Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) Vì biểu thức \(\sqrt{\dfrac{-5}{x^2+6}}\)có -5<0 nên làm cho cả phân số âm
Từ đó suy ra căn thức vô nghiệm
Vậy không có giá trị nào của x để biểu thức trên xác định
b) \(\sqrt{\left(x-1\right)\left(x-3\right)}\)
Để biểu thức trên xác định thì chia ra 4 TH (vì để xác định thì cả x-1 và x-3 cùng dương hoặc cùng âm)
\(\left[\begin {array} {} \begin{cases} x-1\geq0\\ x-3\geq0 \end{cases} \Leftrightarrow \begin{cases} x\geq1\\ x\geq3 \end{cases} \Rightarrow x\geq3 \\ \begin{cases} x-1\leq0\\ x-3\leq0 \end{cases} \Leftrightarrow \begin{cases} x\leq1\\ x\leq3 \end{cases} \Rightarrow x\leq1 \end{array} \right.\)
c) \(\sqrt{x^2-4}\) \(\Leftrightarrow\)\(\sqrt{\left(x-2\right)\left(x+2\right)}\)
Rồi làm như câu b
d) \(\sqrt{\dfrac{2-x}{x+3}}\)
Để biểu thức trên xác định thì
\(\begin{cases}2-x\ge0\\x+3>0\end{cases}\Leftrightarrow\begin{cases}x\ge2\\x>-3\end{cases}\) \(\Rightarrow\) \(x\ge2\) hoặc \(x>-3\)
e) Ở các biểu thức sau này nếu chỉ có căn thức có ẩn và + (hoặc trừ) với 1 số thì chỉ cần biến đổi cái có ẩn còn cái số thì kệ xác nó đi 
)
\(\sqrt{x^2-3x}\Leftrightarrow\sqrt{x\left(x-3\right)}\)
Để biểu thức trên xác định thì \(x\ge0\) và \(x-3\ge0\Leftrightarrow x\ge3\)
Bữa sau mình làm tiếp
1: \(\left\{{}\begin{matrix}\left|x-1\right|+\dfrac{2}{y}=2\\-\left|x-1\right|+\dfrac{4}{y}=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{6}{y}=3\\\left|x-1\right|=2-\dfrac{2}{y}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=2\\\left|x-1\right|=2-\dfrac{2}{2}=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=2\\x\in\left\{2;0\right\}\end{matrix}\right.\)
2: \(\left\{{}\begin{matrix}2\left|x-1\right|-\dfrac{5}{y-1}=-3\\\left|x-1\right|+\dfrac{2}{y-1}=3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2\left|x-1\right|-\dfrac{5}{y-1}=-3\\2\left|x-1\right|+\dfrac{4}{y-1}=6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-\dfrac{9}{y-1}=-9\\\left|x-1\right|+\dfrac{2}{y-1}=3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=2\\\left|x-1\right|=3-\dfrac{2}{2}=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=2\\x\in\left\{3;-1\right\}\end{matrix}\right.\)
3: \(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{2}{x-5}+\dfrac{12}{\sqrt{y}-2}=4\\\dfrac{2}{x-5}-\dfrac{1}{\sqrt{y}-2}=-9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{13}{\sqrt{y}-2}=13\\\dfrac{1}{x-5}=2-\dfrac{6}{\sqrt{y}-2}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=9\\\dfrac{1}{x-5}=2-\dfrac{6}{3-2}=2-\dfrac{6}{1}=-4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=9\\x-5=-\dfrac{1}{4}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{19}{4}\\y=9\end{matrix}\right.\)
a, Để \(\sqrt{\left(x-1\right)\left(x-3\right)}\) xác định thì (x-1)(x-3)\(\ge\)0
TH1: \(\left\{{}\begin{matrix}x-1\ge0\\x-3\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge1\\x\ge3\end{matrix}\right.\Leftrightarrow}x\ge3}\)TH2:\(\left\{{}\begin{matrix}x-1\le0\\x-3\le0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\le1\\x\le3\end{matrix}\right.\Leftrightarrow}x\le1}\) Vậy nếu \(x\ge3\) hoặc \(x\le1\) thì biểu thức có nghĩa
