## 2020年1月10日金曜日

### 数学 - 代数学 - 因数分解と分数式 - 比および比例式 - 性質、等式の証明

1. $\frac{a}{b}=\frac{c}{d}=k$

とおく。

$\begin{array}{l}a=bk\\ c=dk\\ \frac{a+2b}{2a+b}\\ =\frac{bk+2b}{2bk+b}\\ =\frac{k+2}{2k+1}\\ \frac{c+2d}{2c+d}\\ =\frac{dk+2d}{2dk+d}\\ =\frac{k+2}{2k+1}\\ \frac{a+2b}{2a+b}=\frac{c+2d}{2c+d}\end{array}$

2. $\begin{array}{l}\frac{pa+rb}{qa+sb}\\ =\frac{pbk+rb}{qbk+sb}\\ =\frac{pk+r}{qk+s}\\ \frac{pc+rd}{qc+sd}\\ =\frac{pdk+rd}{qdk+sd}\\ =\frac{pk+r}{qk+s}\\ \frac{pa+rb}{qa+sb}=\frac{pc+rd}{qc+sd}\end{array}$

3. $\begin{array}{l}\frac{{a}^{2}+{c}^{2}}{ab+cd}\\ =\frac{{b}^{2}{k}^{2}+{d}^{2}{k}^{2}}{bkb+dkd}\\ =k\\ \frac{ab+cd}{{b}^{2}+{d}^{2}}\\ =\frac{{b}^{2}k+{d}^{2}k}{{b}^{2}+{d}^{2}}\\ =k\\ \frac{{a}^{2}+{c}^{2}}{ab+cd}=\frac{ab+cd}{{b}^{2}+{d}^{2}}\end{array}$

4. $\begin{array}{l}\frac{{\left(a-c\right)}^{2}}{{\left(b-d\right)}^{2}}\\ =\frac{{\left(bk-dk\right)}^{2}}{{\left(b-d\right)}^{2}}\\ ={k}^{2}\\ \frac{{a}^{2}+{c}^{2}}{{b}^{2}+{d}^{2}}\\ =\frac{{b}^{2}{k}^{2}+{d}^{2}{k}^{2}}{{b}^{2}+{d}^{2}}\\ ={k}^{2}\end{array}$