Câu hỏi của Nguyễn Trân

Question

1)

a) So sánh (-99)20 và 9999100

b) Chứng tỏ A = 2720 + 361  + 931 chia hết cho 13

2)

Cho$\dfrac{a}{b}$= $\dfrac{b}{c}$=$\dfrac{c}{2011}$ =$\dfrac{2011}{a}$ và a+b+c khác 0. Tính a+b-c

3)...

Mới vô · Accepted Answer

3, $M=\dfrac{\dfrac{4}{237}-\dfrac{4}{2371}+\dfrac{4}{23711}}{\dfrac{-5}{237}+\dfrac{5}{2371}-\dfrac{5}{23711}}=\dfrac{\left(-4\right)\cdot\left(\dfrac{-1}{237}+\dfrac{1}{2371}-\dfrac{1}{23711}\right)}{5\cdot\left(\dfrac{-1}{237}+\dfrac{1}{2371}-\dfrac{1}{23711}\right)}=\dfrac{-4}{5}$ Vậy $M=\dfrac{-4}{5}$ 2, Áp dụng tính chất dãy tỉ số bằng nhau ta có: $\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{2011}=\dfrac{2011}{a}=\dfrac{a+b+c+2011}{b+c+2011+a}=\dfrac{a+b+c+2011}{a+b+c+2011}=1$ $\dfrac{a}{b}=1\Rightarrow a=b\left(1\right)\ \dfrac{b}{c}=1\Rightarrow b=c\left(2\right)$ Từ (1) và (2) ta có: $a=c$ $\Rightarrow a+b-c=a+a-a=a$ 1) b) $A=27^{20}+3^{61}+9^{31}\ =\left(3^3\right)^{20}+3^{61}+\left(3^2\right)^{31}\ =3^{60}+3^{61}+3^{62}\ =3^{60}\cdot\left(1+3+3^2\right)\ =3^{60}\cdot\left(1+3+9\right)\ =3^{60}\cdot13⋮13$ Vậy $A⋮13$ a, $\left(-99\right)^{20}=\left(-99\right)^{2\cdot10}=\left[\left(-99\right)^2\right]^{10}=9801^{10}\ 9999^{100}=\left(9999^{10}\right)^{10}>\left(9999^{10}\right)^1=9999^{10}$ Vì $9801^{10}< 9999^{10}< \left(9999^{10}\right)^{10}=9999^{100}\Rightarrow\left(-99\right)^{20}< 9999^{100}$ Vậy $\left(-99\right)^{20}< 9999^{100}$Câu trả lời: 3, $M=\dfrac{\dfrac{4}{237}-\dfrac{4}{2371}+\dfrac{4}{23711}}{\dfrac{-5}{237}+\dfrac{5}{2371}-\dfrac{5}{23711}}=\dfrac{\left(-4\right)\cdot\left(\dfrac{-1}{237}+\dfrac{1}{2371}-\dfrac{1}{23711}\right)}{5\cdot\left(\dfrac{-1}{237}+\dfrac{1}{2371}-\dfrac{1}{23711}\right)}=\dfrac{-4}{5}$ Vậy $M=\dfrac{-4}{5}$ 2, Áp dụng tính chất dãy tỉ số bằng nhau ta có: $\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{2011}=\dfrac{2011}{a}=\dfrac{a+b+c+2011}{b+c+2011+a}=\dfrac{a+b+c+2011}{a+b+c+2011}=1$ $\dfrac{a}{b}=1\Rightarrow a=b\left(1\right)\ \dfrac{b}{c}=1\Rightarrow b=c\left(2\right)$ Từ (1) và (2) ta có: $a=c$ $\Rightarrow a+b-c=a+a-a=a$ 1) b) $A=27^{20}+3^{61}+9^{31}\ =\left(3^3\right)^{20}+3^{61}+\left(3^2\right)^{31}\ =3^{60}+3^{61}+3^{62}\ =3^{60}\cdot\left(1+3+3^2\right)\ =3^{60}\cdot\left(1+3+9\right)\ =3^{60}\cdot13⋮13$ Vậy $A⋮13$ a, $\left(-99\right)^{20}=\left(-99\right)^{2\cdot10}=\left[\left(-99\right)^2\right]^{10}=9801^{10}\ 9999^{100}=\left(9999^{10}\right)^{10}>\left(9999^{10}\right)^1=9999^{10}$ Vì $9801^{10}< 9999^{10}< \left(9999^{10}\right)^{10}=9999^{100}\Rightarrow\left(-99\right)^{20}< 9999^{100}$ Vậy $\left(-99\right)^{20}< 9999^{100}$

Huy Thắng Nguyễn · Answer

1/

a) (-99)20 = 9920

Vì 99 < 9999

20 < 100

Nên 9920 < 9999100

Vậy (-99)20 < 9999100

b) $A=27^{20}+3^{61}+9^{31}$

$=\left(3^3\right)^{20}+3^{61}+\left(3^2\right)^{31}$

$=3^{60}+3^{61}+3^{62}$

$=3^{60}\left(1+3+3^2\right)$

$=3^{60}.13⋮13$

Vậy A chia hết cho 13.

2) Áp dụng tính chất dãy tỉ số bằng nhau ta có:

$\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{2011}=\dfrac{2011}{a}=\dfrac{a+b+c+2011}{b+c+2011+a}=1$

$\Rightarrow\dfrac{a}{b}=1;\dfrac{b}{c}=1\Rightarrow a=b=c$ (*)

Thay (*) vào a + b - c: a + a - a = a

Vậy a + b - c = a.

3. $M=\dfrac{\dfrac{4}{237}-\dfrac{4}{2371}+\dfrac{4}{23711}}{-\dfrac{5}{237}+\dfrac{5}{2371}-\dfrac{5}{23711}}$

$=\dfrac{4\left(\dfrac{1}{237}-\dfrac{1}{2371}+\dfrac{1}{23711}\right)}{-5\left(\dfrac{1}{237}-\dfrac{1}{2371}+\dfrac{1}{23711}\right)}$

$=-\dfrac{4}{5}$Câu trả lời: 1/