Tìm GTNN của biểu thức sau:
\(A=x^2-4x+y^2-y+5\)
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- \(\frac{x^3}{8}\) + \(\frac34x^2\) - \(\frac32x\) + 1
= (-\(\frac{x}{2}\))\(^3\) + 3.(-\(\frac{x}{2}\))\(^2\).1 + 3(\(-\frac12\)\(x\) ).1\(^2\) + 1\(^3\)
= (\(\frac{-x}{2}+1\))\(^3\)
\(-\frac{x^3}{8}+\frac34x^2-\frac32x+1\)
\(=\left(-\frac12x\right)^3+3\cdot\left(-\frac12x\right)^2\cdot1+3\cdot\left(-\frac12x\right)\cdot1^2+1^3\)
\(=\left(-\frac12x+1\right)^3\)

1 I usually walk to school.
2 She isn't sleeping now.
3 They was arrive late yesterday.
4 Is he watching TV at the moment?
5 We aren't having class next Monday.
6 My mom cooks dinner every evening.
7 Do you go to the cinema last weeked?
8 Listen! Someone is knocking at the door.
9 I think she pass the test.
10 They don't eat meat on Fridays.

+) The teacher is explaining a new lesson to us.
(–) The teacher is not explaining a new lesson to us.
(?) Is the teacher explaining a new lesson to us?
cho mình tick nhé !
A = x^2 - 4x + y^2 - y + 5
A=(x^2 - 4x + 4) + (y^2 - y + 1/4) + 5 - 4 - 1/4
A= (x - 2)^2 + (y - 1/2)^2 + 3/4
vì (x - 2)^2 \(\ge\) 0; (y - 1/2)^2 \(\ge\) 0 (1)
nên (x - 2)^2 + (y - 1/2)^2 + 3/4 \(\ge\) 3/4 (2)
từ (1) và (2) => GTNN của biểu thức A là 3/4 khi x = 2; y = 1/2
\(A=x^2-4x+y^2-y+5\)
\(=x^2-4x+4+y^2-y+\frac14+\frac34\)
\(=\left(x-2\right)^2+\left(y-\frac12\right)^2+\frac34\ge\frac34\forall x,y\)
Dấu '=' xảy ra khi \(\begin{cases}x-2=0\\ y-\frac12=0\end{cases}\Rightarrow\begin{cases}x=2\\ y=\frac12\end{cases}\)