cho \(\dfrac{a}{b}=\dfrac{c}{d}\)Chứng minh rằng
\(\dfrac{2018a-2019b}{2019c+2020d}\)=\(\dfrac{2018c-2018c}{2019a+2020b}\)
CMR: câu a) 2018a-2019b / 2019c+2020d = 2018c - 2019d / 2019a+2020b
câu b) a^2 + c^2 / b^2 + d^2 = a/bd
Cho các số thực a,b,c,d,e thỏa mãn \(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{d}{e}\)chứng minh rằng: \(\left(\dfrac{2019b+2020c-2021d}{2019c+2020d-2021e}\right)=\dfrac{a^2}{b.c}\)
Sửa: CMR: \(\left(\dfrac{2019b+2020c-2021d}{2019c+2020d-2021e}\right)^3=\dfrac{a^2}{bc}\)
\(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{d}{e}=\dfrac{2019b+2020c-2021d}{2019c+2020d-2021e}\\ \Rightarrow\left(\dfrac{a}{b}\right)^3=\left(\dfrac{2019b+2020c-2021d}{2019c+2020d-2021e}\right)^3\left(1\right)\\ \dfrac{a}{b}=\dfrac{b}{c}=k\Rightarrow a=bk;b=ck\Rightarrow a=ck^2\\ \Rightarrow\dfrac{a^2}{bc}=\dfrac{c^2k^4}{ck\cdot c}=k^3=\left(\dfrac{a}{b}\right)^3\left(2\right)\\ \left(1\right)\left(2\right)\RightarrowĐpcm\)
Cho tỉ lệ thức \(\frac{a}{b}=\frac{c}{d}\). Chứng minh rằng \(\frac{2019a^2+2020b^2}{2019a^2-2020b^2}=\frac{2019c^2+2020d^2}{2019c^2-2020d^2}\)
\(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}\Rightarrow\frac{a^2}{c^2}=\frac{b^2}{d^2}\Rightarrow\frac{2019a^2}{2019c^2}=\frac{2020b^2}{2020d^2}=\)
\(=\frac{2019a^2+2020b^2}{2019c^2+2020d^2}=\frac{2019a^2-2020b^2}{2019c^2-2020d^2}\Rightarrow\frac{2019a^2+2020b^2}{2019a^2-2020b^2}=\frac{2019c^2+2020d^2}{2019c^2-2020d^2}\)
Bạn ơi tham khảo thử cách này nhé !
Từ \(\frac{a}{b}=\frac{c}{d}\)( bài cho )
Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow\hept{\begin{cases}a=bk\\c=dk\end{cases}}\)
Khi đó :
+) \(\frac{2019a^2+2020b^2}{2019a^2-2020b^2}=\frac{2019\left(bk\right)^2+2020b^2}{2019\left(bk\right)^2-2020b^2}=\frac{b^2\left(2019k^2+2020\right)}{b^2\left(2019k^2-2020\right)}=\frac{2019k^2+2020}{2019k^2-2020}\)
+) \(\frac{2019c^2+2020d^2}{2019c^2-2020d^2}=\frac{2019\left(dk\right)^2+2020d^2}{2019\left(dk\right)^2-2020d^2}=\frac{d^2\left(2019k^2+2020\right)}{d^2\left(2019k^2-2020\right)}=\frac{2019k^2+2020}{2019k^2-2020}\)
Cho \(\dfrac{a}{b}=\dfrac{c}{d}\)
cm:\(\dfrac{2018a-2018c}{2018b-2018c}=\dfrac{2017a+2017c}{2017b+2017d}\)
Ta có:
a/b = c/d => 2018a/2018b = 2018c/2018d = 2018a - 2018c / 2018b- 2018d
a/b = c/d => 2017a/2017b = 2017c/2017d =2017a+ 2017c/ 2017b+ 2017d
