a) Ta có: \(\dfrac{2\sqrt{8}-\sqrt{12}}{\sqrt{18}-\sqrt{48}}-\dfrac{\sqrt{5}+\sqrt{27}}{\sqrt{30}+\sqrt{162}}\)

\(=\dfrac{-2\left(\sqrt{3}-\sqrt{8}\right)}{\sqrt{6}\left(\sqrt{3}-\sqrt{6}\right)}-\dfrac{\sqrt{5}+\sqrt{27}}{\sqrt{6}\left(\sqrt{5}+\sqrt{27}\right)}\)

\(=\dfrac{-3}{\sqrt{6}}=\dfrac{-3\sqrt{6}}{6}=\dfrac{-\sqrt{6}}{2}\)

b) Ta có: \(\left(1+\sqrt{2}+\sqrt{3}\right)\left(1-\sqrt{2}-\sqrt{3}\right)\)

\(=1-\left(\sqrt{2}+\sqrt{3}\right)^2\)

\(=1-5-2\sqrt{6}\)

\(=-4-2\sqrt{6}\)

Giúp mình đi mọi người.

Giúp mình nhanh nha mọi người

a)5-2(1/2)^2

=5-2*1

=5-2

=3

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

b) \(x\left(3x-18\right)-3\left(x-4\right)\left(x-2\right)+8=3x^2-18x-3x^2+18x-24+8=-16\)

1. 
a) 503 + 120 = 623
b) 1000 - 120 = 880
c) 2 + 18 : 2 = 2 + 9 = 11
d) 21 : 7 - 3 = 3 - 3 = 0

2. 
a) x - 3 = 21 => x = 24
b) 15 - x . 3 = 6 => 15 - 3x = 6 => 3x = 15 - 6 = 9 => x = 3
c) x + 21 : 7 = 6 => x + 3 = 6 => x = 3
d) 44 + x : 3 = 50 => x : 3 = 50 - 44 = 6 => x = 18

3. 
a) 15 . (21 - 3 . 7) = 15 . (21 - 21) = 15 . 0 = 0
b) (4 : 2 - 2) . 105 = (2 - 2) . 105 = 0 . 105 = 0
c) 376 + 285 + 124 + 715 = 1500
d) 97 + 998 + 9999 + 16 = 11110
e) 252 + 139 - 52 - 39 = 300

4. 
a) b - a = 5 - 3 = 2
b) a + b = 3 + 5 = 8
c) 2a + b = 2*3 + 5 = 6 + 5 = 11
d) a . (b + 1) = 3 . (5 + 1) = 3 . 6 = 18

5. 
a) 37581 - 9999 = 27582
b) 7345 - 1998 = 5347
c) 485321 - 99999 = 385322
d) 7593 - 1997 = 5596

6. 
a) (x - 42) - 110 = 0 => x - 42 = 110 => x = 110 + 42 = 152
b) 2436 : x = 12 => x = 2436 / 12 = 203
c) 74 . (x - 3) = 0 => x - 3 = 0 => x = 3
d) x - 36 : 18 = 2 => x - 2 = 36 => x = 36 + 2 = 38

7. 
a) 67 + 135 + 33 = 235
b) 997 + 86 = 1083
c) 37 . 38 + 62 . 37 = 1406
d) 43 . 11 = 473
e) 67 . 99 = 6633

8. 
a) 71 - (33 + x) = 26 => 71 - 33 - x = 26 => 38 - x = 26 => x = 38 - 26 = 12
b) 97 - (64 - x) = 44 => 97 - 64 + x = 44 => x = 44 - 97 + 64 => x = 11
c) x - 36 : 18 = 12 => x - 2 = 12 => x = 14
d) 3636 : (12 . x - 91) = 36 => 3636 = 36 * (12 . x - 91) => 3636 = 432 . x - 3276 => 432 . x = 3636 + 3276 => 432 . x = 6912 => x = 6912 / 432 => x = 16
e) ( x : 23 + 45) . 67 = 8911 => (x / 23 + 45) . 67 = 8911 => (x / 23 + 45) = 8911 / 67 => (x / 23 + 45) = 133 => x / 23 = 133 - 45 => x / 23 = 88 => x = 88 . 23 => x = 2024

