A=1+1/2+1+1/6+1+1/12+...+1+1/90=

=9+1/2+1/6+1/12+...+1/90

1/2+1/6+1/12+...+1/90=

1/1x2+1/2x3+2/3x4+...+1/9x10=

\(=\dfrac{2-1}{1x2}+\dfrac{3-2}{2x3}+\dfrac{4-3}{3x4}+...+\dfrac{10-9}{9x10}=\)

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{9}-\dfrac{1}{10}=\)

\(=1-\dfrac{1}{10}=\dfrac{9}{10}\)

\(\Rightarrow A=9+\dfrac{9}{10}=9\dfrac{9}{10}\)

\(d_1\) nhận \(\overrightarrow{u_1}=\left(3;1\right)\) là 1vtcp

\(d_2\) nhận (2;-1) là 1 vtpt nên nhận \(\overrightarrow{u_2}=\left(1;2\right)\) là 1 vtcp

\(\Rightarrow cos\widehat{\left(d_1;d_2\right)}=\left|cos\widehat{\left(\overrightarrow{u_1};\overrightarrow{u_2}\right)}\right|=\dfrac{\left|\overrightarrow{u_1}.\overrightarrow{u_2}\right|}{\left|\overrightarrow{u_1}\right|.\left|\overrightarrow{u_2}\right|}=\dfrac{\left|3.1+1.2\right|}{\sqrt{3^2+1^2}.\sqrt{1^2+2^2}}=\dfrac{\sqrt{2}}{2}\)

\(\Rightarrow\widehat{\left(d_1;d_2\right)}=45^0\)

\(\dfrac{x}{2}+\dfrac{3x}{5}-\dfrac{6}{5}=3\Leftrightarrow\dfrac{11x}{10}=3+\dfrac{6}{5}=\dfrac{21}{5}\)

\(\Rightarrow11x=42\Leftrightarrow x=\dfrac{42}{11}\)

\(\frac{1}{2}\)x +  \(\frac{3}{5}\)( x - 2 ) = 3

\(\frac{1}{2}\)x + \(\frac{3}{5}\) x - \(\frac{3}{5}\). 2 = 3

x \((\)\(\frac{1}{2}\)+ \(\frac{3}{5}\) \()\)- \(\frac{6}{5}\) = 3 

x . \(\frac{11}{10}\) - \(\frac{6}{5}\) = 3

x . \(\frac{11}{10}\) = 3 + \(\frac{6}{5}\) = \(\frac{21}{5}\)

x = \(\frac{21}{5}\) : \(\frac{11}{10}\) = \(\frac{42}{11}\)

NẾU CÓ GÌ SAI SÓT MONG BẠN THÔNG CẢM 

1/2 . x + 3/5 . ( x - 2 ) = 3         x + 3/5 . ( x - 2 ) = 3 : 1/2         x + 3/5 . ( x - 2 ) = 6                 x  . ( x - 2 ) = 6 - 3/5                 x  . ( x - 2 ) = 27/5                  x . x + x - 2 = 27/5             x mũ x + x - 2 = 27/5              2 . x mũ x - 2 = 27/5                   x mũ x - 2 = 27/5 : 2                   x mũ x - 2 = 27/10                        x mũ x = 27/10 + 2                        x mũ x = 47/10                            x . x = 47/10                            x . x = 4,7                            x . x = căn 4,7                                 x = căn 4,7 : x                           Vậy x = căn 4,7 : x hoặc là 47/10 : x                      

`30 xx 25 xx 7 xx8`

`=(30 xx7) xx (25xx8)`

`=210 xx200`

`=42000`

`----`

`75 xx 8 xx 12 xx 14`

`=(75xx8) xx(12xx14)`

`=600xx168`

`=100800`

\(a.30\times25\times7\times8=\left(30\times7\right)\times\left(25\times8\right)=210\times200=42000\)

\(b.75\times8\times12\times14=\left(75\times12\right)\times\left(8\times14\right)=900\times112=100800\)

\(a,30.25.7.8\\ =\left(30.7\right).\left(25.8\right)\\ =210.200\\ =42000\\ b,75.8.12.14\\ =\left(75.12\right).\left(8.14\right)\\ =900.112\\ =100800\)

\(3\left(x-1\right)^2-3x\left(2-5\right)=21\)

\(\Leftrightarrow3x^2-6x+3+9x-21=0\)

\(\Leftrightarrow3x^2+3x-18=0\)

\(\Leftrightarrow3\left(x^2+x-6\right)=0\)

\(\Leftrightarrow3\left(x-2\right)\left(x+3\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\x+3=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=2\\x=-3\end{cases}}\)

Vậy \(S=\left\{2;-3\right\}\)

