Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}=\frac{2\left(x-1\right)}{4}=\frac{3\left(y-2\right)}{9}=\frac{2x-2+3y-6-z+3}{4+9-4}=\frac{2x+3y-z-5}{9}=\frac{90}{9}=10\)
=> x-1 = 10.2 = 20 => x= 21
y-2 = 10.3 = 30 => y = 32
z-3 = 10.4 =40 => z = 43
Có: \(\frac{y-2}{3}=\frac{2y-4}{6}\)
\(\frac{z-3}{4}=\frac{3z-9}{12}\)
Suy ra\(\frac{x-1}{2}=\frac{2y-4}{6}=\frac{3z-9}{12}=\frac{\left(x-1\right)-\left(2y-4\right)+\left(3z-9\right)}{2-6+12}\)
\(=\frac{\left(x-2y+3z\right)-6}{8}=\frac{14-6}{8}=1\)
Vậy có \(\frac{x-1}{2};\frac{y-2}{3};\frac{z-3}{4}=1\)Thay vào có x=3; y=5; z=7
#)Giải :
a) Ta có : \(\frac{x}{3}=\frac{y}{4};\frac{y}{5}=\frac{z}{7}\Rightarrow\frac{x}{15}=\frac{y}{20};\frac{y}{20}=\frac{z}{28}\Rightarrow\frac{x}{15}=\frac{y}{20}=\frac{z}{28}\)
Áp dụng tính chất dãy tỉ số bằng nhau :
\(\frac{x}{15}=\frac{y}{20}=\frac{z}{28}=\frac{2x+3y-z}{30+60-28}=\frac{186}{62}=3\)
\(\hept{\begin{cases}\frac{x}{15}=3\\\frac{y}{20}=3\\\frac{z}{28}=3\end{cases}\Rightarrow\hept{\begin{cases}x=45\\y=60\\z=84\end{cases}}}\)
Vậy x = 45; y = 60; z = 84
b) Áp dụng tính chất dãy tỉ số bằng nhau :
\(\frac{y+z+1}{x}=\frac{x+z+2}{y}=\frac{x+y-3}{z}=\frac{\left(y+z+1\right)+\left(x+z+2\right)+\left(x+y-3\right)}{x+y+z}=\frac{2\left(x+y+z\right)}{x+y+z}=2\)
\(\Rightarrow\frac{y+z+1}{x}=\frac{x+z+2}{y}=\frac{x+y-3}{z}=\frac{1}{x+y+z}=2\)
\(\Rightarrow\hept{\begin{cases}y+z+1=2x\left(1\right)\\x+z+2=2y\left(2\right)\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}x+y-3=2z\left(3\right)\\x+y+z=\frac{1}{2}\left(4\right)\end{cases}}\)
\(\left(+\right)x+y+z=\frac{1}{2}\Rightarrow y+z=\frac{1}{2}-z\)
Thay (1) vào (+) ta được :
\(\frac{1}{2}-x+1=2x\Rightarrow\frac{3}{2}=3x\Rightarrow x=\frac{1}{2}\)
\(\left(+_2\right)x+y+z=\frac{1}{2}\Rightarrow x+z=\frac{1}{2}-y\)
Thay (2) và (+2) ta được :
\(\frac{1}{2}-y+2=2y\Rightarrow\frac{5}{2}=3y\Rightarrow y=\frac{5}{6}\)
\(\left(+_3\right)x+y+z=\frac{1}{2}+\frac{5}{6}+z=\frac{1}{2}\Rightarrow\frac{4}{3}+z=\frac{1}{2}\Rightarrow z=\frac{-5}{6}\)
Vậy \(\hept{\begin{cases}x=\frac{1}{2}\\y=\frac{5}{6}\\z=\frac{-5}{6}\end{cases}}\)
\(\frac{x}{2}=\frac{y}{3}=\frac{z}{5}=k\)
\(\Rightarrow x=2k;y=3k;z=5k\)
\(\Rightarrow xyz=2k\cdot3k\cdot5k=30k^3\)
Mà \(xyz=810\Rightarrow30k^3=810\)
\(\Rightarrow k^3=27\)
\(\Rightarrow k=3\)
Thay vào tìm x,,z.
