Rút gọn các Biểu Thức sau
a)\(\frac{a^2m-a^2n-b^2n+b^2m}{a^2+b^2}\)
b)\(\frac{\left(ab+bc+cd+da\right)abcd}{\left(c+d\right)\left(a+b\right)+\left(b-c\right)\left(a-d\right)}\)
Rút gọn rồi tính gt biểu thức :
a ) \(\frac{ax^2\left(a-x\right)-a^2x\left(x-a\right)}{3a^2-3x^2}\) với \(a=\frac{1}{2};x=-3\)
b ) \(\frac{\left(ab+bc+ca+da\right)abcd}{\left(c+d\right)\left(a+b\right)+\left(b-c\right)\left(a-d\right)}\) với \(a=-3;b=-4;c=2;d=3\).
a ) \(A=\frac{ax^2\left(a-x\right)-a^2x\left(x-a\right)}{3a^2-3x^2}=\frac{ax\left(a-x\right)\left(a+x\right)}{3\left(a-x\right)\left(a+x\right)}=\frac{ax}{3}\)
Thay \(a=\frac{1}{2};x=-3\), ta có :
\(A=\frac{\frac{1}{2}.-3}{3}=-\frac{1}{2}\)
b ) \(B=\frac{\left(ab+bc+cd+da\right)abcd}{\left(c+d\right)\left(a+b\right)+\left(b-c\right)\left(a-d\right)}=\frac{\left[\left(ab+ad\right)+\left(bc+cd\right)\right]abcd}{ca+cb+da+db+ba-bd-ca+cd}\)
\(=\frac{\left[a\left(b+d\right)+c\left(b+d\right)\right]abcd}{ba+da+cb+cd}=\frac{\left(b+d\right)\left(a+c\right)abcd}{\left(b+d\right)\left(a+c\right)}=abcd\)
Thay \(a=-3;b=-4;c=2;d=3\), ta có :
\(B=\left(-3\right).\left(-4\right).2.3=72\)
Rút gọn các phân thức :
a, \(\frac{\left(a-b\right)\left(c-d\right)}{\left(b^2-a^2\right)\left(d^2-c^2\right)}\)
b, \(\frac{m^4-m}{2m^2+2m+2}\)
\(\frac{\left(a-b\right)\left(c-d\right)}{\left(b^2-a^2\right)\left(d^2-c^2\right)}=\frac{\left(b-a\right)\left(d-c\right)}{\left(b-a\right)\left(b+a\right)\left(d-c\right)\left(d+c\right)}=\frac{1}{\left(a+b\right)\left(c+d\right)}\)
\(\frac{m^4-m}{2m^2+2m+2}=\frac{m\left(m^3-1\right)}{2m^2+2m+2}=\frac{m\left(m-1\right)\left(m^2+m+1\right)}{2\left(m^2+m+1\right)}=\frac{m\left(m-1\right)}{2}\)
rút gọn biểu thức
\(\frac{a^2-bc}{\left(a+b\right)\left(a+c\right)}+\frac{b^2-ca}{\left(b+c\right)\left(b+a\right)}+\frac{c^2-ab}{\left(c+a\right)\left(c+b\right)}\)
\(\frac{\text{(ab+bc+cd+da)abcd}}{\left(c+d\right)\left(a+b\right)+\left(b-c\right)\left(a-d\right)}\)
\(\frac{\left(ab+bc+cd+ad\right).abcd}{\left(c+d\right)\left(a+b\right)+\left(b-c\right)\left(a-d\right)}\)\(=\frac{\left(ab+bc+cd+ad\right)abcd}{ac+ad+bc+bd+ab+cd-ac-bd}\)
\(=\frac{\left(ab+bc+ad+cd\right).abcd}{ab+bc+cd+ad}\)\(=abcd\)
Cho ab=cd, hãy rút gọn biểu thức
P=\(\frac{\left(a+c\right)\left(a+d\right)\left(b+c\right)\left(b+d\right)}{\left(a+b+c+d\right)^2}\)
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
Rút gọn biểu thức:
A=\(\frac{bc}{\left(a-b\right)\left(a-c\right)}+\frac{ac}{\left(b-c\right)\left(b-a\right)}+\frac{ab}{\left(c-a\right)\left(c-b\right)}\)
Giúp mik tick cho.
