标题

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中考数学复习:解特殊复杂方程的解题技巧及知识梳理

解特殊复杂方程的思维导图

题型一:解含参数的一元二次方程

解含参数的一元二次方程时,只需将参数当做已知数来解方程,需要注意的是二次项系数如果含有参数的话,要考虑一元二次方程的定义保证二次项系数不能为零.

\({\color{red}{【例题】}}\)解关于x的方程\(a{{x}^{2}}+bx+c=0\)
\({\color{red}{【解析】}}\)⑴ 当\(a = 0\),\(b \ne 0\)时,\(x = – \frac{c}{b}\)
\(b = 0\)时,\(c = 0\),方程有无数解;
\(b = 0\)时,\(c \ne 0\),方程无解;
⑵ 当\(a \ne 0\)时,原方程可化为\({x^2} + \frac{b}{a}x + \frac{c}{a} = 0\)
\({\left( {x + \frac{b}{{2a}}} \right)^2} = \frac{{{b^2} – 4ac}}{{4{a^2}}}\)
当\({b^2} – 4ac > 0\)时, \({x_1} = \frac{{ – b + \sqrt {{b^2} – 4ac} }}{{2a}}\),\({x_2} = \frac{{ – b – \sqrt {{b^2} – 4ac} }}{{2a}}\)
当\({b^2} – 4ac = 0\)时,\({x_1} = {x_2} = – \frac{b}{{2a}}\)
当\({b^2} – 4ac < 0\)时,该方程无实数根.

\({\color{red}{【练习】}}\)解关于x的方程\(\left( {a – 1} \right){x^2} – 2ax + a = 0\).
\({\color{red}{【解析】}}\)⑴当\(a = 1\)时,方程的根为\(x = \frac{1}{2}\);
⑵当\(a \ne 1\)时,\(\Delta = {\left( { – 2a} \right)^2} – 4a\left( {a – 1} \right) = 4a\)
当\(\Delta > 0\)时,即当\(a > 0\)且\(a \ne 1\)时,方程有两个不相等的实数根\({{x}_{1}}=\frac{a+\sqrt{a}}{a-1}\ \ {{x}_{2}}=\frac{a-\sqrt{a}}{a-1}\);
当\(\Delta = 0\)时,即当\(a = 0\)时,方程有两个相等的实数根\({x_1} = {x_2} = 0\);
当\(\Delta < 0\)时,即当\(a < 0\)时,方程没有实数根.

题型二:解高次整式方程和方程组

求解高次方程与方程组的基本思路与解一元二次方程的思路一样,都可以采取消元以及降次的方法将原方程进行化简,但是有时候要注意可以采取换元法等进行计算会更加简便.

\({\color{red}{【例题】}}\)解下列方程:
⑴ \({x^3} – 2{x^2} = 3x\)
⑵ \({x^4} – 7{x^2} + 12 = 0\)
\({\color{red}{【解析】}}\)⑴ 移项,得\({x^3} – 2{x^2} – 3x = 0\).提取公因式并应用十字相乘法分解因式,得
\(x(x – 3)(x + 1) = 0\).所以原方程的解是 ,\({x_3} = – 1\).
⑵ 方程中只含有未知数的四次项、二次项与常数项,这是通常所称的双二次方程,
将\({x^2}\)作为一个整体分解因式,得\(({x^2} – 4)({x^2} – 3) = 0\),即\(\left( {x + 2} \right)\left( {x – 2} \right)\left( {x + \sqrt 3 } \right)\left( {x – \sqrt 3 } \right) = 0\),
从而原方程的解是\({x_1} = – 2\),\({x_2} = 2\),\({x_3} = – \sqrt 3 \),\({x_4} = \sqrt 3 \).

