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Nguyễn Dũng
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Nguyễn Lê Phước Thịnh
3 tháng 2 2023 lúc 11:28

uses crt;

var i,n,s:integer;

begin

clrscr;

readln(n);

s:=0;

for i:=1 to n do

if i mod 2=0 then s:=s+i;

write(s);

readln;

end.

Lưu Phương Thảo
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NGUYỄN NHẬT TÂN
1 tháng 3 2023 lúc 20:26

a)17/6                                  b)5

c)25312                               d)101248

Lùi Văn Tiến
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Thương Hoàng
2 tháng 4 2017 lúc 9:45

\(A=\frac{1}{2.5}+\frac{1}{3.5}+\frac{1}{3.7}+\frac{1}{4.7}+...+\frac{1}{9.19}+\frac{1}{10.19}\)

\(\Rightarrow\frac{1}{2}A=\frac{1}{4.5}+\frac{1}{6.5}+\frac{1}{6.7}+\frac{1}{8.7}+...+\frac{1}{18.19}+\frac{1}{20.19}\)

\(\Rightarrow\frac{1}{2}A=\frac{5-4}{4.5}+\frac{6-5}{6.5}+\frac{7-6}{6.7}+...+\frac{20-19}{20.19}\)

\(\Rightarrow\frac{1}{2}A=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{19}-\frac{1}{20}\)

\(\Rightarrow\frac{1}{2}A=\frac{1}{4}-\frac{1}{20}\)

\(\Rightarrow\frac{1}{2}A=\frac{1}{5}\)

\(\Rightarrow A=\frac{2}{5}\)

Marie Curie
22 tháng 4 2017 lúc 13:41

Mình có cách giải khác:

A= \(\frac{1}{2.5}+\frac{1}{3.5}+\frac{1}{3.7}+\frac{1}{4.7}+...+\frac{1}{9.19}+\frac{1}{10.19}\)

A= \(\frac{2.1}{2.2.5}+\frac{2.1}{2.3.5}+\frac{2.1}{2.3.7}+\frac{2.1}{2.4.7}+...+\frac{2.1}{2.9.19}+\frac{2.1}{2.10.19}\)

A= \(\frac{2.1}{4.5}+\frac{2.1}{5.6}+\frac{2.1}{6.7}+\frac{2.1}{7.8}+...+\frac{2.1}{18.19}+\frac{2.1}{19.20}\)

A= \(2.\left(\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+...+\frac{1}{18.19}+\frac{1}{19.20}\right)\)

A=\(2.\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{18}-\frac{1}{19}+\frac{1}{19}-\frac{1}{20}\right)\)

A= \(2.\left(\frac{1}{4}+0+0+0+...+0+0-\frac{1}{20}\right)\)

A=\(2.\left(\frac{1}{4}-\frac{1}{20}\right)\)

A=\(2.\left(\frac{5}{20}-\frac{1}{20}\right)\)

A= \(2.\frac{1}{5}\)

A=\(\frac{2}{5}\)

Xong rùi đó!!!!! :))

:D :D
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『Kuroba ム Tsuki Ryoo...
27 tháng 6 2023 lúc 12:23

`@` `\text {Ans}`

`\downarrow`

`a)`

Thu gọn:

`P(x)=`\(5x^4 + 3x^2 - 3x^5 + 2x - x^2 - 4 +2x^5\)

`= (-3x^5 + 2x^5) + 5x^4 + (3x^2 - x^2) + 2x - 4`

`= -x^5 + 5x^4 + 2x^2 + 2x - 4`

`Q(x) =`\(x^5 - 4x^4 + 7x - 2 + x^2 - x^3 + 3x^4 - 2x^2\)

`= x^5 + (-4x^4 + 3x^4) - x^3 + (x^2 - 2x^2) + 7x - 2`

`= x^5 - x^4 - x^3 - x^2 + 7x - 2`

`@` Tổng:

