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A=-(-2a+3b-5c)+(-5b+4a-2c)-(3c-4b-7a)
=2a-3b+5c-5b+4a-2c-3c+4b+7a
=13a-4b.
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có 2 nghiệm format('truetype')%3Bfont-weight%3Anormal%3Bfont-style%3Anormal%3B%7D%3C%2Fstyle%3E%3C%2Fdefs%3E%3Ctext%20font-family%3D%22Arial%22%20font-size%3D%2224%22%20text-anchor%3D%22middle%22%20x%3D%226.5%22%20y%3D%2222%22%3Ex%3C%2Ftext%3E%3Ctext%20font-family%3D%22Arial%22%20font-size%3D%2212%22%20text-anchor%3D%22middle%22%20x%3D%2215.5%22%20y%3D%2229%22%3E1%3C%2Ftext%3E%3Ctext%20font-family%3D%22math1f7177163c833dff4b38fc8d287%22%20font-size%3D%2216%22%20text-anchor%3D%22middle%22%20x%3D%2221.5%22%20y%3D%2219%22%3E%2C%3C%2Ftext%3E%3Ctext%20font-family%3D%22Arial%22%20font-size%3D%2224%22%20text-anchor%3D%22middle%22%20x%3D%2230.5%22%20y%3D%2222%22%3Ex%3C%2Ftext%3E%3Ctext%20font-family%3D%22Arial%22%20font-size%3D%2212%22%20text-anchor%3D%22middle%22%20x%3D%2239.5%22%20y%3D%2229%22%3E2%3C%2Ftext%3E%3C%2Fsvg%3E){kind=small}
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bằng:
`3.2^x-4^{x-1}-8=0`
`<=>4^{x-1}-3.2^x+8=0`
`<=>1/4*4^x-3.2^x+8=0`
`<=>4^x-12.2^x+32=0`
`<=>(2^x)^2-12.2^x+32=0`
Đặt `t=2^x`
`pt<=>t^2-12t+32=0`
`<=>(t-4)(t-8)=0`
`<=>[(t=4),(t=8):}`
`=>[(x=2),(x=3):}=>|x_1-x_2|=|2-3|=1`
B=(-5c+3a-4b)-(3a-4b+7c)-(-12b-6a+15c)+(-3c+21a-10b)
=-5c+3a-4b-3a+4b-7c+12b+6a-15c
=6a +12b -27c
C=-(-32b-12c+5a)+(2c-4b-23a)-(17a-16c-31b)-(-6b+3c)
=32b+12c-5a+2c-4b-23a-17a+16c+31b+6b-3c
=-45a+65b+9c
Đặt:
\(\dfrac{a}{b}=\dfrac{c}{d}=k\)
\(\Rightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)
\(\Rightarrow\dfrac{2a+5b}{3a-4b}=\dfrac{2bk+5b}{3bk-4b}=\dfrac{b\left(2k+5\right)}{b\left(3k-4\right)}=\dfrac{2k+5}{3k-4}\)
\(\Rightarrow\dfrac{2c+5d}{3c-4d}=\dfrac{2dk+5d}{3dk-4d}=\dfrac{d\left(2k+5\right)}{d\left(3k-4\right)}=\dfrac{2k+5}{3k-4}\)
\(\Rightarrow\dfrac{2a+5b}{3a-4b}=\dfrac{2c+5d}{3c-4d}\)
\(\dfrac{2016a-2017b}{2017c+2018d}=\dfrac{2016bk-2017b}{2017dk+2018d}=\dfrac{b\left(2016k-2017\right)}{d\left(2017k+2018\right)}\)
\(\dfrac{2016c-2017d}{2017a+2018b}=\dfrac{2016dk-2017d}{2017bk+2018b}=\dfrac{d\left(2016k-2017\right)}{b\left(2017k+2018\right)}\)
\(\Rightarrow\dfrac{2016a-2017b}{2017c+2018d}=\dfrac{2016c-2017d}{2017a+2018b}\)
\(\dfrac{7a^2+5ac}{7a^2-5ac}=\dfrac{7bk^2+5bdk^2}{7bk^2-5bdk^2}=\dfrac{k^2\left(7b+5bd\right)}{k^2\left(7b-5bd\right)}=\dfrac{7b+5bd}{7b-5bd}\)
\(\dfrac{7b^2+5ab}{7b^2-5ab}=\dfrac{7b^2+5kb^2}{7b^2-5kb^2}=\dfrac{b^2\left(7+5k\right)}{b^2\left(7-5k\right)}=\dfrac{7+5k}{7-5k}\)
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