`# \text {Kaizu DN}`

`a)`

`(3x + 6) + (7x - 14) = 0?`

\(\Rightarrow3x+6+7x-14=0\\ \Rightarrow\left(3x+7x\right)+\left(6-14\right)=0\\ \Rightarrow10x-8=0\\ \Rightarrow10x=8\Rightarrow x=\dfrac{8}{10}\\ \Rightarrow x=\dfrac{4}{5}\)

Vậy, \(x=\dfrac{4}{5}\) 

`b)`

`17y + 35 + 4x + 17 = 42`

\(\Rightarrow\left(17y+17\right)+\left(35+4x\right)=42\\ \Rightarrow17\left(y+1\right)+\left(35+4x\right)=42\)

Bạn xem lại đề ;-;.

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

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

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

Ta có \(\dfrac{1}{2^2}< \dfrac{1}{1.2};\dfrac{1}{3^2}< \dfrac{1}{2.3};...;\dfrac{1}{2022^2}< \dfrac{1}{2021.2022}\)

cộng vế với vế 

\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{2022^2}< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2021}-\dfrac{1}{2022}\)

\(=1-\dfrac{1}{2022}=\dfrac{2021}{2022}\)

Vậy ta có đpcm 

\(\dfrac{x+4}{3}=\dfrac{x-11}{-6}\)

\(\dfrac{2x+8}{6}=\dfrac{-x+11}{6}\)

\(\Leftrightarrow2x+8=-x+11\)

\(\Leftrightarrow3x=3\)

\(\Leftrightarrow x=1\)

Nhân chéo ta được\(-6(x+4)=3(x-11)=>-6x-24=3x-33=>6x-3x-24+33=0=>3x+9=0=>3x=-9=>x=-3\)

Bài 2:

1: \(x+\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\cdots+\frac{1}{31\cdot33}=3\)

=>\(x+\frac12\left(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\cdots+\frac{2}{31\cdot33}\right)=3\)

=>\(x+\frac12\left(1-\frac13+\frac13-\frac15+\cdots+\frac{1}{31}-\frac{1}{33}\right)=3\)

=>\(x+\frac12\left(1-\frac{1}{33}\right)=3\)

=>\(x+\frac12\cdot\frac{32}{33}=3\)

=>\(x+\frac{16}{33}=3\)

=>\(x=3-\frac{16}{33}=\frac{99}{33}-\frac{16}{33}=\frac{83}{33}\)

2: \(x-\frac{3}{1\cdot5}-\frac{3}{5\cdot9}-\cdots-\frac{3}{61\cdot65}=2\)

=>\(x-\frac34\left(\frac{4}{1\cdot5}+\frac{4}{5\cdot9}+\cdots+\frac{4}{61\cdot65}\right)=2\)

=>\(x-\frac34\left(1-\frac15+\frac15-\frac19+\cdots+\frac{1}{61}-\frac{1}{65}\right)=2\)

=>\(x-\frac34\left(1-\frac{1}{65}\right)=2\)

=>\(x-\frac34\cdot\frac{64}{65}=2\)

=>\(x-\frac{48}{65}=2\)

=>\(x=2+\frac{48}{65}=\frac{130}{65}+\frac{48}{65}=\frac{178}{65}\)

Bài 1:

2: \(B=\frac{1}{6\cdot7}+\frac{1}{7\cdot8}+\cdots+\frac{1}{89\cdot90}\)

\(=\frac16-\frac17+\frac17-\frac18+\cdots+\frac{1}{89}-\frac{1}{90}\)

\(=\frac16-\frac{1}{90}=\frac{14}{90}=\frac{7}{45}\)

3: \(C=\frac{1}{2\cdot4}+\frac{1}{4\cdot6}+\cdots+\frac{1}{60\cdot62}\)

\(=\frac12\left(\frac{2}{2\cdot4}+\frac{2}{4\cdot6}+\cdots+\frac{2}{60\cdot62}\right)\)

\(=\frac12\left(\frac12-\frac14+\frac14-\frac16+\cdots+\frac{1}{60}-\frac{1}{62}\right)\)

\(=\frac12\left(\frac12-\frac{1}{62}\right)=\frac12\cdot\frac{30}{62}=\frac12\cdot\frac{15}{31}=\frac{15}{62}\)

4: \(D=\frac{2}{2\cdot5}+\frac{2}{5\cdot8}+\cdots+\frac{2}{92\cdot95}\)

\(=\frac23\left(\frac{3}{2\cdot5}+\frac{3}{5\cdot8}+\cdots+\frac{3}{92\cdot95}\right)\)

\(=\frac23\left(\frac12-\frac15+\frac15-\frac18+\cdots+\frac{1}{92}-\frac{1}{95}\right)\)

\(=\frac23\left(\frac12-\frac{1}{95}\right)=\frac23\cdot\frac{93}{190}=\frac{1}{95}\cdot31=\frac{31}{95}\)

5:Sửa đề: \(E=\frac{6}{1\cdot3}+\frac{6}{3\cdot5}+\cdots+\frac{61}{63\cdot65}\)

\(=3\left(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\cdots+\frac{2}{63\cdot65}\right)\)

\(=3\left(1-\frac13+\frac13-\frac15+\cdots+\frac{1}{63}-\frac{1}{65}\right)\)

\(=3\left(1-\frac{1}{65}\right)=3\cdot\frac{64}{65}=\frac{192}{65}\)

= x³ + 3³ -x(x² -1) -27 =0 ( tổng các lập phuong)

x =0 

CX100%

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