Đáp số x=-40 em nhé, lúc nãy anh nhầm dạng =)

đk : x khác 10 ; 0 

\(\Rightarrow600x-600\left(x-10\right)=3x\left(x-10\right)\)

\(\Leftrightarrow6000=3x^2-30x\Leftrightarrow x^2-10x-2000=0\Leftrightarrow x=50;x=-40\)(tm)

\(Q=\left(\dfrac{1}{2\sqrt{x}+1}+\dfrac{1}{2\sqrt{x}-1}\right):\dfrac{1}{1-4x}\)

\(=\left(\dfrac{2\sqrt{x}-1}{\left(2\sqrt{x}+1\right)\left(2\sqrt{x}-1\right)}+\dfrac{2\sqrt{x}+1}{\left(2\sqrt{x}+1\right)\left(2\sqrt{x}-1\right)}\right).\left(1-4x\right)\)

\(=\left(\dfrac{2\sqrt{x}-1+2\sqrt{x}+1}{4x-1}\right)\left(1-4x\right)\)

\(=\dfrac{-4\sqrt{x}.\left(4x-1\right)}{4x-1}=-4\sqrt{x}\)

\(Q=\left(\dfrac{1}{2\sqrt{x}+1}+\dfrac{1}{2\sqrt{x}-1}\right):\dfrac{1}{1-4x}\left(dkxd:x\ge0;x\ne\dfrac{1}{4}\right)\)

\(=\left[\dfrac{2\sqrt{x}-1}{\left(2\sqrt{x}-1\right)\left(2\sqrt{x}+1\right)}+\dfrac{2\sqrt{x}+1}{\left(2\sqrt{x}-1\right)\left(2\sqrt{x}+1\right)}\right]\cdot\left(1-4x\right)\)

\(=\dfrac{2\sqrt{x}-1+2\sqrt{x}+1}{4x-1}\cdot\left[-\left(4x-1\right)\right]\)

\(=4\sqrt{x}\cdot\left(-1\right)\)

\(=-4\sqrt{x}\)

bài này bn trục căn thức mẫu nhé

\(P=\dfrac{1-\sqrt{2}}{1-2}+\dfrac{\sqrt{2}-\sqrt{3}}{2-3}+...+\dfrac{\sqrt{99}-\sqrt{100}}{99-100}\)

\(P=\dfrac{1-\sqrt{2}+\sqrt{2}-\sqrt{3}+....+\sqrt{99}-\sqrt{100}}{-1}=\dfrac{1-\sqrt{100}}{-1}=\sqrt{100}-1\)

\(=10-1=9\)

`=(\sqrt1-\sqrt2)/(1-2)+(\sqrt2-\sqrt3)/(2-3)+.....+(\sqrt99-\sqrt100)/(99-100)`

`=\sqrt2-\sqrt1+\sqrt3-\sqrt2+.....+\sqrt100-\sqrt99`

`=-\sqrt1+\sqrt100=-1+10=9`

Ta có: \(\dfrac{1}{\sqrt{1}+\sqrt{2}}+\dfrac{1}{\sqrt{2}+\sqrt{3}}+\dfrac{1}{\sqrt{3}+\sqrt{4}}+...+\dfrac{1}{\sqrt{99}+\sqrt{100}}\)

\(=-1+\sqrt{2}-\sqrt{2}+\sqrt{3}-\sqrt{3}+2-...-\sqrt{99}+10\)

=10-1=9

a) x vô nghĩa

b) x=0,8;x=-0,8

d: Đặt x/2=y/5=k

=>x=2k; y=5k

Ta có: xy=10

nên k²=1

Trường hợp 1: k=1

=>x=2; y=5

Trường hợp 2: k=-1

=>x=-2; y=-5

6

\(\Leftrightarrow5x=6\left(x-1\right)\Leftrightarrow5x=6x-6\Leftrightarrow x=6\)

ĐKXĐ:\(x\ne0\)

\(\dfrac{5}{6}=\dfrac{x-1}{x}\\ \Rightarrow5x=6\left(x-1\right)\\ \Rightarrow5x=6x-6\\ \Rightarrow6x-5x-6=0\\ \Rightarrow x-6=0\\ \Rightarrow x=6\)

a) S=\(\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{2017.2019}\)

2S=\(\dfrac{2}{1.3}+\dfrac{2}{3.5}+...+\dfrac{2}{2017.2019}\)

2S=\(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2017}-\dfrac{1}{2019}\)

2S=\(1-\dfrac{1}{2019}\)

2S=\(\dfrac{2018}{2019}\)

S\(\dfrac{1009}{2019}\)

b) Gọi ƯCLN(14n+3,21n+5) là d

14n+3⋮d ⇒42n+9⋮d

21n+5⋮d ⇒42n+10⋮d

(42n+10)-(42n+9)⋮d

1⋮d ⇒ƯCLN(14n+3,21n+5)=1

Vậy \(\dfrac{14n+3}{21n+5}\) là Ps tối giản

Giải:

a) \(S=\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{2017.2019}\)

\(S=\dfrac{1}{2}.\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{2017.2019}\right)\) 

\(S=\dfrac{1}{2}.\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{2017}-\dfrac{1}{2019}\right)\)

\(S=\dfrac{1}{2}.\left(1-\dfrac{1}{2019}\right)\) 

\(S=\dfrac{1}{2}.\dfrac{2018}{2019}\) 

\(S=\dfrac{1009}{2019}\) 

b) Gọi \(ƯCLN\left(14n+3;21n+5\right)=d\) 

\(\Rightarrow\left\{{}\begin{matrix}14n+3⋮d\\21n+5⋮d\end{matrix}\right.\)   \(\Rightarrow\left\{{}\begin{matrix}3.\left(14n+3\right)⋮d\\2.\left(21n+5\right)⋮d\end{matrix}\right.\)   \(\Rightarrow\left\{{}\begin{matrix}42n+9⋮d\\42n+10⋮d\end{matrix}\right.\) 

\(\Rightarrow\left(42n+10\right)-\left(42n+9\right)⋮d\) 

\(\Rightarrow1⋮d\) 

Vậy \(A=\dfrac{14n+3}{21n+5}\) là p/s tối giản.

Ko nên thức thâu đêm bạn nha!

\(\dfrac{2+\sqrt{2}}{\sqrt{2}+1}+\dfrac{1}{\sqrt{3}+\sqrt{2}}\)

\(=\dfrac{\sqrt{2}\left(\sqrt{2}+1\right)}{\sqrt{2}+1}+\dfrac{\sqrt{3}-\sqrt{2}}{3-2}\)

\(=\sqrt{2}+\sqrt{3}-\sqrt{2}=\sqrt{3}\)

-3/6 = -1/2 = -18/36

\(-\dfrac{3}{6}=\dfrac{x}{2}=-\dfrac{18}{y}\)

Xét \(-\dfrac{3}{6}=\dfrac{x}{2}\)

\(\Rightarrow x=\dfrac{-3.2}{6}=-1\)

Xét \(\dfrac{x}{2}=-\dfrac{18}{y}\)

     \(-\dfrac{1}{2}=-\dfrac{18}{y}\)

   \(y=\dfrac{-18.2}{-1}=36\)

= -1/2 = -18/36