Giải:

a) \(M=21^9+21^8+21^7+...+21+1\) 

Do \(21^n\) luôn có tận cùng là 1

\(\Rightarrow M=21^9+21^8+21^7+...+21+1\) 

Tân cùng của M là:

     \(1+1+1+1+1+1+1+1+1+1=10\) tận cùng là 0

\(\Rightarrow M⋮10\) 

\(\Leftrightarrow M⋮2;5\) 

b) \(N=6+6^2+6^3+...+6^{2020}\) 

\(N=6.\left(1+6\right)+6^3.\left(1+6\right)+...+6^{2019}.\left(1+6\right)\) 

\(N=6.7+6^3.7+...+6^{2019}.7\) 

\(N=7.\left(6+6^3+...+6^{2019}\right)⋮7\) 

\(\Rightarrow N⋮7\) 

Ta thấy: \(N=6+6^2+6^3+...+6^{2020}⋮6\) 

Mà \(6⋮̸9\) 

\(\Rightarrow N⋮̸9\) 

c) \(P=4+4^2+4^3+...+4^{23}+4^{24}\) 

\(P=1.\left(4+4^2\right)+4^2.\left(4+4^2\right)+...+4^{20}.\left(4+4^2\right)+4^{22}.\left(4+4^2\right)\) 

\(P=1.20+4^2.20+...+4^{20}.20+4^{22}.20\) 

\(P=20.\left(1+4^2+...+4^{20}+4^{22}\right)⋮20\) 

\(\Rightarrow P⋮20\) 

\(P=4+4^2+4^3+...+4^{23}+4^{24}\) 

\(P=4.\left(1+4+4^2\right)+...+4^{22}.\left(1+4+4^2\right)\) 

\(P=4.21+...+4^{22}.21\) 

\(P=21.\left(4+...+4^{22}\right)⋮21\) 

\(\Rightarrow P⋮21\) 

d) \(Q=6+6^2+6^3+...+6^{99}\) 

\(Q=6.\left(1+6+6^2\right)+...+6^{97}.\left(1+6+6^2\right)\) 

\(Q=6.43+...+6^{97}.43\) 

\(Q=43.\left(6+...+6^{97}\right)⋮43\) 

\(\Rightarrow Q⋮43\) 

Chúc bạn học tốt!

1)1/6 - - 5/6 = 1/6 + 5/6 = 1

2)6/13 - -14/39 = 6/13 + 14/39= 32/39

3)4/5 - 4/-18 = 4/5 + 4/18= 46/45

4)7/21 - 9/-36 = 7/21 + 9/36 = 7/12

5)-12/18 - -21/35  = -12/18 + 21/35 = -1/15

6)-3/21 - 6/42 = -2/7

7)-18/24 - 15/21 = -41/28

8)1/6 - 2/5 =-7/30

 moi nguoi oi giup voi

(-2) + 6 = 4

(-120) + (-60) = -180

| 21 - ( - 7) | + ( - 6)= 23

( - 5) + 21 = 16

( - 70) + ( - 21) = -91

6 + | ( - 7) + ( - 6)| = 19

  27 + -( 5) = 22

27 + ( - 5 ) = 22

Đúng thì k còn sai thì thui chứ đừng k sai nhé.

(-2)+6=4

(-120)+(-60)=-180

(-5)+21=16

(-70)+(-21)=-91

27+-(5)=22

27+(-5)=22

\(x^2=\left(\sqrt{21+6\sqrt{6}}-\sqrt{21-6\sqrt{6}}\right)^2\)

\(x^2=21+6\sqrt{6}+21-6\sqrt{6}-2\sqrt{441-216}\)

\(x^2=42-2\sqrt{225}\)

\(x^2=42-30=12\)

\(x=2\sqrt{3}\)

nếu có sai bn thông cảm nha

cách khác nhé:

\(\sqrt{21+6\sqrt{6}}-\sqrt{21-6\sqrt{6}}\)

\(=\sqrt{21+2.3\sqrt{2}.\sqrt{3}}-\sqrt{21-2.3\sqrt{2}.\sqrt{3}}\)

