Первое домашнее задание курса школьной математики
- \(\displaystyle\frac{2^4\cdot (12)^4}{9^2\cdot (18)^3}\)
- \(\displaystyle\frac{a^4(b^3c)^{-2}}{(ab^4)^{-3}(c^{-3})^2}\)
- \(\displaystyle\frac{2}{3}+\displaystyle\frac{5}{6}:\displaystyle\frac{15}{12}-\displaystyle\frac{13}{9}\)
- \((x-1)(x-3)-(x+2)(3-x)\)
- \((x-10)(x+10)-(2x+3)(2x-3)\)
- \(6b^2-(2b+5)(3b-7)\)
- \(-1-(2x-(x+1)(2x+1))-2x^2\)
- \((x-2)(3x+4)-x(3x-2)\)
- \((2x^2-4)(3x^2+2)-(x^2-1)(6x^2+6)\)
- \(49x^3-x(7x+5)(7x-5)\)
- \((a+b)(a-b+1)-(a-b)(a+b-1)\)
- \((x+2-x^2)(2-x)+3(1-\frac{x^3}{3})\)
- \((1+x)(2+x)(3+x)-(x+2)(x+1)\)
- \(1-(x^2-x)(x^2+x)+x^3(x+1)\)
- \((2+x-3x^2+x^3)(2-x+3x^2+x^3)\)
- \((x-2)(x^4+2x^3+4x^2+8x+16)\)
- \((x+3y)(x+y+2)-(x+y)(x+3y+2)\)
- \((3+5-6)\cdot(-2-3+6)\)
- \(-4\displaystyle\frac{1}{3}-(-\displaystyle\frac{2}{5}+\displaystyle\frac{3}{7})\)
- \(\displaystyle\frac{1}{2(3-2)}-\displaystyle\frac{2(2-4)}{3-5}\)
- \(\frac{5}{4}-\frac{4}{13}\cdot\frac{26}{8}+8:\frac{44}{33}\)
- \(\displaystyle\frac{49^4\cdot 50^3\cdot 3^5:8}{9^3\cdot 35^3\cdot 25:7^{-5}}\)
- \((2,5+2\frac{1}{2})\cdot\frac{8}{12}-(\frac{7}{2}+\frac{18}{27})\cdot\frac{8}{24}-6,2\)
- \(\frac{x}{5}-\frac{1}{3}+\frac{2}{9}=\frac{3x}{27}-\frac{6}{15}\)
- \(4(-x+2)-7(2x+3)=5\)
- \(2x+\displaystyle\frac{1}{3}(7-4x)=0\)
- \(-\displaystyle\frac{1}{6}(-2x-3)=\displaystyle\frac{1}{6}(3x+3)\)
- \(x+5(2-x+3(x-1)-x)=0\)
- \(-(x-4(x-4))=x-4\)
- \(4-2(8-7(2x-3))=x\)
- \(\displaystyle\frac{x}{3}+\displaystyle\frac{2x-3}{6}=0\)
- \(-\displaystyle\frac{x}{2}+x-\displaystyle\frac{3}{4}=0\)
- \(\displaystyle\frac{x+1}{2}=1-\displaystyle\frac{x-1}{3}\)
- \((x-4)\cdot\displaystyle\frac{7}{8}=\displaystyle\frac{5x}{4}-\displaystyle\frac{35}{4}\)
- \(\displaystyle\frac{x-8}{2}=\displaystyle\frac{8-2x}{3}\)
- \(\displaystyle\frac{2x}{5}-\displaystyle\frac{5x+1}{10}=-0,1\)
- \(6-x-\displaystyle\frac{7x-8}{3}=0\)
- \(\frac{x}{2}+\frac{x}{3}+\frac{x}{4}=2-\frac{x}{8}\)
- \(\displaystyle\frac{2x}{3}\cdot (2-0,1)+\displaystyle\frac{x}{4}:2,2=1-\displaystyle\frac{x}{2}\)
- \(\displaystyle\frac{x}{1\cdot 2\cdot 3}+\displaystyle\frac{x}{2\cdot 3\cdot 4}=\displaystyle\frac{2x}{4\cdot 5\cdot 6}-0,2\)
- \((\frac{41}{18}-\frac{17}{36})\cdot\frac{18}{65}+(\frac{8}{7}-\frac{23}{49}):\frac{99}{49}+\frac{7}{6}\)
- \(\frac{10}{16}+\frac{3}{2}\cdot (\frac{17}{4}:17)+3,75:\frac{5}{6}\)
- \(\frac{217}{31}:1,75-(3,5:\frac{5}{4}+\frac{7}{2})+3,4:\frac{17}{8}\)
- \((\frac{14}{15}+\frac{5}{2}+0,3)\cdot \frac{8}{7}\cdot 0,75+\frac{5}{10}\)
- \((1+x+y)(1-(x+y))\)
- \(a^3-(a-4)(a^2+3a+12)-a(a+3)\)
- \((x+2)(x^2-2x+4)-(x^2-3x)(x+3)\)
- \((3d^3-4)(9d^6+12d^3+16)\)
- \(\displaystyle\frac{ab^3}{a^2b}\)
- \(\displaystyle\frac{(a^2)^3b}{a^5(b^3)^2}\)
- \(\displaystyle\frac{5abc}{25ab^{-1}c^2}\)
- \(\displaystyle\frac{18}{c^4}\cdot\frac{2c^3}{27}\)
- \(\displaystyle\frac{7(-a)^3(b^{-1})^2}{21(ab)^2a^{-2}}\)
- \(\displaystyle\frac{144a^2b}{12b^3a^8}\)
- \(\displaystyle\frac{3}{2a^2}\cdot\frac{4a}{9}\)
- \(\displaystyle\frac{2a^8b^9}{4(ab)^3}\)
- \(\displaystyle\frac{a^{-1}b^2c^4}{(ab)^2(bc)^2(ac)^3}\)
2) a^7b^6c^4
4) 2x^2-5x-3
6) 35-b
8) -8
10) 25x
12) -3x^2+7
14) x^3+x^2+1
16) x^5-32
18) 2
20) -3/2
22) 5/3
24) -25/4
26) -7/2
28) 5/6
30) 2
32) 3/2
34) 14
36) 0
38) 48/29
40) -24/21
41) 2
42) 11/2
43) -7/10
44) 37/10
45) -x^2-y^2-2xy+1
46) -3a+48
47) 9x+8
48) 27d^9-64
49) b^2/a
52) 4/(3c)
54)12/(a^6b^2)