Cho x/a=y/b . CMR
( x2+y2)(a2+b2)=(ax+by)2
GỢI Ý:
Đặt x/a=y/b=k. Suy ra x=ka ; y=kb
Thay x,y vào hai vế rồi so sánh
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Ta có: \(\left(a^2+b^2\right)\left(x^2+y^2\right)=\left(ax+by\right)^2\)
\(\Leftrightarrow a^2x^2+a^2y^2+b^2x^2+b^2y^2=a^2x^2+b^2y^2+2axby\)
\(\Leftrightarrow a^2y^2-2axby+b^2x^2=0\)
\(\Leftrightarrow\left(ay-bx\right)^2=0\)
\(\Leftrightarrow ay=bx\)
hay \(\dfrac{a}{x}=\dfrac{b}{y}\)
Ta có : \(\left(a^2+b^2\right)\left(x^2+y^2\right)=\left(ax+by\right)^2\)
\(\Leftrightarrow a^2x^2+a^2y^2+b^2x^2+b^2y^2=a^2x^2+2abxy+b^2y^2\)
\(\Leftrightarrow a^2y^2-2abxy+b^2x^2=0\)
\(\Leftrightarrow\left(ay-bx\right)^2=0\)
\(\Leftrightarrow ay-bx=0\)
\(\Leftrightarrow ay=bx\Leftrightarrow\dfrac{a}{b}=\dfrac{x}{y}\)
\(=\dfrac{2\left(x+y\right)}{\left(a+b\right)^2}.\dfrac{a\left(x-y\right)+b\left(x-y\right)}{2\left(x^2-y^2\right)}\)
\(=\dfrac{2\left(x+y\right)}{\left(a+b\right)^2}.\dfrac{\left(x-y\right)\left(a+b\right)}{2\left(x-y\right)\left(x+y\right)}\)
\(=\dfrac{1}{a+b}\)
\(=\dfrac{a+b-c}{\left(a+b\right)^2-c^2}.\dfrac{\left(a+b\right)^2+c\left(a+b\right)}{\left(a-b\right)\left(a+b\right)}\)
\(=\dfrac{a+b-c}{\left(a+b-c\right)\left(a+b+c\right)}.\dfrac{\left(a+b\right)\left(a+b+c\right)}{\left(a-b\right)\left(a+b\right)}\)
\(=\dfrac{1}{a-b}\)
\(c,\dfrac{x^3+1}{x^2+2x+1}.\dfrac{x^2-1}{2x^2-2x+2}\)
\(=\dfrac{\left(x+1\right)\left(x^2-x+1\right)}{\left(x+1\right)^2}.\dfrac{\left(x-1\right)\left(x+1\right)}{2\left(x^2-x+1\right)}\) \(=\dfrac{x-1}{2}\) \(d,\dfrac{x^8-1}{x+1}.\dfrac{1}{\left(x^2+1\right)\left(x^4+1\right)}\) \(=\dfrac{\left(x^4\right)^2-1}{x+1}.\dfrac{1}{\left(x^2+1\right)\left(x^4+1\right)}\) \(=\dfrac{\left(x^4-1\right)\left(x^4+1\right)}{x+1}.\dfrac{1}{\left(x^2+1\right)\left(x^4+1\right)}\) \(=\dfrac{\left(x^2+1\right)\left(x^2-1\right)}{x+1}.\dfrac{1}{x^2+1}\) \(=\dfrac{\left(x-1\right)\left(x+1\right)}{x+1}\) \(=x-1\) \(e,\dfrac{x-y}{xy+y^2}-\dfrac{3x+y}{x^2-xy}.\dfrac{y-x}{x+y}\) \(=\dfrac{x-y}{y\left(x+y\right)}-\dfrac{3x+y}{x\left(x-y\right)}.\dfrac{-\left(x-y\right)}{x+y}\) \(=\dfrac{x-y}{y\left(x+y\right)}-\dfrac{3x+y}{x}.\dfrac{-1}{x+y}\) \(=\dfrac{x-y}{y\left(x+y\right)}-\dfrac{-3x-y}{x\left(x+y\right)}\) \(=\dfrac{x\left(x-y\right)+y\left(3x+y\right)}{xy\left(x+y\right)}\) \(=\dfrac{x^2-xy+3xy+y^2}{xy\left(x+y\right)}\) \(=\dfrac{x^2+2xy+y^2}{xy\left(x+y\right)}\) \(=\dfrac{\left(x+y\right)^2}{xy\left(x+y\right)}=\dfrac{x+y}{xy}\)tìm giá trị của m để pt 2x-m=1-x nhận giá trị x=-2 là nghiệm
giải hộ e với :)
a: \(ax+by+cz\)
\(=x^3-xyz+y^3-xyz+z^3-xyz\)
\(=x^3+y^3+z^3-3xyz\)
b: \(ax+by+cz\)
\(=x^3+y^3+z^3-3xyz\)
\(=\left(x+y\right)^3+z^3-3xy\left(x+y\right)-3yxz\)
\(=\left(x+y+z\right)\left(x^2+y^2+2xy-xz-yz+z^2\right)-3xy\left(x+y+z\right)\)
\(=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-xz\right)\)
Áp dụng tính chất các dãy tỉ số bằng nhau ta có:
\(\dfrac{x}{a}=\dfrac{y}{b}=\dfrac{z}{c}=\dfrac{x+y+z}{a+b+c}=\dfrac{x+y+z}{1}\)
\(x=a\left(x+y+z\right)=x^2=a^2.\left(x+y+z\right)^2\)
\(y=b\left(x+y+z\right)=y^2=b^2\left(x+y+z\right)^2\)
\(z=c\left(x+y+z\right)=z^2=c^2.\left(x+y+z\right)^2\)
\(\Rightarrow x^2+y^2+z^2=a^2\left(x+y+z\right)^2+b^2\left(x+y+z\right)^2+c^2\left(x+y+z\right)^2\)
\(=\left(x+y+z\right)^2\left(a^2+b^2+c^2\right)=\left(x+y+z\right)^2\) (do \(a^2+b^2+c^2=1\))
https://lazi.vn/edu/exercise/864720/cho-a-b-c-a2-b2-c2-1-va-x-a-y-b-z-c-chung-minh-rang-x-y-z2-x2-y2-z2
liệt phím? Mù mắt?