\(k,=\dfrac{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)+5\left(\sqrt{a}-\sqrt{b}\right)}{\sqrt{a}+\sqrt{b}+5}\\ =\dfrac{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}+5\right)}{\sqrt{a}+\sqrt{b}+5}=\sqrt{a}-\sqrt{b}\)

\(h,=\dfrac{1}{2a-1}\sqrt{25a^2\left(a^2-4a+4\right)}=\dfrac{1}{2a-1}\sqrt{25a^2\left(a-2\right)^2}\\ =\dfrac{\left|5a\left(a-2\right)\right|}{2a-1}=\left[{}\begin{matrix}\dfrac{5a\left(a-2\right)}{2a-1}\left(a\ge2;a\ne\dfrac{1}{2}\right)\\\dfrac{5a\left(2-a\right)}{2a-1}\left(0\le a< 2;a\ne\dfrac{1}{2}\right)\\\dfrac{-5a\left(2-a\right)}{2a-1}\left(a< 0\right)\end{matrix}\right.\)

a: Ta có: \(\sqrt{75}-\sqrt{5\dfrac{1}{3}}+\dfrac{9}{2}\sqrt{2\dfrac{2}{3}}+2\sqrt{27}\)

\(=5\sqrt{3}+\dfrac{4}{3}\sqrt{3}+3\sqrt{6}+6\sqrt{3}\)

\(=\dfrac{37}{3}\sqrt{3}+3\sqrt{6}\)

c: Ta có: \(\left(\sqrt{12}+2\sqrt{27}\right)\cdot\dfrac{\sqrt{3}}{2}-\sqrt{150}\)

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

\(=12-5\sqrt{6}\)

TH1: CO₂ dư

\(Ca\left(OH\right)_2+CO_2\rightarrow CaCO_3+H_2O\\ CaCO_3+CO_2+H_2O\rightarrow Ca\left(HCO_3\right)_2\\ n_{CO_2}=n_{CaCO_3}+n_{Ca\left(OH\right)_2}=\dfrac{10}{100}+0,2=0,3\left(mol\right)\\ \Rightarrow V=V_{CO_2\left(đktc\right)}=0,3.22,4=6,72\left(l\right)\)

TH2: Ca(OH)₂ dư

\(Ca\left(OH\right)_2+CO_2\rightarrow CaCO_3+H_2O\\ n_{CO_2}=n_{CaCO_3}=\dfrac{10}{100}=0,1\left(mol\right)\\ \Rightarrow V=V_{CO_2\left(đktc\right)}=0,1.22,4=2,24\left(l\right)\)

=> Chọn C

Bài 7: 

- Đơn chất: 

+ N₂; \(PTK_{N_2}=14.2=28\left(đvC\right)\)

+ O₃; \(PTK_{O_3}=16.3=48\left(đvC\right)\)

- Hợp chất:

+ H₂SO₄ (axit sunfuric); \(PTK_{H_2SO_4}=1.2+32+16.4=98\left(đvC\right)\)

+ H₂O₂ (oxi già): \(PTK_{H_2O_2}=1.2+16.2=34\left(đvC\right)\)

+ C₆H₁₂O₆ (glucozơ); \(PTK_{C_6H_{12}O_6}=12.6+1.12+16.6=180\left(đvC\right)\)

+ NaNO₃ (natri nitrat); \(PTK_{NaNO_3}=23.1+14.1+16.3=85\left(đvC\right)\)

Bài 8: 

Gọi CTHH của hợp chất là: A₂O₃

a. Ta có: \(NTK_C=12\left(đvC\right)\)

Theo đề, ta có: \(PTK_{A_2O_3}=8,5.12=102\left(đvC\right)\)

b. Ta có: \(PTK_{A_2O_3}=NTK_A.2+16.3=102\left(đvC\right)\)

=> \(NTK_A=27\left(đvC\right)\)

Dựa vào bảng hóa trị, suy ra:

A là nhôm (Al)

Bài 1:

Theo đề ta có : A₁=G₁; T₁=3A₁; X₁ = 2T₁

=>\(\left\{{}\begin{matrix}X_1=6A_1\\T_1=3A_1\\G_1=A_1\end{matrix}\right.\)=>\(\left\{{}\begin{matrix}A=4A_1\\G=7A_1\end{matrix}\right.\)

Lại có: 2A + 3G= 5800

=> 29A₁ = 5800 => A₁=200

=> G = 1400