cho ab=cd hãy rút gọn biểu thức P=\(\dfrac{\left(a+c\right)\left(a+d\right)\left(b+c\right)\left(b+d\right)}{\left(a+b+c+d\right)^2}\)
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DT
2 tháng 2 2020
P=\(\frac{\left(a+c\right)\left(a+d\right)\left(b+c\right)\left(b+d\right)}{\left(a+b+c+d\right)^2}\)=\(\frac{\left(a^2+ad+ac+cd\right)\left(b^2+bd+bc+cd\right)}{\left(a+b+c+d\right)^2}\)
=\(\frac{\left(a^2+ac+ad+ab\right)\left(b^2+bc+bd+ab\right)}{\left(a+b+c+d\right)^2}\) (do ab=cd)
=\(\frac{a\left(a+b+c+d\right)b\left(a+b+c+d\right)}{\left(a+b+c+d\right)^2}\)
=\(\frac{ab\left(a+b+c+d\right)^2}{\left(a+b+c+d\right)^2}\)=ab
TT
17 tháng 8 2020
a) \(\frac{a^2m-a^2n-b^2n+b^2m}{a^2+b^2}=\frac{a^2\left(m-n\right)+b^2\left(m-n\right)}{a^2+b^2}\)
\(=\frac{\left(m-n\right)\left(a^2+b^2\right)}{a^2+b^2}=m-n\)
b) \(\frac{\left(ab+bc+cd+ad\right)abcd}{\left(c+d\right)\left(a+b\right)+\left(b-c\right)\left(a-b\right)}\)
\(=\frac{\left[b.\left(a+c\right)+d.\left(a+c\right)\right].abcd}{ac+bc+da+db+ab-b^2-ca+bc}\)
\(=\frac{\left(a+c\right)\left(d+b\right)abcd}{2bc+da+db+ab-b^2}\)
20 tháng 5 2022
\(A=\dfrac{-\left(ac+bc+ad+bd\right)-\left(cd-ca-bd+ba\right)}{\left(ab+bc+cd+ad\right)\cdot abcd}\)
\(=\dfrac{-ac-bc-ad-bd-cd+ca+bd-ba}{\left(ab+bc+cd+ad\right)\cdot abcd}\)
\(=\dfrac{-bc-ad-cd-ba}{\left(ab+bc+cd+ad\right)\cdot abcd}=-\dfrac{1}{abcd}\)
18 tháng 7 2019
\(\left(a+b+c+d\right)^2+\left(a+b-c-d\right)^2+\left(a-b+c-d\right)^2+\left(a-b-c+d\right)^2\)(Sửa lại nha bn viết sai để)
Đặt x=a+b , y=c+d , z=a-b , t=c-d
Khi đó biểu thức bằng
\(\left(x+y\right)^2+\left(x-y\right)^2+\left(z+t\right)^2+\left(z-t\right)^2\)
\(=x^2+y^2+2xy+x^2+y^2-2xy+z^2+t^2+2zt+z^2+t^2-2zt\)
\(=2\left(x^2+y^2+z^2+t^2\right)=2\left[\left(a+b\right)^2+\left(a-b\right)^2+\left(c+d\right)^2+\left(c-d\right)^2\right]\)
\(=2(a^2+b^2-2ab+a^2+b^2-2ab+c^2+d^2+2cd+c^2+d^2-2cd)\)
\(=2\left(2a^2+2b^2+2c^2+2d^2\right)=4\left(a^2+b^2+c^2+d^2\right)\)
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
help me please
ab=cd
a=b=c=d=0
P khong xac dinh
de thieu