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Ta có : P + Q hay
\(8xy+5x-2y-5x-2y=8xy-4y\)
Vậy \(P+Q=8xy-4y\)
\(\left(7x-13\right)+\left(6x-18\right)-\left(4x+7\right)=3\)
\(\Rightarrow7x-13+6x-18-4x-7-3=0\)
\(\Rightarrow\left(7x+6x-4x\right)-\left(18+13+3+7\right)=0\)
\(\Rightarrow\left(13x-4x\right)-\left(18+16+7\right)=0\)
\(\Rightarrow9x-\left(18+23\right)=0\)
\(\Rightarrow9x-41=0\)
\(\Rightarrow9x=0+41\)
\(\Rightarrow9x=41\)
\(\Rightarrow x=41:9\)
\(\Rightarrow x=41.\frac{1}{9}=\frac{41}{9}\)
Vậy \(x=\frac{41}{9}\)
(=) 7x-13+6x-18-4x-7=3
(=)7x+6x-4x=3+13+18+7
(=)9x41
(=)x=41/9
\(\left(7xy-9xy+6x\right)+A=5xy+11x\)
\(A=5xy+11x-\left(7xy-9xy+6x\right)\)
\(A=5xy+11x-7xy+9xy-6x\)
\(A=\left(5xy-7xy+9xy\right)+\left(11x-6x\right)\)
\(A=7xy+5x\)
P(x)=-3x+2x+4x-2
P(x)=3x-2
Có : Q(x)+P(x)=0
=>Q(x)+(3x-2)=0
=>Q(x)=-(3x-2)
#H
\(Q\left(x\right)=-P\left(x\right)\)
<=>\(Q\left(x\right)=-\left(-3x+2x+4x-2\right)\)
<=>\(Q\left(x\right)=3x-2x-4x+2\)
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Do ad = bc
=> adcd=bccdadcd=bccd
=> ac=bd(đpcm)
Sai