b, Để \(\sqrt{x^2-4}=\sqrt{\left(x-2\right)\left(x+2\right)}\)có nghĩa thì (x-2)(x+2)\(\ge0\)
TH1: \(\left\{{}\begin{matrix}x-2\ge0\\x+2\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge2\\x\ge-2\end{matrix}\right.\Leftrightarrow x\ge}2}\)TH2:\(\left\{{}\begin{matrix}x-2\le0\\x+2\le0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\le2\\x\le-2\end{matrix}\right.\Leftrightarrow}x\le-2}\)Vậy nếu \(x\ge2\) hoặc \(x\le-2\) thì biểu thức có nghĩa
a: \(A=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{x+\sqrt{x}+1}\cdot\dfrac{\sqrt{x}\left(2\sqrt{x}+1\right)}{\sqrt{x}}+2\sqrt{x}+2\)
\(=\sqrt{x}\left(\sqrt{x}-1\right)\left(2\sqrt{x}+1\right)+2\sqrt{x}+2\)
\(=\left(x-\sqrt{x}\right)\left(2\sqrt{x}+1\right)+2\sqrt{x}+2\)
\(=2x\sqrt{x}+x-2x-\sqrt{x}+2\sqrt{x}+2\)
\(=2x\sqrt{x}-x+\sqrt{x}+2\)
b: \(=\dfrac{\sqrt{x}-4+3\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)}:\dfrac{x-4-x}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(=\dfrac{4\left(\sqrt{x}-1\right)}{-4}=-\sqrt{x}+1\)
c: \(=\dfrac{15\sqrt{x}-11-3x-9\sqrt{x}+2\sqrt{x}+6-\left(2\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{-3x+8\sqrt{x}+5-2x+2\sqrt{x}-3\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{-5x+7\sqrt{x}+8}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
Câu 1:
a: \(P=\dfrac{x+\sqrt{x}-2}{\sqrt{x}\left(\sqrt{x}+2\right)}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)
\(=\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}}=\dfrac{\sqrt{x}+1}{\sqrt{x}}\)
b: Để \(2P=2\sqrt{5}+5\) thì \(P=\dfrac{2\sqrt{5}+5}{2}\)
\(\Leftrightarrow\sqrt{x}\left(2\sqrt{5}+5\right)=2\left(\sqrt{x}+1\right)\)
\(\Leftrightarrow\sqrt{x}\left(2\sqrt{5}+3\right)=2\)
hay \(x=\dfrac{4}{29+12\sqrt{5}}=\dfrac{4\left(29-12\sqrt{5}\right)}{121}\)
Câu 1:
a: \(P=\dfrac{x+\sqrt{x}-2}{\sqrt{x}\left(\sqrt{x}+2\right)}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)
\(=\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}}=\dfrac{\sqrt{x}+1}{\sqrt{x}}\)
b: Để \(2P=2\sqrt{5}+5\) thì \(P=\dfrac{2\sqrt{5}+5}{2}\)
\(\Leftrightarrow\sqrt{x}\left(2\sqrt{5}+5\right)=2\left(\sqrt{x}+1\right)\)
\(\Leftrightarrow\sqrt{x}\left(2\sqrt{5}+3\right)=2\)
hay \(x=\dfrac{4}{29+12\sqrt{5}}=\dfrac{4\left(29-12\sqrt{5}\right)}{121}\)
Đặt \(\sqrt{x-1}=a\), khi đó ta có:
\(P=\left(\dfrac{\sqrt{x-1}}{3+\sqrt{x-1}}+\dfrac{x+8}{10-x}\right):\left(\dfrac{3\sqrt{x-1}+1}{x-3\sqrt{x-1}-1}-\dfrac{1}{\sqrt{x-1}}\right)\)
\(=\left[\dfrac{\sqrt{x-1}}{\sqrt{x-1}+3}+\dfrac{\left(x-1\right)+9}{9-\left(x-1\right)}\right]:\left[\dfrac{3\sqrt{x-1}+1}{\left(x-1\right)-3\sqrt{x-1}}-\dfrac{1}{\sqrt{x-1}}\right]\)
\(=\left(\dfrac{a}{a+3}+\dfrac{a^2+9}{9-a^2}\right):\left(\dfrac{3a+1}{a^2-3a}-\dfrac{1}{a}\right)\)
\(=\dfrac{a\left(3-a\right)+\left(a^2+9\right)}{\left(3+a\right)\left(3-a\right)}:\dfrac{\left(3a-1\right)-\left(a-3\right)}{a\left(a-3\right)}\)
\(=\dfrac{3a-a^2+a^2+9}{\left(3+a\right)\left(3-a\right)}:\dfrac{3a-1-a+3}{a\left(a-3\right)}\)
\(=\dfrac{3a+9}{\left(3+a\right)\left(3-a\right)}:\dfrac{2a+4}{a\left(a-3\right)}\)
\(=\dfrac{3\left(a+3\right)}{\left(a+3\right)\left(a-3\right)}.\dfrac{a\left(a-3\right)}{2\left(a+2\right)}\)
\(=\dfrac{-3a}{2\left(a+2\right)}\).
