la 100

100 nhớ chọn mk nha

là 100

5 mũ 56,nha bạn

kết quả là 5 ^ ⁵⁶

Đặt A=(2+x)⁵(3x-1)⁷

^{khai triển ta có:A=(\(_{k=0}^5\Sigma C_5^k2^{5-k}x^k\)}).(\(^7_{i=0}\Sigma C_7^i\left(3x\right)^i\))

=\(\left(_{k=0}^5\Sigma\right)\left(_{i=0}^7\Sigma\right)\left(C_5^kC^i_7\right)\left(x^k.\left(3x\right)^i\right)\)

=số hạng\(\left(C_5^kC^i_7\right)\left(x^k.\left(3x\right)^i\right)\)chứa x⁵ tại k+i=5

có k\(\in\){0,1,2,...5},i\(\in\){0,1,2,...7}

=>(k,i)={(0,5);(1,4);(2,3);(3,2);(4,1);(5,0)}

=>Hệ số của x⁵ là:\(\left(C_5^0C^5_7\right)3^5\)+\(\left(C_5^1C^4_7\right)\left(3^4\right)\)+\(\left(C_5^2C^3_7\right)\left(3^3\right)\)+\(\left(C_5^3C^2_7\right)\left(3^2\right)\)+

\(\left(C_5^4C^1_7\right)\left(3^1\right)\)+\(\left(C_5^5C^0_7\right)3^0\)=30724

Hok tốt!!!

b) ta có (1+x-x²)⁸=(1+(x-x²))⁸

=\(^8_{k=0}\Sigma.C_8^k\left(x-x^2\right)^k\)=\(^8_{k=0}\Sigma.C_8^k\left(x-1\right)^kx^k\)=\(^8_{k=0}\Sigma.C_8^k\left(x-1\right)^kx^k\)

a, y \(\times\) \(\dfrac{4}{3}\) = \(\dfrac{16}{9}\)

    y         =    \(\dfrac{16}{9}\) : \(\dfrac{4}{3}\)

    y         = \(\dfrac{4}{3}\)

b, ( y - \(\dfrac{1}{2}\)) + 0,5 = \(\dfrac{3}{4}\)

    y - 0,5 + 0,5 = \(\dfrac{3}{4}\)

   y                   = \(\dfrac{3}{4}\)

c, \(\dfrac{4}{5}-\dfrac{2}{5}y\) = 0,2

   0,8 - 0,4y = 0,2

           0,4y = 0,8 - 0,2

           0,4y  = 0,6

               y = 1,5

d, (y + \(\dfrac{3}{4}\)) \(\times\) \(\dfrac{5}{7}\) = \(\dfrac{10}{9}\)

    y + \(\dfrac{3}{4}\)           = \(\dfrac{10}{9}\) : \(\dfrac{5}{7}\)

   y + \(\dfrac{3}{4}\)            = \(\dfrac{14}{9}\)

y                    = \(\dfrac{14}{9}\) - \(\dfrac{3}{4}\)

 y                   =   \(\dfrac{29}{36}\)

e, y : \(\dfrac{5}{4}\)         = \(\dfrac{9}{5}\)  + \(\dfrac{1}{2}\)

   y : \(\dfrac{5}{4}\)         =   \(\dfrac{23}{10}\)

  y                =      \(\dfrac{23}{10}\)

  y               =   \(\dfrac{23}{8}\)

f, y \(\times\) \(\dfrac{1}{2}\) + \(\dfrac{3}{2}\) \(\times\) y   = \(\dfrac{4}{5}\)

   y \(\times\) ( \(\dfrac{1}{2}+\dfrac{3}{2}\))      =  \(\dfrac{4}{5}\)

   2y                       = \(\dfrac{4}{5}\)

    y                        = \(\dfrac{2}{5}\)

\(\left(x-12,5\right):\frac{3}{4}=\frac{8}{5}\)

\(x-12,5=\frac{8}{5}.\frac{3}{4}\)

\(x-12,5=\frac{6}{5}\)

\(x=\frac{6}{5}+12,5\)

\(x=\frac{137}{10}\)

\(\left(x-\frac{2}{5}\right)\times\frac{5}{11}=\frac{75}{22}\)

\(x-\frac{2}{5}=\frac{75}{22}:\frac{5}{11}\)

\(x-\frac{2}{5}=\frac{15}{2}\)

\(x=\frac{15}{2}+\frac{2}{5}\)

\(x=\frac{79}{10}\)

Bài làm

\(\left(X-12,5\right):\frac{3}{4}=\frac{8}{5}\)

\(\left(X-\frac{25}{2}\right)=\frac{8}{5}x\frac{3}{4}\)

\(X-\frac{25}{2}=\frac{6}{5}\)

\(X=\frac{6}{5}+\frac{25}{2}\)

\(X=\frac{30}{10}+\frac{125}{10}\)

\(X=\frac{155}{10}\)

\(X=\frac{31}{2}\)

~ Câu b) tương tự, chỉ đổi dấu thôi ~

# Chúc bạn học tốt #

\(\Rightarrow5^{3x+2}< 10^{18}\div2^{18}\)

\(\Rightarrow5^{3x+2}< 5^{18}\)

\(3x+2< 18\)

\(x< \frac{16}{3}\)

latex

Bn ơi , có phải đề bài như thế này ko?

\(5^{x+3}.5^{4x}-5^{2x}.5^{3x+2}=100\)

a: \(=\dfrac{3}{5}\cdot\dfrac{5}{4}+\dfrac{1}{2}\cdot\dfrac{2}{3}=\dfrac{3}{4}+\dfrac{1}{3}=\dfrac{9+4}{12}=\dfrac{13}{12}\)

b: \(\Leftrightarrow x\cdot\dfrac{5}{4}=\dfrac{5}{8}\)

hay x=1/2