a) Ta có : \(\frac{2}{3}x=\frac{3}{4}y=\frac{5}{6}z\)=> \(\frac{2x}{3}=\frac{3y}{4}=\frac{5z}{6}\)=> \(\frac{x}{\frac{3}{2}}=\frac{y}{\frac{4}{3}}=\frac{z}{\frac{6}{5}}\)
=> \(\frac{x^2}{\frac{9}{4}}=\frac{y^2}{\frac{16}{9}}=\frac{z^2}{\frac{36}{25}}\)
Đặt \(\frac{x^2}{\frac{9}{4}}=\frac{y^2}{\frac{16}{9}}=\frac{z^2}{\frac{36}{25}}=k\Leftrightarrow\hept{\begin{cases}x^2=\frac{9}{4}k\\y^2=\frac{16}{9}k\\z^2=\frac{36}{25}k\end{cases}}\)
=> \(x^2+y^2+z^2=\frac{9}{4}k+\frac{16}{9}k+\frac{36}{25}k\)
=> \(\frac{4921}{900}k=724\)
=> \(k=724:\frac{4921}{900}=\frac{651600}{4921}\)
Do đó : \(\hept{\begin{cases}x^2=\frac{9}{4}\cdot\frac{651600}{4921}\\y^2=\frac{16}{9}\cdot\frac{651600}{4921}\\z^2=\frac{36}{25}\cdot\frac{651600}{4921}\end{cases}}\)
Bài toán đây có sai sót j không vậy?Thấy số dữ quá đi :v
b) Ta có : \(\frac{x-1}{2}=\frac{y+2}{3}=\frac{z-3}{4}\)
=> \(\frac{x-1}{2}=\frac{2y+4}{6}=\frac{3z-9}{12}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có :
\(\frac{x-1}{2}=\frac{2y+4}{6}=\frac{3z-9}{12}=\frac{x-1-2y+4+3z-9}{2-6+12}=\frac{x-2y+3z-6}{8}=\frac{46-6}{8}=\frac{40}{8}=5\)
=> \(\hept{\begin{cases}\frac{x-1}{2}=5\\\frac{y+2}{3}=5\\\frac{z-3}{4}=5\end{cases}}\Rightarrow\hept{\begin{cases}x=11\\y=13\\z=23\end{cases}}\)
c) Đặt \(\frac{x}{3}=\frac{y}{16}=k\Rightarrow\hept{\begin{cases}x=3k\\y=16k\end{cases}}\)
=> xy = 16k . 3k
=> 48k2 = 192
=> k2 = 4
=> k = 2 hoặc k = -2
Do đó \(\left(x,y\right)\in\left\{\left(6,32\right);\left(-6,-32\right)\right\}\)
Bài 2 : a) \(\frac{4^2\cdot25^2+16\cdot125}{2^3\cdot5^2}\)
\(=\frac{\left(2^2\right)^2\cdot\left(5^2\right)^2+16\cdot125}{2^3\cdot5^2}\)
\(=\frac{2^4\cdot5^4+2^4\cdot5^3}{2^3\cdot5^2}\)
\(=\frac{2\cdot2^3\left(5^4+5^3\right)}{2^3\cdot5^2}\)
\(=\frac{2\cdot5^3\left(5+1\right)}{5^2}=\frac{2\cdot5\cdot5^2\cdot6}{5^2}=2\cdot5\cdot6=60\)
b) \(\frac{6^8\cdot2^4-4^5\cdot18^4}{27^3\cdot8^4-3^9\cdot2^{13}}\)
\(=\frac{\left(2\cdot3\right)^8\cdot2^4-\left(2^2\right)^5\cdot\left(2\cdot3^2\right)^4}{\left(3^3\right)^3\cdot\left(2^3\right)^4-3^9\cdot2^{13}}\)
\(=\frac{2^8\cdot3^8\cdot2^4-2^{10}\cdot2^4\cdot3^8}{3^9\cdot2^{12}-3^9\cdot2^{13}}\)
\(=\frac{2^{12}\cdot3^8-2^{14}\cdot3^8}{3^9\left(2^{12}-2^{13}\right)}\)
\(=\frac{3^8\left(2^{12}-2^{14}\right)}{3^9\left(2^{12}-2^{13}\right)}=\frac{3^8\left(2^{12}-2^{14}\right)}{3^8\left(2^{12}-2^{13}\right)\cdot3}=1\)