5:

a: sin x=2*cosx

\(A=\dfrac{6cosx+2cosx-4\cdot8\cdot cos^3x}{cos^3x-2cosx}\)

\(=\dfrac{8-32cos^2x}{cos^2x-2}\)

b: VT=sin^4(pi/2-x)+cos^4(x+pi/2)+6*1/2*sin^22x+1/2*cos4x

=cos^4x+sin^4x+3*sin^2(2x)+1/2*(1-2*sin^2(2x))

=1-2*sin^2x*cos^2x+3*sin^2(2x)+1/2-sin^2(2x)

==3/2=VP

- Hợp chất vô cơ: CO₂, Na₂CO₃ 

- Hợp chất hữu cơ: 

+ Hiđrocacbon: C₄H₁₀, C₆H₆, C₃H₄ 

+ Dẫn xuất của hiđrocacbon: C₃H₈O, CH₃Cl, C₆H₆Cl₆

bn đang thi thì tự lm nhé, kể cả KT hay là Thi thử

thi tự lm nhé

a) \(\dfrac{1}{\sqrt[]{x}-1}+\dfrac{1}{1+\sqrt[]{x}}+1\left(x\ge0;x\ne1\right)\)

\(=\dfrac{\sqrt[]{x}+1+\sqrt[]{x}-1+x-1}{\left(\sqrt[]{x}-1\right)\left(\sqrt[]{x}+1\right)}\)

\(=\dfrac{x+2\sqrt[]{x}-1}{x-1}\)

\(=\dfrac{x-1+2\sqrt[]{x}}{x-1}\)

\(=1+\dfrac{2\sqrt[]{x}}{x-1}\)

b) \(\dfrac{1}{\sqrt[]{x}+2}-\dfrac{2}{\sqrt[]{x}-2}-\dfrac{4}{4-x}\left(x\ge0;x\ne4\right)\)

\(=\dfrac{\sqrt[]{x}-2-2\left(\sqrt[]{x}+2\right)+4}{\left(\sqrt[]{x}+2\right)\left(\sqrt[]{x}-2\right)}\)

\(=\dfrac{\sqrt[]{x}-2-2\sqrt[]{x}-4+4}{\left(\sqrt[]{x}+2\right)\left(\sqrt[]{x}-2\right)}\)

\(=\dfrac{-\sqrt[]{x}-2}{\left(\sqrt[]{x}+2\right)\left(\sqrt[]{x}-2\right)}\)

\(=\dfrac{-\left(\sqrt[]{x}+2\right)}{\left(\sqrt[]{x}+2\right)\left(\sqrt[]{x}-2\right)}\)

\(=\dfrac{-1}{\sqrt[]{x}-2}\)

\(\dfrac{x}{6}=\dfrac{7}{4}\Rightarrow x=\dfrac{6\cdot7}{4}=\dfrac{21}{2}\\ \dfrac{3}{x}=\dfrac{21}{17}\Rightarrow x=\dfrac{3\cdot17}{21}=\dfrac{17}{7}\)