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a) \(\left(b+1\right)+b\left(a+1\right)=\left(a+1\right)\left(b+1\right)\)
\(\Leftrightarrow ab+2b+1=ab+a+b+1\)
\(\Leftrightarrow b=a\)
Câu a sai đề, hình như pk là \(\frac{a}{b}=1\)
b) \(2\left(a+1\right)\left(a+b\right)=\left(a+b\right)\left(a+b+2\right)\)
\(\Leftrightarrow\left(2a+2\right)\left(a+b\right)=\left(a+b\right)\left(a+b+2\right)\)
\(\Leftrightarrow\left(2a+2\right)\left(a+b\right)-\left(a+b\right)\left(a+b+2\right)=0\)
\(\Leftrightarrow\left(2a+2-a-b-2\right)\left(a+b\right)=0\)
\(\Leftrightarrow\left(a-b\right)\left(a+b\right)=0\)
\(\Leftrightarrow a^2-b^2=0\)
Hình như đề cx sai
\(\left(a+b-2c\right)^2+\left(b+c-2a\right)^2+\left(c+a-2b\right)^2=\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\)
\(\Leftrightarrow\hept{\begin{cases}a+b-2c=a-b\\b+c-2a=b-c\\c+a-2b=c-a\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}2b-2c=0\\2c-2a=0\\2a-2b=0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}b-c=0\\c-a=0\\a-b=0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}b=c\\c=a\\a=b\end{cases}}\)
\(\Leftrightarrow a=b=c\)( đpcm )
\(\Rightarrow\hept{\begin{cases}a+b-2c=a-b\\b+c-2a=b-c\\c+a-2b=a-c\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}2b-2c=0\\2c-2a=0\\2a-2b=0\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}b-c=0\\c-a=0\\a-b=0\end{cases}\Rightarrow\hept{\begin{cases}b=c\\c=a\\a=b\end{cases}\Rightarrow}a=b=c\left(dpcm\right)}\)
a, Ta có: \(\frac{a}{b}=\frac{c}{d}=k\left(k\ne0\right)\Rightarrow a=kb;c=kd\)
Thay:
\(\frac{ab}{cd}=\frac{b^2}{d^2}\)
\(\frac{\left(a+b\right)^2}{\left(c+d\right)^2}=\frac{b^2\left(k+1\right)^2}{d^2\left(k+1\right)^2}=\frac{b^2}{d^2}\)
=> đpcm
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 :
\(B=\left(-3\right).\left(-4\right).2.3=72\)
tính a,b,c hả bạn