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| x + 1 | + 7 = 25
<=> | x + 1 | = 18
<=> x + 1 = 18 hoặc x + 1 = -18
<=> x = 17 hoặc x = -19
bạn ngọc thiếu 3 , mình sửa luôn
\(3.|x+1|+7=25\)
\(< =>3|x+1|=25-7\)
\(< =>3|x+1|=18\)
\(< =>|x+1|=\frac{18}{3}=6\)
\(< =>\orbr{\begin{cases}x+1=6\\x+1=-6\end{cases}}\)
\(< =>\orbr{\begin{cases}x=5\\x=-7\end{cases}}\)
Ta thấy: 2/2.3 = 2/2 - 2/3 ; 2/3.4 = 2/3 - 2/4 ; 2/4.5 = 2/4 - 2/5
Tổng quát ta có: 2/x(x+1) = 2/x - 2/x + 1 , như vậy thì bài toán trên( bạn chép lại đề)
= 2/1 - 2/x + 1 = 2008/2009
Ta có: 2/1 - 2/x+1 = 2008/2009
2/x+1 = 2 - 2008/2009
2/x+1= 1/2009
x + 1 = 2009
x = 2009 - 1 = 2008
tk nha
\(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{x}-\dfrac{1}{x+1}=\dfrac{1}{3}-\dfrac{x+1}{324}\)
\(\dfrac{1}{3}-\dfrac{1}{x+1}=\dfrac{1}{3}-\dfrac{x+1}{324}\)
\(\dfrac{1}{x+1}=\dfrac{x+1}{324}\)
\(\left(x+1\right)^2=324=18^2\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=18\\x+1=-18\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=17\\x=-19\end{matrix}\right.\)
Ta có \(\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+...+\dfrac{1}{x\left(x+1\right)}=\dfrac{1}{3}-\dfrac{x+1}{324}\)
\(\Rightarrow\)\(\dfrac{4-3}{3.4}+\dfrac{5-4}{4.5}+\dfrac{6-5}{5.6}+...+\dfrac{\left(x+1\right)-x}{x\left(x+1\right)}=\dfrac{1}{3}-\dfrac{x+1}{324}\)
\(\Rightarrow\)\(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{x}-\dfrac{1}{x+1}=\dfrac{1}{3}-\dfrac{x+1}{324}\)
\(\Rightarrow\)\(\dfrac{1}{3}-\dfrac{1}{x+1}=\dfrac{1}{3}-\dfrac{x+1}{324}\)
\(\Rightarrow\)\(\dfrac{1}{3}-\dfrac{1}{3}=-\dfrac{x+1}{324}+\dfrac{1}{x+1}\)
\(\Rightarrow\)\(\dfrac{1}{x+1}-\dfrac{x+1}{324}=0\)
\(\Rightarrow\)\(\dfrac{1}{x+1}=\dfrac{x+1}{324}\)
\(\Rightarrow\)(x+1).(x+1)=324
\(\Rightarrow\)(x+1)2=324
\(\Rightarrow\)(x+1)2 = 182 = (-18)2
TH1: (x+1)2 = 182
\(\Rightarrow\)x+1 = 18
\(\Rightarrow\)x = 17
TH2: (x+1)2 = (-18)2
\(\Rightarrow\)x+1 = -18
\(\Rightarrow\)x = -19
Vậy x\(\in\)\(\left\{17;-19\right\}\)
\(\Leftrightarrow2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2008}{2010}\)
\(\Leftrightarrow2\left(\frac{3-2}{2.3}+\frac{4-3}{3.4}+\frac{5-4}{4.5}+...+\frac{\left(x+1\right)-x}{x\left(x+1\right)}\right)=\frac{2008}{2010}\)
\(\Leftrightarrow2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2008}{2010}\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{1004}{2010}\)
\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{2010}\)
\(\Leftrightarrow x+1=2010\)
\(\Leftrightarrow x=2009\)
Ta có : 2x + 2x + 1 = 24
=> 2x(1 + 2) = 24
=> 2x.3 = 24
=> 2x = 8
=> 2x = 23
=> x = 3
Ta có : (x + 2)4 = (x + 2)6
=> (x + 2)4 - (x + 2)6 = 0
<=> (x + 2)4 (1 - (x + 2)2) = 0
<=> \(\orbr{\begin{cases}\left(x+2\right)^4=0\\\left(1-\left(x+2\right)^2\right)=0\end{cases}}\)
<=> \(\orbr{\begin{cases}x+2=0\\\left(x+2\right)^2=1\end{cases}}\)
<=> \(\orbr{\begin{cases}x+2=0\\x+2=1\end{cases}}\)
<=> \(\orbr{\begin{cases}x=-2\\x=-1\end{cases}}\)
Ta có:
1/3.4=1/3-1/4
1/4.5=1/4-1/5
Vậy:
1/3.4+1/4.5+...+1/x(x+1)= (1/3-1/4)+(1/4-1/5)+....+(1/x-1/(x+1)=1/3-1/(x+1)=3/10
Giải 1/3-1/(x+1)=3/10 là ra đáp án.
Thân