\(a,50\%x-0,2+x=\dfrac{4}{5}\)

\(\Leftrightarrow\dfrac{1}{2}x-0,2+x=\dfrac{4}{5}\)

\(\Leftrightarrow\dfrac{1}{2}x+x=\dfrac{4}{5}+0,2\)

\(\Leftrightarrow\dfrac{3}{2}x=\dfrac{4}{5}+\dfrac{1}{5}\)

\(\Leftrightarrow\dfrac{3}{2}x=1\)

\(\Leftrightarrow x=\dfrac{2}{3}\)

\(b,\left(x-\dfrac{3}{4}\right):\dfrac{1}{2}+\dfrac{3}{2}=\dfrac{25}{2}\)

\(\Leftrightarrow\left(x-\dfrac{3}{4}\right).2=\dfrac{25}{2}-\dfrac{3}{2}\)

\(\Leftrightarrow\left(x-\dfrac{3}{4}\right).2=\dfrac{22}{2}\)

\(\Leftrightarrow x-\dfrac{3}{4}=11:2\)

\(\Leftrightarrow x=\dfrac{11}{2}+\dfrac{3}{4}\)

\(\Leftrightarrow x=\dfrac{25}{4}\)

\(\dfrac{3}{x-5}=\dfrac{-4}{x+2}\left(x\ne5;-2\right).\\ \Leftrightarrow\dfrac{3}{x-5}+\dfrac{4}{x+2}=0.\\ \Leftrightarrow\dfrac{3x+6+4x-20}{\left(x-5\right)\left(x+2\right)}=0.\\ \Rightarrow7x=14.\\ \Leftrightarrow x=2\left(TM\right).\)

\(a,M=\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}+\dfrac{\sqrt{x}}{\sqrt{x}+1}\right):\left(\dfrac{2}{x}-\dfrac{2-x}{x\sqrt{x}+x}\right)\left(x>0;x\ne1\right)\\ M=\dfrac{x+\sqrt{x}+x-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}:\dfrac{2\sqrt{x}+2-2+x}{x\left(\sqrt{x}+1\right)}\\ M=\dfrac{2x}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\cdot\dfrac{x\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+2\right)}\\ M=\dfrac{2x}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)

\(b,M=-\dfrac{1}{2}\Leftrightarrow\dfrac{2x}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}=-\dfrac{1}{2}\\ \Leftrightarrow-4x=x+\sqrt{x}-2\\ \Leftrightarrow5x+\sqrt{x}-2=0\)

Đặt \(\sqrt{x}=t\)

\(\Leftrightarrow5t^2+t-2=0\\ \Delta=1^2-4\cdot5\left(-2\right)=41\\ \Leftrightarrow\left[{}\begin{matrix}t=\dfrac{-1-\sqrt{41}}{10}\\t=\dfrac{-1+\sqrt{41}}{10}\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-\left(1+\sqrt{41}\right)^2}{100}=\dfrac{-42-2\sqrt{41}}{100}\\x=\dfrac{\left(\sqrt{41}-1\right)^2}{100}=\dfrac{42-2\sqrt{41}}{100}\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-21-\sqrt{41}}{50}\left(L\right)\\x=\dfrac{21-\sqrt{41}}{50}\left(N\right)\end{matrix}\right.\\ \Leftrightarrow x=\dfrac{21-\sqrt{41}}{50}\)

a: Ta có: \(M=\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}+\dfrac{\sqrt{x}}{\sqrt{x}+1}\right):\left(\dfrac{2}{x}+\dfrac{x-2}{x\sqrt{x}+x}\right)\)

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

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

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

\(\left|x+1\right|-\left|-2x-2\right|=2\)

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

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

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

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

\(\Leftrightarrow\left|x+1\right|+2=0\)

Mà: \(\left|x+1\right|\ge0\forall x\Rightarrow\left|x+1\right|+2\ge2>0\)

\(\Leftrightarrow\left|x+1\right|+2=0\) (vô lí)

Vậy phương trình vô nghiệm:

\(x\in\varnothing\)

=>|x+1|-2|x+1|=2

=>-|x+1|=2

=>|x+1|=-2(vô lý)

Vậy: \(x\in\varnothing\)

\(TH_1:\left\{{}\begin{matrix}x+1\ge0\\-2x-2\ge0\end{matrix}\right.\) 

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

\(\Leftrightarrow x+1+2x+2=2\)

\(\Leftrightarrow3x+3=2\)  

\(\Leftrightarrow x=-\dfrac{1}{3}\left(l\right)\)

\(TH_2:\left\{{}\begin{matrix}x+1< 0\\-2x-2< 0\end{matrix}\right.\) 

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

\(\Leftrightarrow-x-1+2x+2=2\) 

\(\Leftrightarrow x+1=2\)

\(\Leftrightarrow x=1\left(l\right)\) 

Vậy tập nghiệm rỗng.

Ta có: X x 456 + X x 544 + X = 12000

          X x ( 456 + 544 + 1) = 12000

          X x 1001 = 12000

          X = 12000/1001

\(X\cdot456+X\cdot544+X=12000\)

\(X\cdot\left(456+544+1\right)=12000\)

\(X\cdot1001=12000\)

\(X=\frac{12000}{1001}\)

\(X=11\frac{989}{1001}\)

\(\cdot\)\(Là\)x (nhân) nha

x . 456 + x . 544 + x = 12000

x . 456 + x . 544 + x . 1 = 12000

x (456 + 544 + 1) = 12000

x . 1001 = 12000

x = 12000 : 1001

x = 11.98801198801

#Học tốt!!!

~NTTH~

help mì 

Đặt \(x+2=t\ne0\Rightarrow x+1=t-1\)

\(A=\dfrac{x+1}{\left(x+2\right)^2}=\dfrac{t-1}{t^2}=-\dfrac{1}{t^2}+\dfrac{1}{t}=-\left(\dfrac{1}{t}-\dfrac{1}{2}\right)^2+\dfrac{1}{4}\le\dfrac{1}{4}\)

\(A_{max}=\dfrac{1}{4}\) khi \(t=2\) hay \(x=0\)

giúp mình đi mọi người

5/6-x=2/3

x=5/6-2/3

x=1/6

4/3:x=2

x=4/3:2

x=2/3

a) x = 1/6

b) x = 2/3

29.31.169.\(13^2=25676339=13^4.29.31\)