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7 tháng 1 2018
\(a,\dfrac{2x+2y}{a^2+2ab+b^2}.\dfrac{ax-ay+bx-by}{2x^2-2y^2}\)

\(=\dfrac{2\left(x+y\right)}{\left(a+b\right)^2}.\dfrac{a\left(x-y\right)+b\left(x-y\right)}{2\left(x^2-y^2\right)}\)

\(=\dfrac{2\left(x+y\right)}{\left(a+b\right)^2}.\dfrac{\left(x-y\right)\left(a+b\right)}{2\left(x-y\right)\left(x+y\right)}\)

\(=\dfrac{1}{a+b}\)


\(b,\dfrac{a+b-c}{a^2+2ab+b^2-c^2}.\dfrac{a^2+2ab+b^2+ac+bc}{a^2-b^2}\)

\(=\dfrac{a+b-c}{\left(a+b\right)^2-c^2}.\dfrac{\left(a+b\right)^2+c\left(a+b\right)}{\left(a-b\right)\left(a+b\right)}\)

\(=\dfrac{a+b-c}{\left(a+b-c\right)\left(a+b+c\right)}.\dfrac{\left(a+b\right)\left(a+b+c\right)}{\left(a-b\right)\left(a+b\right)}\)

\(=\dfrac{1}{a-b}\)

\(c,\dfrac{x^3+1}{x^2+2x+1}.\dfrac{x^2-1}{2x^2-2x+2}\)

\(=\dfrac{\left(x+1\right)\left(x^2-x+1\right)}{\left(x+1\right)^2}.\dfrac{\left(x-1\right)\left(x+1\right)}{2\left(x^2-x+1\right)}\) \(=\dfrac{x-1}{2}\) \(d,\dfrac{x^8-1}{x+1}.\dfrac{1}{\left(x^2+1\right)\left(x^4+1\right)}\) \(=\dfrac{\left(x^4\right)^2-1}{x+1}.\dfrac{1}{\left(x^2+1\right)\left(x^4+1\right)}\) \(=\dfrac{\left(x^4-1\right)\left(x^4+1\right)}{x+1}.\dfrac{1}{\left(x^2+1\right)\left(x^4+1\right)}\) \(=\dfrac{\left(x^2+1\right)\left(x^2-1\right)}{x+1}.\dfrac{1}{x^2+1}\) \(=\dfrac{\left(x-1\right)\left(x+1\right)}{x+1}\) \(=x-1\) \(e,\dfrac{x-y}{xy+y^2}-\dfrac{3x+y}{x^2-xy}.\dfrac{y-x}{x+y}\) \(=\dfrac{x-y}{y\left(x+y\right)}-\dfrac{3x+y}{x\left(x-y\right)}.\dfrac{-\left(x-y\right)}{x+y}\) \(=\dfrac{x-y}{y\left(x+y\right)}-\dfrac{3x+y}{x}.\dfrac{-1}{x+y}\) \(=\dfrac{x-y}{y\left(x+y\right)}-\dfrac{-3x-y}{x\left(x+y\right)}\) \(=\dfrac{x\left(x-y\right)+y\left(3x+y\right)}{xy\left(x+y\right)}\) \(=\dfrac{x^2-xy+3xy+y^2}{xy\left(x+y\right)}\) \(=\dfrac{x^2+2xy+y^2}{xy\left(x+y\right)}\) \(=\dfrac{\left(x+y\right)^2}{xy\left(x+y\right)}=\dfrac{x+y}{xy}\)
19 tháng 2 2018

tìm giá trị của m để pt 2x-m=1-x nhận giá trị x=-2 là nghiệm

giải hộ e với :)

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 : \(\left(a^2+b^2\right)\left(x^2+y^2\right)=\left(ax+by\right)^2\)

\(\Leftrightarrow a^2x^2+a^2y^2+b^2x^2+b^2y^2=a^2x^2+2abxy+b^2y^2\)

\(\Leftrightarrow a^2y^2-2abxy+b^2x^2=0\)

\(\Leftrightarrow\left(ay-bx\right)^2=0\)

\(\Leftrightarrow ay-bx=0\)

\(\Leftrightarrow ay=bx\Leftrightarrow\dfrac{a}{b}=\dfrac{x}{y}\)

23 tháng 4 2018

Giải sách bài tập Toán 7 | Giải sbt Toán 7