=> 2018a-2018c/2018b-2018d = 2017a+2017c/2017b+2017d (=a/b=c/d)
Cho các số thực thỏa mãn\(\dfrac{â}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{d}{e}chứngminh:\left(\dfrac{2019b+2020c-2021d}{2019c+2020d-2021e}\right)^3=\dfrac{a^2}{bc}\)
\(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{d}{e}=\dfrac{2019b+2020c-2021d}{2019c+2020d-2021e}\left(1\right)\\ \text{Đặt }\dfrac{a}{b}=\dfrac{b}{c}=k\Leftrightarrow a=bk;b=ck\Leftrightarrow a=ck^2\\ \Leftrightarrow\dfrac{a^2}{bc}=\dfrac{c^2k^4}{c^2k}=k^3=\left(\dfrac{a}{b}\right)^3\left(2\right)\\ \left(1\right)\left(2\right)\Leftrightarrow\left(\dfrac{2019b+2020c-2021d}{2019c+2020d-2021e}\right)^3=\dfrac{a^2}{bc}\)
\(Cho:\frac{a}{2b}+\frac{b}{2c}+\frac{c}{2d}+\frac{d}{2a}\)\(\left(a,b,c,d>0\right)\)Tính:\(\frac{2019a-2018b}{c+d}+\frac{2019b-2018c}{a+d}+\frac{2019c-2018d}{a+b}+\frac{2019d-2018a}{c+b}\)
Cho tỉ lệ thức \(\frac{a}{b}=\frac{c}{d}\) . Chứng minh rằng ta có tỉ lệ thức sau :
\(\frac{2018a^2+2019b^2}{2018a^2-2019b^2}=\frac{2018c^2+2019d^2}{2018c^2-2019d^2}\)
Dăm ba mấy bài đặt k:v
Đặt \(\frac{a}{b}=\frac{c}{d}=k\)
Ta có:
\(\frac{2018a^2+2019b^2}{2018a^2-2019b^2}=\frac{2018b^2k^2+2019b^2}{2018b^2k^2-2019b^2}=\frac{b^2\left(2018k^2+2019\right)}{b^2\left(2018k^2-2019\right)}=\frac{2018k^2+2019}{2018k^2-2019}\)
\(\frac{2018c^2+2019d^2}{2018c^2-2019d^2}=\frac{2018d^2k^2+2019d^2}{2018d^2k^2-2019d^2}=\frac{d^2\left(2018k^2+2019\right)}{d^2\left(2018k^2-2019\right)}=\frac{2018k^2+2019}{2018k^2-2019}\)
Từ đó \(\frac{2018a^2+2019b^2}{2018a^2-2019b^2}=\frac{2018c^2+2019d^2}{2018c^2-2019d^2}\)
cho 3 số a, b, c > 0 thỏa mãn a + b + c = 1. Chứng minh
\(\dfrac{a}{a+\sqrt{2019a+bc}}+\dfrac{b}{b+\sqrt{2019b+ac}}+\dfrac{c}{c+\sqrt{2019c+ab}}\le1\)
Mong mọi người giúp mình với, lâu không dùng bất đẳng thức nên quên. Cám ơn mọi người nhiều!
Cho tỉ lệ thức \(\frac{a}{b}=\frac{c}{d}.Chứngminh\frac{2018a-2019b}{2018c+2019d}=\frac{2018c-2019d}{2018a+2019b}\)
Từ \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}\)
\(\Rightarrow\frac{2018a}{2018c}=\frac{2019b}{2019d}\)
Áp dụng t/c DTSBN : \(\frac{2018a}{2018c}=\frac{2019b}{2019d}=\frac{2018a-2019b}{2018c-2019d}=\frac{2018a+2019b}{2018c+2019d}\)
Cái này đến đây là đề sai nhé ! Đề phải cho là C/m cái (2018a-2019b).(2018c+2019d) = (2018a-2019b)(2018c+2019d) mới đúng