Số lượng số hạng của dãy số trừ 1 đầu:

\(\left(99-1\right):1+1=99\) (số hạng)

Tổng của dãy số là:

\(\left(99+1\right)\cdot99:2+1=4951\)

Số số hạng của dãy 1 + 2 + 3 + ... + 89:

89 - 1 + 1 = 89 số:

Tổng là:

1 + (89 + 1) . 89 : 2 = 1 + 45 . 89 = 4006

\(1+1+2+3+...+89\)

\(=1+\left[\left(89-1\right):1+1\right]\left(1+89\right):2\)

\(=1+89.90:2\)

\(=1+4005\)

\(=4006\)

\(\dfrac{7}{19}x\dfrac{8}{23}+\dfrac{7}{19}x\dfrac{15}{23}+1\dfrac{7}{19}\)

= \(\dfrac{7}{19}x\left(\dfrac{8}{23}+\dfrac{15}{23}\right)+1+\dfrac{7}{19}\)

=\(\dfrac{7}{19}x1+1+\dfrac{7}{19}\)

= \(\dfrac{7}{19}+1+\dfrac{7}{19}=1\dfrac{14}{19}\) = \(\dfrac{33}{19}\)

\(\dfrac{75}{100}+\dfrac{18}{21}+\dfrac{49}{32}+\dfrac{1}{4}+\dfrac{3}{21}-\dfrac{17}{32}\)

=  \(\dfrac{3}{4}+\dfrac{6}{7}+\dfrac{49}{32}+\dfrac{1}{4}+\dfrac{1}{7}-\dfrac{17}{32}\)

= \(\left(\dfrac{3}{4}+\dfrac{1}{4}\right)+\left(\dfrac{6}{7}+\dfrac{1}{7}\right)+\left(\dfrac{49}{32}-\dfrac{17}{32}\right)\)

= 1 + 1 + 1 = 3

\(\dfrac{8}{9}x\dfrac{15}{16}x\dfrac{24}{25}x\dfrac{35}{36}x\dfrac{48}{49}x\dfrac{63}{64}\)

= \(\dfrac{3}{4}\) *Câu này bạn tự sử dụng gạch nhé!

`1,`

`a,`

`7/19 \times 8/23 + 7/19 \times 15/23 + 1 7/19`

`= 7/19 \times 8/23 + 7/19 \times 15/23 + 1 + 7/19`

`= 7/19 \times (8/23 + 15/23 + 1) + 1`

`= 7/19 \times 2 + 1`

`=14/19 + 1`

`= 33/19`

`b,`

`75/100 + 18/21 + 49/32 + 1/4 + 3/21 - 17/32`

`= 75/100 + (18/21 + 3/21) + (49/32 - 17/32) + 1/4`

`= 0,75 + 1 + 1 + 0,25`

`= (0,75 + 0,25) + 1 + 1`

`= 1+1+1=3`

`c,`

`8/9 \times 15/16 \times 24/25 \times 35/36 \times 48/49 \times 63/64`

`=` \(\dfrac{2\times3}{3\times3}\times\dfrac{3\times5}{4\times4}\times\dfrac{3\times4\times2}{5\times5}\times\dfrac{5\times7}{6\times6}\times\dfrac{6\times8}{7\times7}\times\dfrac{7\times9}{8\times8}\)

`= 3/4` (bạn sử dụng gạch, rút gọn các số là được nhé).

A=(17^18+1)/(17^19+1)

17A=17(17^18+1)/17^19+1=17^19+17/17^19+1

17A=(17^19+1)+16/(17^19+1)=1+16/17^19+1    

B=(17^17+1)/(17^18+1)

17B=17(17^17+1)/17^18+1=17^18+17/17^18+1

17B=(17^18+1)+16/(17^18+1)=1+16/17^18+1

Từ (1) và (2)⇒1+16/17^19+1<1+16/17^18+1

=> 17A<17B

Hay A<B

Vậy A<B

S = 1 + 3¹ + 3² + 3³ + ... + 3³⁰

3S = 3 + 3² + 3³ + 3⁴ + ... + 3³¹

3S - S = (3 + 3² + 3³ + 3⁴ + ... + 3³¹) - (1 + 3¹ + 3² + 3³ + ... + 3³⁰)

2S = 3³¹ - 1

\(S=\frac{3^{31}-1}{2}\)

học lớp 10 chưa

\(\left(\frac{1}{2}-\frac{1}{3}\right).6^x+6^{x+2}\)=\(6^{15}+6^{18}\)

\(\frac{1}{6}.6^x+6^{x+2}=6^{15}+6^{18}\)

\(6^{x-1}+6^{x+2}=6^{15}+6^{18}\)

\(6^{x-1}.\left(1+6^3\right)=6^{15}.\left(1+6^3\right)\)

              \(6^{x-1}=6^{15}\)

     =>        \(x-1=15\)

     =>        \(x\)       \(=16\)

Vậy x=16

chúc bn học tốt!

Lời giải:
Có:
$(a^2+1)(b^2+1)(c^2+1)=(a^2+ab+bc+ac)(b^2+ab+bc+ac)(c^2+ab+bc+ac)$

$=(a+b)(a+c)(b+c)(b+a)(c+a)(c+b)=[(a+b)(b+c)(c+a)]^2$

Và:

$(a+b+c-abc)^2=[(a+b+c)(ab+bc+ac)-abc]^2$

$=[ab(a+b)+bc(b+c)+ca(c+a)+2abc]^2$

$=[ab(a+b+c)+bc(b+c+a)+ca(c+a)]^2$

$=[(a+b+c)(ab+bc)+ca(c+a)]^2=[b(a+b+c)(a+c)+ac(c+a)]^2$

$=[(c+a)(ab+b^2+bc+ac)]^2=[(c+a)(b+a)(b+c)]^2$
Do đó: $P=\frac{[(a+b)(b+c)(c+a)]^2}{[(a+b)(b+c)(c+a)]^2}=1$

program tinhtoan;

uses crt;

var: i;n:interger;

S:real;

writeln(' Nhap n='); readln(n);

S:=0;

For i:=1 to n*(n*1) do S:=S+\(\frac{1}{i};\)

writeln(' S=',S);

End.

(ps: ko chắc )

Trả lời:

\(10+11+12+13+14+15+16+17+18+19+20\)

\(=\left(10+20\right)+\left(11+19\right)+\left(12+18\right)+\left(13+17\right)+\left(14+16\right)+15\)

\(=30+30+30+30+30+15\)

\(=165\)

\(1+2+3+4+5+6+7+8+9\)

\(=\left(1+9\right)+\left(2+8\right)+\left(3+7\right)+\left(4+6\right)+5\)

\(=10+10+10+10+5\)

\(=45\)

10+11+12+13+14+15+16+17+18+19+20

(10+20)+(11+19)+(12+18)+(13+17)+(14+16)+15

30+30+30+30+30+15

165

\(a,\text{10 + 11 + 12 + 13 + 14 + 16 + 17 + 18 + 19 + 20}\)

\(=\left(10+20\right)+\left(11+19\right)+\left(12+18\right)+\left(13+17\right)+\left(14+16\right)\)

\(=30+30+30+30+30\)

\(=30\times5\)

\(=150.\)

\(b,\text{1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9}\)

\(=\left(1+9\right)+\left(2+8\right)+\left(3+7\right)+\left(4+6\right)+5\)

\(=10+10+10+10+5\)

\(=45.\)