\(Dk:x,y\ge\frac{-5}{4}\)

\(\left\{{}\begin{matrix}\left(2x-3\right)^2=4y+5\\\left(2y-3\right)^2=4x+5\end{matrix}\right.\Rightarrow\left(2y-3\right)^2-\left(2x-3\right)^2=4x-4y\Leftrightarrow\left(2y-2x\right)\left(2x+2y-6\right)=4\left(x-y\right)\Leftrightarrow4\left(y-x\right)\left(x+y-3\right)=4\left(x-y\right)\Leftrightarrow-4\left(x-y\right)\left(x+y-3\right)=4\left(x-y\right)\)

\(+,x=y\Rightarrow\left(2x-3\right)^2=4x+5\Leftrightarrow4x^2-12x+9=4x+5\Leftrightarrow4x^2-16x+4=0\Leftrightarrow x^2-4x+1=0\)

\(\Delta=16-4=12>0\Rightarrow\left[{}\begin{matrix}x=2+\sqrt{3}\\x=2-\sqrt{3}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=y=2+\sqrt{3}\left(tm\right)\\x=y=2-\sqrt{3}\left(tm\right)\end{matrix}\right.\)

\(+,x\ne y\Rightarrow-4\left(x+y-3\right)=4\Leftrightarrow x+y-3=-1\Leftrightarrow x+y=2\)

\(\Leftrightarrow x=2-y\Rightarrow\left(1-2y\right)^2=4y+5\Leftrightarrow1-4y+4y^2=4y+5\Leftrightarrow4y^2-8y-4=0\Leftrightarrow y^2-2y-1=0;\Delta=\left(-2\right)^2-\left(-1\right).1.4=4-\left(-4\right)=8>0\Rightarrow\left[{}\begin{matrix}x=1+\sqrt{2}\\x=1-\sqrt{2}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}y=1-\sqrt{2};x=1+\sqrt{2}\left(tm\right)\\x=1-\sqrt{2};y=1+\sqrt{2}\left(tm\right)\end{matrix}\right.\)

\(x=\frac{1}{2}\frac{\sqrt{\left(\sqrt{2}-1\right)^2}}{\sqrt{\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}}=\frac{1}{2}.\left(\sqrt{2}-1\right)\)

\(\Rightarrow2x=\sqrt{2}-1\Rightarrow2x+1=\sqrt{2}\)

\(\Rightarrow4x^2+4x+1=2\Rightarrow4x^2+4x-1=0\)

\(B=\left[x^3\left(4x^2+4x-1\right)-x\left(4x^2+4x-1\right)+4x^2+4x-1-1\right]^{2018}+2018\)

\(=\left(-1\right)^{2018}+2018=2019\)

ĐKXĐ: \(-\dfrac{1}{4}\le x\le3\)

\(\left(\sqrt{4x+1}-3\right)+\left(1-\sqrt{3-x}\right)+\left(4x^2-5x-6\right)=0\)

\(\Leftrightarrow\dfrac{4\left(x-2\right)}{\sqrt{4x+1}+3}+\dfrac{x-2}{1+\sqrt{3-x}}+\left(x-2\right)\left(4x+3\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(\dfrac{4}{\sqrt{4x+1}+3}+\dfrac{1}{1+\sqrt{3-x}}+4x+3\right)=0\)

Do \(\dfrac{4}{\sqrt{4x+1}+3}+\dfrac{1}{1+\sqrt{3-x}}+4x+3>0;\forall x\in\left[-\dfrac{1}{4};3\right]\)

\(\Rightarrow x-2=0\Rightarrow x=2\)

dat x+3=a ta co

\(\sqrt{8-a}+2\sqrt{a}=6\)

=>\(8-a=\left(6-2\sqrt{a}\right)^2\)

=>\(8-a=36-24\sqrt{a}+a\)

=>\(2a-24\sqrt{a}+24=0\)

=>tim a roi tim x

a) ( 4x - 2 )x + 5 = 0

=> \(\hept{\begin{cases}4x-2=0\\x+5=0\end{cases}}\)

+ TH1 : 4x - 2 = 0

             4x      = 2 

               x      = 0,5 ( vô lí nếu x c Z )

+ TH2 : x + 5 = 0

             x       = -5

=> x c \(\hept{ }\)0,5 ; -5 ( đóng ngoặc nhọn )

b) 2x - 9 = -8 -9 

    2x - 9 = -17

    2x      = -8

      x      = -4

c) 5( 3x + 8 ) - 7 . ( 2x + 3 ) = 16

    15x + 40 -  ( 14x + 21 )    = 16

     15x + 40 - 14x - 24         = 16

     ( 15x -14x ) + (40 - 24 )  = 16

     x + 16 = 16

     x = 0

Xin lỗi phần c mk hơi sai sót !!! Mk sẽ làm lại phần c :

c) 5( 3x + 8 ) - 7( 2x + 3 ) = 16

    15x + 40 - ( 14x + 21 ) = 16

    15x + 40 - 14x - 21       = 16

     ( 15x - 14x ) + ( 40 - 21 ) = 16

     x + 19 = 16

     => x = -3