a: \(\left(-256\right)\cdot45-256\cdot56+256\)

\(=256\left(-45-56+1\right)\)

\(=256\left(-100\right)=-25600\)

b: \(\left(-2\right)^3\cdot1975\cdot\left(-4\right)\cdot\left(-5\right)^3\cdot25\)

\(=\left(-8\right)\cdot\left(-125\right)\cdot\left(-4\right)\cdot25\cdot1975\)

\(=1000\cdot\left(-100\right)\cdot1975=-197500000\)

c: \(2076-1976\cdot65-1976\cdot35\)

\(=2076-1976\left(65+35\right)\)

\(=2076-1976\cdot100=2076-197600=-195524\)

d: \(-437-25\cdot78+25\cdot178\)

\(=-437+25\left(178-78\right)\)

\(=-437+2500=2063\)

c/

$C=\frac{11}{2}(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{91.93})$

$=\frac{11}{2}\left(\frac{3-1}{1.3}+\frac{5-3}{3.5}+...+\frac{93-91}{91.93}\right)$

$=\frac{11}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+....+\frac{1}{91}-\frac{1}{93}\right)$

$=\frac{11}{2}(1-\frac{1}{93})$

$=\frac{11}{2}.\frac{92}{93}=\frac{506}{93}$

d/

$D=5\left(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+...+\frac{1}{675}\right)$

$=\frac{5}{2}\left(\frac{2}{3}+\frac{2}{15}+\frac{2}{35}+...+\frac{2}{675}\right)$

$=\frac{5}{2}\left(\frac{3-1}{1.3}+\frac{5-3}{3.5}+\frac{7-5}{5.7}+...+\frac{27-25}{25.27}\right)$

$=\frac{5}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{25}-\frac{1}{27}\right)$
$=\frac{5}{2}\left(1-\frac{1}{27}\right)$
$=\frac{5}{2}.\frac{26}{27}=\frac{65}{27}$

e/

$E=13\left(\frac{1}{15}+\frac{1}{35}+\frac{1}{63}...+\frac{1}{1023}\right)$

$=\frac{13}{2}\left(\frac{2}{15}+\frac{2}{35}+\frac{2}{63}+...+\frac{2}{1023}\right)$

$=\frac{13}{2}\left(\frac{5-3}{3.5}+\frac{7-5}{5.7}+\frac{9-7}{7.9}+...+\frac{33-31}{31.33}\right)$

$=\frac{13}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{31}-\frac{1}{33}\right)$
$=\frac{13}{2}(\frac{1}{3}-\frac{1}{33})=\frac{65}{33}$
$=\frac{5}{2}.\frac{26}{27}=\frac{65}{27}$

S_ABCD = 52cm² => S_AOB = 52/4 = 13cm²

Mà S_AOB = \(\dfrac{1}{2}\cdot OA\cdot OB=\dfrac{1}{2}OA^2\) (OA=OB)

Nên \(\dfrac{1}{2}OA^2=13\Leftrightarrow OA^2=26\Leftrightarrow OA=\sqrt{26}\left(cm\right)\)

Diện tích hình tròn là : \(\pi\cdot r^2=3,14\cdot26=81,64\left(cm^2\right)\)

Vậy diện tích phần gạch chéo là 81,64-52=29,64(cm)²

S_ABCD = 52cm² => S_AOB = 52/4 = 13cm²

Mà S_AOB = 12OA2=13⇔OA2=26⇔OA=√26(cm)12OA2=13⇔OA2=26⇔OA=26(cm)

Diện tích hình tròn là : 

Vậy diện tích phần gạch chéo là 81,64-52=29,64(cm)²

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