`P(x)+Q(x)=`\((-x^5 + 5x^4 + 2x^2 + 2x - 4) + (x^5 - x^4 - x^3 - x^2 + 7x - 2)\)

`= -x^5 + 5x^4 + 2x^2 + 2x - 4 + x^5 - x^4 - x^3 - x^2 + 7x - 2`

`= (-x^5 + x^5) - x^3 + (5x^4 - x^4) + (2x^2 - x^2) + (2x + 7x) + (-4-2)`

`= 4x^4 - x^3 + x^2 + 9x - 6`

`@` Hiệu:

`P(x) - Q(x) =`\((-x^5 + 5x^4 + 2x^2 + 2x - 4) - (x^5 - x^4 - x^3 - x^2 + 7x - 2)\)

`= -x^5 + 5x^4 + 2x^2 + 2x - 4 - x^5 + x^4 + x^3 + x^2 - 7x + 2`

`= (-x^5 - x^5) + (5x^4 + x^4) + x^3 + (2x^2 + x^2) + (2x - 7x) + (-4+2)`

`= -2x^5 + 6x^4 + x^3 + 3x^2 - 5x - 2`

`b)`

`@` Thu gọn:

\(H (x) = ( 3x^5 - 2x^3 + 8x + 9) - ( 3x^5 - x^4 + 1 - x^2 + 7x)\)

`= 3x^5 - 2x^3 + 8x + 9 - 3x^5 + x^4 - 1 + x^2 - 7x`

`= (3x^5 - 3x^5) + x^4 - 2x^3 - x^2 + (8x + 7x) + (9+1)`

`= x^4 - 2x^3 - x^2 + 15x + 10`

\(R( x) = x^4 + 7x^3 - 4 - 4x ( x^2 + 1) + 6x\)

`= x^4 + 7x^3 - 4 - 4x^3 - 4x + 6x`

`= x^4 + (7x^3 - 4x^3) + (-4x + 6x) - 4`

`= x^4 + 3x^3 + 2x - 4`

`@` Tổng:

`H(x)+R(x)=` \((x^4 - 2x^3 - x^2 + 15x + 10)+(x^4 + 3x^3 + 2x - 4)\)

`= x^4 - 2x^3 - x^2 + 15x + 10+x^4 + 3x^3 + 2x - 4`

`= (x^4 + x^4) + (-2x^3 + 3x^3) - x^2 + (15x + 2x) + (10-4)`

`= 2x^4 + x^3 - x^2 + 17x + 6`

`@` Hiệu: 

`H(x) - R(x) =`\((x^4 - 2x^3 - x^2 + 15x + 10)-(x^4 + 3x^3 + 2x - 4)\)

`=x^4 - 2x^3 - x^2 + 15x + 10-x^4 - 3x^3 - 2x + 4`

`= (x^4 - x^4) + (-2x^3 - 3x^3) - x^2 + (15x - 2x) + (10+4)`

`= -5x^3 - x^2 + 13x + 14`

`@` `\text {# Kaizuu lv u.}`

Kak Best
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loann nguyễn
6 tháng 7 2021 lúc 14:42

là sao ????=))

giữa các phân số là cộng hay trừ vậy???

\(\dfrac{1}{1.3}+\dfrac{1}{2.3}+\dfrac{1}{2.5}+\dfrac{1}{3.5}+\dfrac{1}{3.7}+\dfrac{1}{4.7}+\dfrac{1}{4.9}\) 

\(=\dfrac{1}{1.3}+\dfrac{1}{3.2}+\dfrac{1}{2.5}+\dfrac{1}{5.3}+\dfrac{1}{3.7}+\dfrac{1}{7.4}+\dfrac{1}{4.9}\) 

\(=\left(\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+\dfrac{1}{8.9}\right):\dfrac{1}{2}\) 

\(=\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}\right):\dfrac{1}{2}\) 

