Áp dụng tính chất dãy tỉ số bằng nhau ta có :
\(\dfrac{x+1}{3}=\dfrac{y+2}{4}=\dfrac{z-3}{5}\Rightarrow\dfrac{3x+2y+4z+\left(1+2-3\right)}{3.3+2.4+4.5}=\dfrac{3x+2y+4z}{3.3+2.4+4.5}=\dfrac{47}{37}\)
\(\dfrac{x+1}{3}=\dfrac{47}{37}\Rightarrow x+1=\dfrac{141}{37}\Leftrightarrow x=\dfrac{107}{37}\)
\(\dfrac{y+2}{4}=\dfrac{47}{37}\Rightarrow y+2=\dfrac{188}{37}\Leftrightarrow y=\dfrac{114}{37}\)
\(\dfrac{z-3}{5}=\dfrac{47}{37}\Rightarrow z-3=\dfrac{235}{37}\Leftrightarrow z=\dfrac{346}{37}\)
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