Câu 1.

Tờ vé số có dạng \(\overline{a_1a_2a_3a_4a_5a_6}\in A=\left\{0;1;2;3;4;5;6;7;8;9\right\}\)

\(;a_i\ne a_j\)

Chọn \(a_1\ne0\) nên \(a_1\) có 9 cách chọn.

5 số còn lại là chỉnh hợp chập 5 của 8 số còn lại \(\in A\backslash\left\{a_1\right\}\)

\(\Rightarrow\)Có \(A_8^5\) cách.

Vậy có tất cả \(A_8^5\cdot9=60480\) vé số.

c

Giúp em giải với huhu 

a: Ta có: \(\left(x-3\right)^3-\left(x-3\right)\left(x^2+3x+9\right)+9\left(x+1\right)^2=4\)

\(\Leftrightarrow x^3-9x^2+27x-27-x^3+27+9x^2+18x+9=4\)

\(\Leftrightarrow45x=-5\)

hay \(x=-\dfrac{1}{9}\)

b: Ta có: \(x\left(x-5\right)\left(x+5\right)-\left(x+2\right)\left(x^2-2x+4\right)=17\)

\(\Leftrightarrow x^3-25x-x^3-8=17\)

\(\Leftrightarrow-25x=25\)

hay x=-1

Bài 9:

a: Ta có: \(x^2-10x=-25\)

\(\Leftrightarrow x^2-10x+25=0\)

\(\Leftrightarrow x-5=0\)

hay x=5

b: ta có: \(4x^2-4x=-1\)

\(\Leftrightarrow4x^2-4x+1=0\)

\(\Leftrightarrow2x-1=0\)

hay \(x=\dfrac{1}{2}\)

c: Ta có: \(\left(2x-1\right)^2=\left(3x-2\right)^2\)

\(\Leftrightarrow\left(3x-2\right)^2-\left(2x-1\right)^2=0\)

\(\Leftrightarrow\left(3x-2-2x+1\right)\left(3x-2+2x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(5x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{3}{5}\end{matrix}\right.\)

Bài 7a đề sai bạn nhé.

Bài 7:

a. \(\left(x-3\right)^2-\left(5-2x\right)^2=\left(x-3-5+2x\right)\left(x-3+5-2x\right)=\left(3x-8\right)\left(2-x\right)\)

b. \(\left(x+y\right)^2-x^2+4xy-4y^2=\left(x+y\right)^2-\left(x-2y\right)^2=\left(x+y-x+2y\right)\left(x-y+x-2y\right)=3y\left(2x-3y\right)\)

c. \(\left(x+y\right)^3-\left(x-y\right)^3=\left(x+y-x+y\right)\left[\left(x+y\right)^2+\left(x+y\right)\left(x-y\right)+\left(x-y\right)^2\right]=2y\left(3x^2+y^2\right)\)

d. \(x^3+y^3+z^3-3xyz\)

\(=\left(x+y\right)^3-3x^2y-3xy^2+z^3-3xyz\)

\(=\left[\left(x+y\right)^3+z^3\right]-3xy\left(x+y+z\right)\)

\(=\left(x+y+z\right)\left[\left(x+y\right)^2-z\left(x+y\right)+z^2\right]-3xy\left(x+y+z\right)\)

\(=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-xz-yz\right)\)

Bài 7a đề sai bạn nhé

Bài 8:

a: \(73^2-27^2=\left(73-27\right)\left(73+27\right)=4600\)

b: \(63^2-27^2+72^2-18^2\)

\(=\left(63-18\right)\left(63+18\right)+\left(72-27\right)\left(72+27\right)\)

\(=45\cdot\left(63+18+72+27\right)\)

\(=45\cdot180=8100\)

Mọi người ơi 

b: Ta có: \(\left(x+y\right)^2-x^2+4xy-4y^2\)

\(=\left(x+y\right)^2-\left(x-2y\right)^2\)

\(=\left(x+y-x+2y\right)\left(x+y+x-2y\right)\)

\(=3y\cdot\left(2x-y\right)\)

c: Ta có: \(\left(x+y\right)^3-\left(x-y\right)^3\)

\(=x^3+3x^2y+3xy^2+y^3-x^3+3x^2y-3xy^2+y^3\)

\(=2y^3+6x^2y\)

\(=2y\left(3x^2+y^2\right)\)