Sửa lại:A.x=x²+x³+...+x¹⁰¹

=>A.x-A=(x²+x³+...+x¹⁰¹)-(x+x²+...+x¹⁰⁰)

=>A(x-1)=x¹⁰¹-x

=>A=\(\dfrac{x^{101}-x}{x-1}\)

Thay x=\(\dfrac{1}{2}\)vào A ta có:

A=\(\dfrac{\left(\dfrac{1}{2}\right)^{101}-\dfrac{1}{2}}{\dfrac{1}{2}-1}=\dfrac{\left(\dfrac{1}{2}\right)^{101}-\dfrac{1}{2}}{-\dfrac{1}{2}}=1-\left(\dfrac{1}{2}\right)^{100}=\dfrac{2^{100}-1}{2^{100}}\)

Ta có:A.x=x²+x³+...+x¹⁰¹

=>A.x-A=(x²+x³+...+x¹⁰¹)-(x+x²+...+x¹⁰⁰)

=>A(x-1)=x¹⁰¹-x

=>A=\(\dfrac{x^{101}-x}{x-1}\)

Thay x=\(\dfrac{1}{2}\)

=>A=\(\dfrac{\left(\dfrac{1}{2}\right)^{101}-\dfrac{1}{2}}{\dfrac{1}{2}-1}=\dfrac{\left(\dfrac{1}{2}\right)^{101}-\dfrac{1}{2}}{-\dfrac{1}{2}}=1-\left(\dfrac{1}{2}\right)^{101}\)

C=1 A=B=0 khi f(0)= 1

Không được đâu, tự tìm thôi.
Về khoản tìm x thì còn được.

4 x y = 7 giờ 40 phút 

4 x y =  \(7+\dfrac{2}{3}\) giờ

4 x y = \(\dfrac{23}{3}\) giờ

y = \(\dfrac{23}{3}:4\) (giờ)

y = \(\dfrac{23}{12}\) (giờ) 

y = \(\dfrac{23}{12}\times60\) (phút)

y = 115 phút

=>4*y=460 phút

=>y=460/4=115 phút

BPT đã cho vô nghiệm khi và chỉ khi BPT \(f\left(x\right)\le0\) nghiệm đúng với mọi x

TH1: \(\left\{{}\begin{matrix}2m^2+m-6=0\\2m-3=0\end{matrix}\right.\) \(\Rightarrow m=\dfrac{3}{2}\)

TH2: \(\left\{{}\begin{matrix}2m^2+m-6< 0\\\Delta=\left(2m-3\right)^2+4\left(2m^2+m-6\right)\le0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}2m^2+m-6< 0\\12m^2-8m-15\le0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}-3< m< \dfrac{3}{2}\\-\dfrac{5}{6}\le m\le\dfrac{3}{2}\end{matrix}\right.\) \(\Rightarrow-\dfrac{5}{6}\le m< \dfrac{3}{2}\)

Kết hợp 2 trường hợp ta được \(-\dfrac{5}{6}\le m\le\dfrac{3}{2}\)

7 giờ 40 phút = 460 phút

4 x y = 460 phút

      y = 460 phút : 4 = 115 phút

                                = 1 giờ 55 phút

a, (sửa đề )

\(1+\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+.....+\frac{1}{x.\left(x+1\right)}=\frac{1999}{2000}\)

=\(1+\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{x.\left(x+1\right)}\right)=\frac{1999}{2000}\)

=\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{x+\left(x+1\right)}=1-\frac{1999}{2000}=\frac{1}{2000}\)

=\(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{x}-\frac{1}{x+1}=\frac{1}{2000}\)

=\(\frac{1}{1}-\frac{1}{x+1}=\frac{1}{2000}\)

=\(\frac{1}{x+1}=\frac{1}{1}-\frac{1}{2000}=\frac{1999}{2000}\)

=> \(x+1=1:\frac{1999}{2000}=\frac{2000}{1999}\)

=>\(x=\frac{2000}{1999}-1=\frac{1}{1999}\)

Vậy x ∈{ \(\frac{1}{1999}\)}

b, \(\frac{1}{21}+\frac{1}{28}+\frac{1}{36}+.....+\frac{2}{x+\left(x+1\right)}=\frac{2}{9}\)

=> \(\frac{2}{42}+\frac{2}{56}+\frac{2}{72}+.....+\frac{2}{x+\left(x+1\right)}=\frac{2}{9}\)

=>\(\frac{2}{6.7}+\frac{2}{7.8}+\frac{2}{8.9}+.....+\frac{2}{x+\left(x+1\right)}=\frac{2}{9}\)

=>2.(\(\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+....+\frac{1}{x.\left(x+1\right)}\))=\(\frac{2}{9}\)

=>\(\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+....+\frac{1}{x+\left(x+1\right)}=\frac{2}{9}:2=\frac{1}{9}\)

=>\(\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+....+\frac{1}{x}-\frac{1}{x+1}=\frac{1}{9}\)

=>\(\frac{1}{6}-\frac{1}{x+1}=\frac{1}{9}\)

=>\(\frac{1}{x+1}=\frac{1}{6}-\frac{1}{9}=\frac{1}{18}\)

=>\(x+1=18\)

=>\(x=18-1=17\)

=>x∈{17}

\(y=\dfrac{x-1}{x^2-mx+1}\)

\(\lim\limits_{x\rightarrow+\infty}\dfrac{x-1}{x^2-mx+1}=\lim\limits_{x\rightarrow-\infty}\dfrac{x-1}{x^2-mx+1}=0\)

Đồ thị có 3 tiệm cận khi đồ thị có 2 tiệm cận đứng

\(\Rightarrow x^2-mx+1\) có 2 nghiệm phân biệt khác 1

\(\Leftrightarrow\left\{{}\begin{matrix}\Delta=m^2-4>0\\1-m+1\ne0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}m< -2\\m>2\end{matrix}\right.\\m\ne2\end{matrix}\right.\)

X x 1/2 = 2/3

X = (2/3) / (1/2) = 4/3

\(x\) \(\times\) \(\dfrac{1}{2}\) = 1 - \(\dfrac{1}{3}\) 

\(x\) \(\times\)\(\dfrac{1}{2}\) = \(\dfrac{2}{3}\)

\(x\)        = \(\dfrac{2}{3}\) : \(\dfrac{1}{2}\)

\(x\)        = \(\dfrac{4}{3}\)

X x 1/2 = 1 - 1/3

X x 1/2 = 2/3

X = 2/3 : 1/2

X = 4/3

\(\dfrac{x-1}{3}=\dfrac{2-x}{-2}\)

⇔ \(\dfrac{x-1}{3}=\dfrac{x-2}{2}\)

⇔ \(3x-6-2x+2=0\)

⇔ \(x-4=0\)

⇒ \(x=4\)

=>-2x+2=6-3x

=>-2x+3x=6-2

=>x=4