A = x² + 5x - 5 

A = x² + 2. \(\dfrac{5}{2}\) x + ( \(\dfrac{5}{2}\))² - \(\dfrac{45}{4}\)

A = (x + \(\dfrac{5}{2}\))² - \(\dfrac{45}{4}\)

(x + \(\dfrac{5}{2}\))² ≥ 0 ⇔ A = (x + \(\dfrac{5}{2}\))² - \(\dfrac{45}{4}\) ≥ \(\dfrac{-45}{4}\)

A(min) = -45/4   = xảy ra ⇔ x + 5/2 = 0 ⇔ x = -5/2

ĐKXĐ: \(x\ne1;x\ne2\)

\(\dfrac{4}{x-1}-\dfrac{5}{x-2}=-3\)

\(\Leftrightarrow\dfrac{4\left(x-2\right)-5\left(x-1\right)}{\left(x-1\right)\left(x-2\right)}=\dfrac{-3\left(x-1\right)\left(x-2\right)}{\left(x-1\right)\left(x-2\right)}\)

\(\Rightarrow4x-8-5x+5=-3x^2+6x+3x-6\)

\(\Leftrightarrow3x^2-10x+3=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{1}{3}\end{matrix}\right.\)(TM)

Vậy...

= 4/x - 5/x - 1 - 2 = -3

= \(\dfrac{4-5}{x}\) - 3 = -3

=> -1/x = 0 

=> x = -1 : 0

=> x ko có gtr thỏa mãn 

Em nên gõ công thức trực quan để được hỗ trợ tốt hơn nhé

Bạn chịu khó đánh chữ bằng Latex ra nhé.

\(A=1997.1999=\left(1998-1\right)\left(1998+1\right)=1998^2-1< 1998^2=B\)

I'll let you do the drawing. 

a) Consider the 2 triangles AHM and ABH, which both have a common angle at A, have \(\widehat{AMH}=\widehat{AHB}\left(=90^o\right)\). Therefore, \(\Delta AHM~\Delta ABH\left(a.a\right)\). This means \(\dfrac{AH}{AB}=\dfrac{AM}{AH}\) or \(AH^2=AM.AB\)

Similarly, we have \(AH^2=AN.AC\). From these, we get \(AM.AB=AN.AC=AH^2\)

We can easily prove that AMHN is a rectangle (because  \(\widehat{MAN}=\widehat{AMH}=\widehat{ANH}=90^o\)). Thus, \(AH=MN\)(2 diagonals of a rectangle are equal) 

And finally, we get \(AM.AB=AN.AC=MN^2\), and that's what we must prove!

b) We can easily prove \(HN//AB\left(\perp AC\right)\), which means \(\widehat{FHN}=\widehat{B}\)

Consider the right triangle BHM (right at M), it has the median ME. Therefore, \(ME=\dfrac{BH}{2}\). We also have \(BE=\dfrac{BH}{2}\) so \(ME=BE\) or \(\Delta BEM\) is an isosceles triangle, or \(\widehat{BEM}=180^o-2.\widehat{B}\)

Similarly, we have \(\widehat{HFN}=180^o-2.\widehat{FHN}\)

We have already had \(\widehat{B}=\widehat{FHN}\). Thus, \(\widehat{BEM}=\widehat{HFN}\) or \(ME//NF\) (2 equal staggered angles)

Therefore, MEFN is a trapezoid.

In this trapezoid, I is the midmpoint of EF, O is the midpoint of MN (2 diagonal AH, MN of the rectangle AMHN meets at O). Thus, OI is the avergage line of the trapezoid MEFN (ME//NF) or \(OI//NF\)

It's easy to see \(\widehat{FNC}=\widehat{C}\); \(\widehat{MNH}=\widehat{MAH}\)

Also, \(\widehat{C}=\widehat{MAH}\left(=90^o-\widehat{B}\right)\). So, \(\widehat{FNC}=\widehat{MNH}\) or \(\widehat{FNC}+\widehat{FNH}=\widehat{MNH}+\widehat{FNH}\) or \(\widehat{CNH}=\widehat{MNF}\). Because \(\widehat{CNH}=90^o\), it's easy to see \(\widehat{MNF}=90^o\) or \(NF\perp MN\)

We have already prove that \(OI//NF\). Therefore, \(OI\perp MN\), and that's what we must prove!

c) I'm thinking about this question.

sai, parabol úp

cái này đáng ra là tìm giá trị lớn nhất chứ không phải nhỏ nhất

p = 2 + x - x²

P = -x² + x + 2

P = - ( x² - 2. \(\dfrac{1}{2}\)x + \(\dfrac{1}{4}\)) + \(\dfrac{9}{4}\)

P = - (x - \(\dfrac{1}{2}\))² + \(\dfrac{9}{4}\)

- ( x - 1/2 ) ² ≤ 0 ⇔ p ≤  \(\dfrac{9}{4}\)⇔ P(max) = 9/4 dấu = xảy ra khi x = 1/2