cho ba số a, b, c thỏa mãn abc = 27 và 1/a+1/b+1/c = (a+b+c)/9 Chứng minh (a*2020-9*1010)(b*2020-9*1010)(c*2020-9*1010)=0
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\(bx^2=ay^2\Leftrightarrow\dfrac{x^2}{a}=\dfrac{y^2}{b}\Leftrightarrow\left(\dfrac{x^2}{a}\right)^{1010}=\left(\dfrac{y^2}{b}\right)^{1010}\\ \Leftrightarrow\dfrac{x^{2020}}{a^{1010}}=\dfrac{y^{2020}}{a^{1010}}\)
Áp dụng t/c dtsbn:
\(\dfrac{x^{2020}}{a^{1010}}=\dfrac{y^{2020}}{b^{1010}}=\dfrac{x^{2020}+y^{2020}}{a^{1010}+b^{1010}}\left(3\right)\)
Đặt \(\dfrac{x^2}{a}=\dfrac{y^2}{b}=k\Leftrightarrow x^2=ak;y^2=bk\)
\(x^2+y^2=1\Leftrightarrow ak+bk=1\Leftrightarrow k\left(a+b\right)=1\Leftrightarrow a+b=\dfrac{1}{k}\)
\(\Leftrightarrow\dfrac{2}{\left(a+b\right)^{1010}}=\dfrac{2}{\left(\dfrac{1}{k}\right)^{1010}}=2:\dfrac{1}{k^{1010}}=k^{1010}\left(1\right)\)
Mà \(\dfrac{x^{2020}}{a^{1010}}=\dfrac{\left(x^2\right)^{1010}}{a^{1010}}=\dfrac{a^{1010}k^{1010}}{a^{1010}}=k^{1010}\left(2\right)\)
Từ \(\left(1\right)\left(2\right)\left(3\right)\) ta được đpcm
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\(a^{2020}+b^{2020}+c^{2020}=a^{1010}b^{1010}+b^{1010}c^{1010}+c^{1010}a^{1010}\)
\(\Leftrightarrow a^{2020}+b^{2020}+c^{2020}-a^{1010}b^{1010}-b^{1010}c^{1010}-c^{1010}a^{1010}=0\)
\(\Leftrightarrow2a^{2020}+2b^{2020}+2c^{2020}-2a^{1010}b^{1010}-2b^{1010}c^{1010}-2a^{1010}c^{1010}=0\)
\(\Leftrightarrow\left(a^{2020}-2a^{1010}b^{1010}+b^{2020}\right)+\left(b^{2020}-2b^{1010}c^{1010}+c^{2020}\right)+\left(c^{2020}-2a^{1010}c^{1010}+a^{2020}\right)=0\)
\(\Leftrightarrow\left(a^{1010}-b^{1010}\right)^2+\left(b^{1010}-c^{1010}\right)^2+\left(c^{1010}-a^{1010}\right)^2=0\)
\(\Leftrightarrow\hept{\begin{cases}\left(a^{1010}-b^{1010}\right)=0\\b^{1010}-c^{1010}=0\\c^{1010}-a^{1010}=0\end{cases}}\Leftrightarrow a^{1010}=b^{1010}=c^{1010}\Leftrightarrow\pm a=\pm b=\pm c\)
Rồi thay :> Còn thay kiểu nào thì mình cũng hong biết :">
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\(2a^{2020}+2b^{2020}+2c^{2020}-2\left(ab\right)^{1010}-2\left(bc\right)^{1010}-2\left(ca\right)^{1010}=0\)
\(\Leftrightarrow\left(a^{1010}-b^{1010}\right)^2+\left(b^{1010}-c^{1010}\right)^2+\left(c^{1010}-a^{1010}\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}a^{1010}-b^{1010}=0\\b^{1010}-c^{1010}=0\\c^{1010}-a^{1010}=0\end{matrix}\right.\)
\(\Rightarrow\left|a\right|=\left|b\right|=\left|c\right|\)
Nếu đề không cho a;b;c dương thì không tính được cụ thể giá trị A
Nếu a;b;c dương thì \(a=b=c\Rightarrow A=0\)
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a) UCLN(99,9)=9
b) UCLN(100,50)=50
c) UCLN(2020,1010)=1010
d) UCLN(2021,2022,1)=1