Dạng 1: Bài luyện tập 1 có đáp án

$P=\frac{x^2+x+1}{2}=\frac{1}{2}\left(x^2+2.x.\frac{1}{2}+\frac{1}{4}+\frac{3}{4}\right)=\frac{1}{2}{\left(x+\frac{1}{2}\right)}^2+\frac{3}{8}$

$P=\left(\frac{x+1}{1-x}-\frac{1-x}{1+x}-\frac{4x^2}{x^2-1}\right):\frac{4x^2-4}{x^2-2x+1}$$DK:x\ne \pm 1$

$=\left(\frac{-{\left(x+1\right)}^2}{\left(x-1\right)\left(x+1\right)}+\frac{{\left(x-1\right)}^2}{\left(x-1\right)\left(x+1\right)}-\frac{4x^2}{\left(x-1\right)\left(x+1\right)}\right):\frac{4\left(x+1\right)\left(x-1\right)}{{\left(x-1\right)}^2}$

$=\left(\frac{{\left(x-1\right)}^2-{\left(x+1\right)}^2-4x^2}{\left(x-1\right)\left(x+1\right)}\right):\frac{4\left(x+1\right)}{\left(x-1\right)}$

$=\left(\frac{-4x-4x^2}{\left(x-1\right)\left(x+1\right)}\right).\frac{\left(x-1\right)}{4\left(x+1\right)}$

$=\left(\frac{-4x\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}\right).\frac{\left(x-1\right)}{4\left(x+1\right)}$

$=\frac{-x}{x+1}$

$x^2+4x=5\iff x^2+4x-5=0\iff \left(x+5\right)\left(x-1\right)=0\iff \left[\begin{array}{l}x=-5\left(tm\right)\\ x=1\left(l\right)\end{array}\right.$

Thay $x=-5\left(tm\right)$ vào P , ta có$P=\frac{5}{-5+1}=\frac{-5}{4}$

$P=\frac{-x}{x+1}=\frac{-x-1+1}{x+1}=-1+\frac{1}{x+1}$

$P\in Z\iff \frac{1}{x+1}\in Z\iff x+1\in U\left(1\right)=\left\{\pm 1\right\}$

$P=\left(\frac{x^2-3x}{x^2-9}-1\right):\left(\frac{9-x^2}{x^2+x-6}-\frac{x-3}{2-x}-\frac{x-2}{x+3}\right)$$DK:x\ne \pm 3;x\ne 2$

$=\left(\frac{x\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}-1\right):\left(\frac{9-x^2}{\left(x-2\right)\left(x+3\right)}+\frac{x-3}{x-2}-\frac{x-2}{x+3}\right)$

$=\left(\frac{x}{\left(x+3\right)}-1\right):\left(\frac{9-x^2}{\left(x-2\right)\left(x+3\right)}+\frac{\left(x-3\right)\left(x+3\right)}{\left(x-2\right)\left(x+3\right)}-\frac{{\left(x-2\right)}^2}{\left(x-2\right)\left(x+3\right)}\right)$

$=\left(\frac{x-x-3}{\left(x+3\right)}\right):\left(\frac{9-x^2+x^2-9-x^2+4x-4}{\left(x-2\right)\left(x+3\right)}\right)$

$=\left(\frac{-3}{\left(x+3\right)}\right):\left(\frac{-x^2+4x-4}{\left(x-2\right)\left(x+3\right)}\right)$

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

$=\frac{3}{x-2}$

$x^3–4x=0\iff x\left(x+2\right)\left(x-2\right)=0\iff \left[\begin{array}{l}x=0\left(tm\right)\\ x=2\left(l\right)\\ x=-2\left(tm\right)\end{array}\right.$

$P=\left(\frac{x-1}{x-3}+\frac{x-3}{x+3}+\frac{4x-2}{x^2-9}\right):\left(1-\frac{x+1}{x+3}\right)$$DK:x\ne \pm 3$

$=\left(\frac{\left(x-1\right)\left(x+3\right)}{\left(x+3\right)\left(x-3\right)}+\frac{\left(x-3\right)\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}+\frac{4x-2}{\left(x+3\right)\left(x-3\right)}\right):\left(\frac{x+3-x-1}{x+3}\right)$

$=\left(\frac{x^2+2x-3+x^2-6x+9+4x-2}{\left(x+3\right)\left(x-3\right)}\right):\left(\frac{2}{x+3}\right)$

$=\left(\frac{2x^2+4}{\left(x+3\right)\left(x-3\right)}\right).\left(\frac{x+3}{2}\right)$

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

$=\frac{x^2+2}{x-3}$

$\left|2x-1\right|-1=0\iff \left|2x-1\right|=1\iff 2x-1=\pm 1\iff \left[\begin{array}{l}2x-1=1\\ 2x-1=-1\end{array}\right.\iff \left[\begin{array}{l}x=1\left(tm\right)\\ x=0\left(tm\right)\end{array}\right.$

$P=\frac{x^2+2}{x-3}=x+3+\frac{11}{x-3}$$P\in Z\iff \frac{11}{x-3}\in Z\iff x-3\in U\left(11\right)=\left\{\pm 1;\pm 11\right\}$

