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10 tháng 12 2021 lúc 15:50
cho \(\dfrac{a}{b}=\dfrac{c}{d}\)Chứng minh rằng
\(\dfrac{2018a-2019b}{2019c+2020d}\)=\(\dfrac{2018c-2018c}{2019a+2020b}\)
\(Cho:\frac{a}{2b}+\frac{b}{2c}+\frac{c}{2d}+\frac{d}{2a}\)\(\left(a,b,c,d>0\right)\)Tính:\(\frac{2019a-2018b}{c+d}+\frac{2019b-2018c}{a+d}+\frac{2019c-2018d}{a+b}+\frac{2019d-2018a}{c+b}\)
\(Cho\frac{a}{b}=\frac{c}{d}.CMR:\frac{2019a^2+2020b^2}{2019a^2-2020b^2}=\frac{2019c^2+2020d^2}{2019c^2-2019d^2}\)
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#Duongg
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 \(\frac{a}{b}=\frac{c}{d}.Chứngminh\frac{2018a-2019b}{2018c+2019d}=\frac{2018c-2019d}{2018a+2019b}\)
1.Tính:left(frac{1}{4times9}+frac{1}{9times14}+frac{1}{14times19}+...+frac{1}{44times49}right)timesfrac{1-3-5-7-...-49}{89}2.Cho frac{a}{2b}frac{b}{2c}frac{c}{2d}frac{d}{2a}. Tính: Afrac{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-1right)left(x-3right) 0
Đọc tiếp
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:\(\left(x-1\right)\left(x-3\right)< 0\)
Cho \(\frac{a}{b}=\frac{c}{d}\).CMR \(\frac{2017-2018b}{2018a+2019b}=\frac{2017c-2018d}{2018c+2019d}\)
Cho tỉ lệ thức \(\frac{a}{b}=\frac{c}{d}\) . Chứng minh rằng ta co tỉ thức sau :
\(\frac{2018a^{2\:}+2019b^2}{2018b^2-2019b^2}=\frac{2018c^2+2019d^2}{2018c^2-2019d^2}\)
Đặt bằng k nhé các bạn , giúp mình nhanh lên ạ
Nhanh lên ạ
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=$\frac{a^2}{bc}$