a. 10 – 9 + 8 – 7 + 6 – 5 + 4 – 3 + 2 – 1

= (10 – 9) + (8 – 7) + (6 – 5) + (4 – 3) + (2 – 1)

= 1 + 1 + 1 + 1 + 1

= 5

b. 1 + 3 + 5 + 7 + 9 + 10 + 8 + 6 + 4 + 2 + 0

= (0 +10) +(1 + 9) + (2 + 8) + (3 + 7) + (4 + 6) + 5

= 10 + 10 + 10 + 10 + 10 + 5

= 55

a) 10 - 9 + 8 - 7 + 6 - 5 + 4 - 3 + 2 - 1

= (10 - 9) + (8 - 7) + (6 - 5) + (4 - 3) + (2 - 1)

= 1 + 1 + 1 + 1 + 1

= 1 x 5

= 5

b) 1 + 3 + 5 + 7 + 9 + 10 + 8 + 6 + 4 + 2 + 0

= 0 + 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10

= (0 + 10) x 11 : 2

= 10 x 11 : 2

= 5 x 11

= 55

a)5

b)55

chúc bạn học tốt

1) \(x.\left(x+7\right)=0\)
\(=>\left[\begin{matrix}x=0\\x+7=0\end{matrix}\right.=>\left[\begin{matrix}x=0\\x=-7\end{matrix}\right.\)
2) \(\left(x+12\right).\left(x-3\right)=0\)
\(=>\left[\begin{matrix}x+12=0\\x-3=0\end{matrix}\right.=>\left[\begin{matrix}x=-12\\x=3\end{matrix}\right.\)
3) \(\left(-x+5\right).\left(3-x\right)=0\)
\(=>\left[\begin{matrix}-x+5=0\\3-x=0\end{matrix}\right.=>\left[\begin{matrix}x=5\\x=3\end{matrix}\right.\)
4) \(x.\left(2+x\right).\left(7-x\right)=0\)
\(=>\left[\begin{matrix}x=0\\2+x=0\\7-x=0\end{matrix}\right.=>\left[\begin{matrix}x=0\\x=-2\\x=7\end{matrix}\right.\)
5) \(\left(x-1\right).\left(x+2\right).\left(-x-3\right)=0\)
\(=>\left[\begin{matrix}x-1=0\\x+2=0\\-x-3=0\end{matrix}\right.=>\left[\begin{matrix}x=1\\x=-2\\x=-3\end{matrix}\right.\)

Tổng quát: tích ab = 0 thì a=0 hoặc b=0

\(a\cdot b\cdot c=0\Leftrightarrow\left[\begin{matrix}a=0\\b=0\\c=0\end{matrix}\right.\)(nếu có 3 thừa số)

==40 chuc bn hok tốt 

kết quả là 40 nha bạn 

1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 0 = 45

Kết bạn nhé!

Học tốt!!!

đây bạn

a,7.x-8=6

x-8=6:7

x-8=6/7

x=6/7+8

x=62/7

Vậy x=62/7

b, 8.(x-5)=0

x-5=0:8

x-5=0

x=0+5

x=5

Vậy x = 5

c, x -6:2-(48-24.2:6-3)=0

x-6:2-(48-48:6-3)=0

x-6:2-(48-8-3)=0

x-6:2-(40-3)=0

x-6:2-37=0

x-6:2=0+37

x-3=37

x=37+3

x=40

Vẫy=40

Tích đi mình trả lời

Ta có :

\(3^{2n}=\left(3^2\right)^n=9^n\)

\(2^{3n}=\left(2^3\right)^n=8^n\)

Vì \(9>8\)\(\Rightarrow9^n>8^n\)

Hay \(3^{2n}>2^{3n}\)

Vậy  \(3^{2n}>2^{3n}\)

_Chúc bạn học tốt_

a.390 - (x - 7) = 169 : 13

390 - (x - 7) = 13

x - 7 = 390 - 13

x = 377 + 7

x = 384

b.(x - 140) : 7 = 3³ - 2³.3

(x - 140) : 7 = 3

x - 140 = 3 . 7

x = 21 + 140

x = 161

c.x - 6:2 - (48-24):2: 6-3 = 0

x - 3 -2 - 3 = 0

x = 3 + 2 + 3

x = 8

\(a.390-\left(x-7\right)=169:13\)

    \(390-\left(x-7\right)=13\)

                   \(\left(x-7\right)=390-13\)

                         \(x-7=377\)

                                  \(x=377+7\)

                          Vậy \(x=384\)

