chúc bạn học tốt !

chúc bạn học tốt !

chúc bạn học tốt !

chúc bạn học tốt !

(1/1×2 + 1/2×3 + ... + 1/9×10) × x < 2/1×3 + 2/3×5 + ... + 2/9×11

(1 - 1/2 + 1/2 - 1/3 + ... + 1/9 - 1/10) × x < 1 - 1/3 + 1/3 - 1/5 + ... + 1/9 - 1/11

(1 - 1/10) × x < 1 - 1/11

9/10 × x < 10/11

x < 10/11 : 9/10

x < 10/11 × 10/9

x < 100/99

Mà x là số tự nhiên => x = 0 hoặc 1

BẠN LÀ FAN CỦA HARI WON HẢ

\(\left(\frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{9.10}\right).x< \frac{2}{1.3}+\frac{2}{3.5}+.....+\frac{2}{9.11}\)

\(\Rightarrow\left(1-\frac{1}{10}\right).x< 2\left(\frac{1}{2}.\frac{10}{11}\right)\Rightarrow\frac{9}{10}.x< \frac{10}{11}\Rightarrow x< \frac{100}{99}\)

Mà:\(x\in N\)\(\Rightarrow x=0\)hoặc \(x=1\)

\(\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{8.9.10}\right).x=\frac{23}{45}\)

\(\frac{1}{2}.\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{8.9.10}\right).x=\frac{23}{45}\)

\(\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{8.9}-\frac{1}{9.10}\right).x=\frac{23}{45}\)

\(\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{9.10}\right).x=\frac{23}{45}\)

\(\frac{1}{2}.\frac{22}{45}.x=\frac{23}{45}\)

         \(\frac{11}{45}.x=\frac{23}{45}\)

                  \(x=\frac{23}{45}:\frac{11}{45}\)

                 \(x=\frac{23}{11}\)

Gọi A=(1/1.2.3+ 1/2.3.4 +...+ 1/8.9.10) .x=23/45

    2A=3-1/1.2.3+ 4–2/2.3.4+ 5–4/3.4.5+ ... + 10–8/8.9.10

    2A=1/2 —1/2.3+ 1/2.3 — 1/3.4+ 1/3.4– 1/4.5 +...+1/8.9–1/9.10=1/2–1/9.10=44/90

     A=44/90 : 2=22/90

     x=23/45:A= 23/45 : 22/90=23/11= 2 1/1( hỗn số)

Nhầm đáp án là 2 1/11 nhé

ĐKXĐ: \(x\notin\left\{2;5\right\}\)

Ta có: \(\dfrac{3x}{x-2}-\dfrac{2}{x-5}=\dfrac{3x}{\left(x-2\right)\left(5-x\right)}\)

\(\Leftrightarrow\dfrac{3x\left(x-5\right)}{\left(x-2\right)\left(x-5\right)}-\dfrac{2\left(x-2\right)}{\left(x-2\right)\left(x-5\right)}+\dfrac{3x}{\left(x-2\right)\left(x-5\right)}=0\)

Suy ra: \(3x^2-15x-2x+4+3x=0\)

\(\Leftrightarrow3x^2-14x+4=0\)

\(\Delta=196-4\cdot3\cdot4=196-48=148\)

Vì \(\Delta>0\) nên phương trình có hai nghiệm phân biệt là

\(\left\{{}\begin{matrix}x_1=\dfrac{14-\sqrt{148}}{6}=\dfrac{7-\sqrt{37}}{3}\left(nhận\right)\\x_2=\dfrac{14+\sqrt{148}}{6}=\dfrac{7+\sqrt{37}}{3}\left(nhận\right)\end{matrix}\right.\)

Vậy: \(S=\left\{\dfrac{7-\sqrt{37}}{3};\dfrac{7+\sqrt{37}}{3}\right\}\)

Lớp 8 chưa học delta nên mk sẽ trình bày theo cách khác nha!

