ĐKXĐ: \(\left\{{}\begin{matrix}2x+5>=0\\4-2x>=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}2x>=-5\\2x< =4\end{matrix}\right.\Leftrightarrow-\dfrac{5}{2}< =x< =2\)

\(x^2+\sqrt{2x+5}+\sqrt{4-2x}=4x-1\)

=>\(x^2-4+\sqrt{2x+5}-3+\sqrt{4-2x}=4x-1-7\)

=>\(\left(x-2\right)\left(x+2\right)+\dfrac{2x+5-9}{\sqrt{2x+5}+3}+\sqrt{4-2x}=4x-8\)

=>\(\left(x-2\right)\left[\left(x+2\right)+\dfrac{2}{\sqrt{2x+5}+3}-4\right]+\sqrt{4-2x}=0\)

=>\(-\left(2-x\right)\left[\left(x-2\right)+\dfrac{2}{\sqrt{2x+5}+3}\right]+\sqrt{2\left(2-x\right)}=0\)

=>\(\sqrt{2-x}\left[-\sqrt{2-x}\left(x-2+\dfrac{2}{\sqrt{2x+5}+3}\right)+\sqrt{2}\right]=0\)

=>\(\sqrt{2-x}=0\)

=>x=2(nhận)

(2x^2-3x+1)(2x^2+5x+1)=9x^2

<=> (2x^2+5x+1- 8x)(2x^2 +5x+1)=9x^2

<=> (2x^2+5x+1)^2 -8x(2x^2+5x+1)=9x^2

<=>  (2x^2+5x+1)^2 -2*(4x)*(2x^2+5x+1)=9x^2

<=>  (2x^2+5x+1)^2 -2*(4x)*(2x^2+5x+1)+(4x)^2=9x^2+16x^2

<=> (2x^2+5x+1 - 4x)^2=25x^2

<=> (2x^2+x+1)^2=25x^2

<=> (2x^2+x+1)^2 - 25x^2 =0

<=>(2x^2+x+1-5x)(2x^2+x+1+5x)=0

<=>(2x^2-4x+1)(2x^2+6x+1)=0

<=> (2x^2-4x+1)=0 => 2( x^2 - 2x + 1/2)=0

                                <=> x^2-2x +1/2 =0

                                <=> (x^2-2x+1) -1/2 =0

                                <=> (x-1)^2 =1/2     =>  x-1 =căn(1/2)  => x=căn(1/2)+1

                                                              => x-1=-(căn(1/2)) => x=- (căn(1/2)) +1

Hoặc  2x^2 +6x +1=0 

         <=> x^2 + 3x +1/2 =0                

         <=> (x^2 + 2*(1.5)x + (1.5)^2) -(1.5)^2+1/2 =0

         <=> (x+1.5)^2 - 7/4 =0

         <=> (x+1.5)^2 = 7/4    =>        x+1.5 = căn(7/4) => x=căn(7/4) -1.5

                                           =>      x+1.5 =- căn(7/4) => x=-căn(7/4) -1.5

nhớ thanks bạn (+_+)

Ta có: \(\hept{\begin{cases}\left(\frac{1}{x}+y\right)+\left(\frac{1}{x}-y\right)=\frac{5}{8}\\\left(\frac{1}{x}+y\right)-\left(\frac{1}{x}-y\right)=-\frac{3}{8}\end{cases}\Leftrightarrow\hept{\begin{cases}\frac{2}{x}=\frac{5}{8}\\2y=-\frac{3}{8}\end{cases}\Leftrightarrow}\hept{\begin{cases}x=\frac{16}{5}\\y=-\frac{3}{16}\end{cases}}}\)

a: 3x-4=0

=>3x=4

hay x=4/3

b: (x+2)(2x-3)=0

=>x+2=0 hoặc 2x-3=0

=>x=-2 hoặc x=3/2

x={1/3; 3/2; -5}

Giải phương trình:

a) (x+2)³ - (x-2)³ = 12x(x-1) - 8

<=> (x² + 3.x².2 + 3.x.2² + 2³) - (x² - 3.x².2 + 3.x.2² - 2³) - [12x(x-1) - 8] = 0

<=> (x³ + 6x² + 12x + 8) - (x³ - 6x² + 12x - 8) - (12x² - 12x - 8) = 0

<=> x³ + 6x² + 12x + 8 - x³ + 6x^{2 -} 12x + 8 - 12x² + 12x + 8 = 0

<=> 12x +32 = 0

<=> x =  \(\frac{-32}{12}\) = \(-2\frac{2}{3}\)         

                                                 Vậy phương trình có nghiệm duy nhất là  \(-2\frac{2}{3}\)

b) (3x-1)² - 5(2x+1)² + (6x-3)(2x+1) = (x-1)²

<=> (9x² - 6x + 1) - 5(4x² + 4x + 1) + 3(2x - 1)(2x + 1) - (x² - 2x +1) = 0

<=> 9x² - 6x + 1 - 20x² - 20x - 5 + 3(4x² - 1) - x² + 2x -1 = 0

<=> 9x² - 6x + 1 - 20x² - 20x - 5 + 12x² - 3 - x² + 2x -1 = 0

<=> -24x - 8 = 0

<=> x = \(\frac{-8}{24}\) = \(\frac{-1}{3}\)  

                  Vậy phương trình có nghiệm duy nhất là \(\frac{-1}{3}\)

\(a,2x\left(x-5\right)+4\left(x-5\right)=0\\ \Leftrightarrow\left(x-5\right)\left(2x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-5=0\\2x+4=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=5\\2x=-4\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=5\\x=-2\end{matrix}\right.\)

Vậy \(x\in\left\{5;-2\right\}\)

\(b,3x-15=2x\left(x-5\right)\\ \Leftrightarrow3\left(x-5\right)-2x\left(x-5\right)=0\\ \Leftrightarrow\left(x-5\right)\left(-2x+3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-5=0\\-2x+3=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=5\\2x=3\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=5\\x=\dfrac{3}{2}\end{matrix}\right.\)

Vậy \(x\in\left\{5;\dfrac{3}{2}\right\}\)

\(c,\left(2x+1\right)\left(3x-2\right)=\left(5x-8\right)\left(2x+1\right)\\ \Leftrightarrow\left(2x+1\right)\left(3x-2\right)-\left(5x-8\right)\left(2x+1\right)=0\\ \Leftrightarrow\left(2x+1\right)\left(3x-2-5x+8\right)=0\\ \Leftrightarrow\left(2x+1\right)\left(-2x+6\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}2x+1=0\\-2x+6=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}2x=-1\\2x=6\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=3\end{matrix}\right.\)

Vậy \(x\in\left\{-\dfrac{1}{2};3\right\}\)

Câu d xem lại đề

\(2\left(3x-2\right)+\left(x-3\right)^2=0\)

\(\Rightarrow2\left(3x-2\right)=\left(x-3\right)^2\)

\(\Rightarrow6x-4=x^2-9\)

\(\Rightarrow6x-x^2=4-9\)

\(\Rightarrow6x-x^2=-5\)

\(\Rightarrow...\)

pn tự lm nka, mk ms lp 7 ò

\(\Leftrightarrow6x-4+x^2-6x+9=0\)

\(\Leftrightarrow x^2+5=0\)

\(\Leftrightarrow x^2=-5\)(vô lý)

Vậy ptrình vô nghiệm