Rút gọn
D=\(\dfrac{a^2}{\left(a+b\right)\left(1-b\right)}-\dfrac{b^2}{\left(a+b\right)\left(1+a\right)}-\dfrac{a^2b^2}{\left(1+a\right)\left(1-b\right)}\)
K
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26 tháng 11 2021
\(B=\left(\dfrac{a-b}{a^2+ab}-\dfrac{a}{b^2+ab}\right):\left(\dfrac{b^3}{a^3-ab^2}+\dfrac{1}{a+b}\right)\)
\(=\left(\dfrac{a-b}{a\left(a+b\right)}-\dfrac{a}{b\left(a+b\right)}\right):\left(\dfrac{b^3}{a\left(a-b\right)\left(a+b\right)}+\dfrac{1}{a+b}\right)\)
\(=\dfrac{b\left(a-b\right)-a^2}{ab\left(a+b\right)}:\dfrac{b^3+a\left(a-b\right)}{a\left(a-b\right)\left(a+b\right)}\)
\(=\dfrac{ab-b^2-a^2}{ab\left(a+b\right)}\cdot\dfrac{a\left(a-b\right)\left(a+b\right)}{a^2-ab+b^3}\)
\(=\dfrac{\left(a-b\right)\left(ab-b^2-a^2\right)}{b\left(a^2-ab+b^3\right)}\)
\(=\dfrac{-\left(a-b\right)\left(a^2-ab+b^2\right)}{b\left(a^2-ab+b^3\right)}\)
Đề lỗi rồi chứ mình ko rút gọn đc nữa
N
19 tháng 11 2021
Bài 3:
\(a,A=\dfrac{x^2+xy-xy+y^2}{\left(x-y\right)\left(x+y\right)}:\dfrac{x^2+2xy+y^2-2xy}{\left(x-y\right)\left(x+y\right)}\\ A=\dfrac{x^2+y^2}{\left(x-y\right)\left(x+y\right)}\cdot\dfrac{\left(x-y\right)\left(x+y\right)}{x^2+y^2}=1\\ b,=\left[\dfrac{\left(1-\sqrt{a}\right)\left(a+\sqrt{a}+1\right)}{1-\sqrt{a}}+\sqrt{a}\right]\left[\dfrac{1-\sqrt{a}}{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}\right)}\right]^2\\ =\left(a+2\sqrt{a}+1\right)\left(\dfrac{1}{\sqrt{a}+1}\right)^2\\ =\left(\sqrt{a}+1\right)^2\cdot\dfrac{1}{\left(\sqrt{a}+1\right)^2}=1\)
31 tháng 3 2017
cái này tương tự này, do dài quá nên ngại làm, bn tham khảo nhé Câu hỏi của Thiên An - Toán lớp 9 - Học toán với OnlineMath
NV
Nguyễn Việt Lâm
Giáo viên
20 tháng 12 2020
\(B=\left(ab+bc+ca\right)\left(\dfrac{ab+bc+ca}{abc}\right)-abc\left(\dfrac{a^2b^2+b^2c^2+c^2a^2}{a^2b^2c^2}\right)\)
\(=\dfrac{\left(ab+bc+ca\right)^2-\left(a^2b^2+b^2c^2+c^2a^2\right)}{abc}\)
\(=\dfrac{a^2b^2+b^2c^2+c^2a^2+2abc\left(a+b+c\right)-\left(a^2b^2+b^2c^2+c^2a^2\right)}{abc}\)
\(=2\left(a+b+c\right)\)
TK
23 tháng 12 2018
1)\(\dfrac{c-b}{\left(a-b\right)\left(c-b\right)\left(a-c\right)}+\dfrac{a-c}{\left(b-a\right)\left(b-c\right)\left(a-c\right)}+\dfrac{b-a}{\left(b-a\right)\left(c-b\right)\left(c-a\right)}=\dfrac{c-b+a-c+b-c}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=0\)
D\(=\dfrac{a^2\left(a+1\right)+b^2\left(b-1\right)+a^2b^2\left(a+b\right)}{\left(a+b\right)\left(1-b\right)\left(a+1\right)}=\dfrac{a^3+a^2+b^3-b^2+a^3b^2+a^2b^3}{\left(a+b\right)\left(1-b\right)\left(a+1\right)}\)
\(ĐKXĐ:a\ne-b;a\ne-1;b\ne1\)
\(D=\dfrac{a^2}{\left(a+b\right)\left(1-b\right)}-\dfrac{b^2}{\left(a+b\right)\left(1+a\right)}-\dfrac{a^2b^2}{\left(1+a\right)\left(1-b\right)}\)
\(=\dfrac{a^2\left(1+a\right)-b^2\left(1-b\right)-a^2b^2\left(a+b\right)}{\left(a+b\right)\left(1+a\right)\left(1-b\right)}\)
\(=\dfrac{a^3+a^2-b^2+b^3-a^2b^2\left(a+b\right)}{\left(a+b\right)\left(1+a\right)\left(1-b\right)}\)
\(=\dfrac{\left(a+b\right)\left(a^2-ab+b^2\right)+\left(a+b\right)\left(a-b\right)-a^2b^2\left(a+b\right)}{\left(a+b\right)\left(1+a\right)\left(1-b\right)}\)
\(=\dfrac{\left(a+b\right)\left(a^2-ab+b^2+a-b-a^2b^2\right)}{\left(a+b\right)\left(1+a\right)\left(1-b\right)}\)
\(=\dfrac{a\left(a-b\right)+\left(a-b\right)+b^2\left(1-a^2\right)}{\left(1+a\right)\left(1-b\right)}\)
\(=\dfrac{\left(a-b\right)\left(a+1\right)-b^2\left(a-1\right)\left(a+1\right)}{\left(1+a\right)\left(1-b\right)}\)
\(=\dfrac{\left(1+a\right)\left(a-b-ab^2+b^2\right)}{\left(1+a\right)\left(1-b\right)}\)
\(=\dfrac{a-b-ab^2+b^2}{1-b}\)
\(=\dfrac{b\left(b-1\right)-a\left(b^2-1\right)}{1-b}=\dfrac{a\left(1-b\right)\left(1+b\right)-b\left(1-b\right)}{1-b}=\dfrac{\left(1-b\right)\left(a+ab-b\right)}{1-b}=a+ab-b\)