Toán 9 Giải phương trình nghiệm nguyên

lê thị hồng hà said:

a, x2+xy+y2x2+xy+y2x^{2}+xy+y^{2} = 2x+y
<=> 4x24x24x^{2} + 4xy + 4y24y24y^{2} - 8x - 4y = 0
<=> 4x24x24x^{2} + 4x(y-2) + (y−2)2(y−2)2(y-2)^{2} - (y−2)2(y−2)2(y-2)^{2} + 4y24y24y^{2} - 4y =0
<=> (2x+y−2)2(2x+y−2)2(2x+y-2)^{2} - y2y2y^{2} + 4y -4 + 4y24y24y^{2} -4y=0
<=> (2x+y−2)2(2x+y−2)2(2x+y-2)^{2} + 3y2y2y^{2} - 4 =0
<=> (2x+y−2)2(2x+y−2)2(2x+y-2)^{2} = 4-3y2y2y^{2}
vì (2x−y−2)2(2x−y−2)2(2x-y-2)^{2} ≥≥\geq 0 => 4-3y2y2y^{2} ≥≥\geq 0
<=> 3y23y23y^{2} ≤≤\leq 4
<=> y2≤43y2≤43y^{2} \leq \frac{4}{3}
<=> −23√≤y≤23√−23≤y≤23\frac{-2}{\sqrt{3}} \leq y\leq \frac{2}{\sqrt{3}}
vì y∈Zy∈Zy\in Z nên y={−1;0;1}y={−1;0;1}y=\left \{ -1 ; 0 ;1 \right \}
- với y = -1 ta có (2x−3)2=1(2x−3)2=1(2x-3)^{2}=1
<=> 2x-3=1 hoặc 2x-3=-1
<=> 2x=4 hoặc 2x=2
<=> x=2 (TM) hoặc x=1 ( TM )
- với y = 0 ta có (2x−2)2=4(2x−2)2=4(2x-2)^{2}=4
<=> 2x-2=2 hoặc 2x-2=-2
<=> 2x=4 hoặc 2x=0
<=> x=2 (TM) hoặc x=0 ( TM )
- với y = 1 ta có (2x−1)2=1(2x−1)2=1(2x-1)^{2}=1
<=> 2x-1=1 hoặc 2x-1=-1
<=> 2x=2 hoặc 2x=0
<=> x=1 (TM) hoặc x=0 ( TM )
vậy pt có nghiệm nguyên (x;y)=( 2 ; -1) ; (1 ; -1) ; (2 ; 0 ) ; (0;0) ; (1;1) ;(0;1)

b, x2+xy+y2x2+xy+y2x^{2}+xy+y^{2}=x+y=x+y= x+y
<=> 4x2+4xy+4y2=4x+4y4x2+4xy+4y2=4x+4y4x^{2}+4xy+4y^{2}=4x+4y
<=> 4x2+4xy+4y2−4x−4y=04x2+4xy+4y2−4x−4y=04x^{2}+4xy+4y^{2}-4x-4y=0
<=> 4x2+4x(y−1)+(y−1)2−(y−1)2+4y2−4y=04x2+4x(y−1)+(y−1)2−(y−1)2+4y2−4y=04x^{2}+4x(y-1)+(y-1)^{2}-(y-1)^{2}+4y^{2}-4y=0
<=> (2x+y−1)2−y2+2y−1+4y2−4y=0(2x+y−1)2−y2+2y−1+4y2−4y=0(2x+y-1)^{2}-y^{2}+2y-1+4y^{2}-4y=0
<=> (2x+y−1)2+3y2−2y−1=0(2x+y−1)2+3y2−2y−1=0(2x+y-1)^{2}+3y^{2}-2y-1=0
<=> (2x+y−1)2=−3y2+2y+1(2x+y−1)2=−3y2+2y+1(2x+y-1)^{2}=-3y^{2}+2y+1
<=> (2x+y−1)2=3y2+3y−y+1(2x+y−1)2=3y2+3y−y+1(2x+y-1)^{2}=3y^{2}+3y-y+1
<=> (2x+y−1)2=−3y(y−1)−(y−1)(2x+y−1)2=−3y(y−1)−(y−1)(2x+y-1)^{2}=-3y(y-1)-(y-1)
<=> (2x+y−1)2=(y−1)(−3y−1)(2x+y−1)2=(y−1)(−3y−1)(2x+y-1)^{2} = (y-1)(-3y-1)
vì (2x+y−1)2≥0(2x+y−1)2≥0(2x+y-1)^{2} \geq 0
⇒(y−1)(−3y−1)≥0⇒(y−1)(−3y−1)≥0\Rightarrow (y-1)(-3y-1)\geq 0
<=> y−1≥0y−1≥0y-1\geq 0 và −3y−1≥0−3y−1≥0-3y-1\geq 0
hoặc y−1≤0y−1≤0y-1\leq 0 và −3y−1≤0−3y−1≤0-3y-1 \leq 0
( đến đây bn tự giải nhé ^^ )

c, x2−3xy+3y2=3yx2−3xy+3y2=3yx^{2}-3xy+3y^{2}=3y
⇔x2−3xy+3y2−3y=0⇔x2−3xy+3y2−3y=0\Leftrightarrow x^{2}-3xy+3y^{2}-3y=0
⇔36y2−36xy−36y+12=0⇔36y2−36xy−36y+12=0\Leftrightarrow 36y^{2}-36xy-36y+12=0
⇔36y2−12y(3x−3)+(3x−3)2−(3x−3)2+12x2=0⇔36y2−12y(3x−3)+(3x−3)2−(3x−3)2+12x2=0\Leftrightarrow 36y^{2}-12y(3x-3)+(3x-3)^{2}-(3x-3)^{2}+12x^{2}=0
⇔(6y−3x+3)2−9x2+18x−9+12x2=0⇔(6y−3x+3)2−9x2+18x−9+12x2=0\Leftrightarrow (6y-3x+3)^{2}-9x^{2}+18x-9+12x^{2}=0
⇔(6y−3x+3)2+3x2+18x−9=0⇔(6y−3x+3)2+3x2+18x−9=0\Leftrightarrow (6y-3x+3)^{2}+3x^{2}+18x-9=0
⇔(6y−3x+3)2=−3x2−18x+9⇔(6y−3x+3)2=−3x2−18x+9\Leftrightarrow (6y-3x+3)^{2}=-3x^{2}-18x+9
⇔(6y−3x+3)2=−3(x2+6x+9)+36⇔(6y−3x+3)2=−3(x2+6x+9)+36\Leftrightarrow (6y-3x+3)^{2}=-3(x^{2}+6x+9)+36
⇔(6y−3x+3)2=−3(x−3)2+36⇔(6y−3x+3)2=−3(x−3)2+36\Leftrightarrow (6y-3x+3)^{2}=-3(x-3)^{2}+36
vì (6y−3x+3)2≥0(6y−3x+3)2≥0(6y-3x+3)^{2}\geq 0 ⇒−3(x−3)2+36≥0⇒−3(x−3)2+36≥0\Rightarrow -3(x-3)^{2}+36\geq 0
⇔−3(x−3)2≥−36⇔−3(x−3)2≥−36\Leftrightarrow -3(x-3)^{2} \geq -36
⇔(x−3)2≤12⇔(x−3)2≤12\Leftrightarrow (x-3)^{2} \leq 12
(đến đây bn tự lm nha ^^ bài sau cx tương tự )

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