Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(P=\dfrac{14^5.9^4-6^9.49^2}{2^{10}.49^3.3^8+6^8.7^5.13}\)
\(=\dfrac{2^5.7^5.3^8-2^9.3^9.7^4}{2^{10}.7^6.3^8+2^8.3^8.7^5.13}\)
\(=\dfrac{2^5.7^4.3^8\left(7-2^4.3\right)}{2^8.3^8.7^5\left(2^2.7+13\right)}\)
\(=\dfrac{-41}{2^3.7.41}\)
\(=\dfrac{-1}{56}\)
Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\) \(\Rightarrow\) \(\begin{cases} a = bk \\ c = dk \end{cases}\)
Ta có: \(\dfrac{a^2+c^2}{b^2+d^2}=\dfrac{b^2k^2+d^2k^2}{b^2+d^2}=\dfrac{k^2\left(b^2+d^2\right)}{b^2+d^2}=k^2\left(1\right)\)
\(\dfrac{a.c}{b.d}=\dfrac{bk.dk}{b.d}=\dfrac{k^2.b.d}{b.d}=k^2\left(2\right)\)
Từ (1) và (2) suy ra: \(\dfrac{a.c}{b.d}=\dfrac{a^2+c^2}{b^2+d^2}\) \(\rightarrow đpcm\).
\(P=\dfrac{2^5\cdot7^5\cdot3^8-2^9\cdot3^9\cdot7^4}{2^{10}\cdot7^6\cdot3^8+2^8\cdot3^8\cdot7^5\cdot13}\)
\(=\dfrac{2^5\cdot7^4\cdot3^8\left(7-2^4\cdot3\right)}{2^8\cdot3^8\cdot7^5\cdot\left(2^2\cdot7+13\right)}\)
\(=\dfrac{1}{8}\cdot\dfrac{1}{7}\cdot\dfrac{7-16\cdot3}{4\cdot7+13}=\dfrac{1}{56}\cdot\left(-1\right)=-\dfrac{1}{56}\)
>> Mình không chép lại đề bài nhé ! <<
Cách 1 :
\(A=\left(\dfrac{36-4+3}{6}\right)-\left(\dfrac{30+10-9}{6}\right)-\left(\dfrac{18-14+15}{6}\right)=\dfrac{35}{6}-\dfrac{31}{6}-\dfrac{19}{6}=-\dfrac{15}{6}=-\dfrac{5}{2}\)
Cách 2 :
\(A=6-\dfrac{2}{3}+\dfrac{1}{2}-5+\dfrac{5}{3}-\dfrac{3}{2}-3-\dfrac{7}{3}+\dfrac{5}{2}\)
\(A=\left(6-5-3\right)-\left(\dfrac{2}{3}+\dfrac{5}{3}-\dfrac{7}{3}\right)+\left(\dfrac{1}{2}+\dfrac{3}{2}-\dfrac{5}{2}\right)\)
\(A=-2-0-\dfrac{1}{2}=-\dfrac{5}{2}\)
Cách 1 :
\(\left(6-\dfrac{2}{3}+\dfrac{1}{2}\right)-\left(5+\dfrac{5}{3}-\dfrac{3}{2}\right)-\left(3-\dfrac{7}{3}+\dfrac{5}{2}\right)\)
\(=\left(\dfrac{36}{6}-\dfrac{4}{6}+\dfrac{3}{6}\right)-\left(\dfrac{30}{6}+\dfrac{10}{6}-\dfrac{9}{6}\right)-\left(\dfrac{18}{6}-\dfrac{14}{6}+\dfrac{15}{6}\right)\)
\(=\dfrac{35}{6}-\dfrac{31}{6}-\dfrac{19}{6}\)
\(=-\dfrac{5}{2}\)
Cách 2 :
\(\left(6-\dfrac{2}{3}+\dfrac{1}{2}\right)-\left(5+\dfrac{5}{3}-\dfrac{3}{2}\right)-\left(3-\dfrac{7}{3}+\dfrac{5}{2}\right)\)
\(=6-\dfrac{2}{3}+\dfrac{1}{2}-5-\dfrac{5}{3}+\dfrac{3}{2}-3+\dfrac{7}{3}-\dfrac{5}{2}\)
\(=\left(6-5-3\right)+\left(\dfrac{-2}{3}+\dfrac{-5}{3}+\dfrac{7}{3}\right)+\left(\dfrac{1}{2}+\dfrac{3}{2}+\dfrac{-5}{2}\right)\)
\(=\left(-2\right)+0+\dfrac{-1}{2}\)
\(=\dfrac{-5}{2}\)
Đề sai bạn nhé. Đưa dữ kiện 3 ẩn bắt tính biểu thức chứa 2 ẩn làm sao làm được ?
