tim x;y;z biết 3x=4y=5z va x-2y+3x=0
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Bài này dễ mà !?
Hiệu trên bằng :
\(3,26-1,549=1,711\)
Đáp số : 1, 711
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: \(\frac{2}{x}=\frac{3}{y};\frac{5}{y}=\frac{7}{z}\Rightarrow2y=3x;5z=7y.\)
\(\Rightarrow\frac{y}{3}=\frac{x}{2};\frac{z}{7}=\frac{y}{5}\Rightarrow\frac{x}{10}=\frac{y}{15}=\frac{z}{21}\)
\(\Rightarrow\frac{x}{10}=\frac{y}{15}=\frac{z}{21}=\frac{3x+5z-7y}{30+105-105}=\frac{30}{30}=1\)
Vậy x = 1 x 10 = 10
y = 1 x 15 = 15
z = 1 x 21 = 21
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|x+1| + x = 3
=> x+1 + x =3 hoặc x+1+x = -3
2x+1=3 2x+1=-3
2x=3-1 2x=-3-1
2x = 2 2x=-4
x=2:2 x=-4:2
x=1 x=-2
Vậy x=1; x=-2
![](https://rs.olm.vn/images/avt/0.png?1311)
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\(\frac{x-1}{2004}+\frac{x-2}{2003}=\frac{x-3}{2002}+\frac{x-4}{2001}\)
\(\Rightarrow\frac{x-1}{2004}-1+\frac{x-2}{2003}-1=\frac{x-3}{2002}-1+\frac{x-4}{2001}-1\)
\(\Rightarrow\frac{x-2005}{2004}+\frac{x-2005}{2003}=\frac{x-2005}{2002}+\frac{x-2005}{2001}\)
\(\Rightarrow\frac{x-2005}{2001}+\frac{x-2005}{2002}-\frac{x-2005}{2003}-\frac{x-2005}{2004}=0\)
\(\Rightarrow\left(x-2005\right).\left(\frac{1}{2001}+\frac{1}{2002}-\frac{1}{2003}-\frac{1}{2004}\right)=0\)
Vì \(\frac{1}{2001}>\frac{1}{2003};\frac{1}{2002}>\frac{1}{2004}\)
\(\Rightarrow\frac{1}{2001}+\frac{1}{2002}-\frac{1}{2003}-\frac{1}{2004}\ne0\)
\(\Rightarrow x-2005=0\)
\(\Rightarrow x=2005\)
Ta có: 3x = 4y = 5z
\(\Rightarrow\frac{x}{\frac{1}{3}}=\frac{y}{\frac{1}{4}}=\frac{z}{\frac{1}{5}}=\frac{x-2y+3z}{\frac{1}{3}-\frac{1}{2}+\frac{3}{5}}=\frac{0}{\frac{13}{30}}=0\)
=> x = 0 x 1/3 = 0
y = 0 x 1/4 = 0
z = 0 x 1/5 = 0