x1^5 +x2^5 = ? giải hộ vs ạ cảm ơn (ứng dụng viet ) 20/11/2021 Bởi Charlie x1^5 +x2^5 = ? giải hộ vs ạ cảm ơn (ứng dụng viet )
Đáp án+ Giải thích các bước giải: Ta có: $(a+b)^{5}$ =$a^{5}+5.a^{4}.b+10.a^{3}.b^{2}+10.a^{2}.b^{3}+5.a.b^{4}+b^{5}$ $x^{5}_{1}+x^{5}_{2}$ =$(x_{1}+x_{2})^{5}-5.x_{1}.x_{2}.(x^{3}_{1}+x^{3}_{2})-10.x^{2}_{1}.x^{2}_{2}.(x_{1}+x_{2})$ =$(x_{1}+x_{2})^{5}-5.x_{1}.x_{2}.((x_{1}+x_{2})^{3}-3.x_{1}.x_{2}.(x_{1}+x_{2}))-10.x^{2}_{1}.x^{2}_{2}.(x_{1}+x_{2})$ =$(-\frac{b}{a})^{5}-5.\frac{c}{a}.((-\frac{b}{a})^{3}-3.\frac{c}{a}.\frac{b}{a})-10.(\frac{c}{a})^{2}.-\frac{b}{a}$ =$(-\frac{b}{a})^{5}+5.\frac{c}{a}.((\frac{b}{a})^{3}+3.\frac{c}{a}.\frac{b}{a})+10.(\frac{c}{a})^{2}.\frac{b}{a}$ =$(-\frac{b}{a})^{5}+5.\frac{c}{a}.(\frac{b}{a})^{3}+25.(\frac{c}{a})^{2}.\frac{b}{a}$ Bình luận
Đáp án+ Giải thích các bước giải:
Ta có:
$(a+b)^{5}$
=$a^{5}+5.a^{4}.b+10.a^{3}.b^{2}+10.a^{2}.b^{3}+5.a.b^{4}+b^{5}$
$x^{5}_{1}+x^{5}_{2}$
=$(x_{1}+x_{2})^{5}-5.x_{1}.x_{2}.(x^{3}_{1}+x^{3}_{2})-10.x^{2}_{1}.x^{2}_{2}.(x_{1}+x_{2})$
=$(x_{1}+x_{2})^{5}-5.x_{1}.x_{2}.((x_{1}+x_{2})^{3}-3.x_{1}.x_{2}.(x_{1}+x_{2}))-10.x^{2}_{1}.x^{2}_{2}.(x_{1}+x_{2})$
=$(-\frac{b}{a})^{5}-5.\frac{c}{a}.((-\frac{b}{a})^{3}-3.\frac{c}{a}.\frac{b}{a})-10.(\frac{c}{a})^{2}.-\frac{b}{a}$
=$(-\frac{b}{a})^{5}+5.\frac{c}{a}.((\frac{b}{a})^{3}+3.\frac{c}{a}.\frac{b}{a})+10.(\frac{c}{a})^{2}.\frac{b}{a}$
=$(-\frac{b}{a})^{5}+5.\frac{c}{a}.(\frac{b}{a})^{3}+25.(\frac{c}{a})^{2}.\frac{b}{a}$