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Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\Rightarrow\left\{{}\begin{matrix}a=b.k\\c=d.k\end{matrix}\right.\)
Ta có:
\(\dfrac{4.a-5.b}{4.a+5.b}=\dfrac{4.a+5.b-10.b}{4.a+5.b}=1-\dfrac{10.b}{4.a+5.b}=1-\dfrac{10.b}{4.b.k+5b}=1-\dfrac{10}{4.k+5}\) (1)
\(\dfrac{4.c-5.d}{4.c+5.d}=\dfrac{4.c+5.d-10.d}{4.c+5.d}=1-\dfrac{10.d}{4.c+5.d}=1-\dfrac{10.d}{4.d.k+5.d}=1-\dfrac{10}{4.k+5}\) (2)
Từ (1) và (2) suy ra \(\dfrac{4.a-5.b}{4.a+5.b}=\dfrac{4.c-5.d}{4.c+5.d}\left(đpcm\right)\)
Lời giải:
Đặt \(\frac{a}{b}=\frac{c}{d}=t\Rightarrow a=bt; c=dt\)
Khi đó ta có:
\(\frac{4a-5b}{4a+5b}=\frac{4bt-5b}{4bt+5b}=\frac{b(4t-5)}{b(4t+5)}=\frac{4t-5}{4t+5}\)
\(\frac{4c-5d}{4c+5d}=\frac{4dt-5d}{4dt+5d}=\frac{d(4t-5)}{d(4t+5)}=\frac{4t-5}{4t+5}\)
Do đó: \(\frac{4a-5b}{4a+5b}=\frac{4c-5d}{4c+5d}\) (đpcm)
\(\frac{1}{20}\left(x-\frac{8}{15}\right)=-\frac{1}{30}\) \(\left(28+\frac{1}{5}\right).\left(\frac{3}{5}.x+\frac{4}{7}\right)=0\)
\(x-\frac{8}{15}=-\frac{1}{30}:\frac{1}{20}\) \(\frac{141}{5}.\left(\frac{3}{5}.x+\frac{4}{7}\right)=0\)
\(x-\frac{8}{15}=-\frac{2}{3}\) \(\frac{3}{5}.x+\frac{4}{7}=0\)
\(x=-\frac{2}{3}+\frac{8}{15}\) \(\frac{3}{5}.x=-\frac{4}{7}\)
\(x=-\frac{2}{15}\) \(x=-\frac{20}{21}\)
Gọi \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow a=kb;c=kd\)(1)
Thay (1) vào ta có :
\(\frac{5a+3b}{5a-3b}=\frac{5kb+3b}{5kb-3b}=\frac{b\left(5k-3\right)}{b\left(5k-3\right)}=\frac{5k+3}{5k-3}\)(2)
\(\frac{5c+3d}{5c-3d}=\frac{5kd+3d}{5kd-3d}=\frac{d\left(5k+3\right)}{d\left(5k-3\right)}=\frac{5k+3}{5k-3}\)(3)
Từ (2) và (3)
\(\Rightarrow\frac{5a+3b}{5a-3b}=\frac{5c+3d}{5c-3d}\)
\(\RightarrowĐPCM\)
\(A=\left[0,8\cdot7+(0,8)^2\right]\cdot\left[1,25\cdot7-\frac{4}{5}\cdot1,25\right]-47,86\)
\(=0,8\cdot(7+0,8)\cdot1,25\cdot(7-0,8)-47,86\)
\(=0,8\cdot7,8\cdot1,25\cdot6,2-47,86\)
\(=48,36-47,86=0,5\)
\(B=\frac{(1,09-0,29)\cdot\frac{5}{4}}{(18,9-16,65)\cdot\frac{8}{9}}=\frac{0,8\cdot1,25}{2,25\cdot\frac{8}{9}}=\frac{1}{2}\)
\(A:B=0,5:\frac{1}{2}=\frac{1}{2}:\frac{1}{2}=\frac{1}{2}\cdot2=1\)
A gấp 1 lần B