4a-3b/5 = 5b-4c/3 = 3c-5a/4 và a+b+c=180
các bạn giúp mình vsHãy 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)
Ta có \(\frac{4a-3b}{5}=\frac{5b-4c}{3}=\frac{3c-5a}{4}\)
=> \(\frac{20a-15b}{25}=\frac{15b-12c}{9}=\frac{12c-20a}{16}=\frac{20a-15b+15b-12c+12c-20a}{25+9+16}=\frac{0}{50}=0\)
=> \(\hept{\begin{cases}4a-3b=0\\5b-4c=0\\3c-5a=0\end{cases}}\Rightarrow\hept{\begin{cases}4a=3b\\5b=4c\\3c=5a\end{cases}}\Rightarrow\hept{\begin{cases}\frac{a}{3}=\frac{b}{4}\\\frac{b}{4}=\frac{c}{5}\\\frac{c}{5}=\frac{a}{3}\end{cases}}\Rightarrow\frac{a}{3}=\frac{b}{4}=\frac{c}{5}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có
\(\frac{a}{3}=\frac{b}{4}=\frac{c}{5}=\frac{a+b+c}{3+4+5}=\frac{180}{12}=15\)
=> \(\hept{\begin{cases}a=45\\b=60\\c=75\end{cases}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Với a2 + b2 + c2 = 0 , a2 = b2 = c2 = 0 .
a3 + b3 - c3 = 0 + 0 - 0 = 0
\(a^3+b^3-c^2\le0\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Áp dụng tính chất dãy tỉ số bằng nhau, ta có:
\(\frac{1+3y}{12}=\frac{1+5y}{5x}=\frac{1+7y}{4x}=\frac{1+3y+1+7y}{12+4x}=\frac{2+10y}{2\left(6+2x\right)}=\frac{2\left(1+5y\right)}{2\left(6+2x\right)}=\frac{1+5y}{6+2x}\)
\(\Rightarrow5x=6+2x\)
\(5x-2x=6\)
\(3x=6\)
\(x=2\)
Với \(x=2\), ta có:
\(\frac{1+3y}{12}=\frac{1+5y}{10}\)
\(\Rightarrow10\left(1+3y\right)=12\left(1+5y\right)\)
\(10+30y=12+60y\)
\(10-12=60y-30y\)
\(30y=-2\)
\(y=-\frac{1}{15}\)
Vậy \(x=2,y=\frac{-1}{15}\).
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: \(\left(\frac{3x-5}{9}\right)^{2018}\ge0\forall x\); \(\left(\frac{3y+0,4}{3}\right)^{2020}\ge0\forall y\)
\(\Rightarrow\left(\frac{3x-5}{9}\right)^{2018}+\left(\frac{3y+0,4}{3}\right)^{2020}\ge0\forall x,y\)
mà \(\left(\frac{3x-5}{9}\right)^{2018}+\left(\frac{3y+0,4}{3}\right)^{2020}=0\)( giả thuyết )
Dấu " = " xảy ra \(\Leftrightarrow\hept{\begin{cases}\frac{3x-5}{9}=0\\\frac{3y+0,4}{3}=0\end{cases}}\Leftrightarrow\hept{\begin{cases}3x-5=0\\3y+0,4=0\end{cases}}\Leftrightarrow\hept{\begin{cases}3x=5\\3y=-\frac{2}{5}\end{cases}}\Leftrightarrow\hept{\begin{cases}x=\frac{5}{3}\\y=-\frac{2}{15}\end{cases}}\)
Vậy \(x=\frac{5}{3}\)và \(y=-\frac{2}{15}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(x^2-2=0\)
\(x^2\) \(=0+2\)
\(x^2\) \(=2\)
\(\Rightarrow x=2\)hoặc \(x=-2\)
Vậy \(x\in\left\{2;-2\right\}\)
b) \(2\sqrt{x}+3=11\)
\(2\sqrt{x}\) \(=11-3\)
\(2\sqrt{x}\) \(=8\)
\(\sqrt{x}\) \(=8:2\)
\(\sqrt{x}\) \(=4\)
\(\Rightarrow x\) \(=2\)
Vậy \(x=2\)
c) \(\left(3\sqrt{x+5}+5\right):2-20=-13\)
\(\left(3\sqrt{x+5}+5\right):2\) \(=-13+20\)
\(\left(3\sqrt{x+5}+5\right):2\) \(=7\)
\(3\sqrt{x+5}+5\) \(=\frac{7}{2}\)
\(3\sqrt{x+5}\) \(=\frac{7}{2}-5\)
\(3\sqrt{x+5}\) \(=\frac{7-10}{2}\)
\(3\sqrt{x+5}\) \(=-\frac{3}{2}\)
\(\sqrt{x+5}\) \(=-\frac{3}{2}:3\)
\(\sqrt{x+5}\) \(=-\frac{3}{2}\times\frac{1}{3}\)
\(\sqrt{x+5}\) \(=\frac{-3\times1}{2\times3}\)
\(\sqrt{x+5}\) \(=\frac{-1\times1}{2\times1}\)
\(\sqrt{x+5}\) \(=\frac{-1}{2}\)
Vì \(\sqrt{x+5}>0\)
\(\Rightarrow\sqrt{x+5}=\frac{-1}{2}\)(vô lí)
\(\Rightarrow\left(x+5\right)\in\varnothing\)
\(\Rightarrow x\in\varnothing\)
Vậy \(x\in\varnothing\)
a) x2 - 2 = 0
=> x2 = 2
=> \(x=\pm\sqrt{2}\)
Vậy \(x=\pm\sqrt{2}\)
b) \(2\sqrt{x}+3=11\)
=> \(2\sqrt{x}=8\)
=> \(\sqrt{x}=4\)
=> \(\sqrt{x}^2=4^2\)
=> x = 16
Vậy x = 16
c) \(\left(3\sqrt{x+5}+5\right):2-20=-13\)
=> \(\left(3\sqrt{x+5}+5\right):2=7\)
=> \(3\sqrt{x+5}+5=14\)
=> \(3\sqrt{x+5}=9\)
=> \(\sqrt{x+5}=3\)
=> \(\sqrt{x+5}^2=3^2\)
=> x + 5 = 9
=> x = 4
Vậy x = 4