题目内容

12.解方程
(1)30% x=120                  (2)x+$\frac{1}{5}$x=240(3)x-60% x=160                
(4)50%x-33%x=34(5)120x-20x=40               (6)x+130%x=460

分析 (1)依据等式的性质,方程两边同时除以0.3求解;
(2)先化简左边,依据等式的性质,方程两边同时乘$\frac{5}{6}$求解;
(3)先化简左边,依据等式的性质,方程两边同时除以0.4求解;
(4)先化简左边,依据等式的性质,方程两边同时除以0.17求解;
(5)先化简左边,依据等式的性质,方程两边同时除以100求解;
(6)先化简左边,依据等式的性质,方程两边同时除以2.3求解.

解答 解:(1)30% x=120                  
                  0.3x=120
            0.3x÷0.3=120÷0.3
                       x=400

(2)x+$\frac{1}{5}$x=240
            $\frac{6}{5}$x=240
        $\frac{6}{5}$x×$\frac{5}{6}$=240×$\frac{5}{6}$
                x=200

(3)x-60% x=160                
              0.4x=160
        0.4x÷0.4=160÷0.4
                   x=400

(4)50%x-33%x=34
                  0.17x=34
          0.17x÷0.17=34÷0.17
                        x=200

(5)120x-20x=40               
               100x=40
          100x÷10=40÷100
                      x=$\frac{2}{5}$

(6)x+130%x=460
                2.3x=460
          2.3x÷2.6=460÷2.3
                     x=200

点评 此题考查了运用等式的性质解方程,即等式两边同加上或同减去、同乘上或同除以一个数(0除外),两边仍相等,同时注意“=”上下要对齐.

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