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\(ax+ay+by=a\left(1-y\right)+ay+\left(2-a\right)y\)
\(=a-ay+ay+2y-ay\)
\(=a\left(1-y\right)+2y\)
\(=ax+2y=\left(2-b\right)x+2y=2x+2y-bx\)
\(=2\left(x+y\right)-bx=2-bx\)
\(\dfrac{bz-cy}{a}=\dfrac{cx-az}{b}=\dfrac{ay-bx}{c}\)
\(\Rightarrow\dfrac{abz-acy}{a^2}=\dfrac{bcx-abz}{b^2}=\dfrac{acy-bcx}{c^2}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{abz-acy}{a^2}=\dfrac{bcx-abz}{b^2}=\dfrac{acy-bcx}{c^2}=\dfrac{abz-acy+bcx-abz+acy-bcx}{a^2+b^2+c^2}=0\)
Suy ra \(\left\{{}\begin{matrix}bz=cy\Leftrightarrow\dfrac{y}{b}=\dfrac{z}{c}\\cx=az\Leftrightarrow\dfrac{x}{a}=\dfrac{z}{c}\\ay=bx\Leftrightarrow\dfrac{x}{a}=\dfrac{y}{b}\end{matrix}\right.\Leftrightarrow\dfrac{x}{a}=\dfrac{y}{b}=\dfrac{z}{c}\left(đpcm\right)\)
p/s: đã sửa đề
\(bx^2=ay^2\Leftrightarrow\dfrac{x^2}{a}=\dfrac{y^2}{b}\Leftrightarrow\left(\dfrac{x^2}{a}\right)^{1010}=\left(\dfrac{y^2}{b}\right)^{1010}\\ \Leftrightarrow\dfrac{x^{2020}}{a^{1010}}=\dfrac{y^{2020}}{a^{1010}}\)
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
a) a.x + a.y + b.x + b.y
= a.(x + y) + b.(x + y)
= a . 17 + b . 17
= (a +b) . 17
= -2 . 17 = -34
b) a.x - a.y + b.x - b.y
= a.(x - y) + b.(x - y)
= a . (-1) + b.(-1)
= (a + b) . (-1)
= -7 . (-1) = 7
ban tk mk mk se tk lai