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Cho phân thức \(A=\frac{x^5+2x^4+2x^3-4x^2+3x+6}{x^2+2x-8}\)
a) Tìm tập xác định của A
b) Tìm các giá trị của x để A = 0
c) Rút gọn A
a) \(\dfrac{\left(a-b\right)\left(c-d\right)}{\left(b^2-a^2\right)\left(d^2-c^2\right)}=\dfrac{\left(a-b\right)\left(c-d\right)}{\left(a-b\right)\left(a+b\right)\left(c-d\right)\left(c+d\right)}=\dfrac{1}{\left(a+b\right)\left(c+d\right)}\)
b) \(\dfrac{m^4-m}{2m^2+2m+2}=\dfrac{m\left(m^3-1\right)}{2\left(m^2+m+1\right)}=\dfrac{m\left(m-1\right)\left(m^2+m+1\right)}{2\left(m^2+m+1\right)}=\dfrac{m\left(m-1\right)}{2}\)
c) \(\dfrac{ab^2+a^3-a^2b}{a^3+b^3}=\dfrac{a\left(b^2+a^2-ab\right)}{\left(a+b\right)\left(a^2-ab+b^2\right)}=\dfrac{a}{a+b}\)
\(\frac{a^2\left(b-c\right)+b^2\left(c-a\right)+c^2\left(a-b\right)}{^{^{ }}a^4\left(b^2-c^2\right)+b^4\left(c^2-a^2\right)+c^4\left(a^2-b^2\right)}\)
=\(\frac{a^2b-a^2c+b^2c-b^2a+c^2a-c^2b}{a^4b^2-a^4c^2+b^4c^2-b^4a^2+c^4a^2-c^4b^2}\)
*Rút gọn âm và dương đối nhau ( VD: \(a^2\)và\(-a^2\)), còn lại bạn tự tìm thêm nhé :)
\(\frac{b-c+c-a+a-b}{b^2-c^2+c^2-a^2+a^2-b^2}\)
Ta lại rút gọn các cặp đối nhau ( như trên VD)
Kết quả cuối cùng là 0
Đặt biểu thức đã cho là A
Xét tử: \(a^2\left(b-c\right)+b^2\left(c-a\right)+c^2\left(a-b\right)\)
\(=a^2b-a^2c+b^2c-b^2a+c^2\left(a-b\right)\)
\(=\left(a^2b-b^2a\right)-\left(a^2c-b^2c\right)+c^2\left(a-b\right)\)
\(=ab\left(a-b\right)-c\left(a-b\right)\left(a+b\right)+c^2\left(a-b\right)\)
\(=ab\left(a-b\right)-\left(a-b\right)\left(ca+bc\right)+c^2\left(a-b\right)\)
\(=\left(a-b\right)\left(ab-ca-bc+c^2\right)\)\(=\left(a-b\right)\left[a\left(b-c\right)-c\left(b-c\right)\right]=\left(a-b\right)\left(a-c\right)\left(b-c\right)\)
Xét mẫu : làm tương tự như trên ta được
\(a^4\left(b^2-c^2\right)+b^4\left(c^2-a^2\right)+c^4\left(a^2-b^2\right)=\left(a^2-b^2\right)\left(a^2-c^2\right)\left(b^2-c^2\right)\)
\(=\left(a-b\right)\left(a+b\right)\left(a-c\right)\left(a+c\right)\left(b-c\right)\left(b+c\right)\)
\(\Rightarrow A=\frac{1}{\left(a+b\right)\left(a+c\right)\left(b+c\right)}\)
Ta thấy: \(\frac{\left(a-b\right)\left(c-d\right)}{\left(b^2-a^2\right)\left(d^2-c^2\right)}=\frac{\left(b-a\right)\left(d-c\right)}{\left(b-a\right)\left(b+a\right)\left(d-c\right)\left(d+c\right)}=\frac{1}{\left(a+b\right)\left(c+d\right)}\)
\(\frac{\left(a-b\right)\left(c-d\right)}{\left(b^2-a^2\right)\left(d^2-c^2\right)}\)
\(=\frac{\left(a-b\right)\left(c-d\right)}{\left(b-a\right)\left(b+a\right)\left(d-c\right)\left(d+c\right)}\)
\(\frac{1}{\left(a+b\right)\left(c+d\right)}\)
ta có : a+b+c=0=>a+b=-c ; b+c=-a ; a+c=-b
ta có: M= \(\frac{2ab}{a^2+\left(b+c\right)\left(b-c\right)}+\frac{2bc}{b^2+\left(c+a\right)\left(c-a\right)}+\frac{2ca}{c^2+\left(a+b\right)\left(a-b\right)}\)
M=\(\frac{2ab}{a^2-a\left(b-c\right)}+\frac{2bc}{b^2-b\left(c-a\right)}+\frac{2ca}{c^2-c\left(a-b\right)}\)
M=\(\frac{2ab}{a\left(a-b+c\right)}+\frac{2bc}{b\left(b-c+a\right)}+\frac{2ca}{c\left(c-a+b\right)}\)
M=\(\frac{2ab}{-ab+\left(a+c\right)}+\frac{2bc}{-bc+\left(a+b\right)}+\frac{2ac}{-ac+\left(b+c\right)}\)
M=\(\frac{2ab}{-2ab}+\frac{2bc}{-2bc}+\frac{2ca}{-2ca}\)
M=-1-1-1=-3
Vậy với a+b+c=0 thì M=-3
\(\frac{\left(a-b\right)\left(c-d\right)}{\left(b^2-a^2\right)\left(d^2-c^2\right)}=\frac{\left(b-a\right)\left(d-c\right)}{\left(b-a\right)\left(b+a\right)\left(d-c\right)\left(d+c\right)}=\frac{1}{\left(a+b\right)\left(c+d\right)}\)
\(\frac{m^4-m}{2m^2+2m+2}=\frac{m\left(m^3-1\right)}{2m^2+2m+2}=\frac{m\left(m-1\right)\left(m^2+m+1\right)}{2\left(m^2+m+1\right)}=\frac{m\left(m-1\right)}{2}\)