a) $\dfrac{120^3}{40^3}=(\dfrac{120}{40})^3=3^3=27$

b) $\dfrac{390^4}{130^4}=(\dfrac{390}{130})^4=3^4=81$

c) $\dfrac{3^2}{(0,375)^2}=(3:0,375)^2=(3:\dfrac{3}{8})^2=8^2=64$

a, (4².4³ / 2¹.0) .2003

= (4².4³/0) .2003

= 0.2003

= 0

b, 120³/40³

= (120/40)³

= 3³

= 9

c, [(-2,5) .0,38 .0,4] - [(-1,25).0,62.(-0,8)]

= {[(-2,5) .0,4] .0,38} - {[(-1,25) .(-0,8)] .0,62}

= [(-1) .0,38] - (1 .0,62)

= (-0.38) - 0,62

= -1

MK chỉnh dấu ngoặc hơi loằng ngoằng

a,\(\dfrac{120^3}{40^3}=3\)

b,\(\dfrac{390^3}{130^3}=3\)

c,\(\dfrac{3^2}{\left(0,375\right)^2}=8\)

a)120³/40³= (120/40)³= 3³= 27

b)390⁴/130⁴=(390/130)⁴= 3⁴= 81

c)3²/(0,375)²= (3/0,375)²= 8²= 64

a) 120^3/40^3=(120/40)^3=3^3=27
b) 390^4/130^4=(390/130)^4=3^4=81

c) 3^2/(0,375)^2=(3/0.375)^2=64

Xét số hạng tổng quát:\(\dfrac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}\)

\(=\dfrac{1}{\sqrt{n\left(n+1\right)}\left(\sqrt{n+1}+\sqrt{n}\right)}\)

\(=\dfrac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n\left(n+1\right)}}=\dfrac{1}{\sqrt{n}}-\dfrac{1}{\sqrt{n+1}}\)

\(\Rightarrow A=1-\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{120}}-\dfrac{1}{\sqrt{121}}\)

\(A=1-\dfrac{1}{11}=\dfrac{10}{11}\)

a) \(\left(\dfrac{1}{5}\right)^5.5^5=\left(\dfrac{1}{5}.5\right)^5=1^5=1\)

b) \(\left(0,125\right)^3.512=\left(0,512\right)^3.8^3=\left(0,512.8\right)^3=1^3=1\)

c) \(\left(0,25\right)^4.1024=\left[\left(0,25\right)^2\right]^2.32^2=\left(\dfrac{1}{6}\right)^2.32^2=\left(\dfrac{1}{6}.32\right)^2=2^2=4\)

d) \(\dfrac{120^3}{40^3}=\left(\dfrac{120}{40}\right)^3=3^3=64\)

e) \(\dfrac{390^4}{130^4}=\left(\dfrac{390}{130}\right)^4=3^4=81\)

g) \(\dfrac{3^2}{\left(0,375\right)^2}=\left(\dfrac{3}{0,375}\right)^3=8^3=512\)

ĐK: ` x \ne 10; x \ne 0`

`120/(x-10)-3/5=120/x`

`<=>120/(x-10)-120/x=3/5`

`<=>1/(x-10) - 1/x= 1/200`

`<=> (x-x+10)/(x(x-10)) = 1/200`

`<=> 10/(x(x-10))= 1/200`

`<=> x^2-10=2000`

`<=>` \(\left[{}\begin{matrix}x=50\\x=-40\end{matrix}\right.\)

Vậy `S={50;-40}`.

`120/(x-10)-3/5=120/x(x ne 0,x ne 10)`

`<=>40/(x-10)-1/5=40/x`

`<=>200x-x(x-10)=200(x-10)`

`<=>200x-200x+2000-x^2+10x=0`

`<=>x^2-10x-2000=0`

`Delta'=25+2000=2025`

`<=>x_1=50,x_2=-40`

Vậy `S={50,-40}`

ĐKXĐ : \(\left\{{}\begin{matrix}x\ne10\\x\ne0\end{matrix}\right.\)

Ta có : \(\dfrac{120}{x-10}-\dfrac{3}{5}=\dfrac{120}{x}=\dfrac{600-3\left(x-10\right)}{5\left(x-10\right)}\)

\(\Leftrightarrow600\left(x-10\right)=600x-3x\left(x-10\right)\)

\(\Leftrightarrow600x-6000=600x-3x^2+30x\)

\(\Leftrightarrow3x^2-30x-6000=0\)

\(\Leftrightarrow\left(x-50\right)\left(x+40\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=50\\x=-40\end{matrix}\right.\) ( TM )

Vậy ...

Tổng quát:

\(\dfrac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}\)\(=\dfrac{1}{\sqrt{n\left(n+1\right)}\left(\sqrt{n+1}+\sqrt{n}\right)}\)

\(=\dfrac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n\left(n+1\right)}}\)\(=\dfrac{1}{\sqrt{n}}-\dfrac{1}{\sqrt{n+1}}\)

\(\Rightarrow S=\dfrac{10}{11}\)

Ta có công thức tổng quát như sau:

\(\dfrac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}\)

\(=\dfrac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{\left[\left(n+1\right)\sqrt{n}+n\sqrt{n+1}\right]\left[\left(n+1\right)\sqrt{n}-n\sqrt{n+1}\right]}\)

\(=\dfrac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{n\left(n+1\right)^2-n^2\left(n+1\right)}\)

\(=\dfrac{\sqrt{n}}{n}-\dfrac{\sqrt{n+1}}{n+1}\)

\(=\dfrac{1}{\sqrt{n}}+\dfrac{1}{\sqrt{n+1}}\)

Áp dụng vào tổng S ta có:

\(S=\dfrac{1}{2\sqrt{1}+1\sqrt{2}}+\dfrac{1}{3\sqrt{2}+2\sqrt{3}}+...+\dfrac{1}{121\sqrt{120}+120\sqrt{121}}\)

\(S=\dfrac{1}{\sqrt{1}}-\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{120}}+\dfrac{1}{\sqrt{121}}\)

\(S=1-\dfrac{1}{\sqrt{121}}=1-\dfrac{1}{11}=\dfrac{10}{11}\)

\(\dfrac{120}{x}+\dfrac{120}{x-10}=\dfrac{3}{5}\left(dkxd:x>0,x\ne10\right)\)

\(\Leftrightarrow\dfrac{120}{x}+\dfrac{120}{x-10}-\dfrac{3}{5}=0\)

\(\Leftrightarrow\dfrac{120.5\left(x-10\right)+5.120x-3x\left(x-10\right)}{5x\left(x-10\right)}=0\)

\(\Leftrightarrow600x-6000+600x-3x^2+30x=0\)

\(\Leftrightarrow-3x^2+1230x-6000=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\approx405\\x\approx5\end{matrix}\right.\)\(\left(tmdk\right)\)

Vậy ...