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khong co tim x

lộn

Mình là chủ nhân của câu hỏi này lên các bạn hãy bỏ một phân số 133phaanf 10 ra ngoài

   (139139 . 133 - 133133 . 139) : (2 + 4+ 6 + ... + 2002)

= (139 . 1001 . 133 - 133 . 1001 . 139) : (2 + 4 + 6 + ... + 2002)

= 0 : (2 + 4 + 6 + ... + 2002)

= 0

\(\frac{19}{x+y}+\frac{19}{y+z}+\frac{19}{z+x}=\frac{133}{10}\)

\(\Rightarrow19\left(\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}\right)=\frac{133}{10}\)

\(\Rightarrow\left(\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}\right)=\frac{7}{10}\)

\(\frac{7x}{y+z}+\frac{7y}{z+x}+\frac{7z}{x+y}=\frac{133}{10}\)

\(\Rightarrow7\left(\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y}\right)=\frac{133}{10}\)

\(\Rightarrow\left(\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y}\right)=\frac{19}{10}\)

\(\Rightarrow\left(\frac{x}{y+z}+1+\frac{y}{z+x}+1+\frac{z}{x+y}+1\right)=\frac{19}{10}+3\)

\(\Rightarrow\left(\frac{x+y+z}{y+z}+\frac{x+y+z}{z+x}+\frac{x+y+z}{x+y}\right)=\frac{49}{10}\)

\(\Rightarrow\left(x+y+z\right)\left(\frac{1}{y+z}+\frac{1}{z+x}+\frac{1}{x+y}\right)=\frac{49}{10}\)

\(\Rightarrow\left(x+y+z\right).\frac{7}{10}=\frac{49}{10}\)

\(\Rightarrow x+y+z=7\)

Vậy x + y + z = 7

 233.123-123.133+10²

= 123. ( 233 - 133 ) + 100

= 123. 100 + 100

= 12300 + 100

= 12400

=123*(233-133)+100

=123*100+100

=12300+100

=12400

cảm ơn nha

a: \(145+43+\left(-145\right)+57\)

\(=\left(145-145\right)+\left(43+57\right)\)

=0+100

=100

b: \(\left(-123\right)+101+\left(-777\right)+99\)

\(=\left(-123-777\right)+\left(101+99\right)\)

=-900+200

=-700

c: \(\left(-8955\right)+33+\left(-45\right)+\left(-133\right)\)

\(=\left(-8955-45\right)+\left(33-133\right)\)

\(=\left(-9000\right)+\left(-100\right)=-9100\)

d: \(\left(-2\right)+\left(-4\right)+\left(-6\right)+\left(-8\right)+\left(-10\right)+8+10+12\)

\(=\left(12-2\right)+\left(-4-6\right)+\left(-8+8\right)+\left(-10+10\right)\)

=10-10+0+0

=0