PT(1)−PT(2)⟺(y−x)(x2y+x3−5)=0⟺[x=y (3)x2y+x3=5 (4)" style="position: relative;" tabindex="0" id="MathJax-Element-8-Frame" class="MathJax">PT(1)−PT(2)⟺(y−x)(x2y+x3−5)=0⟺[x=y (3)x2y+x3=5 (4) ∙" style="position: relative;" tabindex="0" id="MathJax-Element-9-Frame" class="MathJax">∙ TH1: Với x=y, (1)⟺x4+5x−6=0⟺x=−2 hoặc x=1." style="position: relative;" tabindex="0" id="MathJax-Element-10-Frame" class="MathJax">x=y, (1)⟺x4+5x−6=0⟺x=−2 hoặc x=1. ∙" style="position: relative;" tabindex="0" id="MathJax-Element-11-Frame" class="MathJax">∙ TH2: Từ (1), (2)⇒x, y≤65 (⋆):" style="position: relative;" tabindex="0" id="MathJax-Element-12-Frame" class="MathJax">(1), (2)⇒x, y≤65 (⋆):+ Nếu 0≤x≤65, 0≤y≤65" style="position: relative;" tabindex="0" id="MathJax-Element-13-Frame" class="MathJax">0≤x≤65, 0≤y≤65 VT(4)≤432125&lt;5" style="position: relative;" tabindex="0" id="MathJax-Element-14-Frame" class="MathJax">VT(4)≤432125<5, suy ra PT(4)" style="position: relative;" tabindex="0" id="MathJax-Element-15-Frame" class="MathJax">PT(4) vô nghiệm + Nếux&lt;0, PT(4)⇒y&gt;0, PT(1)⇒x2&lt;6, VT(4)&lt;x2y&lt;665&lt;5" style="position: relative;" tabindex="0" id="MathJax-Element-16-Frame" class="MathJax">x<0, PT(4)⇒y>0, PT(1)⇒x2<6√, VT(4)<x2y<66√5<5, suy ra PT(4)" style="position: relative;" tabindex="0" id="MathJax-Element-17-Frame" class="MathJax">PT(4) vô nghiệm + Nếu y&lt;0, PT(4)⇒x&gt;53&gt;65" style="position: relative;" tabindex="0" id="MathJax-Element-18-Frame" class="MathJax">y<0, PT(4)⇒x>5√3>65 mâu thuẫn với (⋆)" style="position: relative;" tabindex="0" id="MathJax-Element-19-Frame" class="MathJax">(⋆). Do đó PT(4)" style="position: relative;" tabindex="0" id="MathJax-Element-20-Frame" class="MathJax">PT(4) vô nghiệmVậy hệ phương trình có nghiệm(−2;−2), (1;1)." style="position: relative;" tabindex="0" id="MathJax-Element-21-Frame" class="MathJax">(−2;−2), (1;1).
PT(1)
−PT(2)
⟺(y
−x)(x2y+x3
−5)=0
⟺[x=y
(3)x2y+x3=5
(4)" style="position: relative;" tabindex="0" id="MathJax-Element-8-Frame" class="MathJax"
>PT(1)−PT(2)⟺(y−x)(x2y+x3−5)=0⟺[x=y (3)x2y+x3=5 (4)
∙" style="position: relative;" tabindex="0" id="MathJax-Element-9-Frame" class="MathJax"
>∙ TH1: Với x=y,
(1)
⟺x4+5x
−6=0
⟺x=
−2
ho
ặc
x=1." style="position: relative;" tabindex="0" id="MathJax-Element-10-Frame" class="MathJax"
>x=y, (1)⟺x4+5x−6=0⟺x=−2 hoặc x=1.
∙" style="position: relative;" tabindex="0" id="MathJax-Element-11-Frame" class="MathJax"
>∙ TH2: Từ (1),
(2)
⇒x,
y
≤65
(
⋆):" style="position: relative;" tabindex="0" id="MathJax-Element-12-Frame" class="MathJax"
>(1), (2)⇒x, y≤65 (⋆):+ N
ếu 0
≤x
≤65,
0
≤y
≤65" style="position: relative;" tabindex="0" id="MathJax-Element-13-Frame" class="MathJax"
>0≤x≤65, 0≤y≤65 VT(4)
≤432125<5" style="position: relative;" tabindex="0" id="MathJax-Element-14-Frame" class="MathJax"
>VT(4)≤432125<5, suy ra PT(4)" style="position: relative;" tabindex="0" id="MathJax-Element-15-Frame" class="MathJax"
>PT(4) vô nghiệm + Nếux<0,
PT(4)
⇒y>0,
PT(1)
⇒x2<6,
VT(4)<x2y<665<5" style="position: relative;" tabindex="0" id="MathJax-Element-16-Frame" class="MathJax"
>x<0, PT(4)⇒y>0, PT(1)⇒x2<6√, VT(4)<x2y<66√5<5, suy ra PT(4)" style="position: relative;" tabindex="0" id="MathJax-Element-17-Frame" class="MathJax"
>PT(4) vô nghiệm + Nếu y<0,
PT(4)
⇒x>53>65" style="position: relative;" tabindex="0" id="MathJax-Element-18-Frame" class="MathJax"
>y<0, PT(4)⇒x>5√3>65 mâu thuẫn với (
⋆)" style="position: relative;" tabindex="0" id="MathJax-Element-19-Frame" class="MathJax"
>(⋆). Do đó PT(4)" style="position: relative;" tabindex="0" id="MathJax-Element-20-Frame" class="MathJax"
>PT(4) vô nghiệmVậy hệ phương trình có nghiệm(
−2;
−2),
(1;1)." style="position: relative;" tabindex="0" id="MathJax-Element-21-Frame" class="MathJax"
>(−2;−2), (1;1).