Bài 1 : 

\(CT:C_nH_{2n-6}\left(n\ge6\right)\)

\(\%C=\dfrac{12n}{14n-6}\cdot100\%=90.57\%\)

\(\Rightarrow n=8\)

\(CT:C_8H_{10}\)

Bài 2 : 

\(n_{CO_2}=\dfrac{17.6}{44}=0.4\left(mol\right)\)

\(CT:C_nH_{2n+1}OH\)

\(\Rightarrow n_{ancol}=\dfrac{n_{CO_2}}{n}=\dfrac{0.4}{n}\left(mol\right)\)

\(M_A=\dfrac{7.4}{\dfrac{0.4}{n}}=\dfrac{37}{2}n\left(\dfrac{g}{mol}\right)\)

\(\Rightarrow14n+18=\dfrac{37}{2}n\)

\(\Rightarrow n=4\)

\(CT:C_4H_9OH\)

\(CTCT:\)

\(B1:\)

\(CH_3-CH_2-CH_2-CH_2-OH:butan-1-ol\)

\(B2:\)

\(CH_3-CH_2-CH\left(CH_3\right)-OH:butan-2-ol\)

\(B2:\)

\(CH_3-CH\left(CH_3\right)-CH_2-OH:2-metylpropan-1-ol\)

\(B3:\)

\(C\left(CH_3\right)_3-OH:2-metylpropan-2-ol\)

Bài 1 : 

CTPT X:  C_nH_2n-6

Ta có :

\(\%C = \dfrac{12n}{14n-6}.100\% = 90,57\%\\ \Rightarrow n = 8\)

Vậy CTPT của X: C₈H₁₀

Bài 2 :

Ancol : C_nH_2n+2O

Ta có :

\(n_C = n_{CO_2} = \dfrac{17,6}{44}= 0,4(mol)\\ \Rightarrow m_C = 0,4.12 = 4,8(gam)\)

Ta có :

\(\dfrac{m_C}{m_{ancol}} = \dfrac{12n}{14n + 18} = \dfrac{4,8}{7,4}\\ \Rightarrow n = 4\)

Vậy, CTCT : 

\(CH_3-CH_2-CH_2-CH_2-OH\) :  butan-1-ol (Ancol bậc 1)

\(CH_3-CH(CH_3)-CH_2-OH\) : 2-metyl propan-1-ol (Ancol bậc 1)

\(CH_3-C(OH)(CH_3)-CH_3\) : 2-metyl propan-2-ol (Ancol bậc 3)

\(CH_3-CH(OH)-CH_2-CH_3\) : butan-2-ol (Ancol bậc 2)

Chứng minh gì vậy bạn?

a: \(3H_2+Fe_2O_3\rightarrow2Fe+3H_2O\)

b: \(n_{Fe_2O_3}=\dfrac{16}{160}=0.1\left(mol\right)\)

\(\Leftrightarrow n_{H_2O}=n_{H_2}=0.1\cdot3=0.3\left(mol\right)\)

\(v_{H_2}=0.3\cdot22.4=6.72\left(lít\right)\)

Đề ko rõ ràng \(\sqrt{x^2}+x+\dfrac{1}{4}\) hay \(\sqrt{x^2+x+\dfrac{1}{4}}\)??

m??

a. \(ZnCl_2+Zn^{2+}+2Cl^-\)

b. \(FeSO_4\rightarrow Fe^{2+}+SO_4^{2-}\)

c. \(Zn\left(NO_3\right)_2\rightarrow Zn^{2+}+2NO_3^-\)

d. \(MgCl_2\rightarrow Mg^{2+}+2Cl^-\)

\(1,ĐK:x\ge2\\ PT\Leftrightarrow\sqrt{3x-6}+x-2-\left(\sqrt{2x-3}-1\right)=0\\ \Leftrightarrow\dfrac{3\left(x-2\right)}{\sqrt{3x-6}}+\left(x-2\right)-\dfrac{2\left(x-2\right)}{\sqrt{2x-3}+1}=0\\ \Leftrightarrow\left(x-2\right)\left(\dfrac{3}{\sqrt{3x-6}}-\dfrac{2}{\sqrt{2x-3}+1}+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2\left(tm\right)\\\dfrac{3}{\sqrt{3x-6}}-\dfrac{2}{\sqrt{2x-3}+1}+1=0\left(1\right)\end{matrix}\right.\)

Với \(x>2\Leftrightarrow-\dfrac{2}{\sqrt{2x-3}+1}>-\dfrac{2}{1+1}=-1\left(3x-6\ne0\right)\)

\(\Leftrightarrow\left(1\right)>0-1+1=0\left(vn\right)\)

Vậy \(x=2\)

\(2,ĐK:x\ge-1\)

Đặt \(\left\{{}\begin{matrix}\sqrt{x+1}=a\\\sqrt{x^2-x+1}=b\end{matrix}\right.\left(a,b\ge0\right)\Leftrightarrow a^2+b^2=x^2+2\)

\(PT\Leftrightarrow2a^2+2b^2-5ab=0\\ \Leftrightarrow\left(a-2b\right)\left(2a-b\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}a=2b\\b=2a\end{matrix}\right.\)

Với \(a=2b\Leftrightarrow x+1=4x^2-4x+4\left(vn\right)\)

Với \(b=2a\Leftrightarrow4x+4=x^2-x+1\Leftrightarrow x^2-5x-3=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5+\sqrt{37}}{2}\left(tm\right)\\x=\dfrac{5-\sqrt{37}}{2}\left(tm\right)\end{matrix}\right.\)

Vậy ...

\(3,ĐK:x\ge-1\\ PT\Leftrightarrow3\left(x^2-x+1\right)-2\left(x+1\right)=5\sqrt{x^3+1}\) 

Đặt \(\left\{{}\begin{matrix}\sqrt{x+1}=a\\\sqrt{x^2-x+1}=b\end{matrix}\right.\left(a,b\ge0\right)\)

\(PT\Leftrightarrow3b^2-2a^2=5ab\\ \Leftrightarrow2a^2+5ab-3b^2=0\\ \Leftrightarrow\left[{}\begin{matrix}a=2b\\a=-3b\left(vn\right)\end{matrix}\right.\Leftrightarrow a=2b\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5+\sqrt{37}}{2}\\x=\dfrac{5-\sqrt{37}}{2}\end{matrix}\right.\left(\text{giống bài 2}\right)\)

1:

a: Khi m=1 thì (1) sẽ là x^2+2x-5=0

=>\(x=-1\pm\sqrt{6}\)

b: Δ=(2m)^2-4(-2m-3)

=4m^2+8m+12

=4m^2+8m+4+8=(2m+2)^2+8>=8>0

=>Phương trình luôn có hai nghiệm phân biệt

2:

Thay x=-1 và y=2 vào (P), ta được:

a*(-1)^2=2

=>a=2