Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Câu a:
|\(\sqrt2\) - \(x\)| = \(\sqrt2\)
\(\left[\begin{array}{l}\sqrt2-x=\sqrt2\\ \sqrt2-x=-\sqrt2\end{array}\right.\)
\(\left[\begin{array}{l}x=0\\ x=2\sqrt2\end{array}\right.\)
Vậy \(x\in\) {0; \(2\sqrt2\)}
Câu b:
|\(x-1\)| = \(\sqrt3\) + 2
\(\left[\begin{array}{l}x-1=\sqrt3+2\\ x-1=-\sqrt{3-2}\end{array}\right.\)
\(\left[\begin{array}{l}x=\sqrt3+2+1\\ x=-\sqrt3-2+1\end{array}\right.\)
\(\left[\begin{array}{l}x=\sqrt3+\left(2+1\right)\\ x=-\sqrt3-\left(2-1\right)\end{array}\right.\)
\(\left[\begin{array}{l}x=\sqrt3+3\\ x=-\sqrt3-1\end{array}\right.\)
Vậy \(x\in\) {- \(\sqrt3\) - 1; \(\sqrt3\) + 3}
\(\dfrac{1}{x\left(x+1\right)}+\dfrac{1}{\left(x+1\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+3\right)}-\dfrac{1}{x}=2010\)
\(\Rightarrow\dfrac{1}{x}-\dfrac{1}{x+1}+\dfrac{1}{x+1}-\dfrac{1}{x+2}+\dfrac{1}{x+2}-\dfrac{1}{x+3}-\dfrac{1}{x}=\dfrac{1}{2010}\)\(\Rightarrow\dfrac{-1}{x+3}=\dfrac{1}{2010}\)
\(\Rightarrow x+3=-2010\)
\(\Rightarrow x=-2013\)
Vậy x = -2013
\(\dfrac{1}{x\left(x+1\right)}+\dfrac{1}{\left(x+1\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+3\right)}-\dfrac{1}{x}=\dfrac{1}{2010}\)
=> \(\dfrac{1}{x}-\dfrac{1}{x+1}+\dfrac{1}{x+1}-\dfrac{1}{x+2}+\dfrac{1}{x+2}-\dfrac{1}{x+3}-\dfrac{1}{x}=\dfrac{1}{2010}\)
=> \(-\dfrac{1}{x+3}=\dfrac{1}{2010}\)
=> \(-2010=x+3\)
=> \(x=-2013\)
Anh chỉ giải câu a thôi, câu b anh thấy nó bình thường mà.
Cộng vào mỗi phân số thêm 1 đơn vị được:
\(\frac{x+2013}{2009}+\frac{x+2013}{2010}=\frac{x+2013}{2011}+\frac{x+2013}{2012}\).
Tới đây tự làm tiếp nhá.
\(1.\sqrt{x-1}=2\)
\(\Rightarrow x-1=4\)
\(\Rightarrow x=5\)
Vậy \(x=5.\)
\(2.\sqrt{3-x}=1\)
\(\Rightarrow3-x=1\)
\(\Rightarrow x=2\)
\(3.\left|x-1\right|+\left|x^2-1\right|=0\)
\(\Rightarrow\left\{{}\begin{matrix}\left|x-1\right|=0\\\left|x^2-1\right|=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x-1=0\\x^2=1\end{matrix}\right.\)
\(\Rightarrow x=1\)
\(4.\left|2x-3\right|-\left|x-1\right|=0\)
\(\Rightarrow\left|2x-3\right|=\left|x-1\right|\)
\(\Rightarrow\left[{}\begin{matrix}2x-3=x-1\\2x-3=-x+1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x-x=3-1\\2x+x=3+1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{4}{3}\end{matrix}\right.\)
Vậy \(\left[{}\begin{matrix}x=2\\x=\dfrac{4}{3}\end{matrix}\right..\)
=> 1/x - 1/x+1 + 1/x+1 - 1/x+2 + 1/x+2 - 1/x+3 - 1/x = 1/2010
=> -1/x+3 = 1/2010
=> 1/x+3 = 1/-2010
=> x+3 = -2010
=> x = -2010-3 = -2013
k mk nha
1/x - 1/x+1 + 1/x+1 - 1/x+2 + 1/x+2 - 1/x+3 - 1/x = 1/2010
=> -1/x+3 = 1/2010
=> 1/x+3 = 1/-2010
=> x+3 = -2010
=> x = -2010-3 = -2013