Tìm x biết x ⋮ 9 , 21 ≤ x < 28
A. x ∈ 21 ; 22 ; 23 ; 24 ; 25 ; 26
B. x ∈ 22 ; 23 ; 24 ; 25 ; 26 ; 27 ; 28
C. x ∈ 21 ; 22 ; 23 ; 24 ; 25 ; 26 ; 27
D. x ∈ 21 ; 23 ; 24 ; 25 ; 26 ; 27
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a: =>4/x=y/21=4/7
=>x=7; y=21*4/7=12
b: x/7=9/y
=>xy=63
mà x>y
nên \(\left(x,y\right)\in\left\{\left(63;1\right);\left(21;3\right);\left(9;7\right);\left(-7;-9\right);\left(-3;-21\right);\left(-1;-63\right)\right\}\)
c: x/15=3/y
=>xy=45
mà x<y<0
nên \(\left(x,y\right)\in\left\{\left(-45;-1\right);\left(-15;-3\right);\left(-9;-5\right)\right\}\)
d: x/y=21/28=3/4
=>x/3=y/4=k
=>x=3k; y=4k(k\(\in Z\))
\(\frac{1}{21}+\frac{1}{28}+\frac{1}{36}+...+\frac{2}{x\left(x+1\right)}=\frac{2}{9}\)
\(\Rightarrow\frac{2}{42}+\frac{2}{56}+\frac{2}{72}+...+\frac{2}{x\left(x+1\right)}=\frac{2}{9}\)
\(\Rightarrow\frac{2}{6\cdot7}+\frac{2}{7\cdot8}+\frac{2}{8\cdot9}+...+\frac{2}{x\left(x+1\right)}=\frac{2}{9}\)
\(\Rightarrow2\left(\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2}{9}\)
\(\Rightarrow2\left(\frac{1}{6}-\frac{1}{x+1}\right)=\frac{2}{9}\)
\(\Rightarrow\frac{1}{6}-\frac{1}{x+1}=\frac{2}{9}\div2\)
\(\Rightarrow\frac{1}{6}-\frac{1}{x+1}=\frac{1}{9}\)
\(\Rightarrow\frac{1}{x+1}=\frac{1}{6}-\frac{1}{9}\)
\(\Rightarrow\frac{1}{x+1}=\frac{1}{18}\)
\(\Rightarrow x+1=18\Rightarrow x=18-1\Rightarrow x=17\)
Ta có : \(\frac{1}{21}+\frac{1}{28}+...+\frac{2}{x\left(x+1\right)}=\frac{2}{9}\)
\(\Rightarrow\frac{2}{42}+\frac{2}{56}+...+\frac{2}{x\left(x+1\right)}=\frac{2}{9}\)
\(\Rightarrow2\left(\frac{1}{42}+\frac{1}{56}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2}{9}\)
\(\Rightarrow2\left(\frac{1}{6.7}+\frac{1}{7.8}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2}{9}\)
\(\Rightarrow2\left(\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2}{9}\)
\(\Rightarrow2\left(\frac{1}{6}-\frac{1}{x+1}\right)=\frac{2}{9}\)
\(\Rightarrow\frac{1}{6}-\frac{1}{x+1}=\frac{2}{9}:2=\frac{1}{9}\)
\(\Rightarrow\frac{1}{x+1}=\frac{1}{6}-\frac{1}{9}=\frac{1}{18}\)
\(\Rightarrow x+1=18\Rightarrow x=17\)
Vậy x = 17
\(\Rightarrow\dfrac{2}{42}+\dfrac{2}{56}+\dfrac{2}{72}+.....+\dfrac{2}{x\left(x+1\right)}\Rightarrow2\left(\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+.....+\dfrac{1}{x\left(x+1\right)}\right)=\dfrac{2}{9}\\ \Rightarrow2\left(\dfrac{1}{6.7}+\dfrac{1}{7.8}+\dfrac{1}{8.9}+....+\dfrac{1}{x\left(x+1\right)}\right)=\dfrac{2}{9}\\ \Rightarrow2\left(\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}+....+\dfrac{1}{x}-\dfrac{1}{x+1}\right)=\dfrac{2}{9}\\ \Rightarrow2\left(\dfrac{1}{6}-\dfrac{1}{x+1}\right)=\dfrac{2}{9}\\ \Rightarrow\dfrac{1}{6}-\dfrac{1}{x+1}=\dfrac{2}{9}:2\\ \Rightarrow\dfrac{1}{6}-\dfrac{1}{x+1}=\dfrac{2}{9}.\dfrac{1}{2}\\ \Rightarrow\dfrac{1}{6}-\dfrac{1}{x+1}=\dfrac{1}{9}\\ \Rightarrow\dfrac{1}{x+1}=\dfrac{1}{6}-\dfrac{1}{9}\\ \Rightarrow\dfrac{1}{x+1}=\dfrac{3}{54}\\ \Rightarrow x+1=\dfrac{54}{3}\\ \Rightarrow x=\dfrac{54}{3}-1=\dfrac{51}{3}\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \)
\(\frac{1}{21}+\frac{1}{28}+\frac{1}{36}+....+\frac{2}{x\left(x+1\right)}=\frac{2}{9}\)
\(\frac{2}{6.7}+\frac{2}{7.8}+\frac{2}{8.9}+.....+\frac{2}{x\left(x+1\right)}=\frac{2}{9}\)
\(2\left(\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+....+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2}{9}\)
\(\frac{1}{6}-\frac{1}{x+1}=\frac{2}{9}:2=\frac{1}{9}\)
\(\frac{1}{x+1}=\frac{1}{6}-\frac{1}{9}=\frac{1}{18}\)
=> x+1 = 18
=> x = 18 - 1
=> x = 17
\(\dfrac{1}{21}+\dfrac{1}{28}+...+\dfrac{2}{x\left(x+1\right)}=\dfrac{2}{9}\)
\(\Leftrightarrow\dfrac{2}{42}+\dfrac{2}{56}+...+\dfrac{2}{x\left(x+1\right)}=\dfrac{2}{9}\)
\(\Leftrightarrow\dfrac{2}{6.7}+\dfrac{2}{7.8}+...+\dfrac{2}{x\left(x+1\right)}=\dfrac{2}{9}\)
\(\Leftrightarrow2\left(\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+...+\dfrac{1}{x}-\dfrac{1}{x+1}\right)=\dfrac{2}{9}\)
\(\Leftrightarrow\dfrac{1}{6}-\dfrac{1}{x+1}=\dfrac{1}{9}\)
\(\Leftrightarrow\dfrac{1}{x+1}=\dfrac{1}{6}-\dfrac{1}{9}\)
\(\Leftrightarrow\dfrac{1}{x+1}=\dfrac{1}{18}\)
\(\Leftrightarrow x+1=18\)
\(\Leftrightarrow x=17\)