Rút gọn biểu thức :
\(\frac{a^2\left(a+b\right)\left(a+c\right)}{\left(a-b\right)\left(a-c\right)}+\frac{b^2\left(b+a\right)\left(b+c\right)}{\left(b-a\right)\left(b-c\right)}+\frac{c^2\left(c+a\right)\left(c+b\right)}{\left(c-a\right)\left(c-b\right)}\)
Rút gọ biểu thức
\(P=\frac{a^2-bc}{\left(a+b\right)\left(a+c\right)}+\frac{b^2-ac}{\left(b+c\right)\left(b+a\right)}+\frac{c^2-ab}{\left(c+a\right)\left(c+b\right)}\)
\(P=\frac{a^2-bc}{\left(a+b\right)\left(a+c\right)}+\frac{b^2-ac}{\left(b+c\right)\left(b+a\right)}+\frac{c^2-ab}{\left(c+a\right)\left(c+b\right)}\)
\(P=\frac{\left(a^2-bc\right)\left(b+c\right)}{\left(a+b\right)\left(a+c\right)\left(b+c\right)}+\frac{\left(b^2-ac\right)\left(c+a\right)}{\left(b+c\right)\left(b+a\right)\left(c+a\right)}+\frac{\left(c^2-ab\right)\left(b+a\right)}{\left(c+a\right)\left(c+b\right)\left(b+a\right)}\)
\(P=\frac{a^2b+a^2c-b^2c-bc^2}{\left(a+b\right)\left(a+c\right)\left(b+c\right)}+\frac{b^2a+b^2c-a^2c-ac^2}{\left(b+c\right)\left(b+a\right)\left(c+a\right)}+\frac{c^2a+c^2b-a^2b-b^2a}{\left(c+a\right)\left(c+b\right)\left(b+a\right)}\)
\(P=\frac{0}{\left(a+b\right)\left(a+c\right)\left(b+c\right)}\)
\(P=0\)
Xét: \(f\left(x\right)=\frac{x^2-bc}{\left(x+b\right)\left(x+c\right)}+\frac{b^2-xc}{\left(b+c\right)\left(b+x\right)}+\frac{c^2-xb}{\left(c+x\right)\left(c+b\right)}\)
\(\Rightarrow f\left(a\right)=P\)
Ta có: \(f\left(b\right)=\frac{b^2-bc}{2b\left(b+c\right)}+\frac{b^2-bc}{2b\left(b+c\right)}+\frac{c^2-b^2}{\left(c+b\right)\left(c+b\right)}\)
\(\Rightarrow f\left(b\right)=\frac{2b\left(b-c\right)}{2b\left(b+c\right)}+\frac{\left(c-b\right)\left(c+b\right)}{\left(c+b\right)\left(c+b\right)}=\frac{b-c}{b+c}+\frac{c-b}{c+b}=0\left(1\right)\)
Chứng minh tương tự ta cũng có: \(f\left(c\right)=0\left(2\right)\)
Từ (1) và (2) suy ra \(f\left(x\right)=0\left(\forall x\right)\Rightarrow f\left(a\right)=0\left(\forall x\right)\)
Vậy A =0
mình ghi sai rồi \(P=1\)mới đúng chứ kg phải A =1
Rút gọn biểu thức :
1. \(\frac{2^{4m}-2^{4n}}{2^{2n}+2^{2m}}\)
2. \(\frac{a^2\left(b-c\right)+b^2\left(c-a\right)+c^2\left(a-b\right)}{ab^2-ac^2-b^3+bc^2}\)
3. \(\frac{2x^3-7x^2-12x+45}{3x^3-19x^2+33x-9}\)
4. \(\frac{x^3-y^3+z^3+3xyz}{\left(x+y\right)^2+\left(y+x\right)^2+\left(z-x\right)^2}\)
5. \(\frac{x^3+y^3+x^3-3xyz}{\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2}\)
Please, help me!~~~ Pt2