1)Tìm x,y,z,t biết 31(xyzt + xy+xt+zt+1)= 40(yzt+y+t)
2) tìm x,y biết x2+xy-2y-3x=3
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BH
3
LT
1
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 )
\(\cdot y=3\Rightarrow z+\frac{1}{t}=\frac{9}{4}\)
\(\frac{9}{4}< 3\Rightarrow z< 3\)
- \(z=0\Rightarrow t=\frac{4}{9}\notin Z\)
- \(z=1\Rightarrow t=\frac{4}{5}\notin Z\)
- \(z=2\Rightarrow t=4\)( Thỏa mãn )
Vậy \(x=1;y=3;z=2;t=4.\)
A
1
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.
LT
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