Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Áp dụng tính chất dãy tỉ số bằng nhau, ta có:
\(\frac{3z-4y}{5}=\frac{5y-3x}{4}=\frac{4x-5z}{3}=\frac{3z-4y+5y-3x+4x-5z}{5+4+3}=\frac{0}{12}=0\)
\(\frac{3z-4y}{5}=0\Rightarrow3z-4y=0\Rightarrow3z=4y\)\(\Rightarrow\frac{y}{3}=\frac{z}{4}\)(1)
\(\frac{5y-3x}{4}=0\Rightarrow5y-3x=0\Rightarrow5y=3x\Rightarrow\frac{x}{5}=\frac{y}{3}\)(2)
Từ (1) và (2) \(\Rightarrow\frac{x}{5}=\frac{y}{3}=\frac{z}{4}\)
\(\Rightarrow\frac{x^2}{25}=\frac{y^2}{9}=\frac{z^2}{16}\)
Áp dụng tính chất dãy tỉ số bằng nhau, ta có:
\(\frac{x^2}{25}=\frac{y^2}{9}=\frac{z^2}{16}=\frac{x^2-z^2}{25-16}=\frac{36}{9}=4\)
\(\frac{x^2}{25}=4\Rightarrow x^2=100\Rightarrow\orbr{\begin{cases}x=10\\x=-10\end{cases}}\)
\(\frac{y^2}{9}=4\Rightarrow y^2=36\Rightarrow\orbr{\begin{cases}y=6\\y=-6\end{cases}}\)
\(\frac{z^2}{16}=4\Rightarrow z^2=64\Rightarrow\orbr{\begin{cases}z=8\\z=-8\end{cases}}\)
Vậy........................
Theo đề ta có: \(x:y:z=3:4:5\Rightarrow\frac{x}{3}=\frac{y}{4}=\frac{z}{5}\)
Đặt: \(\frac{x}{3}=\frac{y}{4}=\frac{z}{5}=k\left(k\inℕ^∗\right)\)
Suy ra: \(x=3k;y=4k;z=5k\) Thay vào biểu thức P ta có:
\(P=\frac{3k+8k+15k}{6k+12k+20k}+\frac{6k+12k+20k}{9k+16k+25k}+\frac{9k+16k+25k}{12k+20k+30k}\)
\(P=\frac{26k}{38k}+\frac{38k}{50k}+\frac{50k}{62k}=\frac{13}{19}+\frac{19}{25}+\frac{25}{31}=\frac{33141}{14725}\)
\(3x=2y\Rightarrow\frac{x}{2}=\frac{y}{3}\)
\(7y=5z\Rightarrow\frac{y}{5}=\frac{z}{7}\)
\(\hept{\begin{cases}\frac{x}{2}=\frac{x}{3}\\\frac{y}{5}=\frac{x}{7}\end{cases}\Rightarrow}\frac{x}{2}=\frac{5y}{15};\frac{3y}{15}=\frac{z}{7}\)
\(\Rightarrow\frac{x}{10}=\frac{y}{15}=\frac{z}{21}\)
Áp dụng tính chát dãy tỉ số = nhau ta có:
\(\frac{x}{10}=\frac{y}{15}=\frac{z}{21}=\frac{x-y+z}{10-15+21}=\frac{32}{16}=2\)
\(\Rightarrow\frac{x}{10}=2\Rightarrow x=20\)
\(\frac{y}{15}=2\Rightarrow y=30\)
\(\frac{z}{21}=3\Rightarrow z=63\)
b, Tự làm
c, \(5x=2y\Leftrightarrow\frac{x}{2}=\frac{y}{5}\)
\(2x=3z\Leftrightarrow\frac{x}{3}=\frac{z}{2}\)
\(\Leftrightarrow\frac{x}{2}=\frac{y}{5};\frac{x}{3}=\frac{z}{2}\)
\(\Leftrightarrow\frac{x}{6}=\frac{y}{15}=\frac{x}{6}=\frac{z}{10}\)
\(\Leftrightarrow\frac{x}{6}=\frac{y}{15}=\frac{z}{10}\)
Đặt \(\frac{x}{6}=\frac{y}{15}=\frac{z}{10}=k(k\inℤ)\)
\(\Leftrightarrow\hept{\begin{cases}x=6k\\y=15k\\z=10k\end{cases}}\)
