bài 23:

1. \(\frac{-10}{11}\cdot\frac{8}{9}+\frac{7}{18}\cdot\frac{10}{11}=\frac{-10}{11}\cdot\frac{8}{9}+\frac{7}{18}\cdot\frac{2 \cdot5}{11}\)

\(\) \(=\frac{-10}{11}\cdot\frac{8}{9}+\frac{7}{9}\cdot\frac{5}{11}=\frac{-10}{11}\cdot\frac{8}{9}+\frac{35}{99}\)

\(=\frac{-80}{99}+\frac{35}{99}=\frac{-45}{99}=\frac{-5}{11}\)

\(\text{2) }\frac{12}{25}\cdot\frac{23}{7}-\frac{12}{7}\cdot\frac{13}{25}=\frac{12}{25}\cdot\frac{23}{7}-\frac{12}{25}\cdot\frac{13}{7}\)

\(=\frac{12}{25}\cdot\left(\frac{23}{7}-\frac{13}{7}\right)=\frac{12}{25}\cdot\frac{10}{7}=\frac{24}{35}\)

\(\text{3) }\frac{3}{7}\cdot\frac{16}{15}-\frac{2}{15}\cdot\frac{-3}{7}=\frac{3}{7}\cdot\frac{16}{15}+\frac{2}{15}\cdot\frac{3}{7}\)

\(=\frac{3}{7}\cdot\left(\frac{16}{15}+\frac{2}{15}\right)=\frac{3}{7}\cdot\frac{18}{15}=\frac{18}{35}\)

\(\text{4) }\frac{-4}{13}\cdot\frac{5}{17}+\frac{-12}{13}\cdot\frac{4}{17}=\frac{-4}{13}\cdot\frac{5}{17}+\frac{-4}{13}\cdot3\cdot\frac{4}{17}\)

\(=\frac{-4}{13}\cdot\frac{5}{17}+\frac{-4}{13}\cdot\frac{12}{17}=\frac{-4}{13}\cdot\left(\frac{5}{17}+\frac{12}{17}\right)\)

\(=\frac{-4}{13}\cdot\frac{17}{17}=\frac{-4}{13}\)

\(\text{5) }\frac{-6}{13}\cdot\frac{13}{21}+\frac{6}{13}\cdot\frac{-8}{21}=\frac{-6}{13}\cdot\frac{13}{21}-\frac{6}{13}\cdot\frac{8}{21}\)

\(=\frac{-6}{13}\cdot\left(\frac{13}{21}+\frac{8}{21}\right)=\frac{-6}{13}\cdot\frac{21}{21}=\frac{-6}{13}\)

\(\text{6) }\frac{-2}{4}\cdot\frac{-1}{13}-\frac{-2}{24}\cdot\frac{7}{13}=\frac{-1}{2}\cdot\frac{-1}{13}-\frac{-1}{12}\cdot\frac{7}{13}\)

\(=\frac{1}{26}-\frac{-7}{156}=\frac{6}{156}+\frac{7}{156}=\frac{13}{156}=\frac{1}{12}\)

\(A = 2(x^6 + y^6) - 3(x^4 + y^4)\)

\(A = 2[(x^2)^3 + (y^2)^3] - 3[(x^2)^2 + (y^2)^2]\)

\(A = 2(x^2 + y^2)[(x^2 + y^2)^2 - 3x^2y^2] - 3[(x^2 + y^2)^2 - 2x^2y^2]\) (1)

thay \(x^2 + y^2 = 1\) vào (1) ta được:

\(A = 2 \cdot 1 \cdot [1^2 - 3x^2y^2] - 3[1^2 - 2x^2y^2]\)

\(A = 2(1 - 3x^2y^2) - 3(1 - 2x^2y^2)\)

\(A=2-6x^2y^2-3+6x^2y^2=-1\)

vậy A không phục thuộc vào x, y

a. \(3x(x-7) + 2xy - 14y\)

\(= 3x(x-7) + 2y(x-7)\)

\(= (x-7)(3x + 2y)\)

b. \(6x^2 - 12x + 6\)

\(= 6(x^2 - 2x + 1)\)

\(= 6(x-1)^2\)

c. \(9(2x-5)^2 + 15x - 6x^2\)

\(= 9(2x-5)^2 - 3x(2x-5)\)

\(= 3(2x-5)[3(2x-5) - x]\)

\(= 3(2x-5)(6x - 15 - x)\)

\(= 3(2x-5)(5x - 15)\)

\(= 15(2x-5)(x-3)\)

d. \(2xy^3-15x^4=x(2y^3-15x^3)\)

e. \(3x^4-48=3(x^4-16)=3[(x^2)^2-4^2]\)

\(=3(x^2-4)(x^2+4)=3(x-2)(x+2)(x^2+4)\)

f. \(x^2 - z^2 + 4xy + 4y^2\)

\(= (x^2 + 4xy + 4y^2) - z^2\)

\(= (x + 2y)^2 - z^2\)

\(= (x + 2y - z)(x + 2y + z)\)

g. \(x^2 - z^2 + 2xy - 6zt + y^2 - 9t^2\)

\(= (x^2 + 2xy + y^2) - (z^2 + 6zt + 9t^2)\)

\(= (x+y)^2 - (z + 3t)^2\)

\(= [(x+y) - (z+3t)][(x+y) + (z+3t)]\)

\(= (x + y - z - 3t)(x + y + z + 3t)\)

h. \(x^2-12x+20=x^2-2x-10x+20\)

\(=x(x-2)-10(x-2)=(x-2)(x-10)\)

i. \(2x^2-x-15=2x^2-6x+5x-15\)

\(=2x(x-3)+5(x-3)=(x-3)(2x+5)\)

3. ta có a vuông góc với AB, b vuông góc với AB

⇒ a // b

ta có góc C1 = góc ACD (đối đỉnh)

mà góc ACD = góc D1 (đồng vị)

nên góc C1 = góc D1 = 110 độ

\(\left( \frac{x}{x^2-36} - \frac{6}{x^2+6x} \right) : \frac{2x-6}{x^2+6x} + \frac{x}{6-x}\) (đkxđ: \(x \neq 0;\ x \neq \pm 6;\ x \neq 3\) )

\(= \left[ \frac{x}{(x-6)(x+6)} - \frac{6}{x(x+6)} \right] \cdot \frac{x(x+6)}{2(x-3)} - \frac{x}{x-6}\)

\(= \frac{x^2 - 6(x-6)}{x(x-6)(x+6)} \cdot \frac{x(x+6)}{2(x-3)} - \frac{x}{x-6}\)

\(= \frac{x^2 - 6x + 36}{x(x-6)(x+6)} \cdot \frac{x(x+6)}{2(x-3)} - \frac{x}{x-6}\)

\(= \frac{x^2 - 6x + 36}{2(x-6)(x-3)} - \frac{x}{x-6}\)

\(= \frac{x^2 - 6x + 36 - 2x(x-3)}{2(x-6)(x-3)}\)

\(= \frac{x^2 - 6x + 36 - 2x^2 + 6x}{2(x-6)(x-3)}\)

\(=\frac{-x^2 + 36}{2(x-6)(x-3)}=\frac{-(x^2 - 36)}{2(x-6)(x-3)}\)

\(=\frac{-(x-6)(x+6)}{2(x-6)(x-3)}=\frac{-(x+6)}{2(x-3)}\)