10 câu Trắc nghiệm Bất đẳng thức có đáp án (Nhận biết)

$(x+y{)}^2\ge x^2+y^2$.

x + y > 0 thì x > 0 hoặc y > 0.

$x\ge y\Rightarrow x^2\ge y^2$.

x + y > 0 thì x.y > 0.

$-\frac{9}{4}$

$-\frac{3}{2}$

0

$\frac{3}{2}$

a < b ⇒ ac < bc.

$a < b \Rightarrow \frac{1}{a}>\frac{1}{b}$

a < b và c < d ⇒ ac < bd

a < b ⇒ ac < bc, (c > 0)

−3a > −3b.

$a^2>b^2$.

2a > 2b.

$\frac{1}{a}<\frac{1}{b}$.

$\left\{\begin{array}{c}a>b\\ c>d\end{array}\right.\Rightarrow ac>bd$

$\left\{\begin{array}{c}a>b\\ c>d\end{array}\right.\Rightarrow \frac{a}{c}>\frac{b}{d}$

$\left\{\begin{array}{c}a>b\\ c>d\end{array}\right.\Rightarrow a-c>b-d$

$\left\{\begin{array}{c}a>b>0\\ c>d>0\end{array}\right.\Rightarrow ac>bd$

$\left\{\begin{array}{c}0<a<b\\ 0<c<d\end{array}\right.\Rightarrow \frac{a}{d}<\frac{b}{c}$

$\left\{\begin{array}{c}a<b\\ c<d\end{array}\right.\Rightarrow a-c<b-d$

$\left\{\begin{array}{c}a<b\\ c<d\end{array}\right.\Rightarrow a+c<b+d$

$\left\{\begin{array}{c}0<a<b\\ 0<c<d\end{array}\right.\Rightarrow ac<bd$

$\left\{\begin{array}{c}a>b>0\\ c>d>0\end{array}\right.\Rightarrow ac>bd$

$\left\{\begin{array}{c}a>b\\ c>d\end{array}\right.\Rightarrow a-c>b-d$

$\left\{\begin{array}{c}a>b\\ c>d\end{array}\right.\Rightarrow ac>bd$

$\left\{\begin{array}{c}a>b\\ c>d\end{array}\right.\Rightarrow \frac{a}{c}>\frac{b}{d}$

$\left\{\begin{array}{c}a\ge x\\ b\ge y\end{array}\right.\Rightarrow a+b\ge x+y$

$a+\frac{1}{a}\ge 2\forall a>0$

$a+b\ge 2\sqrt{ab}\forall a,b\ge 0$

$a>b\Rightarrow \frac{1}{a}<\frac{1}{b}\forall a,b\ne 0$

$\sqrt{xy}\le 6$

$xy<{\left(\frac{x+y}{2}\right)}^2=36$

$2xy<x^2+y^2$

$\sqrt{xy}\ge 6$