Rút gọn pt trên ta được: 3x² - 14x + 4 = 0 (Theo kết quả của Nguyễn Lê Phước Thịnh CTV)

\(\Leftrightarrow\) 3(x² - \(\dfrac{14}{3}\)x + \(\dfrac{4}{3}\)) = 0

\(\Leftrightarrow\) x² - 2.\(\dfrac{14}{6}\)x + \(\dfrac{196}{36}\) - \(\dfrac{37}{9}\) = 0

\(\Leftrightarrow\) (x - \(\dfrac{14}{6}\))² - \(\left(\dfrac{\sqrt{37}}{3}\right)^2\) = 0

\(\Leftrightarrow\) (x - \(\dfrac{14}{6}\) - \(\dfrac{\sqrt{37}}{3}\))(x - \(\dfrac{14}{6}\) + \(\dfrac{\sqrt{37}}{3}\)) = 0

\(\Leftrightarrow\) (x - \(\dfrac{7}{3}\) - \(\dfrac{\sqrt{37}}{3}\))(x - \(\dfrac{7}{3}\) + \(\dfrac{\sqrt{37}}{3}\)) = 0

\(\Leftrightarrow\) \(\left[{}\begin{matrix}x=\dfrac{7+\sqrt{37}}{3}\\x=\dfrac{7-\sqrt{37}}{3}\end{matrix}\right.\) (TM)

Vậy ...

Chúc bn học tốt!

Ta có: \(\frac{54}{11}.\frac{121}{27}< n< \frac{100}{21}:\frac{25}{126}\)

\(\Rightarrow\frac{2.11}{1.1}< n< \frac{100}{21}.\frac{126}{25}\)

\(\Rightarrow22< n< 24\)

\(\Rightarrow n=23\)

Ta có:

\(\frac{54}{11}\cdot\frac{121}{27}=\frac{54\cdot121}{11\cdot27}=22\)

\(\frac{100}{21}:\frac{25}{126}=\frac{100}{21}\cdot\frac{126}{25}=\frac{100\cdot126}{21\cdot25}=24\)

\(\Rightarrow22< n< 24\)

\(\Rightarrow x=23\)

1/2.(1/3+1/6+1/10+...+1/x(x+1))=1/2.2016/2018

1/6+1/12+1/20+...+1/x(x+1)=504/1009

1/2.3+1/3.4+1/4.5+...+1/x(x+1)=504/1009

1/2-1/3+1/3-1/4+1/4-1/5+...+1/x-1/x+1=504/1009

1/2-1/x+1=504/1009

x-1/2(x+1)=504/1009

-> 1009(x-1)=504.2(x+1)

1009x-1009=1008x+1008

1009x-1008x=1008+1009

->x=2017

\(A=\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{x\left(x+1\right):2}=\frac{2016}{2018}\)
\(A=\frac{1}{2\left(2+1\right):2}+\frac{1}{3\left(3+1\right):2}+...+\frac{1}{x\left(x+1\right):2}\)
\(A=\frac{1}{2\left(2+1\right)}\cdot2+\frac{1}{3\left(3+1\right)}\cdot2+...+\frac{1}{x\left(x+1\right)}.2=\frac{2016}{2018}\)
\(A=2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2016}{2018}\)
\(A=2\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2016}{2018}\)
\(A=1-\frac{1}{x+1}=\frac{2016}{2018}\)
\(\Rightarrow\frac{1}{x+1}=1-\frac{2016}{2018}=\frac{1}{1009}\)
\(\Rightarrow x+1=1009\Rightarrow x=1008\)

a) đk: \(\hept{\begin{cases}x\ge0\\x\ne1\end{cases}}\)

Ta có:

\(P=\left(\frac{3x-\sqrt{9x}-3}{x+\sqrt{x}-2}+\frac{1}{\sqrt{x}-1}+\frac{1}{\sqrt{x}+2}\right)\div\frac{1}{x-1}\)

\(P=\frac{3x-3\sqrt{x}-3+\sqrt{x}+2+\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\cdot\left(x-1\right)\)

\(P=\frac{3x-\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\cdot\left(\sqrt{x}-1\right)\cdot\left(\sqrt{x}+1\right)\)

\(P=\frac{\left(\sqrt{x}-1\right)\left(3\sqrt{x}+2\right)\left(\sqrt{x}+1\right)}{\sqrt{x}+2}\)

\(P=\frac{\left(3\sqrt{x}+2\right)\left(x-1\right)}{\sqrt{x}+2}\)