a,

Số số hạng là : ( 2017 - 1 ) : 2 + 1 = 1009 ( số )

Tổng là : ( 2017 + 1 ) . 1009 : 2 = 1018081

b, 

Số số hạng là : ( 101 - 2 ) : 3 + 1 = 34 ( số )

Tổng là : ( 101 + 2 ) . 34 : 2 = 1751

Ai tk mình mình tk lại

a, Số số hạng :

( 2017 - 1 ) : 2 + 1 = 1009 số 

Tổng của dãy trên:

( 2017 + 1 ) x 1009 : 2 = 1018081

Đáp số: 1018081

b, Số số hạng:

( 101 - 2 ) : 3 + 1 = 34 số 

Tổng của dãy trên: 

( 101 + 2 ) x 34 : 2 = 1751

Đáp số: 1751

bạn áp dụng công thức làm nha : 

( lấy số lớn nhất + số bé nhất ) x số số hạng : 2

3x(2-x)-5=1-(3x²+2)

<=>6x-3x²-5=-3x²-2

<=>6x=3

<=>x=1/2

\(\dfrac{13}{2}\) : 4\(\dfrac{2}{3}\): 2

= \(\dfrac{13}{2}\): \(\dfrac{14}{3}\):2

= \(\dfrac{13}{2}\) \(\times\) \(\dfrac{3}{14}\):2

=  \(\dfrac{39}{28}\) : 2

= \(\dfrac{39}{28}\) \(\times\) \(\dfrac{1}{2}\)

= \(\dfrac{39}{56}\)

a) \(\dfrac{13}{20}+\dfrac{3}{5}+x=\dfrac{5}{6}\)

\(\Rightarrow\dfrac{5}{4}+x=\dfrac{5}{6}\)

\(\Rightarrow x=\dfrac{5}{6}-\dfrac{5}{4}\)

\(\Rightarrow x=\dfrac{-5}{12}\)

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

\(\Rightarrow x+\dfrac{1}{3}=\dfrac{11}{15}\)

\(\Rightarrow x=\dfrac{11}{15}-\dfrac{1}{3}\)

\(\Rightarrow x=\dfrac{2}{5}\)

c)\(\dfrac{-5}{8}-x=\dfrac{-3}{20}-\dfrac{-1}{6}\)

\(\dfrac{-5}{8}-x=\dfrac{1}{60}\)

\(\Rightarrow x=\dfrac{-5}{8}-\dfrac{1}{60}\)

\(\Rightarrow x=\dfrac{-77}{120}\)

d) \(\dfrac{3}{5}-x=\dfrac{1}{4}+\dfrac{7}{10}\)

\(\Rightarrow\dfrac{3}{5}-x=\dfrac{19}{20}\)

\(\Rightarrow x=\dfrac{3}{5}-\dfrac{19}{20}\)

\(\Rightarrow x=\dfrac{-7}{20}\)

e) \(\dfrac{-3}{7}-x=\dfrac{4}{5}+\dfrac{-2}{3}\)

\(\Rightarrow\dfrac{-3}{7}-x=\dfrac{2}{15}\)

\(\Rightarrow x=\dfrac{-3}{7}-\dfrac{2}{15}\)

\(\Rightarrow x=\dfrac{-59}{105}\)

g) \(\dfrac{-5}{6}-x=\dfrac{7}{12}+\dfrac{-1}{3}\)

\(\Rightarrow\dfrac{-5}{6}-x=\dfrac{1}{4}\)

\(\Rightarrow x=\dfrac{-5}{6}-\dfrac{1}{4}\)

\(\Rightarrow x=\dfrac{-13}{12}\)

a: 14/5-7/5=7/5

b: 7/8-1/3+5/4

=21/24-8/24+30/24

=43/24

c; =7/6+5/6+2/15+13/15

=2+1

=3

d: =4*5/3*11=20/33

e: =2/9*1/6*1/4=2/9*1/24=1/108

2:
a: \(=\dfrac{3}{9}\cdot\dfrac{4}{4}\cdot\dfrac{5}{5}\cdot\dfrac{6}{6}\cdot\dfrac{7}{7}=\dfrac{1}{3}\)

b: \(=\dfrac{1}{6}\left(\dfrac{22}{3}-\dfrac{2}{3}\right)=\dfrac{10}{3}\cdot\dfrac{1}{6}=\dfrac{10}{18}=\dfrac{5}{9}\)

c; \(=\dfrac{1}{3}\left(9-\dfrac{2}{5}-\dfrac{3}{5}\right)=\dfrac{8}{3}\)

