Bài 1: 

a: \(=-10x^3+20x^4-5x\)

b: \(=\dfrac{1}{3}a^2b+7a^5-1\)

c: \(=a^3+8+25-a^3=33\)

d: \(=x^2-16+8-x^3=-x^3+x^2-8\)

e: \(=a^3+1+8-a^3=9\)

f: \(=\dfrac{7-2x+4x-8}{2x+3}=\dfrac{2x-1}{2x+3}\)

g: \(=\dfrac{3}{2\left(x+3\right)}-\dfrac{2}{x\left(x+3\right)}\)

\(=\dfrac{3x-4}{2x\left(x+3\right)}\)

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

hơi mờ

Câu 2 :

\(1.Al_2O_3\\ 2.Fe\left(NO_3\right)_3\\ 3.CuCl_2\\ 4.SO_4\)

Câu 4 : 

a) \(n_{O_2}=\dfrac{1,8.10^{23}}{6.10^{23}}=0,3\left(mol\right)\)

b) \(m_{HCl}=0,5.36,5=18,25\left(g\right)\)

c) \(n_{N_2}=\dfrac{12,395}{22,4}\approx0,6\left(mol\right)\\ m_{N_2}=0,6.28=16,8\left(g\right)\)

d) \(d\dfrac{O_2}{kk}=\dfrac{32}{29}=1,1034>1\)

=> Khí Oxi nặng hơn không khí 1,1034 lần