a) Ta có: \(\left(2x-3\right)^2=\left(2x-3\right)\left(x+1\right)\)

\(\Leftrightarrow\left(2x-3\right)^2-\left(2x-3\right)\left(x+1\right)=0\)

\(\Leftrightarrow\left(2x-3\right)\left(2x-3-x-1\right)=0\)

\(\Leftrightarrow\left(2x-3\right)\left(x-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-3=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=3\\x=4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=4\end{matrix}\right.\)

Vậy: \(S=\left\{\dfrac{3}{2};4\right\}\)

b) Ta có: \(x\left(2x-9\right)=3x\left(x-5\right)\)

\(\Leftrightarrow x\left(2x-9\right)-3x\left(x-5\right)=0\)

\(\Leftrightarrow x\left(2x-9\right)-x\left(3x-15\right)=0\)

\(\Leftrightarrow x\left(2x-9-3x+15\right)=0\)

\(\Leftrightarrow x\left(6-x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\6-x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=6\end{matrix}\right.\)

Vậy: S={0;6}

c) Ta có: \(3x-15=2x\left(x-5\right)\)

\(\Leftrightarrow3\left(x-5\right)-2x\left(x-5\right)=0\)

\(\Leftrightarrow\left(x-5\right)\left(3-2x\right)=0\)

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

Vậy: \(S=\left\{5;\dfrac{3}{2}\right\}\)

d) Ta có: \(\dfrac{5-x}{2}=\dfrac{3x-4}{6}\)

\(\Leftrightarrow6\left(5-x\right)=2\left(3x-4\right)\)

\(\Leftrightarrow30-6x=6x-8\)

\(\Leftrightarrow30-6x-6x+8=0\)

\(\Leftrightarrow-12x+38=0\)

\(\Leftrightarrow-12x=-38\)

\(\Leftrightarrow x=\dfrac{19}{6}\)

Vậy: \(S=\left\{\dfrac{19}{6}\right\}\)

e) Ta có: \(\dfrac{3x+2}{2}-\dfrac{3x+1}{6}=2x+\dfrac{5}{3}\)

\(\Leftrightarrow\dfrac{3\left(3x+2\right)}{6}-\dfrac{3x+1}{6}=\dfrac{12x}{6}+\dfrac{10}{6}\)

\(\Leftrightarrow6x+4-3x-1=12x+10\)

\(\Leftrightarrow3x+3-12x-10=0\)

\(\Leftrightarrow-9x-7=0\)

\(\Leftrightarrow-9x=7\)

\(\Leftrightarrow x=-\dfrac{7}{9}\)

Vậy: \(S=\left\{-\dfrac{7}{9}\right\}\)

a:Ta có: \(\sqrt{2x+9}=\sqrt{5-4x}\)

\(\Leftrightarrow2x+9=5-4x\)

\(\Leftrightarrow6x=-4\)

hay \(x=-\dfrac{2}{3}\left(nhận\right)\)

b: Ta có: \(\sqrt{2x-1}=\sqrt{x-1}\)

\(\Leftrightarrow2x-1=x-1\)

hay x=0(loại)

c: Ta có: \(\sqrt{x^2+3x+1}=\sqrt{x+1}\)

\(\Leftrightarrow x^2+3x=x\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(loại\right)\\x=-2\left(loại\right)\end{matrix}\right.\)

a. \(\sqrt{2x+9}=\sqrt{5-4x}\)

<=> 2x + 9 = 5 - 4x 

<=> 2x + 4x = 5 - 9

<=> 6x = -4

<=> x = \(\dfrac{-4}{6}=\dfrac{-2}{3}\)

d: Ta có: \(\sqrt{2x^2-3}=\sqrt{4x-3}\)

\(\Leftrightarrow2x^2-3=4x-3\)

\(\Leftrightarrow2x\left(x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(loại\right)\\x=2\left(nhận\right)\end{matrix}\right.\)

c) \(\dfrac{x}{x-2}+\dfrac{x}{x+2}=\dfrac{4x}{x^2-4}.ĐKXĐ:x\ne2;-2\)

<=>\(\dfrac{x\left(x+2\right)}{x^2-4}+\dfrac{x\left(x-2\right)}{x^2-4}=\dfrac{4x}{x^2-4}\)

<=>x²+2x+x²-2x=4x

<=>2x²-4x=0

<=>2x(x-2)=0

<=>\(\left[{}\begin{matrix}2x=0< =>x=0\\x-2=0< =>x=2\left(loại\right)\end{matrix}\right.\)

