![](https://rs.olm.vn/images/avt/0.png?1311)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(x-5\right)^4.|y^2-25|=0\)
\(\Rightarrow\left[{}\begin{matrix}\left(x-5\right)^4=0\\|y^2-25|=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-5=0\\y^2-25=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0+5=5\\y^2=0+25=25\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5\\y=5;-5\end{matrix}\right.\)
Vậy x=5 và y=5;-5
![](https://rs.olm.vn/images/avt/0.png?1311)
5: \(=-\left(x^2+3x+5\right)\)
\(=-\left(x^2+3x+\dfrac{9}{4}+\dfrac{11}{4}\right)\)
\(=-\left(x+\dfrac{3}{2}\right)^2-\dfrac{11}{4}< 0\)
6: \(=-3\left(x^2+2x+\dfrac{4}{3}\right)=-3\left(x^2+2x+1+\dfrac{1}{3}\right)\)
\(=-3\left(x+1\right)^2-1< 0\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(H=4x^2+4x+2=\left(2x+1\right)^2+1>0\)
\(K=4x^2+3x+2=4\left(x^2+2.\frac{3}{8}x+\frac{9}{64}\right)+\frac{23}{16}\)
\(=4\left(x+\frac{3}{8}\right)^2+\frac{23}{16}>0\)
\(L=2x^2+3x+4=2\left(x^2+2.\frac{3}{4}x+\frac{9}{16}\right)+\frac{23}{8}\)
\(=2\left(x+\frac{3}{4}\right)^2+\frac{23}{8}>0\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left|-2\right|\cdot3^x+3^{x+2}=99\)
\(\Rightarrow2\cdot3^x+3^x\cdot3^2=99\)
\(\Rightarrow3^x\cdot\left(2+3^2\right)=99\)
\(\Rightarrow3^x\cdot11=99\)
\(\Rightarrow3^x=99:11\)
\(\Rightarrow3^x=9\)
\(\Rightarrow3^x=3^2\)
\(\Rightarrow x=2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
hk hừng đẳng thưc chưa
chưa bn