\(\Leftrightarrow x\cdot y=6k\cdot15k=90\)
\(\Leftrightarrow90:k^2=90\Leftrightarrow k^2=1\Leftrightarrow k=\pm1\)
\(\Leftrightarrow\hept{\begin{cases}x=6k\\y=15k\\z=10k\end{cases}}\Leftrightarrow\hept{\begin{cases}x=6\\y=15\\z=10\end{cases}}\)hay \(\hept{\begin{cases}x=-6\\y=-15\\z=-10\end{cases}}\)
Vậy \((x,y)\in(6,15);(-6,-15)\)
\(\frac{4\left(3x-2y\right)}{16}=\frac{3\left(2z-4x\right)}{9}=\frac{2\left(4y-3z\right)}{4}=\frac{12x-8y+6z-12x+8y-6z}{16+9+4}=\frac{0}{29}=0\)
\(\Leftrightarrow3x-2y=0\Leftrightarrow\frac{x}{2}=\frac{y}{3}\)
\(\Leftrightarrow2z-4x=0\Leftrightarrow\frac{x}{2}=\frac{z}{4}\)
\(\Leftrightarrow\frac{x}{2}=\frac{y}{3}=\frac{z}{4}\)
\(\Leftrightarrow\frac{x}{2}=\frac{y}{3}=\frac{z}{4}=\frac{2x+4y+5z}{4+12+20}=\frac{8}{36}=\frac{2}{9}=\frac{2x+3y-z}{4+12-4}\)=> A= 2x+3y -z = 12.2/9 =8/3
TA CÓ \(\frac{3x-5y}{2}=\frac{7y-3z}{3}=\frac{5z-7x}{4}\)\(=\frac{21x-35y}{14}=\frac{35y-15z}{15}=\frac{15z-21x}{12}\)=\(\frac{21x-35+35y-15z+15z-21x}{14+15+12}=\frac{0}{41}=0\)
=> \(\hept{\begin{cases}3x-5y=0\\7y-3z=0\\5z-7x=0\end{cases}\left(=\right)\hept{\begin{cases}3x=5y\\7y=3z\\5z=7x\end{cases}\left(=\right)\hept{\begin{cases}\frac{x}{5}=\frac{y}{3}\\\frac{y}{3}=\frac{z}{7}\\\frac{z}{7}=\frac{x}{5}\end{cases}}}}\)
=> \(\frac{x}{5}=\frac{y}{3}=\frac{z}{7}=\frac{x+y+z}{5+3+7}=\frac{17}{15}\)
=>\(\hept{\begin{cases}x=\frac{17}{3}\\y=\frac{17}{5}\\z=\frac{119}{15}\end{cases}}\)
ai trả lời được câu này mình cho 5 k
tìm x, biết
10+11+12+13+.....x=5106
Ta có : \(\frac{3x-2y}{4}=\frac{4y-3z}{2}=\frac{2z-4x}{3}\)
\(\Leftrightarrow\frac{12x-8y}{16}=\frac{8y-6z}{4}=\frac{6z-12x}{9}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có :
\(\frac{12x-8y}{16}=\frac{8y-6z}{4}=\frac{6z-12x}{9}=\frac{12x-8y+8y-6z+6z-12x}{16+4+9}=0\)
\(\Leftrightarrow\hept{\begin{cases}\frac{3x-2y}{4}=0\\\frac{4y-3z}{2}=0\\\frac{2z-4x}{3}=0\end{cases}}\) \(\Leftrightarrow\hept{\begin{cases}3x=2y\\4y=3z\\2z=4x\end{cases}}\) \(\Leftrightarrow\hept{\begin{cases}\frac{x}{2}=\frac{y}{3}\\\frac{y}{3}=\frac{z}{4}\\\frac{x}{2}=\frac{z}{4}\end{cases}}\) \(\Leftrightarrow\frac{x}{2}=\frac{y}{3}=\frac{z}{4}\)
\(\Leftrightarrow\frac{x}{2}=\frac{2y}{6}=\frac{3z}{12}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có :
\(\frac{x}{2}=\frac{2y}{6}=\frac{3z}{12}=\frac{x-2y+3z}{2-6+12}=\frac{8}{8}=1\)
