CHÚC BẠN HỌC TỐT!!!
Trả lời:
Dạng 4:
$1,$
$x^2-x+2$
$=x^2-2.x.\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{7}{4}$
$=\bigg{(}x-\dfrac{1}{2}\bigg{)}^2+\dfrac{7}{4}>0$
$⇒x^2-x+2>0$
$2,$ Áp dụng BĐT Cauchy cho 2 số dương: $a,b>0$
$\dfrac{a}{b}+\dfrac{b}{a}\geq 2.\sqrt{\dfrac{a}{b}.\dfrac{b}{a}}=2.1=2$
$3,$
$(a+b)\bigg{(}\dfrac{1}{a}+\dfrac{1}{b}\bigg{)}$
$=1+\dfrac{a}{b}+\dfrac{b}{a}+1$
$=\dfrac{a}{b}+\dfrac{b}{a}+2$
$\geq 2+2=4$
$4,$
$a>b⇔2a>2b⇔2a+3>2b+3$
$5,$ Áp dụng bất đẳng thức Cauchy cho 4 số dương:
$a^4+b^4+c^4+d^4\geq 4\sqrt[4]{a^4.b^4.c^4.d^4}=4abcd$.
Dạng 5:
$a,m<n$
$⇔-6m>-6n$
$⇔3-6m>3-6n$
$⇔3-6m>1-6n$
$b,m<n$
$⇔4m<4n$
$⇔4m+1<4n+1$
$⇔4m+1<4n+5$
$c,m<n$
$⇔2m<2n$
$⇔2m+1<2n+1$.
