![](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)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\text{Δ}=\left(-3\right)^2-4\cdot\left(2m+1\right)\)
=9-8m-4=-8m+5
Để phương trình có nghiệm kép thì -8m+5=0
hay m=5/8
Pt trở thành \(x^2-3x+\dfrac{9}{4}=0\)
hay x=3/2
![](https://rs.olm.vn/images/avt/0.png?1311)
2) Ta có: \(B=\dfrac{\sqrt{x}}{\sqrt{x}-2}:\left(\dfrac{x-2}{x-4}-\dfrac{1}{\sqrt{x}+2}\right)\)
\(=\dfrac{\sqrt{x}}{\sqrt{x}-2}:\dfrac{x-2-\sqrt{x}+2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{\sqrt{x}}{\sqrt{x}-2}\cdot\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{x-\sqrt{x}}\)
\(=\dfrac{\sqrt{x}+2}{\sqrt{x}-1}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\dfrac{3}{2}-tana\cdot cos^2a\)
\(=\dfrac{3}{2}-\dfrac{sina}{cosa}\cdot cos^2a\)
\(=\dfrac{3}{2}-sina\cdot cosa\)
\(=\dfrac{3}{2}-\dfrac{1}{2}sin2a\)
\(0^0< a< 90^0\)
=>\(0< =2a< =180^0\)
=>\(sin2a\in\left[-1;1\right]\)
\(-1< =sin2a< =1\)
=>\(\dfrac{1}{2}>=-\dfrac{1}{2}sin2a>=-\dfrac{1}{2}\)
=>\(\dfrac{7}{2}>=-\dfrac{1}{2}sin2a+3>=\dfrac{5}{2}\)
=>\(\dfrac{5}{2}< =y< =\dfrac{7}{2}\)
\(y_{min}=\dfrac{5}{2}\) khi sin2a=1
=>\(2a=\dfrac{\Omega}{2}+k2\Omega\)
=>\(a=\dfrac{\Omega}{4}+k\Omega\)
mà 0<a<90
nên a=45
![](https://rs.olm.vn/images/avt/0.png?1311)
bạn đăng tách ra nhé
1, Lấy vế cộng vế ta được \(\dfrac{4}{x-2}=4\Rightarrow x-2=1\Leftrightarrow x=3\)
Thay vào ta được \(\dfrac{2}{3-2}-\dfrac{3}{y+1}=1\Rightarrow\dfrac{3}{y+1}=1\Rightarrow y+1=3\Leftrightarrow y=2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(=2\sqrt{2}\cdot\sqrt{2}-3\sqrt{2}\cdot\sqrt{2}+\sqrt{10}\cdot\sqrt{2}-2\sqrt{5}\)
=4-6
=-2
b: \(=8+2\sqrt{15}-8+2\sqrt{15}=4\sqrt{15}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a, \(\left\{{}\begin{matrix}35x-28y=21\\35x-45y=40\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}17y=-19\\x=\dfrac{3+4y}{5}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-\dfrac{19}{17}\\x=-\dfrac{13}{17}\end{matrix}\right.\)
b, Đặt x;y khác 0
Đặt 1/x = t ; 1/y = u
\(\left\{{}\begin{matrix}t-8u=18\\5t+4u=51\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}5t-40u=90\\5t+4u=51\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-44u=39\\t=18+8u\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}u=-\dfrac{39}{44}\\t=\dfrac{120}{11}\end{matrix}\right.\)
Theo cách đặt y = -44/39 ; x = 11/120 (tm)
\(a,\\ \Leftrightarrow\left\{{}\begin{matrix}35x-28y=21_{\left(1\right)}\\35x-45y=40_{\left(2\right)}\end{matrix}\right.\\ Lấy\left(1\right)-\left(2\right),ta.đc:\\ -17y=19\Leftrightarrow y=\dfrac{-19}{17}\\ Thay.vào.\left(1\right):\\ 35x-28.\dfrac{-19}{17}=21\Leftrightarrow x=\dfrac{-5}{17}\)
Vậy ......
\(b,\\ \Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x}=18+\dfrac{1}{y}\\5.\left(18+\dfrac{1}{y}\right)+\dfrac{4}{y}=51\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x}=18+\dfrac{1}{y}\\\dfrac{9}{y}=-39\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x}=18-\dfrac{13}{3}\\\dfrac{1}{y}=\dfrac{-13}{3}\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x}=\dfrac{41}{3}\\\dfrac{1}{y}=\dfrac{-13}{3}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{3}{41}\\y=\dfrac{-3}{13}\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Gọi tuổi của Minh là x
=>Tuổi của Ninh là x+2
Tuổi của Lan là 1/2x
Theo đề, ta co: 1/2x+x+2+x=27
=>2,5x=25
=>x=10
![](https://rs.olm.vn/images/avt/0.png?1311)
Bạn gõ hẳn đề ra thì khả năng mọi người sẽ giúp bạn sẽ cao hơn nhé. Đọc như thế này hơi khó.
2) \(\sqrt{x^2-4x+4}=1\\ \Rightarrow x-2=1\\ \Rightarrow x=3\)
3) \(\sqrt{1-4x+4x^2}=5\\ \Rightarrow2x-1=5\\ \Rightarrow x=3\)
4) \(\sqrt{4\left(1-2x+x^2\right)}=0\\ \Rightarrow2\left(x-1\right)=0\\ \Rightarrow x-1=0\\ \Rightarrow x=1\)
5) \(\sqrt{9x^2}=2x+1\\ \Rightarrow2x+1-3x=0\\ \Rightarrow-x+=1\\ \Rightarrow x=1\)
10:ta có: \(\sqrt{4x^2-4x+1}=\sqrt{x^2+8x+16}\)
\(\Leftrightarrow\left|2x-1\right|=\left|x+4\right|\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=x+4\\2x-1=-x-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-1\end{matrix}\right.\)
11: Ta có: \(\sqrt{9x^2+6x+1}=\sqrt{x^2-2\sqrt{6x}+6}\)
\(\Leftrightarrow\left|3x+1\right|=\left|x-\sqrt{6}\right|\)
\(\Leftrightarrow\left[{}\begin{matrix}3x+1=x-\sqrt{6}\\3x+1=-x+\sqrt{6}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-\sqrt{6}-1}{2}\\x=\dfrac{\sqrt{6}-1}{4}\end{matrix}\right.\)