\(\Rightarrow2-4x=6-3x\\ \Rightarrow x=-4\)

\(\dfrac{2x-1}{3}=\dfrac{2-x}{-2}\)

\(\Rightarrow-2\left(2x-1\right)=3\left(2-x\right)\)

\(\Rightarrow-4x+2=6-3x\Rightarrow x=-4\)

Có đúng hông dậy mọi người

x*9,9 + x:10= 12,5

x*9,9 + x*1/10=12,5

x(9,9 +1/10)=12,5

x*10=12,5

x      =12,5:10

x      =1,25

1/4*x+x=36,5-21,5

1/4*x+x=15

x+x=15:1/4

x+x=60

x=60/2=30

\(\frac{1}{4}\)x \(x+x\)+ 21,5 = 36,5

\(\frac{1}{4}\)x \(x+x\)= 35,5 - 21,5

\(\frac{1}{4}\)x \(x+x\)= 14

\(\frac{1\text{x }x}{4}+\frac{x}{1}=14\)

\(\frac{1\text{x }x}{4}+\frac{4\text{ x }x}{4}=14\)

\(\frac{1\text{ x }x+4\text{ x }x}{4}=14\)

\(\frac{x\text{ x }\left(1+4\right)}{4}=14\)

\(\frac{x\text{ x }5}{4}=14\)

\(x\text{ x }5=14\text{ x }4\)

\(x\text{ x }5=56\)

\(x=56:5\)

\(x=11,2\)

Lưu ý: \(x\)và x khác nhau

\(x\)là số \(x\)còn x là nhân

Áp dụng BĐT cauchy, ta có:

\(\sqrt{\left(2y+2z-x\right)\cdot3x}\le\dfrac{2z+2y-x+3x}{2}=\dfrac{2\left(x+y+z\right)}{2}=x+y+z\\ \Leftrightarrow\sqrt{2y+2z-x}\le\dfrac{x+y+z}{\sqrt{3x}}\\ \Leftrightarrow\sqrt{\dfrac{x}{2y+2z-x}}\ge\dfrac{\sqrt{x}}{\dfrac{x+y+z}{\sqrt{3x}}}=\dfrac{x\sqrt{3}}{x+y+z}\)

\(\Leftrightarrow S=\sum\sqrt{\dfrac{x}{2y+2z-x}}\ge\sqrt{3}\left(\dfrac{x}{x+y+z}+\dfrac{y}{x+y+z}+\dfrac{z}{x+y+z}\right)\\ \Leftrightarrow S\ge\sqrt{3}\cdot\dfrac{x+y+z}{x+y+z}=\sqrt{3}\)

Dấu \("="\Leftrightarrow x=y=z\) hay tam giác đều

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)

\(\frac{14}{13}=1+\frac{1}{13}\)

\(\frac{15}{14}=1+\frac{1}{14}\)

Do   \(\frac{1}{13}>\frac{1}{14}\)

nên  \(\frac{14}{13}>\frac{15}{14}\)

\(\text{So sánh }2\text{ phân số : }\)

\(\frac{14}{13}\text{ và }\frac{15}{14}\)                             

\(\text{Ta có : }\)

\(\frac{14}{13}=1+\frac{1}{13}\)

\(\frac{15}{14}=1+\frac{15}{14}\)

\(\text{ Do : }\frac{1}{13}>\frac{1}{14}\)

\(\text{Nên : }\frac{14}{13}>\frac{15}{14}\)

14/13 >15/14