Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Câu 2 :
\(x-y=7\)
\(\Rightarrow x=7+y\)
*)
\(B=\dfrac{3\left(7+y\right)-7}{2\left(7+y\right)+y}-\dfrac{3y+7}{2y+7+y}\)
\(=\dfrac{21+3y-7}{14+3y}-\dfrac{3y+7}{3y+7}\)
\(=\dfrac{14y+3y}{14y+3y}-1\)
\(=1-1\)
\(=0\)
Vậy B = 0
2/ Ta có :
\(B=\dfrac{3x-7}{2x+y}-\dfrac{3y+7}{2y+x}\)
\(=\dfrac{3x-\left(x-y\right)}{2x+y}-\dfrac{3y+\left(x-y\right)}{2y+x}\)
\(=\dfrac{3x-x+y}{2y+x}-\dfrac{3y+x-y}{2y+x}\)
\(=\dfrac{2x+y}{2x+y}-\dfrac{2y+x}{2y+x}\)
\(=1-1=0\)
Câu 2:
Theo đề, ta có: \(\dfrac{10a+b}{a+b}=\dfrac{10b+c}{b+c}\)
=>10ab+10ac+b^2+bc=10ab+10b^2+ac+bc
=>9ac-9b^2=0
=>ac-b^2=0
=>ac=b^2
=>a/b=b/c
1)\(\dfrac{x+1}{-12}=\dfrac{-3}{x+1}\)
\(\Rightarrow\left(x+1\right)^2=36\)
\(\Rightarrow\left[{}\begin{matrix}x+1=6\\x+1=-6\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=5\\x=-7\end{matrix}\right.\)
Vậy....
b)\(\left(\dfrac{1}{2}-2^2:\dfrac{4}{3}\right).\dfrac{6}{5}-7\)
\(=\left(\dfrac{1}{2}-4.\dfrac{3}{4}\right).\dfrac{6}{5}-7\)
\(=\left(\dfrac{1}{2}-3\right).\dfrac{6}{5}-7\)
\(=\dfrac{-5}{2}.\dfrac{6}{5}-7\)
\(=-3-7\)
\(=-10\)
Câu 1:
1/ Tìm x:(mk nghĩ là z)
\(\dfrac{x+1}{-12}=\dfrac{-3}{x+1}\Rightarrow\left(x+1\right)^2=\left(-3\right).\left(-12\right)=36\)
\(\Rightarrow x+1=6;x+1=-6\)
+) \(x+1=6\Rightarrow x=5\)
+) \(x+1=-6\Rightarrow x=-7\)
2/Tính:
\(\left(\dfrac{1}{2}-2^2:\dfrac{4}{3}\right).\dfrac{6}{5}-7=\left(\dfrac{1}{2}-\dfrac{4.3}{4}\right).\dfrac{6}{5}-7\)
\(=\left(\dfrac{1}{2}-3\right).\dfrac{6}{5}-7=\left(\dfrac{1}{2}.\dfrac{6}{5}\right)-\left(3.\dfrac{6}{5}\right)-7\)
\(=0,6-3,6-7=-10\)
Bài 1:
Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\Rightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)
a, Ta có: \(\dfrac{a+c}{c}=\dfrac{bk+dk}{dk}=\dfrac{\left(b+d\right)k}{dk}=\dfrac{b+d}{d}\)
\(\Rightarrowđpcm\)
b, Ta có: \(\dfrac{a+c}{b+d}=\dfrac{bk+dk}{b+d}=\dfrac{k\left(b+d\right)}{b+d}=k\) (1)
\(\dfrac{a-c}{b-d}=\dfrac{bk-dk}{b-d}=\dfrac{k\left(b-d\right)}{b-d}=k\) (2)
Từ (1), (2) \(\Rightarrowđpcm\)
c, Ta có: \(\dfrac{a-c}{a}=\dfrac{bk-dk}{bk}=\dfrac{k\left(b-d\right)}{bk}=\dfrac{b-d}{b}\)
\(\Rightarrowđpcm\)
