a. \(\sqrt{2x+5}=\sqrt{1-x}\)
<=> 2x + 5 = 1 - x
<=> 2x + x = 1 - 5
<=> 3x = -4
<=> x = \(\dfrac{-4}{3}\)
Vậy ...............
b. \(\sqrt{x^2-x}=\sqrt{3-x}\)
<=> x2 - x = 3 - x
<=> x2 - x + x = 3
<=> x2 = 3
<=> x = \(\sqrt{3}\)
Vậy ..................
c. \(\sqrt{2x^2-3}=\sqrt{4x-3}\)
<=> 2x2 - 3 = 4x - 3
<=> 2x2 - 4x = -3 + 3
<=> 2x2 - 4x = 0
<=> x(x - 4) = 0
\(\left[{}\begin{matrix}x=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\)
Vậy .................
a,\(ĐK:-\dfrac{5}{2}\le x\le1\)
Ta có: \(\left(1\right)\Leftrightarrow2x+5=1-x\)
\(\Leftrightarrow3x=-4\Leftrightarrow x=-\dfrac{4}{3}\left(tm\right)\)
b,\(ĐK:1\le x\le3\)
Ta có: \(\left(1\right)\Leftrightarrow x^2-x=3-x\)
\(\Leftrightarrow x^2=3\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{3}\left(tm\right)\\x=-\sqrt{3}\left(loại\right)\end{matrix}\right.\)
c,\(ĐK:\left\{{}\begin{matrix}x\ge\sqrt{\dfrac{3}{2}}\\x\le-\sqrt{\dfrac{3}{2}}\end{matrix}\right.\)
Ta có: \(\left(1\right)\Leftrightarrow2x^2-3=4x-3\)
\(\Leftrightarrow2x^2-4x=0\Leftrightarrow2x\left(x-2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\left(loại\right)\\x=2\left(tm\right)\end{matrix}\right.\)
sorry bn mik quên ĐKXĐ và bn thêm x = \(-\sqrt{3}\) vào câu b giùm mik nha
a. ĐKXĐ: \(\left\{{}\begin{matrix}2x+5\ge0\\1-x\ge0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x\ge-\dfrac{5}{2}\\x\le1\end{matrix}\right.\)\(\Rightarrow-\dfrac{5}{2}\le x\le1\)
\(\sqrt{2x+5}=\sqrt{1-x}\\ \Rightarrow2x+5=1-x\\ \Rightarrow2x+5-1+x=0\\ \Rightarrow3x+4=0\\ \Rightarrow x=-\dfrac{4}{3}\)
b, ĐKXĐ: \(\left\{{}\begin{matrix}x^2-x\ge0\\3-x\ge0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x\left(x-1\right)\ge0\\x\le3\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge0\\x-1\ge0\end{matrix}\right.\\\left\{{}\begin{matrix}x\le0\\x-1\le0\end{matrix}\right.\end{matrix}\right.\\x\le3\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge0\\x\ge1\end{matrix}\right.\\\left\{{}\begin{matrix}x\le0\\x\le1\end{matrix}\right.\end{matrix}\right.\\x\le3\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x\ge1\\x\le0\end{matrix}\right.\\x\le3\end{matrix}\right.\)
\(\sqrt{x^2-x}=\sqrt{3-x}\\ \Rightarrow x^2-x=3-x\\ \Rightarrow x^2-x-3+x=0\\ \Rightarrow x^2-3=0\\ \Rightarrow x=\pm\sqrt{3}\)
c, ĐKXĐ: \(\left\{{}\begin{matrix}2x^2-3\ge0\\4x-3\ge0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x\ge\sqrt{\dfrac{3}{2}}\\x\ge\dfrac{3}{4}\end{matrix}\right.\Rightarrow x\ge\dfrac{3}{4}\)
\(\sqrt{2x^2-3}=\sqrt{4x-3}\\ \Rightarrow2x^2-3=4x-3\\ \Rightarrow2x^2-3-4x+3=0\\ \Rightarrow2x^2-4x=0\\ \Rightarrow2x\left(x-2\right)=0\\ \Rightarrow\left[{}\begin{matrix}2x=0\\x-2=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)
a: Ta có: \(\sqrt{2x+5}=\sqrt{1-x}\)
\(\Leftrightarrow2x+5=1-x\)
\(\Leftrightarrow3x=-4\)
hay \(x=-\dfrac{4}{3}\)
b: Ta có: \(\sqrt{x^2-x}=\sqrt{3-x}\)
\(\Leftrightarrow x^2-x=3-x\)
\(\Leftrightarrow x^2=3\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{3}\\x=-\sqrt{3}\end{matrix}\right.\)
