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Bài 1:
Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\Rightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)
a, Ta có: \(\dfrac{a+c}{c}=\dfrac{bk+dk}{dk}=\dfrac{\left(b+d\right)k}{dk}=\dfrac{b+d}{d}\)
\(\Rightarrowđpcm\)
b, Ta có: \(\dfrac{a+c}{b+d}=\dfrac{bk+dk}{b+d}=\dfrac{k\left(b+d\right)}{b+d}=k\) (1)
\(\dfrac{a-c}{b-d}=\dfrac{bk-dk}{b-d}=\dfrac{k\left(b-d\right)}{b-d}=k\) (2)
Từ (1), (2) \(\Rightarrowđpcm\)
c, Ta có: \(\dfrac{a-c}{a}=\dfrac{bk-dk}{bk}=\dfrac{k\left(b-d\right)}{bk}=\dfrac{b-d}{b}\)
\(\Rightarrowđpcm\)
d, Ta có: \(\dfrac{3a+5b}{2a-7b}=\dfrac{3bk+5b}{2bk-7b}=\dfrac{b\left(3k+5\right)}{b\left(2k-7\right)}=\dfrac{3k+5}{2k-7}\)(1)
\(\dfrac{3c+5d}{2c-7d}=\dfrac{3dk+5d}{2dk-7d}=\dfrac{d\left(3k+5\right)}{d\left(2k-7\right)}=\dfrac{3k+5}{2k-7}\) (2)
Từ (1), (2) \(\Rightarrowđpcm\)
e, Sai đề
f, \(\left(\dfrac{a-b}{c-d}\right)^{2012}=\left(\dfrac{bk-b}{dk-d}\right)^{2012}=\left[\dfrac{b\left(k-1\right)}{d\left(k-1\right)}\right]^{2012}=\dfrac{b^{2012}}{d^{2012}}\)(1)
\(\dfrac{a^{2012}+b^{2012}}{c^{2012}+d^{2012}}=\dfrac{b^{2012}k^{2012}+b^{2012}}{d^{2012}k^{2012}+d^{2012}}=\dfrac{b^{2012}\left(k^{2012}+1\right)}{d^{2012}\left(k^{2012}+1\right)}=\dfrac{b^{2012}}{d^{2012}}\) (2)
Từ (1), (2) \(\Rightarrowđpcm\)
a: f(0)=5
=>a*0^2+b*0+c=5
=>c=5
f(1)=1
=>a*1+b*1+1=5
=>a+b=4
f(5)=0
=>25a+5b+1=0
=>25a+5b=-1
mà a+b=4
nên a=-21/20; b=101/20
(P): y=-21/20x^2+101/20x+5
b: f(-1)=-21/20-101/20+5=-11/10<>3
=>D ko thuộc (P)
f(1/2)=-21/20*1/4-101/20*1/2+5=177/80<>9/4
=>E ko thuộc (P)
c: y=-3
=>-21/20x^2+101/20x+8=0
=>x=6,06 hoặc x=-1,26
Câu 1:
Ta có: \(\left[\dfrac{1}{2.5}+\dfrac{1}{5.8}+...+\dfrac{1}{65.68}\right]x-\dfrac{7}{34}=\dfrac{19}{68}\)
\(\Rightarrow\left[\dfrac{1}{3}\left(\dfrac{3}{2.5}+\dfrac{3}{5.8}+...+\dfrac{3}{65.68}\right)\right]x=\dfrac{33}{68}\)
\(\Rightarrow\left[\dfrac{1}{3}\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+...+\dfrac{1}{65}-\dfrac{1}{68}\right)\right]x=\dfrac{33}{68}\)
\(\Rightarrow\left[\dfrac{1}{3}\left(\dfrac{1}{2}-\dfrac{1}{68}\right)\right]x=\dfrac{33}{68}\)
\(\Rightarrow\dfrac{11}{68}x=\dfrac{33}{68}\)
\(\Rightarrow x=3\)
Vậy \(x=3.\)
a) \(\dfrac{a}{b}=\dfrac{c}{d}=\dfrac{e}{f}=\dfrac{a+c+e}{b+d+f}\)
b) Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{x}{4}=\dfrac{y}{7}=\dfrac{x+y}{4+7}=\dfrac{55}{11}=5\)
\(\Rightarrow\dfrac{x}{4}=5\Rightarrow x=4\cdot5=20\)
\(\Rightarrow\dfrac{y}{7}=5\Rightarrow x=7\cdot5=35\)
a) 𝑎𝑏=𝑐𝑑=𝑒𝑓=𝑎+𝑐+𝑒𝑏+𝑑+𝑓ba=dc=fe=b+d+fa+c+e.
b) 𝑥4=𝑦7=𝑥+𝑦4+7=5511=54x=7y=4+7x+y=1155=5;
Suy ra 𝑥=4.5=20;𝑦=7.5=35x=4.5=20;y=7.5=35.