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\(\frac{2016c-2017b}{2015}=\frac{2017a-2015c}{2016}=\frac{2015b-2016a}{2017}\)
\(\Rightarrow\frac{2016c.2015-2017b.2015}{2015^2}=\frac{2017a.2016-2015c.2016}{2016^2}=\frac{2017.2015b-2017.2016a}{2017^2}\)
\(=\frac{2016c.2015-2017b.2015+2017a.2016-2015a.2016+2017.2015b-2017.2016a}{2015^2+2016^2+2017^2}=0\)
Do đó: \(2016c.2015-2017b.2015=0\Rightarrow2016c=2017b\Rightarrow\frac{b}{2016}=\frac{c}{2017}\)
\(2017a.2016-2015c.2016=0\Rightarrow2017a=2015c\Rightarrow\frac{a}{2015}=\frac{c}{2017}\)
Vậy \(\frac{a}{2015}=\frac{b}{2016}=\frac{c}{2017}\)
Đặt:
\(\dfrac{a}{b}=\dfrac{c}{d}=k\)
\(\Rightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)
\(\Rightarrow\dfrac{2a+5b}{3a-4b}=\dfrac{2bk+5b}{3bk-4b}=\dfrac{b\left(2k+5\right)}{b\left(3k-4\right)}=\dfrac{2k+5}{3k-4}\)
\(\Rightarrow\dfrac{2c+5d}{3c-4d}=\dfrac{2dk+5d}{3dk-4d}=\dfrac{d\left(2k+5\right)}{d\left(3k-4\right)}=\dfrac{2k+5}{3k-4}\)
\(\Rightarrow\dfrac{2a+5b}{3a-4b}=\dfrac{2c+5d}{3c-4d}\)
\(\dfrac{2016a-2017b}{2017c+2018d}=\dfrac{2016bk-2017b}{2017dk+2018d}=\dfrac{b\left(2016k-2017\right)}{d\left(2017k+2018\right)}\)
\(\dfrac{2016c-2017d}{2017a+2018b}=\dfrac{2016dk-2017d}{2017bk+2018b}=\dfrac{d\left(2016k-2017\right)}{b\left(2017k+2018\right)}\)
\(\Rightarrow\dfrac{2016a-2017b}{2017c+2018d}=\dfrac{2016c-2017d}{2017a+2018b}\)
\(\dfrac{7a^2+5ac}{7a^2-5ac}=\dfrac{7bk^2+5bdk^2}{7bk^2-5bdk^2}=\dfrac{k^2\left(7b+5bd\right)}{k^2\left(7b-5bd\right)}=\dfrac{7b+5bd}{7b-5bd}\)
\(\dfrac{7b^2+5ab}{7b^2-5ab}=\dfrac{7b^2+5kb^2}{7b^2-5kb^2}=\dfrac{b^2\left(7+5k\right)}{b^2\left(7-5k\right)}=\dfrac{7+5k}{7-5k}\)
Hình như sai sai
Ta có: \(\dfrac{a}{b}=\dfrac{c}{d}\Rightarrow\dfrac{a}{c}=\dfrac{b}{d}\)
Đặt \(\dfrac{a}{c}=\dfrac{b}{d}=k\)
\(\Rightarrow\left\{{}\begin{matrix}a=ck\\b=dk\end{matrix}\right.\)
\(VT=\dfrac{7a^2+5ac}{7a^2-5ac}=\dfrac{a\left(7a+5c\right)}{a\left(7a-5c\right)}=\dfrac{7ck+5c}{7ck-5c}=\dfrac{c\left(7k+5\right)}{c\left(7k-5\right)}=\dfrac{7k+5}{7k-5}\left(1\right)\)
\(VP=\dfrac{7b^2+5bd}{7b^2-5bd}=\dfrac{b\left(7b+5d\right)}{b\left(7b-5d\right)}=\dfrac{7dk+5d}{7dk-5d}=\dfrac{d\left(7k+5\right)}{d\left(7k-5\right)}=\dfrac{7k+5}{7k-5}\left(2\right)\)
Từ \(\left(1\right)\)và \(\left(2\right)\)
\(\Rightarrow\dfrac{7a^2+5ac}{7a^2-5ac}=\dfrac{7b^2+5bd}{7b^2-5bd}\left(đpcm\right)\)
Vậy \(\dfrac{7a^2+5ac}{7a^2-5ac}=\dfrac{7b^2+5bd}{7b^2-5bd}\)
\(\dfrac{2017}{1}+\dfrac{2016}{2}+...+\dfrac{2}{2016}+\dfrac{1}{2017}\)
\(=\left(\dfrac{2016}{2}+1\right)+\left(\dfrac{2015}{3}+1\right)+...+\left(\dfrac{2}{2016}+1\right)+\left(\dfrac{1}{2017}+1\right)+1\)
\(=\dfrac{2018}{2}+\dfrac{2018}{3}+...+\dfrac{2018}{2017}+\dfrac{2018}{2018}\)
\(=2018\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2018}\right)\)
Theo đề, ta có: \(x=\dfrac{2018\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2018}\right)}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2018}}=2018\)
