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\(\dfrac{2}{2.4}+\dfrac{2}{4.6}+...+\dfrac{2}{x.\left(x+2\right)}=\dfrac{4}{9}\)
=\(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{x}-\dfrac{1}{x-2}\)
=\(\dfrac{1}{2}-\dfrac{1}{x+2}\)=\(\dfrac{4}{9}\)
=>\(\dfrac{1}{x+2}=\dfrac{1}{2}-\dfrac{4}{9}=\dfrac{1}{18}\)
=>\(x+2=18\)
=>x=16
\(\dfrac{2}{2.4}+\dfrac{2}{4.6}+...+\dfrac{2}{x\left(x+2\right)}=\dfrac{4}{9}\)
\(\Rightarrow\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{x}-\dfrac{1}{x+2}=\dfrac{4}{9}\)
\(\Rightarrow\dfrac{1}{2}-\dfrac{1}{x+2}=\dfrac{4}{9}\)
\(\Rightarrow\dfrac{x+2-2}{2\left(x+2\right)}=\dfrac{4}{9}\)
\(\Rightarrow\dfrac{x}{2x+4}=\dfrac{4}{9}\)
\(\Rightarrow9x=8x+16\)
\(\Rightarrow x=16\)
Vậy x = 16
\(\text{Sửa đề:}\)
\(\frac{1}{2.4}+\frac{1}{4.6}+...+\frac{1}{\left(2x-2\right).2x}=\frac{1}{8}\)
\(\text{Đặt biểu thức là A:}\)
\(A=\frac{1}{2.4}+\frac{1}{4.6}+...+\frac{1}{\left(2x-2\right).2x}=\frac{1}{8}\)
\(2A=\frac{2}{2.4}+\frac{2}{4.6}+...+\frac{2}{\left(2x-2\right).2x}=\frac{1}{8}\times2=\frac{1}{4}\)
\(2A=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{2x-2}-\frac{1}{2x}=\frac{1}{4}\)
\(2A=\frac{1}{2}-\frac{1}{2x}=\frac{1}{4}\)
\(A=\frac{1}{2}\times\left(\frac{1}{2}-\frac{1}{2x}\right)=\frac{1}{8}\)
\(A=\frac{1}{2}\times\frac{1}{2}-\frac{1}{2}\times\frac{1}{2x}=\frac{1}{8}\)
\(A=\frac{1}{4}-\frac{1}{4x}=\frac{1}{8}\)
\(\Rightarrow\frac{1}{4x}=\frac{1}{4}-\frac{1}{8}=\frac{2}{8}-\frac{1}{8}=\frac{1}{8}\)
\(\Rightarrow4x=8\)
\(\Rightarrow x=8\div4=2\)
1/(2.4) + 1/(4.6) + … + 1/[(2x – 2).2x] = 1/8
suy ra 2/(2.4) + 2/(4.6) + ...+ 2/[(2x - 2).2x] = 2/8
suy ra 1-1/4+1/4-1/6+...+1/(2x-2) - 1/2x = 2/8
suy ra 1 - 1/2x = 2/8
suy ra 1/2x = 1 - 2/8
suy ra 1/2x = 6/8 = 3/4
suy ra 1.4 = 2.x.3
suy ra 4 = 6x
suy ra x thuộc rỗng
Vậy x thuộc rỗng
k cho mình nha. Chúc bạn học tốt!
\(\frac{2}{2.4}+\frac{2}{4.6}+....+\frac{2}{x\left(x+2\right)}=\frac{4}{9}\)
<=> \(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+....+\frac{1}{x}-\frac{1}{x+2}=\frac{4}{9}\)
<=> \(\frac{1}{2}-\frac{1}{x+2}=\frac{4}{9}\)
<=> \(\frac{1}{x+2}=\frac{1}{18}\)
=> \(x+2=18\)
<=> \(x=16\)
Vậy...
\(\dfrac{2}{2.4}+\dfrac{2}{4.6}+...+\dfrac{2}{x\left(x+2\right)}=\dfrac{4}{9}\)
\(\Rightarrow\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{x}-\dfrac{1}{x+2}=\dfrac{4}{9}\)
\(\Rightarrow\dfrac{1}{2}-\dfrac{1}{x+2}=\dfrac{4}{9}\)
\(\Rightarrow\dfrac{x+2-2}{2\left(x+2\right)}=\dfrac{4}{9}\)
\(\Rightarrow\dfrac{x}{2x+4}=\dfrac{4}{9}\)
\(\Rightarrow9x=8x+16\)
\(\Rightarrow x=16\)
Vậy x = 16
2/2.4+2/4.6+......+2/x.(x+2)=4/9
1/2-1/4+1/4-1/6+....+1/x-1/x+2=4/9
1/2-1/x+2=4/9
1/x+2=1/2-4/9
1/x+2=1/18
=>x+2=18
x=18-2=16
a) \(\left(x+1\right)^2=64\Leftrightarrow\left|x+1\right|=8\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x+1=8\\x+1=-8\end{array}\right.\) \(\Leftrightarrow\left[\begin{array}{nghiempt}x=7\\x=-9\end{array}\right.\)
b) \(\frac{9}{2.4}+\frac{9}{4.6}+...+\frac{9}{96.98}+\frac{9}{98.100}=\frac{9}{4}\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{48.49}+\frac{1}{49.50}\right)\)
\(=\frac{9}{4}\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{48}-\frac{1}{49}+\frac{1}{49}-\frac{1}{50}\right)\)
\(=\frac{9}{4}\left(1-\frac{1}{50}\right)=\frac{441}{200}\)
\(S=\frac{2}{2.4}+\frac{2}{4.6}+...+\frac{2}{x\left(x+2\right)}=\frac{4}{9}\)
\(S=\frac{4-2}{2.4}+\frac{6-2}{4.6}+...+\frac{\left(x+2\right)-x}{x\left(x+2\right)}=\frac{4}{9}\)
\(S=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{x}-\frac{1}{x+2}=\frac{4}{9}\)
\(S=\frac{1}{2}-\frac{1}{x+2}=\frac{4}{9}\)
\(\Rightarrow\frac{1}{x+2}=\frac{1}{2}-\frac{4}{9}=\frac{1}{18}\)
\(\Rightarrow x+2=18\Rightarrow x=18-2=16\)
Vậy x=16
\(\frac{2}{2\cdot4}+\frac{2}{4\cdot6}+...+\frac{2}{x\left(x+2\right)}=\frac{4}{9}\)
\(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{x}-\frac{1}{x+2}=\frac{4}{9}\)
\(\frac{1}{2}-\frac{1}{x+2}=\frac{4}{9}\)
\(\frac{1}{x+2}=\frac{1}{2}-\frac{4}{9}\)
\(\frac{1}{x+2}=\frac{1}{18}\)
\(\Leftrightarrow x+2=18\)
=> x = 18 - 2
x = 16
Vậy x =16