Question

[Lê Hà Lâm]

 Giúp tớ bài 6, 10 nhé. Mình đang gấp

Answer 1

[HT.Phong (9A5)]

8)

a) \(A=1-\dfrac{x}{1-\dfrac{x}{x+1}}\left(x\ne-1\right)\)

\(=1-\dfrac{x}{\dfrac{x+1-x}{x+1}}=1-\dfrac{x}{\dfrac{1}{x+1}}=1-x\left(x+1\right)=-x^2-x+1\) 

b) \(B=\dfrac{\dfrac{x}{y}+\dfrac{y}{x}}{\dfrac{x-y}{x+y}+\dfrac{x+y}{x-y}}=\dfrac{\dfrac{x^2}{xy}+\dfrac{y^2}{xy}}{\dfrac{\left(x-y\right)^2+\left(x+y\right)^2}{\left(x+y\right)\left(x-y\right)}}\left(x\ne\pm y;x\ne0;y\ne0\right)\)

\(=\dfrac{\dfrac{x^2+y^2}{xy}}{\dfrac{x^2-2xy+y^2+x^2+2xy+y^2}{\left(x+y\right)\left(x-y\right)}}=\dfrac{\dfrac{x^2+y^2}{xy}}{\dfrac{2\left(x^2+y^2\right)}{x^2-y^2}}\)

\(=\dfrac{x^2+y^2}{xy}\cdot\dfrac{x^2-y^2}{2\left(x^2+y^2\right)}=\dfrac{x^2-y^2}{2xy}\)

Answer 2

[Nguyễn Lê Phước Thịnh]

10:

a: Thời gian dự định là \(\dfrac{60}{x}\left(giờ\right)\)

b: Thời gian đi nửa quãng đường đầu tiên là: \(\dfrac{60}{2}:\left(x+10\right)=\dfrac{30}{x+10}\left(giờ\right)\)

Thời gian đi nửa quãng đường còn lại là:

\(\dfrac{60-30}{x-6}=\dfrac{30}{x-6}\left(giờ\right)\)

c: Ô tô đến B đúng giờ nên ta có: \(\dfrac{30}{x+10}+\dfrac{30}{x-6}=\dfrac{60}{x}\)

=>\(\dfrac{1}{x+10}+\dfrac{1}{x-6}=\dfrac{2}{x}\)

=>\(\dfrac{x-6+x+10}{\left(x+10\right)\left(x-6\right)}=\dfrac{2}{x}\)

=>\(\dfrac{2x+4}{\left(x+10\right)\left(x-6\right)}=\dfrac{2}{x}\)

=>\(\dfrac{x+2}{x^2+4x-60}=\dfrac{1}{x}\)

=>\(x\left(x+2\right)=x^2+4x-60\)

=>\(x^2+2x=x^2+4x-60\)

=>-2x=-60

=>x=30

Vậy: Vận tốc dự định của ô tô là 30km/h

Bài 6:

a: \(A=\dfrac{x-1}{x}-\dfrac{x+1}{x^2-x}+\dfrac{3\left(x-1\right)}{x^2-2x+1}\)

\(=\dfrac{x-1}{x}-\dfrac{x+1}{x\left(x-1\right)}+\dfrac{3}{x-1}\)

\(=\dfrac{\left(x-1\right)\left(x-1\right)-x-1+3x}{\left(x-1\right)\cdot x}\)

\(=\dfrac{x^2-2x+1+2x-1}{x\left(x-1\right)}=\dfrac{x^2}{x\left(x-1\right)}=\dfrac{x}{x-1}\)

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

\(=\dfrac{\left(x+y\right)^2-\left(x-y\right)^2+4y^2}{2\left(x-y\right)\left(x+y\right)}\)

\(=\dfrac{x^2+2xy+y^2-x^2+2xy-y^2+4y^2}{2\left(x-y\right)\left(x+y\right)}=\dfrac{4y^2+4xy}{2\left(x-y\right)\left(x+y\right)}\)

\(=\dfrac{2y\left(x+y\right)}{2\left(x-y\right)\left(x+y\right)}=\dfrac{y}{x-y}\)