1 C

2 B

3 D

4 D

5 C

6 C

7 B

8 B

9 B

10 D

11 D

12 C

13 B

14 B

15 A

16 D

17 A

18 B

19 C

20 B

1/ C

2 /B

3 /D

4 /D

5 /C

6 /C

7 /B

8/ B

9 /B

10/ D

11/ D

12/ C

13 /B

14 /B

15 /A

16/ D

17/ A

18 /B

19 /C

20/ B

Bài 1:

a. $-27+(-154)-(-27)+54$

$=(-27)-(-27)+(-154)+54=0-154+54=0-(154-54)=0-100=-100$

b.

$-35.127+(-35).(-27)+700$

$=(-35)(127-27)+700=-35.100+700=-3500+700=-2800$

c.

$-3^4-2[(-2023)^0+(-5)^2]=-81-2(1+25)=-81-2.26=-81-52$

$=-(81+52)=-133$

Bài 2: 

a. $-34-2(7-x)=-10$

$2(7-x)=-34-(-10)=-24$

$7-x=-24:2=-12$

$x=7-(-12)=19$
b.

$x=ƯC(36,54,90)$

$\Rightarrow ƯCLN(36,54,90)\vdots x$

$\Rightarrow 18\vdots x$

$\Rightarrow x\in \left\{\pm 1; \pm 2; \pm 3; \pm 6; \pm 9; \pm 18\right\}$

Mà $x>5$ nên $x\in \left\{6; 9; 18\right\}$

\(PTK_{Cu\left(OH\right)_2}=64+\left(16+1\right).2=98\left(đvC\right)\)

\(\%m_{Cu}=\dfrac{64}{98}.100=65,30\%\)

\(\%m_O=\dfrac{16.2}{98}.100=32,65\%\)

\(\%m_H=\dfrac{1.2}{98}.100=2,04\%\)

\(PTK_{H_2SO_4}=2.1+32+4.16=98\left(đvC\right)\)

\(\%m_H=\dfrac{2.1}{98}.100=2,04\%\)

\(\%m_S=\dfrac{32}{98}.100=32,65\%\)

\(\%m_O=\dfrac{4.16}{98}.100=65,30\%\)

các ý còn lại làm tương tự

Bài 3: 

a: \(A=\dfrac{3\left(x+4\right)}{\left(x+4\right)\left(x-4\right)}=\dfrac{3}{x-4}\)

b: Để A nguyên thì \(x-4\in\left\{1;-1;3;-3\right\}\)

hay \(x\in\left\{5;3;7;1\right\}\)

Bài 2: 

a: \(=y\left(6x-y\right)\)

b: \(=x\left(y-2\right)+2\left(y-2\right)=\left(y-2\right)\left(x+2\right)\)

c: =(x+2)(x+5)

Tách ra lào