\(=\left(\dfrac{1}{2}-\dfrac{1}{9}\right):\dfrac{1}{2}\) 

\(=\dfrac{7}{18}:\dfrac{1}{2}\) 

\(=\dfrac{7}{9}\)

Vân anh
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Đào Thu Huyền
7 tháng 2 2017 lúc 19:33

a=511/256

b=647/20

c=mình đang suy nghĩ,nhưng nếu bạn k cho mình thì bạn sẽ có câu trả lời

Nguyen Thu Ha
7 tháng 2 2017 lúc 19:47

a. 1 + 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64 + 1/128 + 1/256

= 1 + ( 1 - 1/2) + ( 1/2 - 1/4) + ( 1/4 - 1/8) + ( 1/8 - 1/16) + ( 1/16 - 1/32) + (1/32 - 1/64) + ( 1/64 - 1/128) + (1/128 - 1/256)

= 1 + 1 - 1/2 + 1/2 - 1/4 + 1/4 - 1/8 + 1/8 - 1/16 + 1/16 - 1/32 + 1/32 - 1/64 + 1/64 - 1/128 + 1/128 - 1/256

= 2 - 1/256

= 511/256

Câu b bạn có viết sai đề không vậy?

Nguyen Thu Ha
7 tháng 2 2017 lúc 19:57

b, 3/1x4 + 3/2x5 + 3/3x6 + 3/4x7 + 1/5x8

= 3/4 + 3/10 + 3/18 + 3/28 + 1/40

= 1133/840

c,2/1x3 + 2/3x5 + 2/5x7+..+ 2/2001x2003 + 2/2003x2005

= ( 1 - 1/3) + ( 1/3 - 1/5) + ( 1/5 - 1/7) +...+ ( 1/2001 - 1/2003) + (1/2003 - 1/2005)

= 1 - 1/3 +  1/3 - 1/5 + 1/5 - 1/7 +...+  1/2001 - 1/2003 + 1/2003 - 1/2005

= 1 - 1/2005

= 2004/2005

Dung
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Mới vô
22 tháng 4 2017 lúc 20:29

\(A=\dfrac{1}{2.5}+\dfrac{1}{3.5}+\dfrac{1}{3.7}+...+\dfrac{1}{9.19}+\dfrac{1}{10.19}\)

\(A=\dfrac{2}{4.5}+\dfrac{2}{5.6}+\dfrac{2}{6.7}+...+\dfrac{2}{18.19}+\dfrac{2}{19.20}\)

\(A=2.\left(\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+...+\dfrac{1}{18.19}+\dfrac{1}{19.20}\right)\)

\(A=2.\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+...+\dfrac{1}{19}-\dfrac{1}{20}\right)\)

\(A=2.\left(\dfrac{1}{4}-\dfrac{1}{20}\right)\)

\(A=2.\dfrac{1}{5}\)

\(A=\dfrac{2}{5}\)

Nguyễn Hoàng Tùng
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『Kuroba ム Tsuki Ryoo...
28 tháng 4 2023 lúc 17:06

\(\cdot\) `\text {dnammv}`

`7,`

`a,`

`M(x)=\(-5x^4+3x^5+x\left(x^2+5\right)+14x^4-6x^5-x^3+x-1\)

