 8 ‐x 18 + x 17 ‐...‐ x 4 + x 3 ‐x 3 + x 2 ‐x 2 + x‐x + 1
= 1
Vậy với x=‐24 thì E=1
Học tốt nha Nguyễn Quang Linh
cảm ơn bạn...nhưng câu hỏi đó mik gửi từ ngày 22/7/2015 đến giờ sao bạn thấy mà trả lời vậy???
28) (x+1/5)2 + 17/25=26/25
\(\left(x+\frac{1}{5}\right)^2+\frac{17}{25}=\frac{26}{25}\)
\(\Rightarrow\left(x+\frac{1}{5}\right)^2=\frac{26}{25}-\frac{17}{25}\)
\(\Rightarrow\left(x+\frac{1}{5}\right)^2=\frac{9}{25}=\frac{3}{5}^2\)
\(\Rightarrow x^2=\frac{3}{5}^2-\frac{1}{5}^2=\frac{2}{5}^2\)
\(\Rightarrow x=\frac{2}{5}\)
Mk nhanh nhất k mk nka!!!^_^^_^^_^
(x+1/5)2+17/25=26/25
tìm x
(x+1/5)^2+17/35=26/25
\(\left(x+\frac{1}{5}\right)^2+\frac{17}{25}=\frac{26}{25}\)
\(\left(x+\frac{1}{5}\right)^2=\frac{26}{25}-\frac{17}{25}\)
\(\left(x+\frac{1}{5}\right)^2=\frac{9}{25}=\left(\frac{3}{5}\right)^2=\left(-\frac{3}{5}\right)^2\)
\(\Rightarrow x+\frac{1}{5}=\frac{3}{5}\) \(\Rightarrow x+\frac{1}{5}=-\frac{3}{5}\)
\(x=\frac{3}{5}-\frac{1}{5}\) \(x=-\frac{3}{5}-\frac{1}{5}\)
\(x=\frac{2}{5}\) \(x=\frac{-4}{5}\)
\(vậy...\)