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22 tháng 7 2021

31(xyzt+xy+xt+zt+1)=40(yzt+y+t)31(xyzt+xy+xt+zt+1)=40(yzt+y+t)

⇒xyzt+xy+xt+zt+1yzt+y+t=4031⇒xyzt+xy+xt+zt+1yzt+y+t=4031

⇒x(yzt+y+t)+zt+1yzt+y+t=4031⇒x(yzt+y+t)+zt+1yzt+y+t=4031

⇒x+zt+1yzt+y+t=4031⇒x+zt+1yzt+y+t=4031

⇒x+1(yzt+y+tzt+1)=4031⇒x+1(yzt+y+tzt+1)=4031

⇒x+1(y+tzt+1)=4031⇒x+1(y+tzt+1)=4031

⇒x+1y+1(zt+1t)=4031⇒x+1y+1(zt+1t)=4031

⇒x+1y+1z+1t=4031⇒x+1y+1z+1t=4031

4031<6231=2⇒x<24031<6231=2⇒x<2

Với x = 0; có :

1y+1z+1t=40311y+1z+1t=4031

⇒y+1z+1t=3140⇒y+1z+1t=3140

Mà 3140<1⇒y<1⇒y=03140<1⇒y<1⇒y=0

⇒1z+1t=3140⇒1z+1t=3140

⇒z+1t=4031⇒z+1t=4031

⋅z=0⇒t=3140∉Z⋅z=0⇒t=3140∉Z(Loại )

⋅z=1⇒t=319∉Z⋅z=1⇒t=319∉Z(Loại )

Với x=1;x=1;ta có :

1y+1z+1t=4031−11y+1z+1t=4031−1

⇒1y+1z+1t=931⇒1y+1z+1t=931

⇒y+1z+1t=319⇒y+1z+1t=319

319<369=4⇒y<4319<369=4⇒y<4

⋅y=0⇒z+1t=931⇒z=0⇒t=319∉Z⋅y=0⇒z+1t=931⇒z=0⇒t=319∉Z(Loại)

⋅y=1⇒z+1t=922⇒z=0⇒t=229∉Z⋅y=1⇒z+1t=922⇒z=0⇒t=229∉Z(Loại)

⋅y=2⇒z+1t=913⇒z=0⇒t=139∉Z⋅y=2⇒z+1t=913⇒z=0⇒t=139∉Z(Loại )

⋅y=3⇒z+1t=94⋅y=3⇒z+1t=94

94<3⇒z<394<3⇒z<3

z=0⇒t=49∉Zz=0⇒t=49∉Zz=1⇒t=45∉Zz=1⇒t=45∉Zz=2⇒t=4z=2⇒t=4( Thỏa mãn )

Vậy x=1;y=3;z=2;t=4.

28 tháng 8 2016

Ta có :

\(31\left(xyzt+xy+xt+zt+1\right)=40\left(yzt+y+t\right)\)

\(\Rightarrow\frac{xyzt+xy+xt+zt+1}{yzt+y+t}=\frac{40}{31}\)

\(\Rightarrow\frac{x\left(yzt+y+t\right)+zt+1}{yzt+y+t}=\frac{40}{31}\)

\(\Rightarrow x+\frac{zt+1}{yzt+y+t}=\frac{40}{31}\)

\(\Rightarrow x+\frac{1}{\left(\frac{yzt+y+t}{zt+1}\right)}=\frac{40}{31}\)

\(\Rightarrow x+\frac{1}{\left(y+\frac{t}{zt+1}\right)}=\frac{40}{31}\)

\(\Rightarrow x+\frac{1}{y+\frac{1}{\left(\frac{zt+1}{t}\right)}}=\frac{40}{31}\)

\(\Rightarrow x+\frac{1}{y+\frac{1}{z+\frac{1}{t}}}=\frac{40}{31}\)

\(\frac{40}{31}< \frac{62}{31}=2\Rightarrow x< 2\)

Với x = 0; có :

\(\frac{1}{y+\frac{1}{z+\frac{1}{t}}}=\frac{40}{31}\)

\(\Rightarrow y+\frac{1}{z+\frac{1}{t}}=\frac{31}{40}\)

Mà \(\frac{31}{40}< 1\Rightarrow y< 1\Rightarrow y=0\)

\(\Rightarrow\frac{1}{z+\frac{1}{t}}=\frac{31}{40}\)

\(\Rightarrow z+\frac{1}{t}=\frac{40}{31}\)

\(\cdot z=0\Rightarrow t=\frac{31}{40}\notin Z\)(Loại )

\(\cdot z=1\Rightarrow t=\frac{31}{9}\notin Z\)(Loại )

Với \(x=1;\)ta có :

\(\frac{1}{y+\frac{1}{z+\frac{1}{t}}}=\frac{40}{31}-1\)

\(\Rightarrow\frac{1}{y+\frac{1}{z+\frac{1}{t}}}=\frac{9}{31}\)

\(\Rightarrow y+\frac{1}{z+\frac{1}{t}}=\frac{31}{9}\)

\(\frac{31}{9}< \frac{36}{9}=4\Rightarrow y< 4\)

\(\cdot y=0\Rightarrow z+\frac{1}{t}=\frac{9}{31}\Rightarrow z=0\Rightarrow t=\frac{31}{9}\notin Z\)(Loại)

\(\cdot y=1\Rightarrow z+\frac{1}{t}=\frac{9}{22}\Rightarrow z=0\Rightarrow t=\frac{22}{9}\notin Z\)(Loại)

\(\cdot y=2\Rightarrow z+\frac{1}{t}=\frac{9}{13}\Rightarrow z=0\Rightarrow t=\frac{13}{9}\notin Z\)(Loại )

\(\cdot y=3\Rightarrow z+\frac{1}{t}=\frac{9}{4}\)

\(\frac{9}{4}< 3\Rightarrow z< 3\)

  1. \(z=0\Rightarrow t=\frac{4}{9}\notin Z\)
  2. \(z=1\Rightarrow t=\frac{4}{5}\notin Z\)
  3. \(z=2\Rightarrow t=4\)( Thỏa mãn )

Vậy \(x=1;y=3;z=2;t=4.\)

4 tháng 2 2022

Answer:

\(P=\frac{1}{1+x+xy+xyz}+\frac{1}{1+y+yz+yzt}+\frac{1}{1+z+zt+ztx}+\frac{1}{1+t+tx+txy}\)

\(=\frac{1}{1+x+xy+xyz}+\frac{x}{x+xy+xyz+xyzt}+\frac{xy}{xy+xyz+xyzt+xyzt.x}+\frac{xyz}{xyz+xyzt+xyzt.x+xyzt.xy}\)

\(=\frac{1}{1+x+xy+xyz}+\frac{x}{x+xy+xyz+1}+\frac{xy}{xy+xyz+1+x}+\frac{xyz}{xyz+1+x+xy}\)

\(=\frac{1+x+xy+xyz}{1+x+xy+xyz}\)

\(=1\)

1 tháng 4 2018

ai  mà biết hả

1 tháng 4 2018

Bạn vô duyên quá đấy!