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a) \(\dfrac{2a+3c}{2b+3d}\) = \(\dfrac{2a-3c}{2b-3d}\)
Từ \(\dfrac{a}{b}\) = \(\dfrac{c}{d}\) = k ( k \(\in\) Q, k \(\ne\) 0 )
=> \(\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)
VP = \(\dfrac{2a+3c}{2b+3d}\) = \(\dfrac{2.b.k+3.d.k}{2b+3d}\) = \(\dfrac{k.\left(2b+3d\right)}{2b+3d}\) = k (1)
VT = \(\dfrac{2a-3c}{2b-3d}\) = \(\dfrac{2.b.k-3.d.k}{2b-3d}\) = \(\dfrac{k.\left(2b-3d\right)}{2b-3d}\) = k (2)
Từ (1) và (2) ta có: \(\dfrac{2a+3c}{2b+3d}\) = \(\dfrac{2a-3c}{2b-3d}\)
hay: (2a+3c).(3b-3d) = (2a-3c).(2b+3d)
a: H=5|3x-6|+100>=100
Dấu = xảy ra khi x=2
b: Đặt a/b=c/d=k
=>a=bk; c=dk
\(\dfrac{ac}{bd}=\dfrac{bk\cdot dk}{bd}=k^2\)
\(\left(\dfrac{a+2018c}{b+2018d}\right)^2=\left(\dfrac{bk+2018dk}{b+2018d}\right)^2=k^2\)
=>ĐPCM
Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)
Ta có
\(VT:\frac{a^{2018}+c^{2018}}{b^{2018}+d^{2018}}=\frac{b^{2018}\cdot k^{2018}+d^{2018}\cdot k^{2018}}{b^{2018}+d^{2018}}=\frac{k^{2018}\left(b^{2018}+d^{2018}\right)}{b^{2018}+d^{2018}}=k^{2018}\)
\(VP:\frac{\left(a+c\right)^{2018}}{\left(b+d\right)^{2018}}=\frac{\left(bk+dk\right)^{2018}}{\left(b+d\right)^{2018}}=\frac{k^{2018}\cdot\left(b+d\right)^{2018}}{\left(b+d\right)^{2018}}=k^{2018}\)
\(\Rightarrow VT=VP\)
Hay \(\frac{a^{2018}+c^{2018}}{b^{2018}+d^{2018}}=\frac{\left(a+c\right)^{2018}}{\left(b+d\right)^{2018}}\left(đpcm\right)\)
Áp dụng tính chất dãy tỉ số bằng nhau:
\(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{a}=\dfrac{a+b+c}{b+c+a}=1\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=b\\b=c\\c=a\end{matrix}\right.\Leftrightarrow a=b=c\)
\(D=\left(\dfrac{a}{c}\right)^{2016}+\left(\dfrac{c}{b}\right)^{2017}+\left(\dfrac{b}{a}\right)^{2018}=1+1+1=3\)
Áp dụng tính chất dãy tỉ số bằng nhau, ta có:
\(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{a}=\dfrac{a+b+c}{b+c+a}=1\)
\(\Rightarrow\dfrac{b}{a}=\dfrac{c}{b}=\dfrac{a}{c}=1\)
\(\Rightarrow D=\left(\dfrac{a}{b}\right)^{2016}+\left(\dfrac{c}{b}\right)^{2017}+\left(\dfrac{b}{a}\right)^{2018}=1^{2016}+1^{2017}+1^{2018}=1+1+1=3\)
Vậy, D = 3
a: \(=\dfrac{3}{2}\left(-21-\dfrac{1}{3}+1+\dfrac{1}{3}\right)=\dfrac{3}{2}\cdot\left(-20\right)=-30\)
b: \(=\dfrac{2018}{2019}\left(13-13-\dfrac{2018}{2019}-\dfrac{1}{2019}\right)=-\dfrac{2018}{2019}\)
Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\Rightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\) (1)
a) Từ (1) ta có:
\(\dfrac{a}{a+b}=\dfrac{bk}{bk+b}=\dfrac{bk}{b\left(k+1\right)}=\dfrac{k}{k+1}\) (2)
\(\dfrac{c}{c+d}=\dfrac{dk}{dk+d}=\dfrac{dk}{d\left(k+1\right)}=\dfrac{k}{k+1}\) (3)
Từ (2) và (3) suy ra \(\dfrac{a}{a+b}=\dfrac{c}{c+d}\)
b) Từ (1) ta có:
\(\dfrac{a^{2018}+c^{2018}}{b^{2018}+d^{2018}}=\dfrac{b^{2018}.k^{2018}+d^{2018}.k^{2018}}{b^{2018}+d^{2018}}=\dfrac{k^{2018}\left(b^{2018}+d^{2018}\right)}{b^{2018}+d^{2018}}=k^{2018}\) (4)
\(\dfrac{\left(a+c\right)^{2018}}{\left(b+d\right)^{2018}}=\dfrac{\left(bk+dk\right)^{2018}}{\left(b+d\right)^{2018}}=\dfrac{\left[k\left(b+d\right)\right]^{2018}}{\left(b+d\right)^{2018}}=k^{2018}\) (5)
Từ (4) và (5) suy ra \(\dfrac{a^{2018}+c^{2018}}{b^{2018}+d^{2018}}=\dfrac{\left(a+c\right)^{2018}}{\left(b+d\right)^{2018}}\)