ĐKXĐ: x khác + -2

 A=( 1/(x-2) + 2x/(x-2)(x+2) +1/(x+2)) . (x-1)/2

   =((x+2+2x+x-2)/(x-2)(x+2)).((x-1)/2)

   =(4x/(x-2)(x+2)).(x-1)/2 =2x/ (x-1)(x-2)(x+2)
 

f(x)g(x)=0<=>f(x)=0 hoặc g(x)=0

<=>(x²+x+1)²+x²+x-11=(x-1)(x+2)(x²+x+5) (=>vế phải có 3 TH)

TH1=>x=1

TH2=>x=-2

x²+x+5=0

1²-4(1.5)=-19

=>pt ko có nghiệm thực

=>x=1 hoặc -2

đặt x^2+x+1=a

a^2+a-12=0  <=>(a-3)(a+4)=0 

<=>a-3=0 hoặc a+4=0

Nếu a-3 =0 suy ra x^2+x-2=0 <=>(x+2)(x-1)=0 suy ra x=-2 hoặc x=1

Nếu a+4=0 thì x^2+x+5=0 vô nghiệm do x^2+x+1/4=(x+1/2)>=0

\(=x^2-1+x-1-x^2-x+x^2-x+2x-2=-x^2\)

\(=-4+x^2+x=-x^2\)

\(=-4+x^2+x+x^2=0\)

\(=-4+2x^2+x=0\Rightarrow x=-1,687\)

1/1 x 2 + 1/2 x 3 + 1/3 x 4 + ... + 1/999 x 1000 + 1

= 1/1 - 1/1000 + 1 

= 999/1000 + 1

= 1999/1000

Chuc ban may man

1,999 nha

Câu 1 :

a. \(4x-5=23\\ \Leftrightarrow4x=23+5\\ \Leftrightarrow4x=28\\ \Leftrightarrow x=7\)

b. 

|-2x|=5x+14

 Nếu - 2x > 0 => x < 0 thì |-2x|= - 2x, ta có pt: -2x = 5x+14

 <=> - 2x = 5x + 14

 <=> - 2x - 5x = 14

 <=> - 7x = 14

 <=> x = - 2 (thoã mãn)

 Nếu - 2x < 0 => x > 0 thì |-2x|= = -(- 2x) = 2x.

Ta có pt: 2x = 5x + 14

 <=> - 3x = 14

<=> x = \(-\dfrac{14}{3}\)
 Vậy pt có nghiệm x = - 2

c) \(\dfrac{x+1}{x-1}-\dfrac{1}{x+1}=\dfrac{x^2+2}{x^2-1}\\ ĐKXĐ:x\ne1;x\ne-1\\ \Leftrightarrow\dfrac{\left(x+1\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}-\dfrac{1\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}=\dfrac{x^2+2}{\left(x-1\right)\left(x+1\right)}\\ \Leftrightarrow x^2+x+x+1-x+1=x^2+2\\ \Leftrightarrow x^2+x+x-x-x^2=2-1-1\\ \Leftrightarrow x=0\left(nhận\right)\)

\(a,4x-5=23\)

\(\Leftrightarrow4x=23+5\)

\(\Leftrightarrow4x=28\)

\(\Leftrightarrow x=7\)

\(b,\left|-2x\right|=5x+14\)

\(\Leftrightarrow\left[{}\begin{matrix}2x=5x+14\\2x=-5x-14\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}-3x-14=0\\7x+14=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}-3x=14\\7x=-14\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{14}{3}\\x=-2\end{matrix}\right.\)

Vậy \(S=\left\{-\dfrac{14}{3};-2\right\}\)

\(c,\Leftrightarrow\dfrac{\left(x+1\right)\left(x+1\right)-x+1-x^2-2}{x^2-1}=0\)

\(\Leftrightarrow x^2+x+x+1-x+1-x^2-2=0\)

\(\Leftrightarrow x=0\)

Vậy \(S=\left\{0\right\}\)

a) \(4x-5=23\)

    \(4x=23+5\)

      \(4x=28\)

        \(x=7\)

b) \(\left|-2x\right|=5x+14\)

 \(\Leftrightarrow\)   \(-2x-5=14\)

\(\Leftrightarrow\)    \(-7x=14\)

\(\Leftrightarrow\)         \(x=-2\)

\(\Leftrightarrow\)    \(-2x=-\left(5x+14\right)\)

\(\Leftrightarrow\)    \(-2x=-\left(5x-14\right)\)

\(\Leftrightarrow\)  \(-2x+5x=-14\)

 \(\Leftrightarrow\)    \(3x=-14\)

 \(\Leftrightarrow\) \(x=-\dfrac{14}{3}\) \(\left(\text{vô lí}\right)\)

  \(\Leftrightarrow x=-2\)   

c) \(\dfrac{x+1}{x-1}-\dfrac{1}{x+1}=\dfrac{x^2+2}{x^2-1}\)

 \(\Leftrightarrow\) \(\dfrac{x+1}{x-1}+\dfrac{-1}{x+1}=\dfrac{x^2+2}{\left(x+1\right)\left(x-1\right)}\)

 \(\Leftrightarrow\left(x+1\right)\left(x+1\right)+\left(-1\right)\left(x-1\right)=x^2+2\)

\(\Leftrightarrow x^2+x+2=x^2+2\)

\(\Leftrightarrow x+2=2\)

\(\Leftrightarrow x=0\)

đk : x >= 0 ; x khác 4 

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

xin lỗi, bn cóa thể bấm ∑ cái nài để lm lại đề đc hăm :v?

\(\dfrac{2x-3}{4-x}+\dfrac{1}{3}>\dfrac{1}{2}-\dfrac{3-x}{5}\) 

đúng ko ???

( x - 2 )^x + 1 = ( x - 2 )^x + 3

( x - 2 )^x + 1 - [ ( x - 2 )^x + 3 ] = 0

( x - 2 )^x + 1 - ( x - 2 )^x - 3 = 0

[ ( x - 2 )^x - ( x - 2 )^x ] + ( 1 - 3 ) = 0

-2 = 0 ( vô lý ) 

Vạy không tìm đc x thỏa mãn

P/s : Không chắc :P

Mơn nha

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

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

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

\(\Rightarrow A=\dfrac{x+1}{\sqrt{x}\left(\sqrt{x}-1\right)}:\dfrac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

\(\Rightarrow A=\dfrac{x+1}{\sqrt{x}\left(\sqrt{x}-1\right)}.\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}+1}\)

\(\Rightarrow A=\dfrac{x+1}{\sqrt{x}}\)