\(x^2-xy+y^2=3\)

\(\Leftrightarrow2x^2-2xy+2y^2=6\)

\(\Leftrightarrow\left(x-y\right)^2+x^2+y^2=1+1+4\)

\(\Rightarrow\left(\left(x-y\right)^2,x^2,y^2\right)=\left(1,1,4;1,4,1;4,1,1\right)\)

\(\Rightarrow\left(x,y\right)=\left(-1,-2;1,2;2,1;-2,-1;-1,1;1,-1\right)\)

Ta có

\(x^4+y^4=7z^4+5\Leftrightarrow x^4+y^4+z^4=8z^4+5\)

Áp dụng tính chất lũy thừa bậc 4 của số nguyên a khi chia cho 8 dư 0 hoặc 1

tức là \(a^4\equiv0,1\left(mod8\right)\)

\(\Rightarrow a^4+b^4+c^4\equiv0,1,2,3\left(mod8\right)\)

Mà \(8z^4+5\equiv5\left(mod8\right)\)

vậy pt k có nghiệm nguyên

a) \(\frac{2-x}{3}< \frac{3-2x}{5}\)

<=> \(10-5x< 9-6x\)

<=> x < - 1 

Vậy S = { x| x < -1 }

b)

 

a, \(5\left(m+3x\right)\left(x+1\right)-4\left(1+2x\right)=80\)

Phương trình nhận \(x=2\)làm nghiệm nên :

\(5\left(m+3.2\right)\left(2+1\right)-4\left(1+2.2\right)=80\)

\(\Leftrightarrow15m+90-20=80\)

\(\Leftrightarrow15m=80+20-90\)

\(\Leftrightarrow15m=10\Leftrightarrow m=1,5\)

....

b, \(3\left(2x+m\right)\left(3x+2\right)-2\left(3x+1\right)^2=43\)

Phương trình nhận \(x=1\)làm nghiệm nên :

\(3\left(2.1+m\right)\left(3.1+2\right)-2\left(3.1+1\right)^2=43\)

\(\Leftrightarrow30+15m-32=43\)

\(\Leftrightarrow15m=43+32-30\)

\(\Leftrightarrow15m=45\Leftrightarrow m=3\)

....

\(\frac{315-x}{101}+\frac{313-x}{103}+\frac{311-x}{105}+\frac{309-x}{107}+4=0\)

\(\Leftrightarrow\frac{315-x}{101}+1+\frac{313-x}{103}+1+\frac{311-x}{105}+1+\frac{309-x}{107}+1=0\)

\(\Leftrightarrow\frac{416-x}{101}+\frac{416-x}{103}+\frac{416-x}{105}+\frac{416-x}{107}=0\)

\(\Leftrightarrow\left(416-x\right)\left(\frac{1}{101}+\frac{1}{103}+\frac{1}{105}+\frac{1}{107}\right)=0\)

\(\Leftrightarrow416-x=0\)

\(\Leftrightarrow x=416\)

a) 5(m + 3x)(x + 1) - 4(1 + 2x) = 80

Phương trình có nghiệm x = 2:

5(m + 3.2)(2 + 1) - 4(1 + 2.2) = 80

<=> 5(m + 6).3 - 4.5 = 80

<=> 15(m + 6) - 4.5 = 80

<=> 15(m + 6) - 20 = 80

<=> 15(m + 6) = 80 + 20

<=> 15(m + 6) = 100

<=> m + 6 = 100 : 15

<=> m + 6 = 20/3

<=> m = 20/3 - 6

<=> m = 2/3

b) 3(2x + m)(3x + 2) - 2(3x + 1)² = 43

Phương trình có nghiệm x = 1:

3(2.1 + m)(3.1 + 2) - 2(3.1 + 1)² = 43

<=> 3(2 + m).5 - 2.16 = 43

<=> 15(2 + m) - 32 = 43

<=> 15(2 + m) = 43 + 32

<=> 15(2 + m) = 75

<=> 2 + m = 75 : 15

<=> 2 + m = 5

<=> m = 5 - 2

<=> m = 3

Phương trình trên 

\(< =>\frac{315-x}{101}+1+\frac{313-x}{103}+1+\frac{311-x}{105}+1+\frac{319-x}{107}+1=0\)

\(< =>\frac{315+101-x}{101}+\frac{313+103-x}{103}+\frac{311+105-x}{105}+\frac{309+107-x}{107}=0\)

\(< =>\frac{416-x}{101}+\frac{416-x}{103}+\frac{416-x}{105}+\frac{416-x}{107}=0\)

\(< =>\left(416-x\right).\left(\frac{1}{101}+\frac{1}{103}+\frac{1}{105}+\frac{1}{107}\right)=0\)

Do \(\frac{1}{101}+\frac{1}{103}+\frac{1}{105}+\frac{1}{107}\ne0\)

\(< =>416-x=0\)

\(< =>x=416\)

\(\Leftrightarrow16-3\left(x+1\right)< 24+2\left(x-1\right)\)

=>16-3x-3<24+2x-2

=>-3x+13<2x+22

=>-5x<9

hay x>-9/5