3.tìm x biết :
\(a.\sqrt[3]{2+x}+\sqrt[3]{2-x}=1\)
\(b.2x^3=\left(x-1\right)^3\)
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31 tháng 10 2021
\(1,\\ a,ĐK:\left\{{}\begin{matrix}x\ge0\\x+5\ge0\end{matrix}\right.\Leftrightarrow x\ge0\\ b,Sửa:B=\left(\sqrt{3}-1\right)^2+\dfrac{24-2\sqrt{3}}{\sqrt{2}-1}\\ B=4-2\sqrt{3}+\dfrac{2\sqrt{3}\left(\sqrt{2}-1\right)}{\sqrt{2}-1}\\ B=4-2\sqrt{3}+2\sqrt{3}=4\\ 3,\\ =\left[1-\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{1+\sqrt{x}}\right]\cdot\dfrac{\sqrt{x}-3+2-2\sqrt{x}}{\left(1-\sqrt{x}\right)\left(\sqrt{x}-3\right)}-2\\ =\left(1-\sqrt{x}\right)\cdot\dfrac{-\sqrt{x}-1}{\left(1-\sqrt{x}\right)\left(\sqrt{x}-3\right)}-2\\ =\dfrac{-\sqrt{x}-1}{\sqrt{x}-3}-2=\dfrac{-\sqrt{x}-1-2\sqrt{x}+6}{\sqrt{x}-3}=\dfrac{-3\sqrt{x}+5}{\sqrt{x}-3}\)
27 tháng 8 2021
a) \(\sqrt{2x-1}=3\left(đk:x\ge\dfrac{1}{2}\right)\)
\(\Leftrightarrow2x-1=9\Leftrightarrow2x=10\Leftrightarrow x=5\)(thỏa đk)
b) \(\sqrt{1-3x}=\dfrac{1}{2}\left(đk:x\le\dfrac{1}{3}\right)\)
\(\Leftrightarrow1-3x=\dfrac{1}{4}\Leftrightarrow3x=\dfrac{3}{4}\Leftrightarrow x=\dfrac{1}{4}\)(thỏa đk)
c) \(\sqrt{\left(x-1\right)^2}=\dfrac{1}{2}\)
\(\Leftrightarrow\left|x-1\right|=\dfrac{1}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=\dfrac{1}{2}\\x-1=-\dfrac{1}{2}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=\dfrac{1}{2}\end{matrix}\right.\)
d) \(\sqrt{\left(1+2x\right)^2}=\dfrac{\sqrt{3}}{2}\)
\(\Leftrightarrow\left|1+2x\right|=\dfrac{\sqrt{3}}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}1+2x=\dfrac{\sqrt{3}}{2}\\1+2x=-\dfrac{\sqrt{3}}{2}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-2+\sqrt{3}}{4}\\x=-\dfrac{2+\sqrt{3}}{4}\end{matrix}\right.\)
e) \(\sqrt{\left(1-2x\right)^2}=\left|x-1\right|\)
\(\Leftrightarrow\left|1-2x\right|=\left|x-1\right|\)
\(\Leftrightarrow\left[{}\begin{matrix}1-2x=x-1\\1-2x=1-x\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=0\end{matrix}\right.\)
HT
20 tháng 2 2021
a/ \(=\lim\limits_{x\rightarrow-\infty}\dfrac{\dfrac{x\sqrt{x^2+1}}{x}-\dfrac{2x}{x}+\dfrac{1}{x}}{\sqrt[3]{\dfrac{2x^3}{x^3}-\dfrac{2x}{x^3}}+\dfrac{1}{x}}=0\)
b/ \(=\lim\limits_{x\rightarrow-\infty}\dfrac{\dfrac{8x^7}{x^7}}{\dfrac{\left(-2x^7\right)}{x^7}}=-\dfrac{8}{2^7}\)
c/ \(=\lim\limits_{x\rightarrow+\infty}\dfrac{\sqrt{\dfrac{4x^2}{x^2}+\dfrac{x}{x^2}}+\sqrt[3]{\dfrac{8x^3}{x^3}+\dfrac{x}{x^3}-\dfrac{1}{x^3}}}{\sqrt[4]{\dfrac{x^4}{x^4}+\dfrac{3}{x^4}}}=\dfrac{2+2}{1}=4\)
HP
30 tháng 8 2021
a, \(\sqrt{\left(2x+3\right)^2}=x+1\)
\(\Leftrightarrow\left|2x+3\right|=x+1\)
TH1: \(\left\{{}\begin{matrix}2x+3=x+1\\2x+3\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-2\\x\ge-\dfrac{3}{2}\end{matrix}\right.\Rightarrow\) vô nghiệm.
Vậy phương trình vô nghiệm.
TH2: \(\left\{{}\begin{matrix}-2x-3=x+1\\2x+3< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{4}{3}\\x< -\dfrac{3}{2}\end{matrix}\right.\Rightarrow\) vô nghiệm.
