北师大版八年级下册数学《24第1课时一元一次不等式的解法》教案

北师大版八年级下册数学《2.4第1课时一元一次不等式的解法》教案_范文写作网

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1．理解一元一次不等式、不等式的解集、解不等式等概念；

2．把握一元一次不等式的解法．(重点，难点)

一、情境导入

1．什么叫一元一次方程？

2．解一元一次方程的一般步骤是什么？要留意什么？

3．假如把一元一次方程中的等号改为不等号，怎样求解？

二、合作探究

探究点一：一元一次不等式的概念

【类型一】一元一次不等式的识别

以下不等式中，是一元一次不等式的是()

A．5x－2＞0B．－3＜2＋

C．6x－3y≤－2D．y2＋1＞2

解析：选项A是一元一次不等式，选项B中含未知数的项不是整式，选项C中含有两个未知数，选项D中未知数的次数是2，应选项B，C，D都不是一元一次不等式，所以选A.

方法总结：假如一个不等式是一元一次不等式，必需满意三个条件：①含有一个未知数，②未知数的最高次数为1，③不等号的两边都是整式．

【类型二】依据一元一次不等式的概念求值

已知－x2a－1＋5＞0是关于x的一元一次不等式，则a的值是________．

解析：由－x2a－1＋5＞0是关于x的一元一次不等式得2a－1＝1，计算即可求出a的值，故a＝1.

方法总结：利用一元一次不等式的概念列出相应的方程求解即可．留意：假如未知数的系数中有字母，要检验此系数可不行能为零．

探究点二：一元一次不等式的解法

【类型一】一元一次不等式的解或解集

以下说法：①x＝0是2x－1＜0的一个解；②x＝－3不是3x－2＞0的解；③－2x＋1＜0的解集是x＞2.其中正确的个数是()

A．0个B．1个

C．2个D．3个

解析：①x＝0时，2x－1＜0成立，所以x＝0是2x－1＜0的一个解；②x＝－3时，3x－2＞0不成立，所以x＝－3不是3x－2＞0的解；③－2x＋1＜0的解集是x＞，所以不正确．应选C.

方法总结：推断一个数是不是不等式的解，只要把这个数代入不等式，看是否成立．推断一个不等式的解集是否正确，可把这个不等式化为“x＞a”或“x＜a”的形式，再进展比拟即可．

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解以下一元一次不等式，并在数轴上表示：

(1)2(x＋)－1≤－x＋9；

(2)－1＞.

解析：根据解一元一次不等式的根本步骤求解：去分母、去括号、移项、合并同类项、两边都除以未知数的系数．

解：(1)去括号，得2x＋1－1≤－x＋9，

移项、合并同类项，得3x≤9，

两边都除以3，得x≤3；

(2)去分母，得3(x－3)－6＞2(x－5)，

去括号，得3x－9－6＞2x－10，

移项，得3x－2x＞－10＋9＋6，

合并同类项，得x＞5.

方法总结：解一元一次不等式的根本步骤：去分母、去括号、移项、合并同类项、两边都除以