Biến đổi các biểu thức sau thành phân thức Bài 44 trang 36 SBT Toán 8 Tập 1

Giải SBT Toán 8 Bài 9: Biến đổi các biểu thức hữu tỉ. Giá trị của phân thức

Bài 44 trang 36 SBT Toán 8 Tập 1: Biến đổi các biểu thức sau thành phân thức:

a) $\frac{1}{2}+\text{\hspace{0.17em}\hspace{0.17em}}\frac{x}{1-\frac{x}{x+\text{\hspace{0.17em}\hspace{0.17em}}2}}$;

b) $\frac{x-\frac{1}{x^2}}{1+\text{\hspace{0.17em}\hspace{0.17em}}\frac{1}{x}+\frac{1}{x^2}}$;

c) $\frac{1-\frac{2y}{x}\text{\hspace{0.17em}\hspace{0.17em}}+\text{\hspace{0.17em}}\frac{y^2}{x^2}}{\frac{1}{x}-\frac{1}{y}}$;

d) $\frac{\frac{x}{4}-1+\text{\hspace{0.17em}}\frac{3}{4x}}{\frac{x}{2}-\frac{6}{x}+\frac{1}{2}}$.

Lời giải:

a) $\frac{1}{2}+\text{\hspace{0.17em}\hspace{0.17em}}\frac{x}{1-\frac{x}{x+\text{\hspace{0.17em}\hspace{0.17em}}2}}=\frac{1}{2}+\text{\hspace{0.17em}\hspace{0.17em}}\frac{x}{\frac{x+2-x}{x+\text{\hspace{0.17em}\hspace{0.17em}}2}}$

$=\frac{1}{2}+\frac{x}{\frac{2}{x+2}}=\frac{1}{2}+\frac{x(x+2)}{2}$

$=\frac{1+\text{}x(x+\text{}2)}{2}=\frac{1+x^2+2x}{2}=\frac{{\left(x+1\right)}^2}{2}$

b) $\frac{x-\frac{1}{x^2}}{1+\text{\hspace{0.17em}\hspace{0.17em}}\frac{1}{x}+\frac{1}{x^2}}$

$=\left(x-\frac{1}{x^2}\right):\left(1+\text{\hspace{0.17em}\hspace{0.17em}}\frac{1}{x}+\frac{1}{x^2}\right)$

$=\frac{x^3-1}{x2}:\frac{x^2+x+\text{\hspace{0.17em}}1}{x^2}$

$\begin{array}{l}=\frac{x^3-1}{x2}.\frac{x^2}{x^2+x+\text{\hspace{0.17em}}1}=\frac{(x^3-1)x^2}{x(x^2+x+\text{\hspace{0.17em}}1)_2}\\ =\frac{(x-1)(x^2+x+1)x^2}{x(x^2+x+\text{\hspace{0.17em}}1)_2}=x-1\end{array}$

c) $\frac{1-\frac{2y}{x}\text{\hspace{0.17em}\hspace{0.17em}}+\text{\hspace{0.17em}}\frac{y^2}{x^2}}{\frac{1}{x}-\frac{1}{y}}$

$=\left(1-\frac{2y}{x}\text{\hspace{0.17em}\hspace{0.17em}}+\text{\hspace{0.17em}}\frac{y^2}{x^2}\right):\left(\frac{1}{x}-\frac{1}{y}\right)$

$=\frac{x^2-2xy+y^2}{x^2}:\frac{y-x}{xy}$

$\begin{array}{l}=\frac{{(x-y)}^2}{x^2}.\frac{xy}{y-x}=\frac{{(x-y)}^{2}xy}{x^2.\text{[}-(x-y)\text{]}}\\ =\frac{(x-y)y}{x.(-1)}=\frac{-(x-y)y}{x}=\frac{y\left(y-x\right)}{x}\end{array}$

d) $\frac{\frac{x}{4}-1+\text{\hspace{0.17em}}\frac{3}{4x}}{\frac{x}{2}-\frac{6}{x}+\frac{1}{2}}$

$\begin{array}{l}=\left(\frac{x}{4}-1+\text{\hspace{0.17em}}\frac{3}{4x}\right):\left(\frac{x}{2}-\frac{6}{x}+\frac{1}{2}\right)\\ =\frac{x^2-4x+\text{\hspace{0.17em}}3}{4x}:\frac{x^2-12+x}{2x}\end{array}$

$\begin{array}{l}=\frac{x^2-4x+\text{\hspace{0.17em}}3}{4x}.\frac{2x}{x^2-12+x}\\ =\frac{(x^2-4x+\text{\hspace{0.17em}}3).2x}{4x(x^2-12+x)}=\frac{(x^2-x)-(3x-3)}{2\left.\left(x-_{2}9\right)\text{}+(x-3)\right]}\end{array}$

$=\frac{x(x-1)-3(x-1)}{2\left.(x+\text{}3)(x-3)+(x-3)\right]}=\frac{(x-3)(x-1)}{2(x+3+1)(x-3)}$

$=\frac{(x-3)(x-1)}{2(x+\text{}4)(x-3)}=\frac{x-1}{2(x\text{}+4)}$