HP
30 tháng 8 2021
b,
a, \(\sqrt{\left(2x-1\right)^2}=x+1\)
\(\Leftrightarrow\left|2x-1\right|=x+1\)
TH1: \(\left\{{}\begin{matrix}2x-1=x+1\\2x-1\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\x\ge\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow x=2\)
TH2: \(\left\{{}\begin{matrix}-2x+1=x+1\\2x-1< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\x< \dfrac{1}{2}\end{matrix}\right.\Leftrightarrow x=0\)
25 tháng 8 2021
a)√x−1=2(x≥1)
\(x-1=4
\)
x=5
b)
\(\sqrt{3-x}=4\) (x≤3)
\(\left(\sqrt{3-x}\right)^2=4^2\)
x-3=16
x=19
26 tháng 8 2021
a: Ta có: \(\sqrt{x-1}=2\)
\(\Leftrightarrow x-1=4\)
hay x=5
b: Ta có: \(\sqrt{3-x}=4\)
\(\Leftrightarrow3-x=16\)
hay x=-13
c: Ta có: \(2\cdot\sqrt{3-2x}=\dfrac{1}{2}\)
\(\Leftrightarrow\sqrt{3-2x}=\dfrac{1}{4}\)
\(\Leftrightarrow-2x+3=\dfrac{1}{16}\)
\(\Leftrightarrow-2x=-\dfrac{47}{16}\)
hay \(x=\dfrac{47}{32}\)
d: Ta có: \(4-\sqrt{x-1}=\dfrac{1}{2}\)
\(\Leftrightarrow\sqrt{x-1}=\dfrac{7}{2}\)
\(\Leftrightarrow x-1=\dfrac{49}{4}\)
hay \(x=\dfrac{53}{4}\)
e: Ta có: \(\sqrt{x-1}-3=1\)
\(\Leftrightarrow\sqrt{x-1}=4\)
\(\Leftrightarrow x-1=16\)
hay x=17
f:Ta có: \(\dfrac{1}{2}-2\cdot\sqrt{x+2}=\dfrac{1}{4}\)
\(\Leftrightarrow2\cdot\sqrt{x+2}=\dfrac{1}{4}\)
\(\Leftrightarrow\sqrt{x+2}=\dfrac{1}{8}\)
\(\Leftrightarrow x+2=\dfrac{1}{64}\)
hay \(x=-\dfrac{127}{64}\)
Y
21 tháng 7 2023
\(a,\sqrt{72x}\) xác định \(\Leftrightarrow72x\ge0\Leftrightarrow x\ge0\)
\(b,\dfrac{2x+3}{\sqrt{x^2-4}}\) xác định \(\Leftrightarrow x^2-4>0\Leftrightarrow\left(x-2\right)\left(x+2\right)>0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x-2>0\\x+2>0\end{matrix}\right.\\\left[{}\begin{matrix}x-2< 0\\x+2< 0\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x>2\\x>-2\end{matrix}\right.\\\left[{}\begin{matrix}x< 2\\x< -2\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x>2\\x< -2\end{matrix}\right.\)
\(c,\sqrt{\left(2x+1\right)\left(x+2\right)}\) xác định \(\Leftrightarrow\left(2x+1\right)\left(x+2\right)\ge0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}2x+1\ge0\\x+2\ge0\end{matrix}\right.\\\left[{}\begin{matrix}2x+1\le0\\x+2\le0\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x\ge-\dfrac{1}{2}\\x\ge-2\end{matrix}\right.\\\left[{}\begin{matrix}x\le-\dfrac{1}{2}\\x\le-2\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x\ge-\dfrac{1}{2}\\x\le-2\end{matrix}\right.\)
\(d,3-\sqrt{16x^2-1}\) xác định \(\Leftrightarrow16x^2-1\ge0\Leftrightarrow\left(4x-1\right)\left(4x+1\right)\ge0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}4x-1\ge0\\4x+1\ge0\end{matrix}\right.\\\left[{}\begin{matrix}4x-1\le0\\4x+1\le0\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x\ge\dfrac{1}{4}\\x\ge-\dfrac{1}{4}\end{matrix}\right.\\\left[{}\begin{matrix}x\le\dfrac{1}{4}\\x\le-\dfrac{1}{4}\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x\ge\dfrac{1}{4}\\x\le-\dfrac{1}{4}\end{matrix}\right.\)
\(e,\sqrt{\dfrac{3+x}{4-x}}\) xác định \(\Leftrightarrow\left[{}\begin{matrix}3+x\ge0\\4-x>0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x\ge-3\\x>4\end{matrix}\right.\) \(\Leftrightarrow x>4\)
20 tháng 7 2016
từ dòng cuối là sai rồi bạn à
Bạn bỏ dòng cuối đi còn lại đúng rồi
Ở tử đặt nhân tử chung căn x chung rồi lại đặt căn x +1 chung
Ở mẫu tách 3 căn x ra 2 căn x +căn x rồi đặt nhân tử 2 căn x ra
rút gọn được \(\frac{3\sqrt{x}-5}{2\sqrt{x}+1}\)
