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

Bài 12 : 

a) $M_{hợp\ chất} = R + 1.4 = M_O = 16(đvC) \Rightarrow R = 12$

Vậy R là nguyên tố cacbon, KHHH : C

b) $\%C = \dfrac{12}{16}.100\% = 75\%$

Bài 13 : 

a) $PTK = 32M_{H_2} = 32.2 = 64(đvC)$

b) $M_{hợp\ chất} = X + 16.2 = 64 \Rightarrow X = 32$
Vậy X là lưu huỳnh, KHHH : S

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

đang thi mik ko giúp đc, xin lỗi nha

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 \(x\ge0,x\in Z\)

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

ĐKXĐ: \(x\notin\left\{0;-9\right\}\)

Ta có: \(\dfrac{1}{x+9}-\dfrac{1}{x}=\dfrac{1}{5}+\dfrac{1}{4}\)

\(\Leftrightarrow\dfrac{20x}{20x\left(x+9\right)}-\dfrac{20\left(x+9\right)}{20x\left(x+9\right)}=\dfrac{4x\left(x+9\right)+5x\left(x+9\right)}{20x\left(x+9\right)}\)

Suy ra: \(4x^2+36x+5x^2+45x=20x-20x-180\)

\(\Leftrightarrow9x^2+81x+180=0\)

\(\Leftrightarrow x^2+9x+20=0\)

\(\Leftrightarrow x^2+4x+5x+20=0\)

\(\Leftrightarrow x\left(x+4\right)+5\left(x+4\right)=0\)

\(\Leftrightarrow\left(x+4\right)\left(x+5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+4=0\\x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-4\left(nhận\right)\\x=-5\left(nhận\right)\end{matrix}\right.\)

Vậy: S={-4;-5}

a: Xét ΔHAC vuông tại H và ΔABC vuông tại A có

\(\widehat{C}\) chung

Do đó: ΔHAC~ΔABC

b: ΔABC vuông tại A

=>\(AB^2+AC^2=BC^2\)

=>\(BC^2=15^2+20^2=625\)

=>BC=25

Xét ΔABC vuông tại A có AH là đường cao

nên \(\left\{{}\begin{matrix}BH\cdot BC=BA^2\\AH\cdot BC=AB\cdot AC\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}BH\cdot25=15^2=225\\AH\cdot25=15\cdot20=300\end{matrix}\right.\)

=>BH=9; AH=12

a: \(A=\dfrac{x}{x+3}-1=\dfrac{x-x-3}{x+3}=\dfrac{-3}{x+3}\)

\(B=\dfrac{9-x^2+x^2-9-x^2+4x-4}{\left(x+3\right)\left(x-2\right)}=\dfrac{-\left(x-2\right)^2}{\left(x-2\right)\left(x+3\right)}=\dfrac{-x+2}{x+3}\)

b: B nguyên

=>-x-3+5 chia hết cho x+3

=>x+3 thuộc {1;-1;5;-5}

=>x thuộc {-2;-4;-8}

c: P=A:B

=(-3/x+3):(-x+2)/(x+3)

=3/(x-2)

\(a,=x^2+x+4x+4=\left(x+1\right)\left(x+4\right)\\ b,=x^2+2x-3x-6=\left(x-3\right)\left(x+2\right)\\ c,=x^2-2x-3x+6=\left(x-2\right)\left(x-3\right)\\ d,=3\left(x^2-2x+5x-10\right)=3\left(x-2\right)\left(x+5\right)\\ e,=-3x^2+6x-x+2=\left(x-2\right)\left(1-3x\right)\\ f,=x^2-x-6x+6=\left(x-1\right)\left(x-6\right)\\ h,=4\left(x^2-3x-6x+18\right)=4\left(x-3\right)\left(x-6\right)\\ i,=3\left(3x^2-3x-8x+5\right)=3\left(x-1\right)\left(3x-8\right)\\ k,=-\left(2x^2+x+4x+2\right)=-\left(2x+1\right)\left(x+2\right)\\ l,=x^2-2xy-5xy+10y^2=\left(x-2y\right)\left(x-5y\right)\\ m,=x^2-xy-2xy+2y^2=\left(x-y\right)\left(x-2y\right)\\ n,=x^2+xy-3xy-3y^2=\left(x+y\right)\left(x-3y\right)\)

Bào quan riboxom trong chất tế bào có chức năng gì? 

a) \(=\left(x^2+x\right)+\left(4x+4\right)=x\left(x+1\right)+4\left(x+1\right)=\left(x+1\right)\left(x+4\right)\)

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

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

d) \(3x^2+9x-30=3\left(x^2+3x-10\right)=3\left[\left(x^2+5x\right)-\left(2x+10\right)\right]=3\left[x\left(x+5\right)-2\left(x+5\right)\right]=3\left(x-2\right)\left(x+5\right)\)

e) \(=-\left(3x^2-5x-2\right)=-\left[\left(3x^2-6x\right)+\left(x-2\right)\right]=-\left[3x\left(x-2\right)+\left(x-2\right)\right]=-\left(3x+1\right)\left(x-2\right)\)

f) \(x^2-7x+6=\left(x^2-x\right)-\left(6x-6\right)=x\left(x-1\right)-6\left(x-1\right)=\left(x-1\right)\left(x-6\right)\)

h) \(=4\left(x^2-9x+14\right)=4\left[\left(x^2-7x\right)-\left(2x-14\right)\right]=4\left[x\left(x-7\right)-2\left(x-7\right)\right]=4\left(x-2\right)\left(x-7\right)\)

i) \(=3\left(3x^2-8x+5\right)=3\left[\left(3x^2-3x\right)-\left(5x-5\right)\right]=3\left[3x\left(x-1\right)-5\left(x-1\right)\right]=3\left(x-1\right)\left(3x-5\right)\)

k) \(=-\left(2x^2+5x+2\right)=-\left[\left(2x^2+4x\right)+\left(x+2\right)\right]=-\left[2x\left(x+2\right)+\left(x+2\right)\right]=-\left(x+2\right)\left(2x+1\right)\)

l) \(=\left(x^2-5xy\right)-\left(2xy-10y^2\right)=x\left(x-5y\right)-2y\left(x-5y\right)=\left(x-5y\right)\left(x-2y\right)\)

m) \(=\left(x^2-2xy\right)-\left(xy-2y^2\right)=x\left(x-2y\right)-y\left(x-2y\right)=\left(x-2y\right)\left(x-y\right)\)

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