\({\color{red}{【练习】}}\)解方程组:
⑴\(\left\{ \begin{array}{l}x – y = 1\\{\left( {x + 1} \right)^2} + {y^2} = 10\end{array} \right.\) ⑵\(\left\{ {\begin{array}{l}2x + 3y – 5 = 0\\4{x^2} – 9{y^2} = 5.\end{array}} \right.\)

\({\color{red}{【解析】}}\)⑴对于由一个二元一次方程和一个二元二次方程所组成的方程组,基本解法是代入消元法.
标记①②,由①得\(x = y + 1\)③,代入②,得\({(y + 1 + 1)^2} + {y^2} = 10\),整理得\({y^2} + 2y – 3 = 0\),
解得\({y_1} = – 3\),\({y_2} = 1\).将\({y_1} = – 3\),\({y_2} = 1\)分别代入③,得\({x_1} = – 2\),\({x_2} = 2\),
所以原方程组的解是\(\left\{ \begin{array}{l}{x_1} = – 2\\{y_1} = – 3\end{array} \right.\),\(\left\{ {\begin{array}{l}{x_2} = 2\\{y_2} = 1\end{array}} \right.\)
⑵原方程组可化为\(\left\{ \begin{align}& 2x+3y=5\cdots①\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ & (2x+3y)(2x-3y)=5\cdots②\ \ \ \ \ \\ \end{align} \right.\),
将①代入②,得\(2x – 3y = 1 \cdots \)③,解由①、③组成的方程组,得原方程组的解是\(\left\{ {\begin{array}{l}x = \dfrac{3}{2}\\y = \dfrac{2}{3}\end{array}} \right.\)

题型三:解分式方程和无理方程

这个部分的题型要注意分式方程无理方程都需要注意的就是必须进行检验;一般来说,根号下含有未知数的方程叫做无理方程,解此类方程的思想是将无理方程转化为我们熟悉的有理方程进行求解,再一次强调要注意:无理方程和分式方程一样均需要进行检验,必须带入原方程进行检验是否满足题意.

\({\color{red}{【例题】}}\)\(\sqrt {3x + 2} = 2\)
\({\color{red}{【解析】}}\)两边同时完全平方得:\(3x + 2 = 4\),\(x = \dfrac{2}{3}\),将\(x = \dfrac{2}{3}\)代入\(3x + 2 > 0\),
所以原方程的解为\(x = \dfrac{2}{3}\)

\({\color{red}{【练习】}}\)解方程
(1)\(\dfrac{2}{{{x^2} – 4}} + \dfrac{{x – 4}}{{{x^2} + 2x}} = \dfrac{1}{{{x^2} – 2x}}\)
(2)\(\dfrac{{{x^2} – 5}}{{x – 1}} + \dfrac{{10x – 10}}{{{x^2} – 5}} = 7\)
(3)\(\dfrac{{x + 1}}{{x + 2}} + \dfrac{{x + 6}}{{x + 7}} = \dfrac{{x + 2}}{{x + 3}} + \dfrac{{x + 5}}{{x + 6}}\)
\({\color{red}{【解析】}}\)(1)原方程变形为\(\dfrac{2}{{\left( {x + 2} \right)\left( {x – 2} \right)}} + \dfrac{{x – 4}}{{x\left( {x + 2} \right)}} = \dfrac{1}{{x\left( {x – 2} \right)}}\).
两边同乘以\(x\left( {x + 2} \right)\left( {x – 2} \right)\),并整理得\({x^2} – 5x + 6 = 0\).
解得\({x_1} = 2\),\({x_2} = 3\).
经检验,\(x = 2\)是增根.
∴原方程的解为\(x = 3\).
(2)设\(y = \dfrac{{{x^2} – 5}}{{x – 1}}\).代入原方程,并整理得:
\({y^2} – 7y + 10 = 0\).
解得\({y_1} = 2\),\({y_2} = 5\).
由\(\dfrac{{{x^2} – 5}}{{x – 1}} = 2\)解得\({x_1} = 3\),\({x_2} = – 1\).
由\(\dfrac{{{x^2} – 5}}{{x – 1}} = 5\)解得\({x_3} = 0\),\({x_4} = 5\).
经检验,原方程的解是\({x_1} = 3\),\({x_2} = – 1\),\({x_3} = 0\),\({x_4} = 5\).
(3)原方程化为\(\dfrac{1}{{\left( {x + 6} \right)\left( {x + 7} \right)}} = \dfrac{1}{{\left( {x + 2} \right)\left( {x + 3} \right)}}\),即\(2x + 9 = 0\),解得\(x = – \dfrac{9}{2}\).

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