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 \(\dfrac{ac}{bd}=\dfrac{a^2+c^2}{b^2+d^2}\)

Bài 1:

Ta có:

+) \(3.4=2.6\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{3}{2}=\dfrac{6}{4}\\\dfrac{3}{6}=\dfrac{2}{4}\\\dfrac{4}{2}=\dfrac{6}{3}\\\dfrac{4}{6}=\dfrac{2}{3}\end{matrix}\right.\)

+) \(3.6=2.9\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{3}{2}=\dfrac{9}{6}\\\dfrac{3}{9}=\dfrac{2}{6}\\\dfrac{6}{2}=\dfrac{9}{3}\\\dfrac{6}{9}=\dfrac{2}{3}\end{matrix}\right.\)

Bài 2:

a) Ta có: \(\dfrac{x}{11}=\dfrac{y}{13}\) và \(x-y=6\)

Áp dụng tính chất của dãy tỉ số bằng nhau ta có:

\(\dfrac{x}{11}=\dfrac{y}{13}=\dfrac{x-y}{11-13}=\dfrac{6}{-2}=-3\)

\(\Rightarrow\left\{{}\begin{matrix}x=11.\left(-3\right)=-33\\y=13.\left(-3\right)=-39\end{matrix}\right.\)

Vậy \(x=-33;y=-39\)

b) Theo bài ra ta có:

\(x:y:z=1:2:3\)

\(\Rightarrow\dfrac{x}{1}=\dfrac{y}{2}=\dfrac{z}{3}\)

\(\Rightarrow\dfrac{4x}{4}=\dfrac{3y}{6}=\dfrac{2z}{6}\)

và \(4x-3y+2z=36\)

Áp dụng tính chất của dãy tỉ số bằng nhau ta có:

\(\dfrac{4x}{4}=\dfrac{3y}{6}=\dfrac{2z}{6}=\dfrac{4x-3y+2z}{4-6+6}=\dfrac{36}{4}=9\)

\(\Rightarrow\left\{{}\begin{matrix}4x=4.9=36\Rightarrow x=9\\3y=6.9=54\Rightarrow y=18\\2z=6.9=54\Rightarrow z=27\end{matrix}\right.\)

Vậy \(x=9;y=18;z=27\)

c) Ta có: \(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{z}{-2}\)

\(\Rightarrow\dfrac{5x}{15}=\dfrac{y}{5}=\dfrac{3z}{-6}\)

và \(5x-y+3z=124\)

Áp dụng tính chất của dãy tỉ số bằng nhau ta có:

\(\dfrac{5x}{15}=\dfrac{y}{5}=\dfrac{3z}{-6}=\dfrac{5x-y+3z}{15-5+\left(-6\right)}=\dfrac{124}{4}=31\)

\(\Rightarrow\left\{{}\begin{matrix}5x=15.31=465\Rightarrow x=93\\y=5.31=155\\3z=\left(-6\right).31=-186\Rightarrow z=-62\end{matrix}\right.\)

Vậy \(x=93;y=155;z=-62\)

Ta có:

\(B=17^7-13^{13}\)

\(\Rightarrow B=17^{16}.17-13^{12}.13\)

\(\Rightarrow B=\left(17^4\right)^4.17-\left(13^4\right)^3.13\)

\(\Rightarrow B=\overline{\left(...1\right)}^4.17-\overline{\left(...1\right)}^3.13\)

\(\Rightarrow B=\overline{\left(...1\right)}.17-\overline{\left(...1\right)}.13\)

\(\Rightarrow B=\overline{\left(...7\right)}-\overline{\left(...3\right)}\)

\(\Rightarrow B=\overline{\left(...4\right)}\)

Vì B có chữ số tận cùng là 4

\(\Rightarrow B⋮2\left(đpcm\right)\)

\(\Rightarrow B\) không chia hết cho 5 (đpcm)

Áp dụng tính chất của dãy tỉ số bằng nhau ta có:

\(\dfrac{y+z+1}{x}=\dfrac{x+z+2}{y}=\dfrac{x+y-3}{z}=\dfrac{1}{x+y+z}=\dfrac{y+z+1+x+z+2+x+y-3}{x+y+z}=\dfrac{2.\left(x+y+z\right)}{x+y+z}=2\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{y+z+1}{x}=2\Rightarrow y+z+1=2x\\\dfrac{x+z+2}{y}=2\Rightarrow x+z+2=2y\\\dfrac{x+y-3}{z}=2\Rightarrow x+y-3=2z\\\dfrac{1}{x+y+z}=2\Rightarrow x+y+z=\dfrac{1}{2}\end{matrix}\right.\)

+) \(x+y+z=\dfrac{1}{2}\Rightarrow y+z=\dfrac{1}{2}-x\). Thay vào \(y+z+1=2x\) ta được \(\dfrac{1}{2}-x+1=2x\Rightarrow3x=\dfrac{3}{2}\Rightarrow x=\dfrac{1}{2}\)

+) \(x+y+z=\dfrac{1}{2}\Rightarrow x+z=\dfrac{1}{2}-y\). Thay vào \(x+z+2=2y\) ta được \(\dfrac{1}{2}-y+2=2y\Rightarrow3y=\dfrac{5}{2}\Rightarrow y=\dfrac{5}{6}\)

\(\Rightarrow x+y+z=\dfrac{1}{2}+\dfrac{5}{6}+z=\dfrac{1}{2}\Rightarrow\dfrac{4}{3}+z=\dfrac{1}{2}\Rightarrow z=\dfrac{1}{2}-\dfrac{4}{3}=\dfrac{-5}{6}\)

Vậy \(x=\dfrac{1}{2};y=\dfrac{5}{6};z=\dfrac{-5}{6}\)

