Câu 3:

a) 

CTPT xủa X là C_nH_2n+2O

\(n_{CO_2}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\Rightarrow n_{C_nH_{2n+2}O}=\dfrac{0,4}{n}\left(mol\right)\)

=> \(n_{H_2O}=\dfrac{\dfrac{0,4}{n}.\left(2n+2\right)}{2}=\dfrac{0,4}{n}\left(n+1\right)\left(mol\right)\)

Mà \(n_{H_2O}=\dfrac{9}{18}=0,5\left(mol\right)\)

=> n = 4

=> CTPT: C₄H₁₀O

b) \(n_{C_4H_{10}O}=\dfrac{0,4}{4}=0,1\left(mol\right)\)

=> m = 0,1.74 = 7,4 (g)

c)

(1) \(CH_3-CH_2-CH_2-CH_2OH\)

(2) \(CH_3-CH_2-CH\left(OH\right)-CH_3\)

(3) \(CH_3-C\left(CH_3\right)\left(OH\right)-CH_3\)

(4) \(CH_3-CH\left(CH_3\right)-CH_2OH\)

(5) \(CH_3-CH_2-CH_2-O-CH_3\)

(6) \(CH_3-CH\left(CH_3\right)-O-CH_3\)

(7) \(CH_3-CH_2-O-CH_2-CH_3\)

d)

X là \(CH_3-C\left(CH_3\right)\left(OH\right)-CH_3\) (2-metylpropan-2-ol)

a: Xét tứ giác OBAC có 

\(\widehat{OBA}+\widehat{OCA}=180^0\)

Do đó: OBAC là tứ giác nội tiếp

\(P=\left(\frac{1}{\sqrt{x}}+\frac{\sqrt{x}}{\sqrt{x}+1}\right):\frac{\sqrt{x}}{x+\sqrt{x}}\)ĐK : x > 0 

\(=\left(\frac{\sqrt{x}+1+x}{\sqrt{x}\left(\sqrt{x}+1\right)}\right):\frac{1}{\sqrt{x}+1}=\frac{x+\sqrt{x}+1}{\sqrt{x}}\)

\(P=\frac{\sqrt{x}}{\sqrt{x}-1}+\frac{3}{\sqrt{x}+1}-\frac{6\sqrt{x}-4}{x-1}\)

\(=\frac{x+\sqrt{x}+3\sqrt{x}-3-6\sqrt{x}+4}{x-1}=\frac{x-2\sqrt{x}+1}{x-1}=\frac{\sqrt{x}-1}{\sqrt{x}+1}\)

bạn bổ sung đk hộ mình ý 2 là : \(x\ge0;x\ne1\)nhé 

S_ABCD = 52cm² => S_AOB = 52/4 = 13cm²

Mà S_AOB = \(\dfrac{1}{2}\cdot OA\cdot OB=\dfrac{1}{2}OA^2\) (OA=OB)

Nên \(\dfrac{1}{2}OA^2=13\Leftrightarrow OA^2=26\Leftrightarrow OA=\sqrt{26}\left(cm\right)\)

Diện tích hình tròn là : \(\pi\cdot r^2=3,14\cdot26=81,64\left(cm^2\right)\)

Vậy diện tích phần gạch chéo là 81,64-52=29,64(cm)²

dạ! em cảm ơn nhiều lắm ạ!!!

6: \(=x^3\left(x-2\right)-\left(x-2\right)\)

\(=\left(x-2\right)\left(x-1\right)\left(x^2+x+1\right)\)

7: =(x-4)(x+2)

2/

\(2x^3-8x=2x\left(x^2-4\right)=2x\left(x-2\right)\left(x+2\right)\)

3/

\(9x^2-\left(x-1\right)^2=\left(3x\right)^2-\left(x-1\right)^2=\left(3x-x+1\right)\left(3x+x-1\right)\)

4/

\(x^2-3x+6y-4y^2=x^2-4y^2-3x+6y=\left(x^2-4y^2\right)-\left(3x-6y\right)\)

\(=\left(x-2y\right)\left(x+2y\right)-3\left(x-2y\right)=\left(x-2y\right)\left(x+2y-3\right)\)

7: =(x-4)(x+2)

4: \(=\left(x-2y\right)\left(x+2y\right)-3\left(x-2y\right)\)

\(=\left(x-2y\right)\left(x+2y-3\right)\)