d, Ta có: \(\dfrac{3a+5b}{2a-7b}=\dfrac{3bk+5b}{2bk-7b}=\dfrac{b\left(3k+5\right)}{b\left(2k-7\right)}=\dfrac{3k+5}{2k-7}\)(1)
\(\dfrac{3c+5d}{2c-7d}=\dfrac{3dk+5d}{2dk-7d}=\dfrac{d\left(3k+5\right)}{d\left(2k-7\right)}=\dfrac{3k+5}{2k-7}\) (2)
Từ (1), (2) \(\Rightarrowđpcm\)
e, Sai đề
f, \(\left(\dfrac{a-b}{c-d}\right)^{2012}=\left(\dfrac{bk-b}{dk-d}\right)^{2012}=\left[\dfrac{b\left(k-1\right)}{d\left(k-1\right)}\right]^{2012}=\dfrac{b^{2012}}{d^{2012}}\)(1)
\(\dfrac{a^{2012}+b^{2012}}{c^{2012}+d^{2012}}=\dfrac{b^{2012}k^{2012}+b^{2012}}{d^{2012}k^{2012}+d^{2012}}=\dfrac{b^{2012}\left(k^{2012}+1\right)}{d^{2012}\left(k^{2012}+1\right)}=\dfrac{b^{2012}}{d^{2012}}\) (2)
Từ (1), (2) \(\Rightarrowđpcm\)
4.a
\(\dfrac{3x-y}{x+y}=\dfrac{3}{4}\\ \Leftrightarrow\left(3x-y\right).4=3\left(x+y\right)\\ \Rightarrow12x-4y=3x+3y\\ \Rightarrow12x-3x=4y+3y\\ \Rightarrow9x=7y\\ \Rightarrow\dfrac{x}{y}=\dfrac{7}{9}\)
Bài 1:
Giải:
Ta có: \(\dfrac{4x}{6y}=\dfrac{2x+8}{3y+11}\)
\(\Rightarrow\dfrac{2x}{3y}=\dfrac{2x+8}{3y+11}\)
\(\Rightarrow\left(3y+11\right)2x=\left(2x+8\right)3y\)
\(\Rightarrow6xy+22x=6xy+24y\)
\(\Rightarrow22x=24y\)
\(\Rightarrow\dfrac{x}{y}=\dfrac{24}{22}\)
\(\Rightarrow\dfrac{x}{y}=\dfrac{12}{11}\)
Vậy \(\dfrac{x}{y}=\dfrac{12}{11}.\)
Câu 4:
Giải:
Gọi số h/s lớp 7A, 7B lần lượt là a,b (a,b \(\in N\)*)
Theo bài ra ta có: \(a+b=65\) và \(\dfrac{a}{6}=\dfrac{b}{7}\)
Áp dụng t/c dãy tỉ số bằng nhau ta có:
\(\dfrac{a}{6}=\dfrac{b}{7}=\dfrac{a+b}{6+7}=\dfrac{65}{13}=5\)
Khi đó \(\left[{}\begin{matrix}\dfrac{a}{6}=5\\\dfrac{b}{7}=5\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}a=30\\b=35\end{matrix}\right.\)
Vậy số h/s lớp \(\left[{}\begin{matrix}7A:30\\7B:35\end{matrix}\right.\).
Từ đề bài:A=\(\dfrac{bc}{a}+\dfrac{ac}{b}+\dfrac{ab}{c}=\dfrac{abc}{a^2}+\dfrac{abc}{b^2}+\dfrac{abc}{c^2}=abc\left(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\right)=8\cdot\dfrac{3}{4}=6\)
\(A=\dfrac{bc}{a}+\dfrac{ac}{b}+\dfrac{ab}{c}\)
\(=\dfrac{abc}{a^2}+\dfrac{abc}{b^2}+\dfrac{abc}{c^2}\\ =abc\left(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\right)\\ =8\cdot\dfrac{3}{4}\\ =6\)