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bài 1 : a,ta có 3/x-1 =4/y-2=5/z-3 => x-1/3=y-2/4=z-3/5
áp dụng .... => x-1+y-2+z-3 / 3+4+5 = x+y+z-1-2-3/3+4+5 = 12/12=1
do x-1/3 = 1 => x-1 = 3 => x= 4 ( tìm y,z tương tự
Bài 1:
a) Ta có: 3/x - 1 = 4/y - 2 = 5/z - 3 => x - 1/3 = y - 2/4 = z - 3/5 áp dụng ... =>x - 1 + y - 2 + z - 3/3 + 4 + 5 = x + y + z - 1 - 2 - 3/3 + 4 + 5 = 12/12 = 1 do x - 1/3 = 1 => x - 1 = 3 => x = 4 ( tìm y, z tương tự )
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\(\dfrac{1}{2}x+2\dfrac{1}{2}=3\dfrac{1}{2}x.\left(-\dfrac{1}{3}\right)\\ \Rightarrow\dfrac{1}{2}x+\dfrac{5}{2}=\dfrac{7}{2}x.\left(-\dfrac{1}{3}\right)\\ \Rightarrow\dfrac{1}{2}x+\dfrac{5}{2}+\dfrac{7}{2}x=-\dfrac{1}{3}\\ \Rightarrow\left(\dfrac{1}{2}+\dfrac{7}{2}\right)x+\dfrac{5}{2}=-\dfrac{1}{3}\\ \Rightarrow4x=-\dfrac{17}{6}\\ \Rightarrow x=-\dfrac{17}{24}.\)
\(\dfrac{1}{2}x+2\dfrac{1}{2}=3\dfrac{1}{2}x-\dfrac{1}{3}\\ \Rightarrow\dfrac{1}{2}x-3\dfrac{1}{2}x=-\dfrac{1}{3}-2\dfrac{1}{2}\\ \Rightarrow\left(\dfrac{1}{2}-\dfrac{7}{2}\right)x=-\dfrac{1}{3}-\dfrac{5}{2}\\ \Rightarrow\dfrac{-6}{2}x=-\dfrac{17}{6}\\ \Rightarrow-3x=-\dfrac{17}{6}\\ \Rightarrow x=\left(-\dfrac{17}{6}\right):\left(-3\right)\\ \Rightarrow x=\dfrac{17}{18}\)
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Bài 2:
a) Ta có: \(\left|x-2\right|=\left|4-x\right|\)
\(\Leftrightarrow x-2=4-x\)
\(\Leftrightarrow2x=6\)
hay x=3
b) Ta có: \(\left(\left|2x-1\right|-3\right)\cdot\left(-2\right)+\left(-5\right)=6\)
\(\Leftrightarrow\left(\left|2x-1\right|-3\right)\cdot\left(-2\right)=11\)
\(\Leftrightarrow\left|2x-1\right|-3=\dfrac{-11}{2}\)
\(\Leftrightarrow\left|2x-1\right|=\dfrac{-11}{2}+\dfrac{6}{2}=\dfrac{-5}{2}\)(Vô lý)