`M(x)=-5x^4+3x^5+x^3+5x+14x^4-6x^5-x^3+x-1`

`=(3x^5-6x^5)+(-5x^4+14x^4)+(x^3-x^3)+(5x+x)-1`

`=-3x^5+9x^4+6x-1`

`N(x)=x^4(x - 5) - 3x^3 + 3x + 2x^5 - 4x^4 + 3x^3 - 5`

`= x^5-5x^4-3x^3+3x+2x^5-4x^4+3x^3-5`

`= 3x^5-9x^4+3x-5`

`b,`

`H(x)= N(x)+ M(x)`

`-> H(x)=(-3x^5+9x^4+6x-1)+(3x^5-9x^4+3x-5)`

`= -3x^5+9x^4+6x-1+3x^5-9x^4+3x-5`

`= (-3x^5+3x^5)+(9x^4-9x^4)+(6x+3x)+(-1-5)`

`= 9x-6`

`G(x)=M(x)-N(x)`

`-> G(x)= (-3x^5+9x^4+6x-1)-(3x^5-9x^4+3x-5)`

`= -3x^5+9x^4+6x-1-3x^5+9x^4-3x+5`

`= (-3x^5-3x^5)+(9x^4+9x^4)+(6x-3x)+(-1+5)`

`= -6x^5+18x^4+3x+4`

`c,`

`H(x)=9x-6`

Hệ số cao nhất: `9`

Hệ số tự do: `-6`

`G(x)= -6x^5+18x^4+3x+4`

Hệ số cao nhất: `-6`

Hệ số tự do: `4`

`d,`

`H(1)=9*1-6=9-6=3`

`H(-1)=9*(-1)-6=-9-6=-15`

 

`G(1)=-6*1^5+18*1^4+3*1+4=-6+18+3+4=12+3+4=15+4=19`

`G(0)=-6*0^5+18*0^4+3*0+4=0+0+0+4=4`

 

`H(x)=9x-6=0`

`-> 9x=0+6`

`-> 9x=6`

`-> x= 6 \div 9`

`-> x=`\(\dfrac{2}{3}\)

Vậy, nghiệm của đa thức là `x=`\(\dfrac{2}{3}\)

nguyễn hồng trang
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HT.Phong (9A5)
12 tháng 8 2023 lúc 8:24

a) \(\left(2x^3-x^2+5x\right):x\)

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

\(=\dfrac{x\left(2x^2-x+5\right)}{x}\)

\(=2x^2-x+5\)

b) \(\left(3x^4-2x^3+x^2\right):\left(-2x\right)\)

\(=\dfrac{3x^4-2x^3+x^2}{-2x}\)

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

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

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

c) \(\left(-2x^5+3x^2-4x^3\right):2x^2\)

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

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

\(=-x^3-2x+\dfrac{3}{2}\)

HT.Phong (9A5)
12 tháng 8 2023 lúc 8:29

d) \(\left(x^3-2x^2y+3xy^2\right):\left(-\dfrac{1}{2}x\right)\)

\(=\dfrac{x^3-2x^2y+3xy^2}{-\dfrac{1}{2}x}\)

\(=\dfrac{\dfrac{1}{2}x\left(2x^2-4xy+6y^2\right)}{-\dfrac{1}{2}x}\)

\(=-\left(2x^2-4xy+6y^2\right)\)

\(=-2x^2+4xy-6y^2\)

e) \(\left[3\left(x-y\right)^5-2\left(x-y\right)^4+3\left(x-y\right)^2\right]:5\left(x-y\right)^2\)

\(=\dfrac{3\left(x-y\right)^5-2\left(x-y\right)^4+3\left(x-y\right)^2}{5\left(x-y\right)^2}\)

\(=\dfrac{5\left(x-y\right)^2\left[\dfrac{3}{5}\left(x-y\right)^3-\dfrac{2}{5}\left(x-y\right)^2+\dfrac{3}{5}\right]}{5\left(x-y\right)^2}\)

\(=\dfrac{3}{5}\left(x-y\right)^3-\dfrac{2}{5}\left(x-y\right)^2+\dfrac{3}{5}\)

f) \(\left(3x^5y^2+4x^3y^3-5x^2y^4\right):2x^2y^2\)

\(=\dfrac{3x^5y^2+4x^3y^3-5x^2y^4}{2x^2y^2}\)

\(=\dfrac{2x^2y^2\left(\dfrac{3}{2}x^3+2xy-\dfrac{5}{2}y^2\right)}{2x^2y^2}\)

\(=\dfrac{3}{2}x^3+2xy-\dfrac{5}{2}y^2\)