2019年12月3日火曜日

数学 - 代数学 - 整式の計算 - 式の展開、工夫

1. $4{x}^{3}+13{x}^{2}+9x-2$

2. $\begin{array}{l}\left({x}^{2}-1\right)\left({x}^{2}+1\right)\\ ={x}^{4}-1\end{array}$

3. $\begin{array}{l}{\left(a+b\right)}^{2}+a+b-12\\ {a}^{2}+2ab+{b}^{2}+a+b-12\end{array}$

4. $\begin{array}{l}{\left(2a+3b\right)}^{2}-2\left(2a+3b\right)+{c}^{2}\\ =4{a}^{2}+12ab+9{b}^{2}-4a-6b+{c}^{2}\end{array}$

5. $\begin{array}{l}{x}^{3}+3{x}^{2}\left(-\frac{1}{3}\right)+3x·\frac{1}{9}-\frac{1}{27}\\ ={x}^{3}-{x}^{2}+\frac{1}{3}x-\frac{1}{27}\end{array}$

6. $\begin{array}{l}\left(2a+b\right)\left({\left(2a+b\right)}^{2}-6ab\right)\\ {\left(2a+b\right)}^{3}-12{a}^{2}b-6a{b}^{2}\\ =8{a}^{3}+3·4{a}^{2}b+3·2a·{b}^{2}+{b}^{3}-12{a}^{2}b-6a{b}^{2}\\ =8{a}^{3}+{b}^{3}\end{array}$

7. $\begin{array}{l}{\left({x}^{2}-{y}^{2}\right)}^{2}\\ ={x}^{4}-2{x}^{2}{y}^{2}+{y}^{4}\end{array}$

8. $\begin{array}{l}{\left({a}^{2}+2ab+{b}^{2}\right)}^{2}\\ ={\left({a}^{2}+{b}^{2}\right)}^{2}+4ab\left({a}^{2}+{b}^{2}\right)+4{a}^{2}{b}^{2}\\ ={a}^{4}+2{a}^{2}{b}^{2}+{b}^{4}+4{a}^{3}b+4a{b}^{3}+4{a}^{2}{b}^{2}\\ ={a}^{4}+4{a}^{3}b+6{a}^{2}{b}^{2}+4a{b}^{3}+{b}^{4}\end{array}$

9. $\begin{array}{l}\left({a}^{4}+4{a}^{3}b+6{a}^{2}{b}^{2}+4a{b}^{3}+{b}^{4}\right)\left(a+b\right)\\ ={a}^{5}+4{a}^{4}b+6{a}^{3}{b}^{2}+4{a}^{2}{b}^{3}+a{b}^{4}\\ +{a}^{4}b+4{a}^{3}{b}^{2}+6{a}^{2}{b}^{3}+4a{b}^{4}+{b}^{5}\\ ={a}^{5}+5{a}^{4}b+10{a}^{3}{b}^{2}+10{a}^{2}{b}^{3}+5a{b}^{4}+{b}^{5}\end{array}$

10. $\begin{array}{l}\left({x}^{2}-2x-8\right)\left({x}^{2}-2x-15\right)\\ ={\left({x}^{2}-2x\right)}^{2}-23\left({x}^{2}-2x\right)+120\\ ={x}^{4}-4{x}^{3}+4{x}^{2}-23{x}^{2}+46x+120\\ ={x}^{4}-4{x}^{3}-19{x}^{2}+46x+120\end{array}$

11. $\begin{array}{l}\left({a}^{2}-{b}^{2}\right)\left({a}^{2}+{b}^{2}\right)\left({a}^{4}+{b}^{4}\right)\\ =\left({a}^{4}-{b}^{4}\right)\left({a}^{4}+{b}^{4}\right)\\ ={a}^{8}-{b}^{8}\end{array}$

12. $\begin{array}{l}2{a}^{5}+\left(8b-b\right){a}^{4}+\left(-4{b}^{2}-4{b}^{2}+5{b}^{2}\right){a}^{3}+\left(2{b}^{3}+20{b}^{3}-3{b}^{3}\right){a}^{2}\\ +\left(-10{b}^{4}-12{b}^{4}\right)a+6{b}^{5}\\ =2{a}^{5}+7{a}^{4}b-3{a}^{3}{b}^{2}+19{a}^{2}{b}^{3}-22a{b}^{4}+6{b}^{5}\end{array}$