Từ E kẻ tia En đi qua E và \(En\) // \(AB\)

Lại có: AB // CD

\(\Rightarrow En\) // \(CD\)

+) Vì AB // En

mà \(\widehat{ABE}\) và \(\widehat{AEn}\) là 2 góc so le trong

\(\Rightarrow\widehat{ABE}=\widehat{AEn}\)

mà \(\widehat{ABE}=40^0\)

\(\Rightarrow\widehat{AEn}=40^0\)

Lại có:

\(\widehat{AEC}=\widehat{AEn}+\widehat{nEC}\)

\(\Rightarrow\widehat{nEC}=\widehat{AEC}-\widehat{AEn}\)

\(\Rightarrow\widehat{nEC}=60^0-40^0\)

\(\Rightarrow\widehat{nEC}=20^0\)

Vì En // CD

mà \(\widehat{nEC}\) và \(\widehat{ECD}\) là 2 góc so le trong

\(\Rightarrow\widehat{nEC}=\widehat{ECD}\)

mà \(\widehat{nEC}=20^0\)

\(\Rightarrow\widehat{ECD}=20^0\)

Vậy \(\widehat{ECD}=20^0\)

\(\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2014}\right)x=\dfrac{2013}{1}+\dfrac{2012}{2}+...+\dfrac{2}{2012}+\dfrac{1}{2013}\)

\(\Leftrightarrow\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2014}\right)x=\left(1+\dfrac{2012}{2}\right)+\left(1+\dfrac{2011}{3}\right)+...+\left(1+\dfrac{2}{2012}\right)+\left(1+\dfrac{1}{2013}\right)+1\)

\(\Leftrightarrow\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2014}\right)x=\dfrac{2014}{2}+\dfrac{2014}{3}+...+\dfrac{2014}{2012}+\dfrac{2014}{2013}+\dfrac{2014}{2014}\)

\(\Leftrightarrow\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2014}\right)x=2014.\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2012}+\dfrac{1}{2013}+\dfrac{1}{2014}\right)\)

\(\Leftrightarrow x=\dfrac{2014.\left(\dfrac{1}{2}+\dfrac{1}{3}+....+\dfrac{1}{2014}\right)}{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2014}}\)

\(\Leftrightarrow x=2014\)

Vậy \(x=2014\)

Số 36=(2^2).(3^2)

Số này có 9 ước là:1;2;3;4;6;9;12;18;36

Tick mk nha bạn 

\(\left(-4,6\right)x-1,53-5,4x=8,47\)

\(\Rightarrow\left(-4,6\right)x-5,4x-1,53=8,47\)

\(\Rightarrow x\left[\left(-4,6\right)-5,4\right]=8,47+1,53\)

\(\Rightarrow x.\left(-10\right)=10\)

\(\Rightarrow x=10:\left(-10\right)\)

\(\Rightarrow x=-1\)

Vậy \(x=-1\)

a) \(\left(x-2\right)^3=-27\)

\(\Rightarrow\left(x-2\right)^3=\left(-3\right)^3\)

\(\Rightarrow x-2=-3\)

\(\Rightarrow x=-1\)

Vậy \(x=-1\)

b) \(\left(2x+1\right)^4=81\)

\(\Rightarrow\left(2x+1\right)^4=3^4=\left(-3\right)^4\)

\(\left\{{}\begin{matrix}\left(2x+1\right)^4=3^4\Rightarrow2x+1=3\Rightarrow x=1\\\left(2x+1\right)^4=\left(-3\right)^4\Rightarrow2x+1=-3\Rightarrow x=-2\end{matrix}\right.\)

Vậy \(x=1;x=-2\)

c) Bạn xem lại đề bài nhé!

d) \(\left(5x-2\right)^{10}=\left(5x-2\right)^{100}\)

\(\Rightarrow\left(5x-2\right)^{10}-\left(5x-2\right)^{100}=0\)

\(\Rightarrow\left(5x-2\right)^{10}.\left[1-\left(5x-2\right)^{90}\right]=0\)

+) TH1: \(\left(5x-2\right)^{10}=0\)

\(\Rightarrow5x-2=0\)

\(\Rightarrow x=\dfrac{2}{5}\)

+) TH2: \(1-\left(5x-2\right)^{90}=0\)

\(\Rightarrow\left(5x-2\right)^{90}=1\)

\(\Rightarrow\left(5x-2\right)^{90}=1^{90}=\left(-1\right)^{90}\)

\(\Rightarrow\left\{{}\begin{matrix}\left(5x-2\right)^{90}=1^{90}\Rightarrow5x-2=1\Rightarrow x=\dfrac{3}{5}\\\left(5x-2\right)^{90}=\left(-1\right)^{90}\Rightarrow5x-2=-1\Rightarrow x=\dfrac{1}{5}\end{matrix}\right.\)

Vậy \(x\in\left\{\dfrac{1}{5};\dfrac{2}{5};\dfrac{3}{5}\right\}\)

Ta có:

\(\dfrac{a+b}{6}=\dfrac{b+c}{7}=\dfrac{c+a}{8}\)

Áp dụng tính chất của dãy tỉ số bằng nhau ta có:

\(\dfrac{a+b}{6}=\dfrac{b+c}{7}=\dfrac{c+a}{8}=\dfrac{a+b+b+c+c+a}{6+7+8}=\dfrac{2.\left(a+b+c\right)}{21}=\dfrac{2.14}{21}=\dfrac{4}{3}\)

\(\Rightarrow a+b=6.\dfrac{4}{3}=8\) (1)

Lại có: \(a+b+c=14\) (2)

Từ (1) và (2)

\(\Rightarrow8+c=14\)

\(\Rightarrow c=14-8=6\)

Vậy c = 6