Suy ra: P \(=\dfrac{-3\sqrt{x-1}}{2\left(\sqrt{x-1}+2\right)}\).
Ta lại có: \(x=\sqrt[4]{\dfrac{3+2\sqrt{2}}{3-2\sqrt{2}}}-\sqrt[4]{\dfrac{3-2\sqrt{2}}{3+2\sqrt{2}}}\)
\(=\sqrt[4]{\dfrac{\left(\sqrt{2}+1\right)^2}{\left(\sqrt{2}-1\right)^2}}-\sqrt[4]{\dfrac{\left(\sqrt{2}-1\right)^2}{\left(\sqrt{2}+1\right)^2}}\)
\(=\sqrt{\dfrac{\sqrt{2}+1}{\sqrt{2}-1}}-\sqrt{\dfrac{\sqrt{2}+1}{\sqrt{2}-1}}\)
\(=\sqrt{\dfrac{\left(\sqrt{2}+1\right)^2}{2-1}}-\sqrt{\dfrac{\left(\sqrt{2}-1\right)^2}{2-1}}\)
\(=\left(\sqrt{2}+1\right)-\left(\sqrt{2}-1\right)\)
\(=2\).
Suy ra: \(P=\dfrac{-3\sqrt{2-1}}{2\left(\sqrt{2-1}+2\right)}=\dfrac{-3}{2.3}=-\dfrac{1}{2}\).
TRẦN MINH HOÀNGKhôi Bùi tran nguyen bao quanMysterious PersoDƯƠNG PHAN KHÁNH DƯƠNGn ghi ket qua thoi cx dc
\(\left(\dfrac{x-2\sqrt{x}}{x-4}-1\right):\left(\dfrac{4-x}{x-\sqrt{x}-6}-\dfrac{\sqrt{x}-2}{3-\sqrt{x}}-\dfrac{\sqrt{x}-3}{\sqrt{x}+2}\right)=\dfrac{x-2\sqrt{x}-x+4}{x-4}:\left(\dfrac{4-x}{x-\sqrt{x}-6}+\dfrac{\sqrt{x}-2}{\sqrt{x}-3}-\dfrac{\sqrt{x}-3}{\sqrt{x}+2}\right)=\dfrac{-2\sqrt{x}+4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}:\left[\dfrac{4-x}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}+\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}-\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}\right]=\dfrac{-2\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}:\left[\dfrac{4-x}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}+\dfrac{x-4}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}-\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}\right]=\dfrac{-2}{\sqrt{x}+2}:\dfrac{4-x+x-4-\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}=\dfrac{-2}{\sqrt{x}+2}:\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}=\dfrac{-2}{\sqrt{x}+2}:\dfrac{\sqrt{x}+3}{\sqrt{x}+2}=\dfrac{-2}{\sqrt{x}+2}.\dfrac{\sqrt{x}+2}{\sqrt{x}+3}=\dfrac{-2}{\sqrt{x}+3}\)
Đk: \(\left\{{}\begin{matrix}\sqrt{x}xđ\\x-4\ne0\\\sqrt{x}-2\ne0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x\ge0\\x\ne\pm2\\\sqrt{x}\ne2\end{matrix}\right.\)\(\Rightarrow x\ge0;x\ne2\)