\(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}\Rightarrow\frac{a^2}{c^2}=\frac{b^2}{d^2}\)
\(\Rightarrow\frac{2018a^2}{2018c^2}=\frac{2019b^2}{2019d^2}=\frac{2018a^2+2019b^2}{2018c^2+2019d^2}=\frac{2018a^2-2019b^2}{2018c^2-2019d^2}\)
\(\Rightarrow\frac{2018a^2+2019b^2}{2018a^2-2019b^2}=\frac{2018c^2+2019d^2}{2018c^2-2019d^2}\left(dpcm\right)\)
Cho tỉ lệ thức \(\frac{a}{b}=\frac{c}{d}\) . Chứng minh rằng ta có tỉ lệ thức sau :
\(\frac{2018a^2+2019b^2}{2018a^2-2019b^2}=\frac{2018c^2+2019d^2}{2018c^2-2019d^2}\)
Cho tỉ lệ thức \(\frac{a}{b}=\frac{c}{d}.Chứngminh\frac{2018a-2019b}{2018c+2019d}=\frac{2018c-2019d}{2018a+2019b}\)
cho tỉ lệ thức: \(\frac{a}{b}=\frac{c}{d}\). chúng minh rằng: \(\left(\frac{a+b}{c+d}\right)^2=\frac{2018a^2+2019b^2}{2018c^2+2019d^2}\)
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Cho \(\frac{a}{b}=\frac{c}{d}\).CMR \(\frac{2017-2018b}{2018a+2019b}=\frac{2017c-2018d}{2018c+2019d}\)
\(Cho:\frac{a}{2b}+\frac{b}{2c}+\frac{c}{2d}+\frac{d}{2a}\)\(\left(a,b,c,d>0\right)\)Tính:\(\frac{2019a-2018b}{c+d}+\frac{2019b-2018c}{a+d}+\frac{2019c-2018d}{a+b}+\frac{2019d-2018a}{c+b}\)
CMR: câu a) 2018a-2019b / 2019c+2020d = 2018c - 2019d / 2019a+2020b
câu b) a^2 + c^2 / b^2 + d^2 = a/bd
Cho tỉ lệ thức: \(\frac{a}{b}=\frac{c}{d}\) Chứng minh:
a) \(\frac{a+2019b}{a-2019b}=\frac{c+2019d}{c-2019d}\)
b)\(\frac{2019\left(a+c\right)}{2019a}=\frac{b+d}{b}\)
Cho \(\frac{a}{b}\)= \(\frac{c}{d}\)CMR \(\frac{2017a+2018}{2018a-2019b}\)= \(\frac{2017c+2018d}{2018c-2019d}\)
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:\(\left(x-1\right)\left(x-3\right)< 0\)