9. 
a) 1 + 2 + 3 + ... + 1998 + 1999 = (1999 . (1999 + 1)) / 2 = 1999 . 2000 / 2 = 1999 . 1000 = 1,999,000
b) Tính tổng tất cả các số tự nhiên có 3 chữ số: Tổng các số từ 100 đến 999 = (100 + 999) / 2 * (999 - 100 + 1) = 1099 / 2 * 900 = 549.5 * 900 = 494550
c) Tính tổng tất cả các số lẻ có 3 chữ số: Các số lẻ từ 101 đến 999 là 101, 103, 105, ..., 999. Số lượng các số này là 450 (900 / 2). Tổng các số này là (101 + 999) / 2 * (450) = 550 * 450 = 247,500

10. 
a) 53 . 39 + 47 . 39 - 53 . 21 - 47 . 21 = 2079 + 1833 - 1113 - 987 = 2912
b) 2 . 53 . 12 + 4 . 6 . 87 - 3 . 8 . 40 = 1272 + 1044 - 960 = 1356
c) 47 . 29 - 13 . 29 - 14 . 29 = 1363 - 377 - 406 = 580
d) 1754 : 17 - 74 : 17 + 20 : 17 = 103 - 4 + 1 = 100
e) 26 . 7 - 17 . 9 + 13 . 26 - 17 . 11 = 182 - 153 + 338 - 187 = 180

bài 1: 

a: Ta có: \(2\sqrt{18}-9\sqrt{50}+3\sqrt{8}\)

\(=6\sqrt{2}-45\sqrt{2}+6\sqrt{2}\)

\(=-33\sqrt{2}\)

b: Ta có: \(\left(\sqrt{7}-\sqrt{3}\right)^2+7\sqrt{84}\)

\(=10-2\sqrt{21}+14\sqrt{21}\)

\(=12\sqrt{21}+10\)

Bài 2: 

a: Ta có: \(\sqrt{\left(2x+3\right)^2}=8\)

\(\Leftrightarrow\left|2x+3\right|=8\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+3=8\\2x+3=-8\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-\dfrac{11}{2}\end{matrix}\right.\)

b: Ta có: \(\sqrt{9x}-7\sqrt{x}=8-6\sqrt{x}\)

\(\Leftrightarrow4\sqrt{x}=8\)

hay x=4

c: Ta có: \(\sqrt{9x-9}+1=13\)

\(\Leftrightarrow3\sqrt{x-1}=12\)

\(\Leftrightarrow x-1=16\)

hay x=17

a: \(=20x^2-10x^3+25x^2\)

b: \(=12x^3+18x^2-10x-15\)

a) Ta có: \(A=\dfrac{2}{3}-\left(-\dfrac{5}{7}+\dfrac{2}{3}\right)\)

\(=\dfrac{2}{3}+\dfrac{5}{7}-\dfrac{2}{3}\)

\(=\dfrac{5}{7}\)

Bài 1.

\(a,\left(2^4\cdot3\cdot5^2\right):\left\{450:\left[450-\left(4\cdot5^3-2^3\cdot5^2\right)\right]\right\}\)

\(=\left(16\cdot3\cdot25\right):\left\{450:\left[450- \left(4\cdot125-8\cdot25\right)\right]\right\}\)

\(=\left(48\cdot25\right):\left\{450:\left[450-\left(500-200\right)\right]\right\}\)

\(=1200:\left[450:\left(450-300\right)\right]\)

\(=1200:\left(450:150\right)\)

\(=1200:3\)

\(=400\)

\(---\)

\(b,3^3\cdot5^2-20\left\{90-\left[164-2\cdot\left(7^8:7^6+7^0\right)\right]\right\}\)

