Toán Lớp 6: Tìm số nguyên x biết
a. ( x-5)(x+2) <0
b. ( x^2-5)(x^2-15)<0
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Toán Lớp 6: Tìm số nguyên x biết
a. ( x-5)(x+2) <0
b. ( x^2-5)(x^2-15)<0
Comments ( 1 )
a, (x-5)(x+2) < 0
<=> TH1 : $\left \{ {{x-5 > 0} \atop {x+2 < 0}} \right.$ <=>$\left \{ {{x > 5} \atop {x < -2 }} <=> 5 < x < -2 \right.$ (loại)
TH2 $\left \{ {{x-5<0} \atop {x+2>0}} \right.$ <=> $\left \{ {{x < 5} \atop {y > -2}} \right.$ <=> -2 < x < 5 ( TM)
Vậy -2 < x < 5 <=> (x-5)(x+2)<0
b, (x^2 – 5) ( x^2 -15) < 0
<=> $\left \{ {{x^2 – 5 > 0} \atop {x^2 – 15 < 0}} \right.$ <=> $\left \{ {{x^2 > 5} \atop {x^2 < 15}} \right.$ <=> $\left \{ {{x > 5^2 } \atop {x < 15^2}} \right.$ <=> $5^{2}$ < x < $15^{2}$ ( TM )
TH2 $\left \{ {{x^2 – 5 < 0} \atop {x^2 – 15 > 0}} \right.$ <=>$\left \{ {{x^2 < 5} \atop {x^2 > 15}} \right.$ <=> $\left \{ {{x < 5^2 } \atop {x > 15^2}} \right.$ <=> $15^{2}$ < x < $5^{2}$ ( loại )
Vậy $5^{2}$ < x < $15^{2}$ <=> ($x^{2}$ – 5)($x^{2}$ – 15) < 0