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1 was repaired

2 Was - bought

3 are watered

4 wasn't learned

5 was watched

6 Is - read

7 was stolen

8 was broken

9 wasn't cut

10 is cleaned

was repaired

was-bought

are watered

động vật

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tui ko bíc

\(=\dfrac{1}{2}-\dfrac{1}{2}cos\dfrac{2\pi}{9}+\dfrac{1}{2}-\dfrac{1}{2}cos\dfrac{4\pi}{9}+\dfrac{1}{2}cos\dfrac{\pi}{9}-\dfrac{1}{2}cos\dfrac{3\pi}{9}\)

\(=1-\dfrac{1}{2}\left(cos\dfrac{2\pi}{9}+cos\dfrac{4\pi}{9}\right)+\dfrac{1}{2}cos\dfrac{\pi}{9}-\dfrac{1}{4}\)

\(=\dfrac{3}{4}-cos\dfrac{\pi}{9}.cos\dfrac{3\pi}{9}+\dfrac{1}{2}cos\dfrac{\pi}{9}\)

\(=\dfrac{3}{4}-\dfrac{1}{2}cos\dfrac{\pi}{9}+\dfrac{1}{2}cos\dfrac{\pi}{9}\)

\(=\dfrac{3}{4}\)

12: \(3\left(2^2+1\right)\left(2^4+1\right)\cdot\ldots\cdot\left(2^{64}+1\right)+1\)

\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\cdot\ldots\cdot\left(2^{64}+1\right)+1\)

\(=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\cdot\ldots\cdot\left(2^{64}+1\right)+1\)

\(=\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\cdot\left(2^{32}+1\right)\cdot\left(2^{64}+1\right)+1\)

\(=\left(2^{16}-1\right)\left(2^{16}+1\right)\cdot\left(2^{32}+1\right)\cdot\left(2^{64}+1\right)+1\)

\(=\left(2^{32}-1\right)\cdot\left(2^{32}+1\right)\cdot\left(2^{64}+1\right)+1\)

\(=\left(2^{64}-1\right)\cdot\left(2^{64}+1\right)+1\)

\(=2^{128}-1+1=2^{128}\)

11: \(100^2-99^2+98^2-97^2+\cdots+2^2-1^2\)

=(100-99)(100+99)+(98-97)(98+97)+...+(2-1)(2+1)

=100+99+...+2+1

\(=\frac{100\left(100+1\right)}{2}=5050\)

10: \(\left(a+b+c\right)^2+\left(a+b-c\right)^2-2\left(a+b\right)^2\)

\(=\left(a+b\right)^2+2c\left(a+b\right)+c^2+\left(a+b\right)^2-2c\left(a+b\right)+c^2-2\left(a+b\right)^2\)

\(=2c^2\)

9: \(\left(x+1\right)^3+\left(x-1\right)^3+x^3-3x\left(x+1\right)\left(x-1\right)\)

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

\(=3x^3+6x-3x^3+3x=9x\)

7: \(\left(6x+1\right)^2+\left(6x-1\right)^2-2\left(6x+1\right)\left(6x-1\right)\)

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

\(=\left(6x+1-6x+1\right)^2=2^2=4\)

6: \(\left(2x+1\right)^2+\left(3x-1\right)^2+2\left(2x+1\right)\left(3x-1\right)\)

\(=\left(2x+1+3x-1\right)^2=\left(5x\right)^2=25x^2\)

5: \(\left(4x-5\right)\left(2x+3\right)-4\left(x+2\right)\left(2x-1\right)+10x+7\)

\(=8x^2+12x-10x-15-4\left(2x^2-x+4x-2\right)+10x+7\)

\(=8x^2+2x-15-8x^2-12x+8+10x+7\)

=-15+8+7

=0

4: (x+2)(x-2)-(x-3)(x+1)

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

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

=2x-1

3: \(3x\left(x-2\right)-5x\left(1-x\right)-8\left(x^2-3\right)\)

\(=3x^2-6x-5x+5x^2-8x^2+24\)

=-11x+24

2: \(3\left(2x-1\right)-5\left(x-3\right)+6\left(3x-4\right)-19x\)

=6x-3-5x+15+18x-24-19x

=(6x-5x+18x-19x)+(-3+15-24)

=0+12-24

=12-24

=-12

1: \(x\left(3x+12\right)-\left(7x-20\right)+x^2\left(2x-3\right)-x\left(2x^2+5\right)\)

\(=3x^2+12x-7x+20+2x^3-3x^2-2x^3-5x\)

=20

\(n_{AgNO_3}=1.0,5=0,5\left(mol\right);n_{HCl}=2.0,3=0,6\left(mol\right)\)

PTHH: AgNO₃ + HCl → AgCl ↓ + HNO₃

Mol:      0,5          0,5                          0,5

Ta có: \(\dfrac{0,5}{1}< \dfrac{0,6}{1}\)⇒ AgNO₃ pứ hết, HCl dư

* V_{dd sau pứ} = 0,5+0,6 = 1,1 (l)

\(\Rightarrow C_{M_{ddNO_3}}=\dfrac{0,5}{1,1}=0,4545M\)

\(C_{M_{ddHCldư}}=\dfrac{0,6-0,5}{1,1}=0,0909M\)

\(m_{ddAgNO_3}=500.1,2=600\left(g\right);m_{ddHCl}=300.1,5=450\left(g\right)\)

* m_{dd sau pứ} = 600+450 = 1050 (g)

\(C\%_{ddHNO_3}=\dfrac{0,5.63.100\%}{1050}=3\%\)

\(C\%_{ddHCl}=\dfrac{\left(0,6-0,5\right).36,5.100\%}{1050}=0,348\%\)

Vdd sau pứ phải là 0,5 + 0,3 = 0,8 (l) (dòng 4) chứ bn.

D

B