a. \(327 \times (126 + x) = 67035\)

\(126 + x = 67035 : 327\)

\(126 + x = 205\)

\(x=205-126=79\)

b. \(10208 : (108 + x) = 58\)

\(108 + x = 10208 : 58\)

\(108 + x = 176\)

\(x=176-108=68\)

c. \(68 \times (236 - x) = 9860\)

\(236 - x = 9860 : 68\)

\(236 - x = 145\)

\(x=236-145=91\)

d. \(17856 : (405 - x) = 48\)

\(405 - x = 17856 : 48\)

\(405 - x = 372\)

\(x=405-372=33\)

bài 25:

\(\text{1) }35\frac{1}{6}:\frac{-4}{5}-45\frac{1}{6}:\frac{-4}{5}=\left(35\frac{1}{6}-45\frac{1}{6}\right):\frac{-4}{5}\)

\(=(-10)\cdot\frac{-5}{4}=\frac{50}{4}=\frac{25}{2}\)

\(\text{2) }4\frac{5}{9}:\frac{-5}{7}+5\frac{4}{9}:\frac{-5}{7}=\left(4\frac{5}{9}+5\frac{4}{9}\right):\frac{-5}{7}\)

\(=\left(\frac{41}{9}+\frac{49}{9}\right):\frac{-5}{7}=10\cdot\frac{-7}{5}=-14\)

\(\text{3) }\frac{-7}{5}\cdot2\frac{2}{3}-1\frac{2}{5}\cdot\frac{-2}{3}=\frac{-7}{5}\cdot\frac{8}{3}-\frac{7}{5}\cdot\frac{-2}{3}\)

\(=\frac{-7}{5}\cdot\frac{8}{3}-\frac{-7}{5}\cdot\frac{2}{3}=\frac{-7}{5}\cdot\left(\frac{8}{3}-\frac{2}{3}\right)\)

\(=\frac{-7}{5}\cdot\frac{6}{3}=\frac{-7}{5}\cdot2=\frac{-14}{5}\)

\(\text{4) }3\frac{1}{3}:\frac{-9}{4}+3\frac{1}{3}:\frac{9}{13}=3\frac{1}{3}\cdot\frac{-4}{9}+3\frac{1}{3}\cdot\frac{13}{9}\)

\(=3\frac{1}{3}\cdot\left(\frac{-4}{9}+\frac{13}{9}\right)=\frac{10}{3}\cdot\frac{9}{9}\)

\(=\frac{10}{3}\cdot1=\frac{10}{3}\)

\(\text{5) }\frac{1}{9}:\frac{2}{145}-4\frac{1}{3}:\frac{2}{145}=\left(\frac{1}{9}-4\frac{1}{3}\right):\frac{2}{145}\)

\(=\left(\frac{1}{9}-\frac{13}{3}\right)\cdot\frac{145}{2}=\frac{-38}{9}\cdot\frac{145}{2}=\frac{-2755}{9}\)

\(\text{6) }19\frac{1}{3}:\frac{4}{3}-39\frac{1}{3}:\frac{4}{3}=\left(19\frac{1}{3}-39\frac{1}{3}\right):\frac{4}{3}\)

\(=(-20)\cdot\frac{3}{4}=-15\)

\(\text{7) }5\frac{2}{3}:\frac{4}{7}-17\frac{2}{3}:\frac{4}{7}=\left(5\frac{2}{3}-17\frac{2}{3}\right):\frac{4}{7}\)

\(=(-12)\cdot\frac{7}{4}=-3\cdot7=-21\)

bài 24:

\(\text{1) }\frac{5}{6}\cdot17\frac{1}{3}-\frac{5}{6}\cdot47\frac{1}{3}=\frac{5}{6}\cdot\left(17\frac{1}{3}-47\frac{1}{3}\right)\)

\(=\frac{5}{6}\cdot(-30)=-25\)

\(\text{2) }\frac{4}{5}\cdot19\frac{1}{3}-\frac{4}{5}\cdot39\frac{1}{3}=\frac{4}{5}\cdot\left(19\frac{1}{3}-39\frac{1}{3}\right)\)

\(=\frac{4}{5}\cdot(-20)=-16\)

\(\text{3) }\frac{2}{5}\cdot15\frac{1}{3}-\frac{2}{5}\cdot10\frac{1}{3}=\frac{2}{5}\cdot\left(15\frac{1}{3}-10\frac{1}{3}\right)\)

\(=\frac{2}{5}\cdot5=2\)

\(\text{4) }\frac{3}{5}\cdot13\frac{1}{4}-\frac{3}{5}\cdot33\frac{1}{4}=\frac{3}{5}\cdot\left(13\frac{1}{4}-33\frac{1}{4}\right)\)

\(=\frac{3}{5}\cdot(-20)=-12\)

\(\text{5) }\frac{3}{4}\cdot26\frac{1}{5}-\frac{3}{4}\cdot44\frac{1}{5}=\frac{3}{4}\cdot\left(26\frac{1}{5}-44\frac{1}{5}\right)\)

\(=\frac{3}{4}\cdot(-18)=\frac{-27}{2}\)

\(\text{6) }\frac{4}{13}\cdot15\frac{3}{41}-\frac{4}{13}\cdot2\frac{3}{41}=\frac{4}{13}\cdot\left(15\frac{3}{41}-2\frac{3}{41}\right)\)

\(=\frac{4}{13}\cdot13=4\)

\(\text{7) }12\frac{1}{5}\cdot\frac{25}{4}-10\frac{1}{5}\cdot\frac{25}{4}=\frac{25}{4}\cdot\left(12\frac{1}{5}-10\frac{1}{5}\right)\)

\(=\frac{25}{4}\cdot2=\frac{25}{2}\)

\(\text{8) }43\frac{1}{4}\cdot\frac{-2}{3}-13\frac{1}{4}\cdot\frac{-2}{3}=\frac{-2}{3}\cdot\left(43\frac{1}{4}-13\frac{1}{4}\right)\)

\(=\frac{-2}{3}\cdot30=-20\)

\(\text{9) }16\frac{3}{5}\cdot\frac{-1}{3}-13\frac{3}{5}\cdot\frac{-1}{3}=\frac{-1}{3}\cdot\left(16\frac{3}{5}-13\frac{3}{5}\right)\)

\(=\frac{-1}{3}\cdot3=-1\) \(\)

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