14/5-7/5=7/5 

7/8-1/3+5/4=13/24+5/4=43/24 

7/6+2/15+5/6+13/15=13/10+5/6+13/15=32/15+13/15=3

4/3*5/11=20/35

2/9:6*1/4=1/27*1/4=1/108

Có gì khó đâu bạn -..-

( 2x + 5 )( 2x - 7 ) - ( -4x - 3 )² = 16

<=> 2x( 2x - 7 ) + 5( 2x - 7 ) - [ (-4x)² - 2.3.(-4x) + 3² ] = 16

<=> 4x² - 14x + 10x - 35 - [ 16x² + 24x + 9 ] = 16

<=> 4x² - 4x - 35 - 16x² - 24x - 9 = 16

<=> -12x² - 28x - 44 - 16 = 0

<=> -12x² - 28x - 60 = 0

<=> -4( 3x² + 7x + 15 ) = 0

<=> 3x² + 7x + 15 = 0

Ta có : 3x² + 7x + 15 = 3( x² + 7/3x + 49/36 ) + 131/12 = 3( x + 7/6 )² + 131/12 ≥ 131/12 > 0 ∀ x

=> Vô nghiệm 

\(4x^2-14x+10x-35-\left(16x^2+24x+9\right)=16\) 

\(4x^2-4x-35-16x^2-24x-9-16=0\)           

\(-12x^2-28x-60=0\) 

\(-4\left(3x^2+7x+15\right)=0\) 

\(3x^2+7x+15=0\) 

\(3\left(x^2+\frac{7}{3}x+5\right)=0\) 

\(x^2+\frac{7}{3}x+5=0\) 

\(x^2+2\cdot x\cdot\frac{7}{6}+\left(\frac{7}{6}\right)^2-\left(\frac{7}{6}\right)^2+5=0\) 

\(\left(x+\frac{7}{6}\right)^2+\frac{131}{36}=0\)  

\(\left(x+\frac{7}{6}\right)^2=-\frac{131}{36}\) ( vô lí vì \(\left(x+\frac{7}{6}\right)^2\ge0\forall x\)  ) 

Vậy phương trình vô nghiệm 

\(\left(2x+5\right)\left(2x-7\right)-\left(-4x-3\right)^2=16\)

\(\Leftrightarrow4x^2-14x+10x-35-16x^2-24x-9-16=0\)

\(\Leftrightarrow-12x^2-28x-60=0\)

\(\Leftrightarrow12x^2+28x+60=0\)

\(\Leftrightarrow12\left(x^2+\frac{7}{3}x+\frac{49}{36}\right)+\frac{131}{3}=0\)

\(\Leftrightarrow12\left(x+\frac{7}{6}\right)^2+\frac{131}{3}=0\)

Mà \(12\left(x+\frac{7}{6}\right)^2+\frac{131}{3}\ge\frac{131}{3}\)

=> Pt vô nghiệm 

ta có 

\(S_2=\left(1-3\right)+\left(5-7\right)+..+\left(1997-1999\right)+2001\)

ha y \(S_2=-2-2-2..+2001=-2.500+2001=1001\)

\(S_3=\left(1-2-3+4\right)+\left(5-6-7+8\right)+..+\left(1997-1998-1999+2002\right)\)

hay \(S_3=0+0+..+0=0\)

\(S_2=\left(1-3\right)+\left(5-7\right)+...+\left(1997-1999\right)+2001\)

\(=\left(-2\right)+\left(-2\right)+....+\left(-2\right)+2001=\left(-2\right).500+2001=-1000+2001=1001\)

\(S_3=\left(0+1-2-3\right)+\left(4+5-6-7\right)+...+\left(1996+1997-1998-1999\right)+2000\)

\(=-4+\left(-4\right)+...+\left(-4\right)+2000=\left(-4\right).500+2000=0\)

1/2

= (-5/7 + -2/7) + (8/11 + 3/11) + 1/2  = -1 + 1 + 1/2 = 1/2

\(\dfrac{-5}{7}+\dfrac{8}{11}+\left(\dfrac{-2}{7}\right)+\dfrac{1}{2}+\dfrac{3}{11}\\ =\left[\dfrac{-5}{7}+\left(\dfrac{-2}{7}\right)\right]+\left(\dfrac{8}{11}+\dfrac{3}{11}\right)+\dfrac{1}{2}=\left(-1\right)+1+\dfrac{1}{2}=\dfrac{1}{2}\)