Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow\hept{\begin{cases}a=bk\\c=dk\end{cases}}\)
\(\Rightarrow\frac{2019a^2+2020b^2}{2019a^2-2020b^2}=\frac{2019b^2k^2+2020b^2}{2019b^2k^2-2020b^2}\)
\(=\frac{2019k^2+2020}{2019k^2-2020}\)(1)
và\(\Rightarrow\frac{2019c^2+2020d^2}{2019c^2-2020d^2}=\frac{2019d^2k^2+2020d^2}{2019d^2k^2-2020d^2}\)
\(=\frac{2019k^2+2020}{2019k^2-2020}\)(2)
Từ (1) và (2) suy ra \(\frac{2019a^2+2020b^2}{2019a^2-2020b^2}\)\(=\frac{2019c^2+2020d^2}{2019c^2-2020d^2}\left(đpcm\right)\)
Cho tỉ lệ thức \(\frac{a}{b}=\frac{c}{d}\). Chứng minh rằng \(\frac{2019a^2+2020b^2}{2019a^2-2020b^2}=\frac{2019c^2+2020d^2}{2019c^2-2020d^2}\)
CMR: câu a) 2018a-2019b / 2019c+2020d = 2018c - 2019d / 2019a+2020b
câu b) a^2 + c^2 / b^2 + d^2 = a/bd
cho \(\dfrac{a}{b}=\dfrac{c}{d}\)Chứng minh rằng
\(\dfrac{2018a-2019b}{2019c+2020d}\)=\(\dfrac{2018c-2018c}{2019a+2020b}\)
B12: Cho tỷ lệ thức \(\frac{a}{b}\)= \(\frac{c}{d}\)(b, d khác 0). Chứng minh rằng:
a)\(\frac{2019a+3b}{2019a-5b}\)= \(\frac{2019c+3d}{2019c-5d}\) b) \(\frac{ab}{cd}\)= \(\frac{a^2-4b^2}{c^2-4d^2}\)
\(Cho:\frac{a}{2b}+\frac{b}{2c}+\frac{c}{2d}+\frac{d}{2a}\)\(\left(a,b,c,d>0\right)\)Tính:\(\frac{2019a-2018b}{c+d}+\frac{2019b-2018c}{a+d}+\frac{2019c-2018d}{a+b}+\frac{2019d-2018a}{c+b}\)
Cho a,b,c,d thỏa mãn $\frac{a}{b}$ =$\frac{b}{c}$ =$\frac{c}{d}$ =$\frac{d}{a}$
CMR:($\frac{2019b+2020c-2021d}{2019c+2020d-2021e}$)^3=$\frac{a^2}{bc}$
cho dãy tỉ số bằng nhau
\(\frac{2019a+b+c+d}{a}=\frac{a+2019b+c+d}{b}=\frac{a+b+2019c+d}{c}=\frac{a+b+c+2019d}{d}\)
tính giá trị biểu thức M=\(\frac{a+b}{c+d}+\frac{b+c}{a+d}+\frac{c+d}{a+b}+\frac{d+a}{b+c}\)
Cho dãy tỉ số bằng nhau :
\(\frac{2019a+b+c+d}{a}=\frac{a+2019b+c+d}{b}=\frac{a+b+2019c+d}{c}=\frac{a+b+c+2019d}{d}\)
Tính : \(M=\frac{a+b}{c+d}+\frac{b+c}{d+a}+\frac{c+d}{b+a}+\frac{d+a}{b+c}\)
giúp mình với mình cần gấp
1.Tính:
\(\left(\frac{1}{4\times9}+\frac{1}{9\times14}+\frac{1}{14\times19}+...+\frac{1}{44\times49}\right)\times\frac{1-3-5-7-...-49}{89}\)
2.Cho \(\frac{a}{2b}=\frac{b}{2c}=\frac{c}{2d}=\frac{d}{2a}\). Tính: \(A=\frac{2019a-2018b}{c+d}+\frac{2019b-2018c}{a+d}+\frac{2019c-2018d}{a+b}+\frac{2019d-2018a}{b+c}\)
3.Tìm x biết:\(\left(x-1\right)\left(x-3\right)< 0\)
