VT=(1/5+1/6+...+1/9)+(1/10+1/11+...+1/14)+(1/15+1/16+1/17)

VT<(1/5+...+1/5)+(1/10+...+1/10)+(1/15+1/15+1/15)=

=5/5+5/10+3/15=1+1/2+1/5<2

1) \(\left(2x+3\right)^2=4x^2+12x+9\)

\(\left(3x+2\right)^2=9x^2+12x+4\)

\(\left(2x+5\right)^2=4x^2+20x+25\)

\(\left(2x+\dfrac{1}{3}\right)^2=4x^2+\dfrac{4}{3}x+\dfrac{1}{9}\)

\(\left(3x+\dfrac{1}{3}\right)^2=9x^2+2x+\dfrac{1}{9}\)

2)  \(\left(2x-3\right)^2=4x^2-12x+9\)

\(\left(3x-2\right)^2=9x^2-12x+4\)

\(\left(2x-5\right)^2=4x^2-20x+25\)

\(\left(2x-\dfrac{1}{3}\right)^2=4x^2-\dfrac{4}{3}x+\dfrac{1}{9}\)

\(\left(3x-\dfrac{1}{3}\right)^2=9x^2-2x+\dfrac{1}{9}\)

3) \(\left(2x-3\right)\left(2x+3\right)=4x^2-9\)

\(\left(3x-4\right)\left(3x+4\right)=9x^2-16\)

\(\left(2x-5\right)\left(2x+5\right)=4x^2-25\)

\(\left(x-\dfrac{1}{2}\right)\left(x+\dfrac{1}{2}\right)=x^2-\dfrac{1}{4}\)

\(\left(2x-\dfrac{1}{3}\right)\left(2x+\dfrac{1}{3}\right)=4x^2-\dfrac{1}{9}\)

1: \(\left(2x+3\right)^2=4x^2+12x+9\)

\(\left(3x+2\right)^2=9x^2+12x+4\)

\(\left(2x+5\right)^2=4x^2+20x+25\)

\(\left(2x+\dfrac{1}{3}\right)^2=4x^2+\dfrac{4}{3}x+\dfrac{1}{9}\)

\(\left(3x+\dfrac{1}{3}\right)^2=9x^2+2x+\dfrac{1}{9}\)

Bài 2: 

a: \(\dfrac{-5}{9}+\dfrac{4}{9}=\dfrac{-1}{9}\)

b: \(=\dfrac{5}{17}+\dfrac{-1}{17}\cdot3=\dfrac{5}{17}-\dfrac{3}{17}=\dfrac{2}{17}\)

c: \(=\dfrac{1}{5}\left(\dfrac{4}{7}+\dfrac{3}{7}\right)-\dfrac{1}{5}=\dfrac{1}{5}-\dfrac{1}{5}=0\)

d: =5,17-2,24-5,17+3,24=1

bài đou ?

    25,1 : 0,25 + 2,7 x 4 - 27,8 x 3,9

= 25,1 x 4 + 2,7 x 4 - 27,8 x 3,9

= 4 x ( 25,1 + 2,7) - 27,8 x 3,9

= 4 x 27,8 - 27,8 x 3,9

= 27,8 x ( 4 - 3,9)

= 27,8 x 0,1

= 2,78 

57:

a: \(x^2-4x+3=\left(x-1\right)\left(x-3\right)\)

b: \(x^2+5x+4=\left(x+1\right)\left(x+4\right)\)

c: \(x^2-x-6=\left(x-3\right)\left(x+2\right)\)

d: \(x^4+4=\left(x^2+2x+2\right)\left(x^2-2x+2\right)\)

Câu 4: 

a: Xét ΔABH vuông tại H và ΔACH vuông tại H có

AB=AC

AH chung

Do đó: ΔABH=ΔACH

b: Xét ΔADH vuông tại D và ΔAEH vuông tại E có

AH chung

\(\widehat{DAH}=\widehat{EAH}\)

Do đó: ΔADH=ΔAEH

Suy ra:HD=HE