\(210+\left(y-15\right)\times3=195\)

\(210+\left(y-15\right)=195:3=65\)

\(y-15=210-65=145\)

\(\Rightarrow x=145+15=160\)

Vậy \(x=160\)

210 + (y - 15) x 3 = 195

          (y - 15) x 3  = 195 - 210

          (y - 15) x 3  = - 15

            y - 15        = -15 : 3

            y - 15        = - 5

            y               = -5 + 15

            y               = 10

120 - 4x(y - 3) = 136

=> 4x(y - 3)     = 136 - 120

=> 4x(y - 3)     = 16

=>   x(y - 3)     = 16 : 4

=>   x(y - 3)     = 4

Mà 4 = 1.4 = 2.2 = (-1).(-4) = (-2).(-2)

Lập bảng xét tất cả 6 trường hợp : 

Vậy các cặp (x,y) thỏa mãn là : (1;7) ; (2;5) ; (4;4) ; (-2 ; 1) ; (-1 ; -1) ; (- 4;2)

120 - 4(y - 3) = 136

<=> -4(y - 3) = 136 - 120

<=> -4(y - 3) = 16

<=> y - 3 = 16 : (-4)

<=> y - 3 = -4

<=> y = -4 + 3

<=> y = -1

=> y = -1

a) \(2^3\cdot4^5=2^3\cdot2^{10}=2^{13}\)

b) \(3^4\cdot27^2=3^4\cdot3^6=3^{10}\)

c) \(16^{10}:8^2=2^{40}:2^6=2^{34}\)

d) \(12^3:3^3=\left(12:3\right)^3=4^3\)

e) \(3^4\cdot2^4=\left(3\cdot2\right)^4=6^4\)

g) \(9^{13}\cdot5^{26}=9^{13}\cdot25^{13}=\left(9\cdot25\right)^{13}=225^{13}\)

=))

\(A=124\left(\frac{1}{1.1985}+\frac{1}{2.1986}+\frac{1}{3.1987}+...+\frac{1}{16.2000}\right)\)

      \(=\frac{124}{1984}.\left(1-\frac{1}{1985}+\frac{1}{2}-\frac{1}{1986}+...+\frac{1}{16}-\frac{1}{2000}\right)\)

       \(=\frac{1}{16}\left[\left(1+\frac{1}{2}+...+\frac{1}{16}\right)-\left(\frac{1}{1985}+\frac{1}{1986}+...+\frac{1}{2000}\right)\right]\)

  Và \(B=\frac{1}{1.17}+\frac{1}{2.18}+...+\frac{1}{1984.2000}\)

             \(=\frac{1}{16}\left[\left(1-\frac{1}{17}+\frac{1}{2}-\frac{1}{18}+...+\frac{1}{1984}-\frac{1}{2000}\right)\right]\)

              \(=\frac{1}{16}\left[\left(1+\frac{1}{2}+...+\frac{1}{1984}\right)-\left(\frac{1}{17}+\frac{1}{18}+...+\frac{1}{2000}\right)\right]\)

=      \(\frac{1}{16}\)  .    \(\left[\left(1+...+\frac{1}{16}\right)+\left(\frac{1}{17}+...+\frac{1}{1984}-\frac{1}{17}-...-\frac{1}{1984}\right)-\left(\frac{1}{1985}+...+\frac{1}{2000}\right)\right]\)

 = \(=\frac{1}{16}\left[\left(1+\frac{1}{2}+...+\frac{1}{16}\right)-\left(\frac{1}{1985}+\frac{1}{1986}+...+\frac{1}{2000}\right)\right]\)

Vậy A = B

Làm mẫu 1 phần ko hiểu thì bảo mình làm típ

a) Ta có: \(\left(x+3\right)^2\ge0;\forall x\)

\(\left(x+3\right)^2-2\ge0-2;\forall x\)

Hay \(A\ge-2;\forall x\)

Dấu "=" xảy ra \(\Leftrightarrow\left(x+3\right)^2=0\)

                         \(\Leftrightarrow x=-3\)

VẬY MIN A=-2 \(\Leftrightarrow x=-3\)

\(\frac{1}{3^x}.\frac{1}{9}.\frac{1}{81}=\frac{1}{3^{20}}\)

\(\Rightarrow\frac{1}{3^x}=\frac{1}{3^{20}}:\left(\frac{1}{3^2}.\frac{1}{3^4}\right)\)

\(\Rightarrow\frac{1}{3^x}=\frac{1}{3^{14}}\)

\(\Rightarrow x=14\)

\(\frac{1}{3^x}.\frac{1}{3^2}.\frac{1}{3^4}=\frac{1}{3^{14}}.\frac{1}{3^2}.\frac{1}{3^4}\)

Vậy x = 14

Study well######

Ta có: 

C = \(\frac{7^2}{2.9}+\frac{7^2}{9.16}+\frac{7^2}{16.23}+...+\frac{7^2}{65.72}\)

=> C = \(7.\left(\frac{7}{2.9}+\frac{7}{9.16}+\frac{7}{16.23}+...+\frac{7}{65.72}\right)\)

=> C = \(7.\left(\frac{1}{2}-\frac{1}{9}+\frac{1}{9}-\frac{1}{16}+\frac{1}{16}-\frac{1}{23}+...+\frac{1}{65}-\frac{1}{72}\right)\)

=> C = \(7.\left(\frac{1}{2}-\frac{1}{72}\right)\)

=> C = \(7.\frac{35}{72}=\frac{245}{72}\)

Nhìn kĩ là ra thôi :

 \(\frac{7^2}{2.9}+\frac{7^2}{9.16}+...+\frac{7^2}{65.72}\)

= \(7\left(\frac{7}{2.9}+\frac{7}{9.16}+...+\frac{7}{65.72}\right)\)

= \(7\left(\frac{1}{2}-\frac{1}{9}+\frac{1}{9}-\frac{1}{16}+...+\frac{1}{65}-\frac{1}{72}\right)\)

= \(7\left(\frac{1}{2}-\frac{1}{72}\right)\)

= \(7.\frac{35}{72}=3\frac{29}{72}\)

Bài làm

     ( x - 6 )² = 9

=> ( x - 6 )² = 3² 

=>   x - 6     = 3 

       x           = 3 + 6

       x           = 9

Vậy x = 9

# Học tốt #

(x - 6 )^2 = 9

( x - 6 )^2=3^2

=>x - 6= 3

x = 3+6

x= 9

a, * thuộc { 0; 2; 4 ; 6 : 8 }

b., * thuộc { 0; 5 }

c, * thuộc { 0 }

a) thuộc số {0,2,4,6,8}

b) thuộc {0,5}

c) số chia hết cho 2vs5 là: {0}

hok tốt