3.x2 -15.x=0
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\(\dfrac{2x-1}{3}=\dfrac{2-x}{-2}\)
\(\Rightarrow-2\left(2x-1\right)=3\left(2-x\right)\)
\(\Rightarrow-4x+2=6-3x\Rightarrow x=-4\)
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
\(\frac{1}{4}\)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 CnH2n+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 \(\frac{14}{13}>\frac{15}{14}\)
\(3x^2-15x=0\)
\(\Leftrightarrow3x\left(x-5\right)=0\Leftrightarrow x=0;5\)