\(=\sqrt{18+2.\sqrt{18}.\sqrt{3}+3}-\sqrt{18-2.\sqrt{18}.\sqrt{3}+3}\)

\(=\sqrt{\left(\sqrt{18}+\sqrt{3}\right)^2}-\sqrt{\left(\sqrt{18}-\sqrt{3}\right)^2}\)

\(=\left(\sqrt{18}+\sqrt{3}\right)-\left(\sqrt{18}-\sqrt{3}\right)\)

\(=2\sqrt{3}\)

p/s: mk đã phân tích kĩ ra cho bn rồi đó

\(A=\sqrt{\left(3\sqrt{2}\right)^2+2.3\sqrt{2}.\sqrt{3}+\left(\sqrt{3}\right)^2}+\sqrt{\left(3\sqrt{2}\right)^2-2.3.\sqrt{2}.\sqrt{3}+\left(\sqrt{3}\right)^2}\)
\(A=\sqrt{\left(3\sqrt{2}+\sqrt{3}\right)^2}+\sqrt{\left(3\sqrt{2}-\sqrt{3}\right)^2}\)
\(A=3\sqrt{2}+\sqrt{3}+3\sqrt{2}-\sqrt{3}=6\sqrt{2}\)

Cảm ơn ạ

Lê Hùng Hải không có gì :))

\(D=\frac{6}{15\times18}+\frac{6}{18\times21}+\frac{6}{21\times24}+...+\frac{6}{87\times90}\)

\(=2\times\left(\frac{3}{15\times18}+\frac{3}{18\times21}+...+\frac{3}{87\times90}\right)\)

\(=2\times\left(\frac{18-15}{15\times18}+\frac{21-18}{18\times21}+...+\frac{90-87}{87\times90}\right)\)

\(=2\times\left(\frac{1}{15}-\frac{1}{18}+\frac{1}{18}-\frac{1}{21}+...+\frac{1}{87}-\frac{1}{90}\right)\)

\(=2\times\left(\frac{1}{15}-\frac{1}{90}\right)=\frac{1}{9}\)

\(\Leftrightarrow\dfrac{6}{16}\cdot2< x< \dfrac{20}{21}\cdot\dfrac{21}{4}=5\)

=>3/4<x<5

hay \(x\in\left\{1;2;3;4\right\}\)

kết quả bằng \(\frac{1}{4}\)

Lời giải:

Đặt \(\sqrt[3]{27+6\sqrt{21}}=a; \sqrt[3]{27-6\sqrt{21}}=b\) thì ta cần tính tổng $A=a+b$.

Ta có:

$a^3+b^3=54$

\(ab=\sqrt[3]{(27+6\sqrt{21})(27-6\sqrt{21})}=-3\)

$A^3=(a+b)^3=a^3+b^3=3ab(a+b)=54+3(-3)A$

$\Leftrightarrow A^3=54-9A$

$\Leftrightarrow A^3+9A-54=0$

$\Leftrightarrow A^2(A-3)+3A(A-3)+18(A-3)=0$

$\Leftrightarrow (A^2+3A+18)(A-3)=0$

$\Leftrightarrow A-3=0$ (do $A^2+3A+18>0$)

$\Leftrightarrow A=3$

Cách khác :

\(\sqrt{21+6\sqrt{6}}-\sqrt{21-6\sqrt{6}}=\dfrac{\sqrt{36+2.6.\sqrt{6}+6}-\sqrt{36-2.6.\sqrt{6}+6}}{\sqrt{2}}=\dfrac{6+\sqrt{6}-6+\sqrt{6}}{\sqrt{2}}=\dfrac{2\sqrt{6}}{\sqrt{2}}=2\sqrt{3}\)

Đặt: \(A=\sqrt{21+6\sqrt{6}}-\sqrt{21-6\sqrt{6}}\)

\(A^2=21+6\sqrt{6}-2\sqrt{\left(21+6\sqrt{6}\right)\left(21-6\sqrt{6}\right)}+21-6\sqrt{6}\)

\(A^2=42-2\sqrt{441-216}\)

\(A^2=42-2\sqrt{225}\)

\(A^2=42-30=12\)

\(\Rightarrow A=\sqrt{12}=2\sqrt{3}\)

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