\(\Leftrightarrow\hept{\begin{cases}\frac{x}{2}=1\\\frac{y}{3}=1\\\frac{z}{4}=1\end{cases}\Leftrightarrow}\hept{\begin{cases}x=2\\y=3\\z=4\end{cases}}\)
Vậy : \(\left(x,y,z\right)=\left(2,3,4\right)\)
Từ giả thiết suy ra : \(\frac{5\left(3z-4y\right)}{25}=\frac{4\left(5y-3x\right)}{16}=\frac{3\left(4x-5z\right)}{9}=\frac{0}{25+16+9}=0\)
( Tính chất dãy tỉ số bằng nhau )
Vì vậy có : \(\left\{{}\begin{matrix}3z-4y=0\\5y-3x=0\\4x-5z=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}3z=4y\\5y=3x\\4x=5z\end{matrix}\right.\) \(\Leftrightarrow\frac{z}{4}=\frac{y}{3}=\frac{x}{5}\)
\(\Leftrightarrow\frac{z^2}{16}=\frac{y^2}{9}=\frac{x^2}{25}=\frac{x^2-z^2}{25-16}=\frac{36}{9}=4\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\pm25\\y=\pm6\\z=\pm8\end{matrix}\right.\)
Ta có: \(\frac{3z-4y}{5}=\frac{5y-3x}{4}=\frac{4x-5z}{3}.\)
\(\Rightarrow\frac{5.\left(3z-4y\right)}{25}=\frac{4.\left(5y-3x\right)}{16}=\frac{3.\left(4x-5z\right)}{9}.\)
\(\Rightarrow\frac{15z-20y}{25}=\frac{20y-12x}{16}=\frac{12x-15z}{9}.\)
Áp dụng tính chất dãy tỉ số bằng nhau ta được:
\(\frac{15z-20y}{25}=\frac{20y-12x}{16}=\frac{12x-15z}{9}=\frac{15z-20y+20y-12x+12x-15z}{25+16+9}=\frac{\left(15z-15z\right)-\left(20y-20y\right)-\left(12x-12x\right)}{50}=\frac{0}{50}=0.\)
\(\Rightarrow\left\{{}\begin{matrix}\frac{3z-4y}{5}=0\\\frac{5y-3x}{4}=0\\\frac{4x-5z}{3}=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}3z-4y=0\\5y-3x=0\\4x-5z=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}3z=4y\\5y=3x\\4x=5z\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\frac{z}{4}=\frac{y}{3}\\\frac{y}{3}=\frac{x}{5}\\\frac{x}{5}=\frac{z}{4}\end{matrix}\right.\Rightarrow\frac{x}{5}=\frac{y}{3}=\frac{z}{4}.\)
\(\Rightarrow\frac{x^2}{25}=\frac{y^2}{9}=\frac{z^2}{16}\) và \(x^2-z^2=36.\)
Áp dụng tính chất dãy tỉ số bằng nhau ta được:
\(\frac{x^2}{25}=\frac{y^2}{9}=\frac{z^2}{16}=\frac{x^2-z^2}{25-16}=\frac{36}{9}=4.\)
\(\Rightarrow\left\{{}\begin{matrix}\frac{x^2}{25}=4\Rightarrow x^2=100\Rightarrow\left[{}\begin{matrix}x=10\\x=-10\end{matrix}\right.\\\frac{y^2}{9}=4\Rightarrow y^2=36\Rightarrow\left[{}\begin{matrix}y=6\\y=-6\end{matrix}\right.\\\frac{z^2}{16}=4\Rightarrow z^2=64\Rightarrow\left[{}\begin{matrix}z=8\\z=-8\end{matrix}\right.\end{matrix}\right.\)
Vậy \(\left(x;y;z\right)=\left(10;6;8\right),\left(-10;-6;-8\right).\)
Chúc bạn học tốt!