北師大版八年級(jí)下冊(cè)數(shù)學(xué)《24第1課時(shí)一元一次不等式的解法》教案

北師大版八年級(jí)下冊(cè)數(shù)學(xué)《2.4第1課時(shí)一元一次不等式的解法》教案_范文寫作網(wǎng)

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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．(重點(diǎn)，難點(diǎn))

一、情境導(dǎo)入

1．什么叫一元一次方程？

2．解一元一次方程的一般步驟是什么？要留意什么？

3．假如把一元一次方程中的等號(hào)改為不等號(hào)，怎樣求解？

二、合作探究

探究點(diǎn)一：一元一次不等式的概念

【類型一】一元一次不等式的識(shí)別

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：選項(xiàng)A是一元一次不等式，選項(xiàng)B中含未知數(shù)的項(xiàng)不是整式，選項(xiàng)C中含有兩個(gè)未知數(shù)，選項(xiàng)D中未知數(shù)的次數(shù)是2，應(yīng)選項(xiàng)B，C，D都不是一元一次不等式，所以選A.

方法總結(jié)：假如一個(gè)不等式是一元一次不等式，必需滿意三個(gè)條件：①含有一個(gè)未知數(shù)，②未知數(shù)的最高次數(shù)為1，③不等號(hào)的兩邊都是整式．

【類型二】依據(jù)一元一次不等式的概念求值

已知－x2a－1＋5＞0是關(guān)于x的一元一次不等式，則a的值是________．

解析：由－x2a－1＋5＞0是關(guān)于x的一元一次不等式得2a－1＝1，計(jì)算即可求出a的值，故a＝1.

方法總結(jié)：利用一元一次不等式的概念列出相應(yīng)的方程求解即可．留意：假如未知數(shù)的系數(shù)中有字母，要檢驗(yàn)此系數(shù)可不行能為零．

探究點(diǎn)二：一元一次不等式的解法

【類型一】一元一次不等式的解或解集

以下說(shuō)法：①x＝0是2x－1＜0的一個(gè)解；②x＝－3不是3x－2＞0的解；③－2x＋1＜0的解集是x＞2.其中正確的個(gè)數(shù)是()

A．0個(gè)B．1個(gè)

C．2個(gè)D．3個(gè)

解析：①x＝0時(shí)，2x－1＜0成立，所以x＝0是2x－1＜0的一個(gè)解；②x＝－3時(shí)，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＞，所以不正確．應(yīng)選C.

方法總結(jié)：推斷一個(gè)數(shù)是不是不等式的解，只要把這個(gè)數(shù)代入不等式，看是否成立．推斷一個(gè)不等式的解集是否正確，可把這個(gè)不等式化為“x＞a”或“x＜a”的形式，再進(jìn)展比擬即可．

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解以下一元一次不等式，并在數(shù)軸上表示：

(1)2(x＋)－1≤－x＋9；

(2)－1＞.

解析：根據(jù)解一元一次不等式的根本步驟求解：去分母、去括號(hào)、移項(xiàng)、合并同類項(xiàng)、兩邊都除以未知數(shù)的系數(shù)．

解：(1)去括號(hào)，得2x＋1－1≤－x＋9，

移項(xiàng)、合并同類項(xiàng)，得3x≤9，

兩邊都除以3，得x≤3；

(2)去分母，得3(x－3)－6＞2(x－5)，

去括號(hào)，得3x－9－6＞2x－10，

移項(xiàng)，得3x－2x＞－10＋9＋6，

合并同類項(xiàng)，得x＞5.

方法總結(jié)：解一元一次不等式的根本步驟：去分母、去括號(hào)、移項(xiàng)、合并同類項(xiàng)、兩邊都除以