Question

[NguyễnMai]

Tìm x , y ,z biết :

a) \(\frac{2}{3}x\)= \(\frac{3}{4}y\)=\(\frac{5}{6}z\)và x² + y² + z² = 724

b) \(\frac{x-1}{2}\)= \(\frac{y+2}{3}\)= \(\frac{z-3}{4}\)và x - 2y + 3z = 46

c) \(\frac{x}{3}=\frac{y}{16}\)và x .y = 192

Bài 2 . Tính

a) \(\frac{4^2.25^2+16.125}{2^3.5^2}\)

b) \(\frac{6^8.2^4-4^5.18^4}{27^3.8^4-3^9.2^{13}}\)

P/S : cần gấp !!!!!!!!    Giải thích chi tiết .

Answer 1

[Huỳnh Quang Sang]

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 \(\left(x,y\right)\in\left\{\left(6,32\right);\left(-6,-32\right)\right\}\)

Answer 2

[Huỳnh Quang Sang]

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) \(\frac{6^8\cdot2^4-4^5\cdot18^4}{27^3\cdot8^4-3^9\cdot2^{13}}\)

\(=\frac{\left(2\cdot3\right)^8\cdot2^4-\left(2^2\right)^5\cdot\left(2\cdot3^2\right)^4}{\left(3^3\right)^3\cdot\left(2^3\right)^4-3^9\cdot2^{13}}\)

\(=\frac{2^8\cdot3^8\cdot2^4-2^{10}\cdot2^4\cdot3^8}{3^9\cdot2^{12}-3^9\cdot2^{13}}\)

\(=\frac{2^{12}\cdot3^8-2^{14}\cdot3^8}{3^9\left(2^{12}-2^{13}\right)}\)

\(=\frac{3^8\left(2^{12}-2^{14}\right)}{3^9\left(2^{12}-2^{13}\right)}=\frac{3^8\left(2^{12}-2^{14}\right)}{3^8\left(2^{12}-2^{13}\right)\cdot3}=1\)