a)\(\dfrac{x+1}{10}+\dfrac{x+1}{11}+\dfrac{x+1}{12}=\dfrac{x+1}{13}+\dfrac{x+1}{14}\)
\(\Rightarrow\left(x+1\right)\left(\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}\right)=\left(x+1\right)\left(\dfrac{1}{13}+\dfrac{1}{14}\right)\)
\(\Rightarrow\left(x+1\right)\left(\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}-\dfrac{1}{13}-\dfrac{1}{14}\right)=0\)
\(\Rightarrow x+1=0\)
\(\Rightarrow x=-1\)
b)\(\dfrac{x+4}{2014}+\dfrac{x+3}{2015}=\dfrac{x+2}{2016}+\dfrac{x+1}{2017}\)
\(1+\dfrac{x+4}{2014}+1+\dfrac{x+3}{2015}=1+\dfrac{x+2}{2016}+1+\dfrac{x+1}{2017}\)
\(\Rightarrow\dfrac{x+2018}{2014}+\dfrac{x+2018}{2015}=\dfrac{x+2018}{2016}+\dfrac{x+2018}{2017}\)
Giải tương tự câu a ta được \(x=-2018\)
a) \(\dfrac{x+1}{10}+\dfrac{x+1}{11}+\dfrac{x+1}{12}=\dfrac{x+1}{13}+\dfrac{x+1}{14}\)
\(\Rightarrow6006\left(x+1\right)+5460\left(x+1\right)+5005\left(x+1\right)=4620\left(x+1\right)+4290\left(x+1\right)\)
\(\Leftrightarrow\left(6006+5460+5005\right)\cdot\left(x+1\right)=\left(4620+4290\right)\cdot\left(x+1\right)\)
\(\Leftrightarrow16471\left(x+1\right)=8910\left(x+1\right)\)
\(\Leftrightarrow16471x+16471=8910x+8910\)
\(\Leftrightarrow16471x-8910x=8910-16471\)
\(\Leftrightarrow7561x=-7561\)
\(\Rightarrow x=-1\)
Vậy \(x=-1\)
b) \(\dfrac{x+4}{2014}+\dfrac{x+3}{2015}=\dfrac{x+2}{2016}+\dfrac{x+1}{2017}\)
\(\Rightarrow4096749040\left(x+4\right)+4094735904\left(x+3\right)=4092704785\left(x+2\right)+4090675680\left(x+1\right)\)
\(\Leftrightarrow4096769040x+16387076160+4094735904x+12284207712=4092704785x+8185409570+4090675680x+4090675680\)
\(\Leftrightarrow8191504944x+28671283872=8183380465x+12276085250\)
\(\Leftrightarrow8191504944x-8183380465x=12276085250-28671283872\)
\(\Leftrightarrow8124479x=-16395198622\)
\(\Rightarrow x=-2018\)
Vậy \(x=-2017\)
P/s: đây không phải cách làm tối ưu, vì vậy mình nghĩ bạn nên tham khảo từ các bài làm khác nhé!
\(\Leftrightarrow\left(\dfrac{x+4}{2015}+1\right)+\left(\dfrac{x+3}{2016}+1\right)=\left(\dfrac{x+2}{2017}+1\right)+\left(\dfrac{x+1}{2018}+1\right)\)
=>x+2019=0
=>x=-2019
b) Ta có: [tex]\frac{a^{2} + c^{2}}{b^{2} + a^{2}}[/tex]= [tex]\frac{bc + c^{2}}{b^{2} + bc}= \frac{c(b +c)}{b(b + c)}= \frac{c}{b}[/tex] (đpcm)
Lời giải:
Ta có \(\frac{2016c-2017b}{2015}=\frac{2017a-2015c}{2016}=\frac{2015b-2016a}{2017}\)
\(\Rightarrow \frac{2015.2016c-2015.2017b}{2015^2}=\frac{2016.2017a-2016.2015c}{2016^2}=\frac{2017.2015b-2017.2016a}{2017^2}\)
Áp dụng tính chất dãy tỉ số bằng nhau:
\( \frac{2015.2016c-2015.2017b}{2015^2}=\frac{2016.2017a-2016.2015c}{2016^2}=\frac{2017.2015b-2017.2016a}{2017^2}\)
\(=\frac{2015.2016c-2015.2017b+2016.2017a-2016.2015c+2017.2015b-2017.2016a}{2015^2+2016^2+2017^2}=0\)
\(\Rightarrow \left\{\begin{matrix} 2015.2016c-2015.2017b=0\\ 2016.2017a-2016.2015c=0\\ 2017.2015b-2016.2016a=0\end{matrix}\right.\)
\(\Rightarrow \left\{\begin{matrix} 2016c=2017b\\ 2017a=2015c\\ 2015b=2016a\end{matrix}\right.\Rightarrow \frac{a}{2015}=\frac{b}{2016}=\frac{c}{2017}\)
Ta có đpcm.