Bạn kiểm tra lại nha
\(\left(\dfrac{-5}{13}\right)^{2017}\cdot\left(\dfrac{13}{5}\right)^{2016}=\left(\dfrac{-5}{13}\right)\cdot\left(-\dfrac{5}{13}\right)^{2016}\cdot\left(\dfrac{13}{5}\right)^{2016}=\left(\dfrac{-5}{13}\right)\cdot\left(\dfrac{5}{13}\right)^{2016}\cdot\left(\dfrac{13}{5}\right)^{2016}=\left(-\dfrac{5}{13}\right)\cdot\left[\left(\dfrac{5}{13}\right)^{2016}\cdot\left(\dfrac{13}{5}\right)^{2016}\right]=\left(-\dfrac{5}{13}\right)\cdot1^{2016}=\left(-\dfrac{5}{13}\right)\cdot1=-\dfrac{5}{13}\)
Bài 2:
a) Ta có : Từ \(\dfrac{a}{b}=\dfrac{c}{d}\Rightarrow\dfrac{a}{c}=\dfrac{b}{d}\)
\(\Rightarrow\dfrac{5a}{5c}=\dfrac{7b}{7d}\)
Theo tính chất dãy tỉ số bằng nhau, ta có :
\(\dfrac{5a}{5c}=\dfrac{7b}{7d}=\dfrac{5a+7b}{5c+7d}\left(1\right)\)
Và \(\dfrac{5a}{5c}=\dfrac{7b}{7d}=\dfrac{5a-7b}{5c-7d}\left(2\right)\)
Từ (1) và (2)=> \(\dfrac{5a+7b}{5c+7d}=\dfrac{5a-7b}{5c-7d}\Rightarrow\dfrac{5a+7b}{5a-7b}=\dfrac{5c+7d}{5c-7d}\)Vậy...
b) Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\Rightarrow a=bk;c=dk\)
Thay các đẳng thức vừa tìm được , ta có :
\(\dfrac{ac}{bd}=\dfrac{bk.dk}{bd}=k^2\left(1\right)\)
\(\dfrac{a^2+c^2}{b^2+d^2}=\dfrac{\left(bk\right)^2+\left(dk\right)^2}{b^2+d^2}=\dfrac{b^2k^2+d^2k^2}{b^2+d^2}\)
\(=\dfrac{k^2\left(b^2+d^2\right)}{b^2+d^2}=k^2\left(2\right)\)
từ (1) và (2)=> đpcm
tik mik nha !!!
1. Bạn xem lại đề bài nhé! Mình nghĩ là \(2x=3y=5z\) thì đúng hơn!
2.
a) Ta có: \(\dfrac{a}{b}=\dfrac{c}{d}\)
\(\Rightarrow\dfrac{a}{c}=\dfrac{b}{d}\)
\(\Rightarrow\dfrac{5a}{5c}=\dfrac{7b}{7d}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{5a}{5c}=\dfrac{7b}{7d}=\dfrac{5a+7b}{5c+7d}=\dfrac{5a-7b}{5c-7d}\)
Từ \(\dfrac{5a+7b}{5c+7d}=\dfrac{5a-7b}{5c-7d}\Rightarrow\dfrac{5a+7b}{5a-7b}=\dfrac{5c+7d}{5c-7d}\)(đpcm)
Vậy \(\dfrac{5a+7b}{5a-7b}=\dfrac{5c+7d}{5c-7d}\)
b) Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)
\(\Rightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)
Ta có:
\(VT=\dfrac{ac}{bd}=\dfrac{bk.dk}{bd}=\dfrac{bd.k^2}{bd}=k^2\left(1\right)\)
\(VP=\dfrac{a^2+c^2}{b^2+d^2}=\dfrac{\left(bk\right)^2+\left(dk\right)^2}{b^2+d^2}=\dfrac{b^2.k^2+d^2.k^2}{b^2+d^2}=\dfrac{k^2.\left(b^2+d^2\right)}{b^2+d^2}=k^2\left(2\right)\)
Từ \(\left(1\right)\) và \(\left(2\right)\)
\(\Rightarrow\dfrac{ac}{bd}=\dfrac{a^2+c^2}{b^2+d^2}\left(đpcm\right)\)
Vậy \(\dfrac{ac}{bd}=\dfrac{a^2+c^2}{b^2+d^2}\)