\(=27\cdot25-20\left\{90-\left[164-2\cdot\left(7^2+1\right)\right]\right\}\)

\(=675-20\left\{90-\left[164-2\cdot\left(49+1\right)\right]\right\}\)

\(=675-20\left[90-\left(164-2\cdot50\right)\right]\)

\(=675-20\left[90-\left(164-100\right)\right]\)

\(=675-20\left(90-64\right)\)

\(=675-20\cdot26\)

\(=675-520\)

\(=155\)

\(---\)

\(c,\left[\left(18^7:18^6-17\right)\cdot2022-1986\right]\cdot5\cdot1^{2022}-13^2\cdot2020^0\)

\(=\left[\left(18-17\right)\cdot2022-1986\right]\cdot5\cdot1-169\cdot1\)

\(=\left(1\cdot2022-1986\right)\cdot5-169\)

\(=\left(2022-1986\right)\cdot5-169\)

\(=36\cdot5-169\)

\(=180-169\)

\(=11\)

Bài 2.

\(a) (2^x+1)^2+3\cdot(2^2+1)=2^2\cdot10\\\Rightarrow (2^x+1)^2+3\cdot(4+1)=4\cdot10\\\Rightarrow (2^x+1)^2+3\cdot5=40\\\Rightarrow (2^x+1)^2+15=40\\\Rightarrow (2^x+1)^2=40-15\\\Rightarrow (2^x+1)^2=25\\\Rightarrow (2^x+1)^2= (\pm 5)^2\\\Rightarrow \left[\begin{array}{} 2^x+1=5\\ 2^x+1=-5 \end{array} \right.\\ \Rightarrow \left[\begin{array}{} 2^x=4\\ 2^x=-6 (vô.lí) \end{array} \right. \\ \Rightarrow 2^x=2^2\\\Rightarrow x=2\)

Vậy \(x=2\).

\(---\)

\(b)3\cdot(x-7)+2\cdot(x+5)=41\\\Rightarrow 3\cdot x+3\cdot(-7)+2\cdot x+2\cdot5=41\\\Rightarrow 3x-21+2x+10=41\\\Rightarrow (3x+2x)+(-21+10)=41\\\Rightarrow 5x-11=41\\\Rightarrow 5x=41+11\\\Rightarrow 5x=52\\\Rightarrow x=\dfrac{52}{5}\)

Vậy \(x=\dfrac{52}{5}\).

\(Toru\)

a) \(2xy\left(3x^2-5xy+4y^2\right)=6x^3y-10x^2y^2+8xy^3\)

b) \(\left(x-3\right)^2+\left(x+5\right)\left(5-x\right)=x^2-6x+9+25-x^2=34-6x\)

a: \(2xy\left(3x^2-5xy+4y^2\right)=6x^3y-10x^2y^2+8xy^3\)

b: \(\left(x-3\right)^2+\left(x+5\right)\left(5-x\right)\)

\(=x^2-6x+9+25-x^2\)

=-6x+34

\(a,\left(x-2\right)\left(x+3\right)-x\left(x-5\right)=x^2-2x+3x-6-x^2+5x=6x-6\)

\(b,\dfrac{1}{x-2}+\dfrac{-2}{x+2}+\dfrac{2x-8}{x^2-4}=\dfrac{x+2}{\left(x-2\right)\left(x+2\right)}-\dfrac{2\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}+\dfrac{2x-8}{\left(x+2\right)\left(x-2\right)}=\dfrac{x+2}{\left(x-2\right)\left(x+2\right)}-\dfrac{2x-4}{\left(x+2\right)\left(x-2\right)}+\dfrac{2x-8}{\left(x+2\right)\left(x-2\right)}=\dfrac{x+2-2x+4+2x-8}{\left(x+2\right)\left(x-2\right)}=\dfrac{x-2}{\left(x+2\right)\left(x-2\right)}=\dfrac{1}{x+2}\)