\(b.\left(x-140\right):7=3^3--2^3.3\)

    \(\left(x-140\right):7=3^3+2^3.3\)

    \(\left(x-140\right):7=27+8.3\)

    \(\left(x-140\right):7=27+24\)

    \(\left(x-140\right):7=51\)

        \(\left(x-140\right)=51.7\)

              \(x-140=357\)

                             \(x=357+140\)

                      Vậy \(x=497\)

\(c.x-6:2-\left(48-24\right):2:6-3=0\)

                          \(x-3-24:2:6-3=0\)

                              \(x-3-12:6-3=0\)

                                     \(x-3-2-3=0\)

                                             \(x-\left(-2\right)=0\)

                                                       \(x+2=0\)

                                                                \(x=0-2\)

                                                        Vậy \(x=-2\)

Chúc bạn học tốt nha!

Cách tui lẹ hơn cách bạn Nguyễn Duy Khánh nè!

Ta có: \(C=\frac{7^{28}+7^{24}+....+7^4+7^0}{\left(7^{30}+7^{26}+...+7^6+7^2\right)+\left(7^{28}+7^{24}+...+7^4+7^0\right)}\)

\(=\frac{7^{28}+7^{24}+...+7^4+7^0}{7^2\left(7^{28}+7^{24}+...+7^4+7^0\right)+\left(7^{28}+7^{24}+...+7^4+7^0\right)}\)

\(=\frac{7^{28}+7^{24}+...+7^4+7^0}{\left(7^{28}+7^{24}+...+7^4+7^0\right)\left(7^2+1\right)}=\frac{1}{7^2+1}=\frac{1}{50}\)

P/s: Easy đúng không?

\(C=\frac{7^{28}+7^{2\text{4}}+...+7^{\text{4}}+7^0}{7^{30}+7^{28}+...+7^2+7^0}\)

Đặt A là tử số ,B là mẫu số.Ta có:

\(7^{\text{4}}A=7^{32}+7^{28}+...+7^8+7^{\text{4}}+7^0\)

\(20\text{4}1A-A=\left(7^{32}+7^{28}+7^{2\text{4}}+...+7^8+7^{\text{4}}\right)-\left(7^{28}+7^{2\text{4}}+...+7^{\text{4}}+7^0\right)\)

\(2\text{4}00A=7^{32}-7^0=7^{32}-1\)

\(\Rightarrow A=\left(7^{32}-1\right):2\text{4}00\)

\(7^2B=\left(7^{32}+7^{30}+7^{28}+...+7^{\text{4}}+7^2\right)\)

49B-B= ....tự..điền......như A nhé.....

48B=7³²-1 =>B=[72³²-1]:48

=>\(C=\frac{A}{B}=\frac{\left(7^{32}-1\right):2\text{4}00}{\left(7^{32}-1\right):\text{4}8}\)

Tui nghĩ vậy đc r á

p/s:ko chắc

.

Đặt:  \(A=7^{28}+7^{24}+...+7^4+7^0\)

=>  \(7^4.A=7^{32}+7^{28}+...+7^8+7^4\)

=>  \(7^4.A-A=\left(7^{32}+7^{28}+...+7^8+7^4\right)-\left(7^{28}+7^{24}+...+7^4+7^0\right)\)

=>  \(\left(7^4-1\right).A=7^{32}-1\)

=>  \(A=\frac{7^{32}-1}{7^4-1}\)

Đặt:  \(B=7^{30}+7^{28}+...+7^2+7^0\)

tương tự tính được:  \(B=\frac{7^{32}-1}{7^2-1}\)

\(C=\frac{A}{B}=\frac{7^{32}-1}{7^4-1}:\frac{7^{32}-1}{7^2-1}=\frac{7^2-1}{7^4-1}\)

\(=\frac{7^2-1}{\left(7^2-1\right)\left(7^2+1\right)}=\frac{1}{7^2+1}=\frac{1}{50}\)

x(x + 7) = 0

=> x = 0 hoac x + 7 = 0

=> x = 0 hoac x = -7

cac phan sau tt

c) (-x + 5)(3 - x) = 0

=> \(\orbr{\begin{cases}-x+5=0\\3-x=0\end{cases}}\)

=> \(\orbr{\begin{cases}-x=-5\\x=3\end{cases}}\)

=> \(\orbr{\begin{cases}x=5\\x=3\end{cases}}\)

Vậy...

a x=0,x=-7

b x=-12,x=3

c x=5,x=3

d x=0,x=-2,x=7

e x=1,x=-2,x=-3