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Bài 1:
a) \(\frac{1}{5}x^4y^3-3x^4y^3\)
= \(\left(\frac{1}{5}-3\right)x^4y^3\)
= \(-\frac{14}{5}x^4y^3.\)
b) \(5x^2y^5-\frac{1}{4}x^2y^5\)
= \(\left(5-\frac{1}{4}\right)x^2y^5\)
= \(\frac{19}{4}x^2y^5.\)
Mình chỉ làm 2 câu thôi nhé, bạn đăng nhiều quá.
Chúc bạn học tốt!
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 :
\(B=\left(-3\right).\left(-4\right).2.3=72\)
a) \(x\ge\frac{2}{3}\Rightarrow3x-2\ge0\Rightarrow\left|3x-2\right|=3x-2\)
\(\Rightarrow P=\frac{1}{2}-\frac{1}{2}\left(x:\frac{1}{6}-\frac{1}{4}\right)-2\left(3x-2\right)\)
\(\Rightarrow P=\frac{1}{2}-\frac{1}{2}\left(6x-\frac{1}{4}\right)-6x+4\)
\(\Rightarrow P=\frac{4}{8}-3x+\frac{1}{8}-6x+\frac{32}{8}\)
\(\Rightarrow P=\frac{37}{8}-9x\)
b) \(x< \frac{2}{3}\Rightarrow3x-2< 0\Rightarrow\left|3x-2\right|=2-3x\)
\(\Rightarrow P=\frac{1}{2}-\frac{1}{2}\left(x:\frac{1}{6}-\frac{1}{4}\right)-2\left(2-3x\right)\)
\(\Rightarrow P=\frac{1}{2}-\frac{1}{2}\left(6x-\frac{1}{4}\right)-4+6x\)
\(\Rightarrow P=\frac{4}{8}-3x+\frac{1}{8}-\frac{32}{8}+6x\)
\(\Rightarrow P=\frac{-27}{8}+3x\)