Bài 2 : (1) liên kết ; (2) electron ; (3) liên kết ; (4) : electron ; (5) sắp xếp electron

Bài 4 : 

$\dfrac{M_X}{4} = \dfrac{M_K}{3} \Rightarrow M_X = 52$

Vậy X là crom,KHHH : Cr

Bài 5 : 

$M_X = 3,5M_O = 3,5.16 = 56$ đvC

Tên : Sắt

KHHH : Fe

Bài 9 : 

$M_Z = \dfrac{5,312.10^{-23}}{1,66.10^{-24}} = 32(đvC)$

Vậy Z là lưu huỳnh, KHHH : S

Bài 10  :

a) $PTK = 22M_{H_2} = 22.2 = 44(đvC)$

b) $M_{hợp\ chất} = X + 16.2 = 44 \Rightarrow X = 12$
Vậy X là cacbon, KHHH : C

Bài 11 : 

a) $PTK = 32.5 = 160(đvC)$

b) $M_{hợp\ chất} = 2A + 16.3 = 160 \Rightarrow A = 56$
Vậy A là sắt

c) $\%Fe = \dfrac{56.2}{160}.100\% = 70\%$

b: Để A nguyên thì \(x+2\in\left\{1;-1\right\}\)

hay \(x\in\left\{-1;-3\right\}\)

Để B nguyên thì \(\sqrt{x}-1\in\left\{-1;1;2;3;6\right\}\)

\(\Leftrightarrow\sqrt{x}\in\left\{0;2;3;4;7\right\}\)

hay \(x\in\left\{0;4;9;16;49\right\}\)

a) Ta có: \(\dfrac{a}{3b+c}=\dfrac{b}{a+3c}=\dfrac{c}{3a+b}=\dfrac{a+b+c}{3b+c+a+3c+3a+b}=\dfrac{a+b+c}{4\left(a+b+c\right)}=\dfrac{1}{4}\)

\(\Rightarrow\left\{{}\begin{matrix}3b+c=4a\\a+3c=4b\\3a+b=4c\end{matrix}\right.\)

\(\Rightarrow\dfrac{3b+c}{a}+\dfrac{a+3c}{b}+\dfrac{3a+b}{c}=\dfrac{4a}{a}+\dfrac{4b}{b}+\dfrac{4c}{c}=4+4+4=12\)

b) \(A=\dfrac{x+1}{x+2}=\dfrac{x+2}{x+2}-\dfrac{1}{x+2}=1-\dfrac{1}{x+2}\in Z\)

\(\Rightarrow\left(x+2\right)\inƯ\left(1\right)=\left\{-1;1\right\}\)

\(\Rightarrow x\in\left\{-3;-1\right\}\)

\(B=\dfrac{\sqrt{x}+5}{\sqrt{x}-1}\left(đk:x\ge0\right)=1+\dfrac{6}{\sqrt{x}-1}\in Z\)

\(\Rightarrow\sqrt{x}-1\inƯ\left(6\right)=\left\{-6;-3;-2;-1;1;2;3;6\right\}\)

Do \(x\ge0,x\in Z\)

\(\Rightarrow x\in\left\{0;4;9;16;49\right\}\)

\(26,\\ a,\sin45^0=\cos45^0< \sin50^025'< \sin57^048'=\cos32^012'< \sin72^0=\cos18^0< \sin75^0\\ b,\tan37^026'< \tan47^0< \tan58^0=\cot32^0< \tan63^0< \tan66^019'=\cot23^041'\\ 27,\\ A=\dfrac{\left(\sin^226^0+\sin^264^0\right)+2\left(\cos^215^0+\cos^275^0\right)}{\left(\sin^255^0+\cos^255^0\right)+\left(\sin^242^0+\cos^242^0\right)}-\dfrac{\tan81^0}{2\tan81^0}\\ A=\dfrac{\left(\sin^226^0+\cos^226^0\right)+2\left(\sin^215^0+\cos^215^0\right)}{1+1}-\dfrac{1}{2}\\ A=\dfrac{1+2}{2}-\dfrac{1}{2}=2-\dfrac{1}{2}=\dfrac{3}{2}\)

\(28,\\ \sin^2\alpha=1-\cos^2\alpha=1-\dfrac{1}{2}=\dfrac{1}{2}\\ \Leftrightarrow\sin\alpha=\dfrac{\sqrt{2}}{2}\)

Bài 1: 

a: \(=\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)=2\)

b: \(=\sqrt{11}+\sqrt{2}-\sqrt{11}-\sqrt{7}-\sqrt{2}=-\sqrt{7}\)

Em ơi đăng tách bài ra mỗi lượt đăng 1-2 bài thôi nha!

đang thi mik ko giúp đc, xin lỗi nha

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