\(2,1\left(\dfrac{m}{s}\right)=\dfrac{189}{25}\left(\dfrac{km}{h}\right),3000m=3km\)

Thời gian người đó đi hết quãng đường thứ nhất:

\(t_1=\dfrac{S_1}{v_1}=\dfrac{4,5}{\dfrac{189}{25}}=\dfrac{25}{42}\left(h\right)\)

Vận tốc trung bình trên cả 2 quãng đường:

\(v_{tb}=\dfrac{S_1+S_2}{t_1+t_2}=\dfrac{4,5+3}{\dfrac{25}{42}+0,75}\approx5,58\left(\dfrac{km}{h}\right)\)

Đổi 2,1m/s=7,56km/s, 3000m=3km

Thời gian người đó đi trên quãng đường đầu 

\(t=s:v=4,5:7,56=0,59\left(h\right)\)

Vận tốc trung bình của người đó trên cả hai quãng đường là

\(v_{tb}=\dfrac{s_1+s}{t_1+t}=\dfrac{4,5+3}{0,59+0,75}=\dfrac{7,5}{1,34}=5,59\left(kmh\right)\)

khó vl

a: Xét ΔDOE vuông tại O và ΔKOE vuông tại O có

EO chung

\(\widehat{DEO}=\widehat{KEO}\)

Do đó: ΔDOE=ΔKOE

b: Xét ΔEDI vàΔEKI có

ED=EK

\(\widehat{DEI}=\widehat{KEI}\)

EI chung

Do đó: ΔEDI=ΔEKI

Suy ra: \(\widehat{EDI}=\widehat{EKI}=90^0\)

hay IK\(\perp\)FE

c: Xét ΔDIQ vuông tại D và ΔKIF vuông tại K có

ID=IK

\(\widehat{DIQ}=\widehat{KIF}\)

Do đó: ΔDIQ=ΔKIF

Suy ra: IQ=IF

\(a,\left(\dfrac{31}{35}-\dfrac{4}{7}\right)\times\dfrac{8}{7}:2\\ =\left(\dfrac{31}{35}-\dfrac{4\times5}{35}\right)\times\dfrac{8}{7}:2\\ =\dfrac{11}{35}\times\dfrac{8}{7}:2\\ =\dfrac{88}{245}:2\\ =\dfrac{44}{245}\\ b,\left(1-\dfrac{1}{2}\right)\times\left(1-\dfrac{1}{3}\right)\times\left(1-\dfrac{1}{4}\right)\times\left(1-\dfrac{1}{5}\right)\\ =\left(\dfrac{2-1}{2}\right)\times\left(\dfrac{3-1}{3}\right)\times\left(\dfrac{4-1}{4}\right)\times\left(\dfrac{5-1}{5}\right)\\ =\dfrac{1}{2}\times\dfrac{2}{3}\times\dfrac{3}{4}\times\dfrac{4}{5}\\ =\dfrac{1}{3}\times\dfrac{3}{4}\times\dfrac{4}{5}\\ =\dfrac{1}{4}\times\dfrac{4}{5}=\dfrac{1}{5}\)

a, ( \(\dfrac{31}{35}\) - \(\dfrac{4}{7}\)) \(\times\) \(\dfrac{8}{7}\): 2

= \(\left(\dfrac{31}{35}-\dfrac{20}{35}\right)\) \(\times\) \(\dfrac{8}{7}\) : 2

= \(\dfrac{11}{35}\) \(\times\) \(\dfrac{8}{7}\) \(\times\) \(\dfrac{1}{2}\)

= \(\dfrac{44}{35}\) \(\times\) \(\dfrac{4}{7}\)

= \(\dfrac{44}{245}\)

b, ( 1 - \(\dfrac{1}{2}\)) \(\times\) ( 1 - \(\dfrac{1}{3}\)) \(\times\) ( 1 - \(\dfrac{1}{4}\)) \(\times\) ( 1 - \(\dfrac{1}{5}\))

= \(\dfrac{1}{2}\) \(\times\) \(\dfrac{2}{3}\) \(\times\) \(\dfrac{3}{4}\) \(\times\) \(\dfrac{4}{5}\)

= \(\dfrac{1}{5}\) \(\times\) \(\dfrac{2\times3\times4}{2\times3\times4}\)

= \(\dfrac{1}{5}\)

\(\dfrac{x}{\dfrac{4}{2}}=\dfrac{4}{\dfrac{x}{2}}\)

\(\Rightarrow\dfrac{x}{2}=\dfrac{8}{x}\)

\(\Rightarrow x^2=8\cdot2\)

\(\Rightarrow x^2=16\)

\(\Rightarrow x^2=4^2\)

\(\Rightarrow\left[{}\begin{matrix}x=4\\x=-4\end{matrix}\right.\)