Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\sqrt{x^2-x+16}=4\)
\(\Rightarrow x^2-x+16=16\\ \Rightarrow x^2-x=0\\ \Rightarrow x\left(x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=0\\x-1=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
Ta có: \(\sqrt{x^2-x+16}=4\)
\(\Leftrightarrow x^2-x=0\)
\(\Leftrightarrow x\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
c: Ta có: \(\sqrt{2x}=\sqrt{5}\)
\(\Leftrightarrow2x=5\)
hay \(x=\dfrac{5}{2}\)
d: Ta có: \(\sqrt{3x-1}=4\)
\(\Leftrightarrow3x-1=16\)
\(\Leftrightarrow3x=17\)
hay \(x=\dfrac{17}{3}\)
Ta có: \(\sqrt{4\cdot\left(1-x\right)^2}=6\)
\(\Leftrightarrow2\left|x-1\right|=6\)
\(\Leftrightarrow\left|x-1\right|=3\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=3\\x-1=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-2\end{matrix}\right.\)
\(\sqrt{4x^2-4x+9}=3\\ \Rightarrow4x^2-4x+9=9\\ \Rightarrow4x\left(x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}4x=0\\x-1=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
Ta có: \(\sqrt{4x^2-4x+9}=3\)
\(\Leftrightarrow4x^2-4x=0\)
\(\Leftrightarrow4x\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
1)
ĐKXĐ: x>4
Ta có: \(\dfrac{\sqrt{x+5}}{\sqrt{x-4}}=\dfrac{\sqrt{x-2}}{\sqrt{x+3}}\)
\(\Leftrightarrow x^2+8x+15=x^2-6x+8\)
\(\Leftrightarrow8x+6x=8-15\)
\(\Leftrightarrow14x=-7\)
hay \(x=-\dfrac{1}{2}\)(loại)
2) Ta có: \(\sqrt{4x^2-9}=3\sqrt{2x-3}\)
\(\Leftrightarrow\sqrt{2x-3}\left(\sqrt{2x+3}-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-3=0\\2x+3=9\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=3\\2x=6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=3\end{matrix}\right.\)
Giaỉ phương trình sau ;
\(\sqrt[]{4\left(1-x\right)^2}-6=0\)
5phút nữa em pk nộp r mn giúp e nha =((((
Ta có: \(\sqrt{4\cdot\left(1-x\right)^2}=6\)
\(\Leftrightarrow2\left|x-1\right|=6\)
\(\Leftrightarrow\left|x-1\right|=3\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=3\\x-1=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-2\end{matrix}\right.\)
\(\Leftrightarrow\left|2\left(1-x\right)\right|=6\)
\(\Leftrightarrow\left[{}\begin{matrix}2\left(1-x\right)=6\\2\left(1-x\right)=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}1-x=3\\1-x=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=4\end{matrix}\right.\)
\(25\sqrt{\dfrac{x-3}{25}}-7\sqrt{\dfrac{4x-12}{9}}-7\sqrt{x^2-9}+18\sqrt{\dfrac{9x^2-81}{81}}=0\left(x\ge3\right)\)
\(=25\sqrt{\dfrac{1}{25}.\left(x-3\right)}-7\sqrt{\dfrac{4}{9}.\left(x-3\right)}-7\sqrt{x^2-9}+18\sqrt{\dfrac{1}{9}.\left(x^2-9\right)}=0\)
\(=5\sqrt{x-3}-\dfrac{14}{3}\sqrt{x-3}-7\sqrt{x^2-9}+6\sqrt{x^2-9}=0\)
\(\Rightarrow\dfrac{1}{3}\sqrt{x-3}-\sqrt{\left(x-3\right)\left(x+3\right)}=0\Rightarrow\sqrt{x-3}-3\sqrt{\left(x-3\right)\left(x+3\right)}=0\)
\(\Rightarrow\sqrt{x-3}\left(1-3\sqrt{x+3}\right)=0\Rightarrow\left[{}\begin{matrix}\sqrt{x-3}=0\\1=3\sqrt{x+3}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{26}{9}\left(l\right)\end{matrix}\right.\)
f: Ta có: \(\left(x+1\right)\left(x-2\right)-\left(2-x\right)\left(3-x\right)>0\)
\(\Leftrightarrow x^2-2x+x-2-\left(x-2\right)\left(x-3\right)>0\)
\(\Leftrightarrow x^2-x-2-x^2+5x-6>0\)
\(\Leftrightarrow4x>8\)
hay x>2
g: Ta có: \(\left(2x-1\right)^2\le2\left(x-1\right)^2\)
\(\Leftrightarrow4x^2-4x+1-2x^2+4x-2\le0\)
\(\Leftrightarrow2x^2\le1\)
\(\Leftrightarrow x^2\le\dfrac{1}{2}\)
\(\Leftrightarrow-\dfrac{\sqrt{2}}{2}\le x\le\dfrac{\sqrt{2}}{2}\)
Ta có: \(\sqrt{4x^2-4x+9}=3\)
\(\Leftrightarrow4x^2-4x=0\)
\(\Leftrightarrow4x\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)