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mình cần phải nộp gấp

Em cần làm gì với biểu thức này???

a, \(Zn+2HCl\rightarrow ZnCl_2+H_2\)

\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)

b, \(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\)

\(n_{HCl}=0,5.2,1=1,05\left(mol\right)\)

Theo PT: \(n_{HCl}=2n_{Zn}+3n_{Al}\Rightarrow n_{Zn}=0,225\left(mol\right)\)

⇒ m_Zn = 0,225.65 = 14,625 (g)

\(n_{H_2}=\dfrac{1}{2}n_{HCl}=0,525\left(mol\right)\Rightarrow V_{H_2}=0,525.24,79=13,01475\left(l\right)\)

c, Theo PT: \(\left\{{}\begin{matrix}n_{ZnCl_2}=n_{Zn}=0,225\left(mol\right)\\n_{AlCl_3}=n_{Al}=0,2\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}C_{M_{ZnCl_2}}=\dfrac{0,225}{0,5}=0,45\left(M\right)\\C_{M_{AlCl_3}}=\dfrac{0,2}{0,5}=0,4\left(M\right)\end{matrix}\right.\)

Ta có:

\(a^2+a+1=\left(a^2+2.a.\dfrac{1}{2}+\dfrac{1}{4}\right)+\dfrac{3}{4}=\left(a+\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0\forall a\)

\(\Rightarrow\)PT đã cho vô nghiệm

Vậy không có giá trị \(a\) thỏa mãn \(P=a^{2014}+\dfrac{1}{a^{2014}}\)

Bài 1:

AB//CD

=>\(\widehat{A}+\widehat{D}=180^0\)

=>\(2\widehat{D}+\widehat{D}=180^0\)

=>\(3\cdot\widehat{D}=180^0\)

=>\(\widehat{D}=60^0\)

\(\widehat{A}=2\cdot60^0=120^0\)

AB//CD

=>\(\widehat{B}+\widehat{C}=180^0\)

=>\(\widehat{C}+\widehat{C}+40^0=180^0\)

=>\(2\cdot\widehat{C}=180^0-40^0=140^0\)

=>\(\widehat{C}=70^0\)

\(\widehat{B}=70^0+40^0=110^0\)

Bài 2:

Xét ΔAHD vuông tại H và ΔBKC vuông tại K có

AD=BC

\(\widehat{ADH}=\widehat{BCK}\)

Do đó: ΔAHD=ΔBKC

=>DH=CK

Bài này có Thịnh làm cho em rồi nhé. 

Đa thức này mũ 3 nhé bạn

a: ĐKXĐ: \(x\notin\left\{1;-1\right\}\)

\(C=\left(\dfrac{1}{x^2+1}-\dfrac{x+1}{x^4-1}\right):\dfrac{x+1}{x^5+x^4-x-1}\)

\(=\dfrac{x^2-1-x-1}{\left(x^2+1\right)\left(x^2-1\right)}:\dfrac{x+1}{x^4\left(x+1\right)-\left(x+1\right)}\)

\(=\dfrac{x^2-x-2}{\left(x^2+1\right)\left(x-1\right)\left(x+1\right)}\cdot\dfrac{\left(x+1\right)\left(x^4-1\right)}{x+1}\)

\(=\dfrac{\left(x-2\right)\left(x+1\right)}{\left(x^2+1\right)\left(x-1\right)\left(x+1\right)}\cdot\dfrac{x^4-1}{1}\)

=(x-2)(x+1)

b: Để C=0 thì (x-2)(x+1)=0

\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\left(nhận\right)\\x=-1\left(loại\right)\end{matrix}\right.\)

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

\(=x^2-x+\dfrac{1}{4}-\dfrac{9}{4}\)

\(=\left(x-\dfrac{1}{2}\right)^2-\dfrac{9}{4}>=-\dfrac{9}{4}\forall x\)

Dấu '=' xảy ra khi \(x-\dfrac{1}{2}=0\)

=>\(x=\dfrac{1}{2}\)

a,b,c là các số thực đôi một phân biệt

=>\(a-b;b-c;a-c\) đều khác 0

\(a^3+b^3+c^3=3bac\)

=>\(\left(a+b\right)^3+c^3-3ab\left(a+b\right)-3abc=0\)

=>\(\left(a+b+c\right)\left[\left(a+b\right)^2-c\left(a+b\right)+c^2\right]-3ab\left(a+b+c\right)=0\)

=>\(\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2-3ab\right)=0\)

=>\(\left(a+b+c\right)\left(a^2+b^2+c^2-ab-ac-bc\right)=0\)

=>\(\left(a+b+c\right)\left[2a^2+2b^2+2c^2-2ab-2ac-2bc\right]=0\)

=>\(\left(a+b+c\right)\left[\left(a-b\right)^2+\left(b-c\right)^2+\left(a-c\right)^2\right]=0\)

=>\(\left[{}\begin{matrix}a+b+c=0\\\left(a-b\right)^2+\left(b-c\right)^2+\left(a-c\right)^2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}a+b+c=0\\a=b=c\left(loại\right)\end{matrix}\right.\)

=>a+b+c=0

=>a+b=-c; a+c=-b; b+c=-a

\(P=\dfrac{a+b}{c}\cdot\dfrac{b+c}{a}\cdot\dfrac{c+a}{b}=\dfrac{-c}{c}\cdot\dfrac{-a}{a}\cdot\dfrac{-b}{b}=-1\)

`a^3 + b^3 + c^3 = 3abc`

`=> a^3 + b^3 + c^3 - 3abc = 0`

`=> (a+b)^3 - 3ab(a+b) + c^3 - 3abc = 0`

`=> ((a+b)^3  + c^3) - (3ab(a+b) + 3abc) = 0`

`=> (a+b+c) ((a+b)^2 - (a+b)c + c^2) - 3ab(a+b+c) = 0`

`=> (a+b+c)(a^2 + 2ab + b^2 - ac - bc + c^2) - 3ab(a+b+c) = 0`

`=> (a+b+c)(a^2 + 2ab + b^2 - ac - bc + c^2 - 3ab) = 0`

`=> (a+b+c)(a^2 - ab + b^2 - ac - bc + c^2) = 0`

Trường hợp 1: 

`a+b+c = 0 (đpcm)`

Trường hợp 2: 

`a^2 - ab + b^2 + ac + bc + c^2 = 0`

`<=> 2a^2 - 2ab + 2b^2 - 2bc +2c^2 - 2ca = 0`

`<=> a^2 - 2ab + b^2 + b^2 - 2bc +c^2 + c^2 - 2ac + a^2 = 0`

`<=> (a-b)^2 + (b-c)^2 + (c-a)^2 = 0`

Do `{((a-b)^2 >=0),((b-c)^2 >=0),((c-a)^2 >=0):}`

`=> (a-b)^2 + (b-c)^2 + (c-a)^2 >= 0`

Dấu = có khi: 

`{(a=b),(b=c),(c=a):}`

Hay `a=b=c  (đpcm)`

Ta có :a^3+b^3+c^3=3abc⇒a^3+b^3+c^3-3abc=0

⇒(a+b+c)(a^2+b^2+c^2-ab-bc-ac)=0

TH1: a+b+c=0

TH2:a^2+b^2+c^2-ab-ac-bc=0

⇒2a^2+2b^2+2c^2-2ab-2bc-2ac=0

(a-b)^2+(b-c)^2+(c-a)^2=0

⇒a=b=c