6:

\(2^{225}=\left(2^3\right)^{75}=8^{75}\)

\(3^{150}=\left(3^2\right)^{75}=9^{75}\)

mà 8<9

nên \(2^{225}< 3^{150}\)

4: \(\left|5x+3\right|>=0\forall x\)

=>\(-\left|5x+3\right|< =0\forall x\)

=>\(-\left|5x+3\right|+5< =5\forall x\)

Dấu = xảy ra khi 5x+3=0

=>x=-3/5

1:

\(\left(2x+1\right)^4>=0\)

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

=>\(M=\dfrac{3}{\left(2x+1\right)^4+2}< =\dfrac{3}{2}\)

Dấu = xảy ra khi 2x+1=0

=>x=-1/2

mình làm câu dưới r nha

Tất cả k dưới đây đều là \(k\in Z\)

6.

\(\Leftrightarrow\sqrt{3}cot\left(3x-\dfrac{\pi}{3}\right)=1\)

\(\Leftrightarrow cot\left(3x-\dfrac{\pi}{3}\right)=\dfrac{1}{\sqrt{3}}\)

\(\Leftrightarrow cot\left(3x-\dfrac{\pi}{3}\right)=cot\left(\dfrac{\pi}{3}\right)\)

\(\Leftrightarrow3x-\dfrac{\pi}{3}=\dfrac{\pi}{3}+k\pi\)

\(\Leftrightarrow3x=\dfrac{2\pi}{3}+k\pi\)

\(\Leftrightarrow x=\dfrac{2\pi}{9}+\dfrac{k\pi}{3}\) 

7.

\(\Leftrightarrow\sqrt{3}tan\left(3x-15^0\right)=-1\)

\(\Leftrightarrow tan\left(3x-15^0\right)=-\dfrac{1}{\sqrt{3}}\)

\(\Leftrightarrow tan\left(3x-15^0\right)=tan\left(-30^0\right)\)

\(\Leftrightarrow3x-15^0=-30^0+k180^0\)

\(\Leftrightarrow3x=-15^0+k180^0\)

\(\Leftrightarrow x=-3^0+k60^0\)

8.

\(\Leftrightarrow\sqrt{3}cot\left(3x-30^0\right)=1\)

\(\Leftrightarrow cot\left(3x-30^0\right)=\dfrac{1}{\sqrt{3}}\)

\(\Leftrightarrow cot\left(3x-30^0\right)=cot\left(60^0\right)\)

\(\Leftrightarrow3x-30^0=60^0+k180^0\)

\(\Leftrightarrow3x=90^0+k180^0\)

\(\Leftrightarrow x=30^0+k60^0\)

bài 3: 

tổng số giờ đã chảy đc từ 2 vòi : 1+1=2(giờ)

tổng số phần bể đã chảy được từ 2 vòi : \(\dfrac{1}{5}+\dfrac{1}{7}=\dfrac{7}{35}+\dfrac{5}{35}=\dfrac{12}{35}\left(ph\text{ần} b\text{ể}\right)\)

nếu chảy cùng lúc mỗi giờ chảy được : \(\dfrac{12}{35}:2=\dfrac{12}{35\cdot2}=\dfrac{6}{35}\left(ph\text{ần}b\text{ể}\right)\)

bài 4:

cách 1:

độ dài đoạn AB là : \(\dfrac{3}{4}+\dfrac{9}{8}=\dfrac{18}{24}+\dfrac{27}{24}=\dfrac{45}{24}\left(m\right)\)

diện tích ABCD là : \(\dfrac{45}{27}\cdot\dfrac{4}{7}=\dfrac{15}{14}\left(m^2\right)\)

cách 2: 

diện tích AEFD là : \(\dfrac{3}{4}\cdot\dfrac{4}{7}=\dfrac{3}{7}\left(m^2\right)\)

diện tích EBCF là : \(\dfrac{9}{8}\cdot\dfrac{4}{7}=\dfrac{9}{14}\left(m^2\right)\)

diện tích ABCD là : \(\dfrac{3}{7}+\dfrac{9}{14}=\dfrac{15}{14}\left(m^2\right)\)

Tách ra nhé bn !!

5.watching

6.Did Nam play...(ko chắc)

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6.Did Nam play

7.không biết chỗ điền

enjoy nha bn 

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6.Did/play

Theo ĐLBTKL: m_CaCO3 = m_CaO + m_CO2

=> m_CO2 = 20-11,2 = 8,8 (g)

\(Bài.1:\\ a,n_{CO_2}=\dfrac{m}{M}=\dfrac{11}{44}=0,25\left(mol\right)\\ b,n_{H_2O}=\dfrac{m}{M}=\dfrac{12.8}{18}=0,7\left(1\right)\left(mol\right)\\ Bài.2:\\ a,m_{H_2SO_4}=n.M=0,25.98=24,5\left(g\right)\\ b,ko.hiểu.đề.cho.đơn.vị.đại.lượng.gì.hết\)

Bài 1 :

a) \(n_{CO_2}=\dfrac{m}{M}=\dfrac{11}{44}=0,25\left(mol\right)\)

b) \(n_{H_2O}=\dfrac{m}{M}=\dfrac{12,8}{18}=0,71\left(mol\right)\)

Bài 2 : 

a) \(m_{H_2SO_4}=n.M=0,25.98=24,5\left(g\right)\)

b) \(n_{CO_2}=\dfrac{m}{M}=\dfrac{0,88}{44}=0,02\left(mol\right)\\ V_{CO_2}=n.22,4=0,2.22,4=0,448\left(l\right)\)

\(n_{H_2}=\dfrac{m}{M}=\dfrac{0,02}{2}=0,01\left(mol\right)\\ V_{H_2}=n.22,4=0,01.22,4=0,224\left(l\right)\)

\(n_{O_2}=\dfrac{m}{M}=\dfrac{0,32}{32}=0,01\left(mol\right)\\ V_{O_2}=0,01:22,4=0,224\left(l\right)\)

\(1) 4Al + 3O_2 \xrightarrow{t^o} 2Al_2O_3\\ 2) 4K + O_2 \xrightarrow{t^o} 2K_2O\\ 3) 2Al(OH)_3 \xrightarrow{t^o} Al_2O_3 + 3H_2O\\ 4) Al_2O_3 + 6HCl \to 2AlCl_3 + 3H_2O\\5)2Al + 6HCl \to 2AlCl_3 + 3H_2\\ 6) FeO + 2HCl \to FeCl_2 + H_2O\\ 8) 2NaOH + H_2SO_4 \to Na_2SO_4 + 2H_2O\\ 9) BaCl_2 + H_2SO_4 \to BaSO_4 + 2HCl\)

1)4Al+ 3O₂-->2Al₂O₃

 2)4K+O₂-->2K₂O

3)2Al(OH)₃-->Al₂O₃+3H₂O

4)Al₂O₃+6HCl-->2AlCl₃+3H₂O

5)2Al+6HCl-->2AlCl₃+3H₂

6)FeO+2HCl-->FeCl₂+H₂O

7) Fe₂O₃+3H₂SO₄-->Fe₂(SO₄)₃ + 3H₂O

8)H₂SO₄+2NaOH-->2H₂O+Na₂SO₄

 9)BaCl₂+H₂SO₄-->2HCl+BaSO₄