Toán 9 CM BĐT

tutuyet2018@gmail.com said:

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câu 5: [tex]GT <=> (x+z).(y+z)=1\\\\ +, (x+z; y+z)=(a;b)\\\\ => x-y=a-b >0 \\\\ => L=\frac{1}{4(x-y)^2}+\frac{1}{(x+z)^2}+\frac{1}{(y+z)^2}\\\\ =\frac{1}{4(a-b)^2}+\frac{1}{a^2}+\frac{1}{b^2}\\\\ =\frac{1}{4(a-b)^2}+a^2+b^2\\\\ =\frac{1}{4(a-b)^2}+(a-b)^2+2\geq 4\geq 3[/tex]
câu 3:
với y=1 => không có x nguyên thỏa mãn
=> y khác 1 khi đó:
[tex]9B=(3x)^2+9y^2+12.|3x|-18.|y|\\\\ =(5-4y)^2+9y^2+12.|5-4y|-18|y|\\\\ =25y^2-40y+25+12|5-4y|-18.|y|\\\\ +, y>0 => y\geq 2\\\\ => 9B=25y^2-40y+25+12.(4y-5)-18y\\\\ =25y^2-58y+48y+25-60\\\\ =25y^2-10y-35\\\\ =25y^2-50y+40y-80+45\\\\ =(y-2).(25y+40)+45\geq 45\\\\ +, y<0 \\\\ => 9B=25y^2-40y+25+12.(5-4y)+18y\\\\ =25y^2-40y+25+60-48y+18y\\\\ =25y^2-70y+85>85\\\\ => 9B\geq 45 => B\geq 5[/tex]
dấu "=" <=> y=2 và x=-1