$\begin{array}{l}\left(\frac{2ab}{a^2-b^2}+\frac{a-b}{2a+2b}\right).\frac{2a}{a+b}+\frac{b}{b-a}\\ =\left(\frac{2ab}{\left(a+b\right)\left(a-b\right)}+\frac{a-b}{2(a+b)}\right).\frac{2a}{a+b}+\frac{b}{b-a}\\ =\left(\frac{4ab}{2\left(a+b\right)\left(a-b\right)}+\frac{{\left(a-b\right)}^2}{2\left(a+b\right)\left(a-b\right)}\right).\frac{2a}{a+b}+\frac{b}{b-a}\\ =\left(\frac{4ab+a^2-2ab+b^2}{2\left(a+b\right)\left(a-b\right)}\right).\frac{2a}{a+b}+\frac{b}{b-a}\\ =\left(\frac{a^2+2ab+b^2}{2\left(a+b\right)\left(a-b\right)}\right).\frac{2a}{a+b}+\frac{b}{b-a}\\ =\left(\frac{{\left(a+b\right)}^2}{2\left(a+b\right)\left(a-b\right)}\right).\frac{2a}{a+b}+\frac{b}{b-a}\\ =\frac{\left(a+b\right)}{2\left(a-b\right)}.\frac{2a}{a+b}+\frac{b}{b-a}\\ =\frac{a}{\left(a-b\right)}+\frac{b}{b-a}\\ =\frac{a-b}{a-b}\\ =1\end{array}$

$\begin{array}{l}\frac{x}{x-y}-\frac{x^3-x{y}^2}{x^2+y^2}.\left[\frac{x}{{\left(x-y\right)}^2}-\frac{y}{x^2-y^2}\right]\\ =\frac{x}{x-y}-\frac{x\left(x+y\right)\left(x-y\right)}{x^2+y^2}.\left[\frac{x}{{\left(x-y\right)}^2}-\frac{y}{\left(x+y\right)\left(x-y\right)}\right]\\ =\frac{x}{x-y}-\frac{x\left(x+y\right)\left(x-y\right)}{x^2+y^2}.\left[\frac{x\left(x+y\right)}{{\left(x-y\right)}^2\left(x+y\right)}-\frac{y\left(x-y\right)}{\left(x+y\right){\left(x-y\right)}^2}\right]\\ =\frac{x}{x-y}-\frac{x\left(x+y\right)\left(x-y\right)}{x^2+y^2}.\left[\frac{x^2+xy-xy+y^2}{{\left(x-y\right)}^2\left(x+y\right)}\right]\\ =\frac{x}{x-y}-\frac{x\left(x+y\right)\left(x-y\right)}{x^2+y^2}.\left[\frac{x^2+y^2}{{\left(x-y\right)}^2\left(x+y\right)}\right]\\ =\frac{x}{x-y}-\frac{x}{x-y}\\ =0\end{array}$

$\begin{array}{l}\frac{y}{3-y}+\frac{y^2+3y}{2y+3}.\left(\frac{y+3}{y^2-3y}-\frac{y}{y^2-9}\right)\\ =\frac{y}{3-y}+\frac{y^2+3y}{2y+3}.\left(\frac{y+3}{y\left(y-3\right)}-\frac{y}{\left(y+3\right)\left(y-3\right)}\right)\\ =\frac{y}{3-y}+\frac{y\left(y+3\right)}{2y+3}.\left(\frac{{\left(y+3\right)}^2}{y\left(y-3\right)\left(y+3\right)}-\frac{y^2}{y\left(y+3\right)\left(y-3\right)}\right)\\ =\frac{y}{3-y}+\frac{y\left(y+3\right)}{2y+3}.\left(\frac{\left(y+3+y\right)\left(y+3-y\right)}{y\left(y-3\right)\left(y+3\right)}\right)\\ =\frac{y}{3-y}+\frac{y\left(y+3\right)}{2y+3}.\left(\frac{3\left(2y+3\right)}{y\left(y-3\right)\left(y+3\right)}\right)\\ =\frac{y}{3-y}+\frac{3}{y-3}\\ =\frac{y-3}{3-y}\\ =-1\end{array}$

$\begin{array}{l}\left(\frac{x}{x^2-36}-\frac{x-6}{x^2+6x}\right):\frac{2x-6}{x^2+6x}+\frac{x}{6-x}\\ =\left(\frac{x}{\left(x+6\right)\left(x-6\right)}-\frac{x-6}{x\left(x+6\right)}\right):\frac{2\left(x-3\right)}{x\left(x+6\right)}+\frac{x}{6-x}\\ =\left(\frac{x^2}{x\left(x+6\right)\left(x-6\right)}-\frac{{\left(x-6\right)}^2}{x\left(x+6\right)\left(x-6\right)}\right).\frac{x\left(x+6\right)}{2\left(x-3\right)}+\frac{x}{6-x}\\ =\left(\frac{\left(x+x-6\right)\left(x-x+6\right)}{x\left(x+6\right)\left(x-6\right)}\right).\frac{x\left(x+6\right)}{2\left(x-3\right)}+\frac{x}{6-x}\\ =\frac{12\left(x-3\right)}{x\left(x+6\right)\left(x-6\right)}.\frac{x\left(x+6\right)}{2\left(x-3\right)}+\frac{x}{6-x}\\ =\frac{6}{\left(x-6\right)}+\frac{x}{6-x}\\ =\frac{6-x}{x-6}\\ =-1\end{array}$