Vậy pt trên có nghiệm là S={0}

d) 11x-9=5x+3

<=>11x-5x=9+3

<=>6x=12

<=>x=2

Vậy pt trên có nghiệm là S={2}

e) (2x+3)(3x-4) =0

<=> \(\left[{}\begin{matrix}2x+3=0< =>x=\dfrac{-3}{2}\\3x-4=0< =>x=\dfrac{4}{3}\end{matrix}\right.\)

Vậy pt trên có tập nghiệm là S={\(\dfrac{-3}{2};\dfrac{4}{3}\)}

a) 5x+9 =2x

<=> 5x-2x=9

<=> 3x=9

<=> x=3

Vậy pt trên có nghiệm là S={3}

b) (x+1)(4x-3)=(2x+5)(x+1)

<=> (x+1)(4x-3)-(2x+5)(x+1)=0

<=>(x+1)(2x-8)=0

<=>\(\left[{}\begin{matrix}x+1=0< =>x=-1\\2x-8=0< =>2x=8< =>x=4\end{matrix}\right.\)

Vậy pt trên có tập nghiệm là S={-1;4}

c) 

<=>

<=>x²+2x+x²-2x=4x

<=>2x²-4x=0

<=>2x(x-2)=0

<=>

Vậy pt trên có nghiệm là S={0}

d) 11x-9=5x+3

<=>11x-5x=9+3

<=>6x=12

<=>x=2

Vậy pt trên có nghiệm là S={2}

e) (2x+3)(3x-4) =0

<=> 

Vậy pt trên có tập nghiệm là S={}

a) \(\left(x-2\right)^2=\left(x-4\right)\left(x+4\right)\) 

\(\Leftrightarrow x^2-4x+4-x^2+16=0\)

\(\Leftrightarrow20-4x=0\)

\(\Leftrightarrow4x=20\)

\(\Leftrightarrow x=5\)

Vậy S = {5}

b) ĐKXĐ: \(x\ne0;x\ne-2\)

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

\(\Leftrightarrow\dfrac{x+2}{x}=\dfrac{x^2+4x+x+4+x^2}{x\left(x+2\right)}\)

\(\Leftrightarrow\dfrac{x+2}{x}=\dfrac{2x^2+5x+4}{x\left(x+2\right)}\)

\(\Rightarrow x\left(x+2\right)^2=x\left(2x^2+5x+4\right)\)

\(\Leftrightarrow x^3+4x^2+4x=2x^3+5x^2+4x\)

\(\Leftrightarrow x^3+x^2=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x^2=0\\x+1=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=0\left(loại\right)\\x=-1\left(TM\right)\end{matrix}\right.\)

Vậy S = {-1}

c) Câu này mình không chắc về đề lắm! Bạn dùng ô chữ M bị ngược để viết lại đề nhé!

a) Ta có: \(\left(x-2\right)^2=\left(x-4\right)\left(x+4\right)\)

\(\Leftrightarrow x^2-4x+4=x^2-16\)

\(\Leftrightarrow x^2-4x+4-x^2+16=0\)

\(\Leftrightarrow-4x+20=0\)

\(\Leftrightarrow-4x=-20\)

hay x=5

Vậy: S={5}

a: \(27^{2-x}< =9\)

=>\(\left(3^3\right)^{2-x}< =3^2\)

=>\(3^{6-3x}< =3^2\)

=>6-3x<=2

=>-3x<=-4

=>\(x>=\dfrac{4}{3}\)

b: \(7^{3-x}< 49\)

=>\(7^{3-x}< 7^2\)

=>3-x<2

=>-x<2-3=-1

=>x>1

c: \(27^{3-x}>9\)

=>\(\left(3^3\right)^{3-x}>3^2\)

=>\(3^{9-3x}>3^2\)

=>9-3x>2

=>-3x>-7

=>\(x< \dfrac{7}{3}\)

d: \(2^{3-x}< 2^3\)

=>3-x<3

=>-x<0

=>x>0

e: \(27^{3-x^2}< 27^{x+1}\)

=>\(3-x^2< x+1\)

=>\(-x^2-x+2< 0\)

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

=>(x+2)(x-1)>0

=>\(\left[{}\begin{matrix}x>1\\x< -2\end{matrix}\right.\)

a. \(\dfrac{-3}{x^2-9}+\dfrac{5}{3-x}=\dfrac{2}{x+3}\)

<=> \(\dfrac{-3}{x^2-9}+\dfrac{-5}{x-3}=\dfrac{2}{x+3}\)

<=> \(\dfrac{-3}{x^2-9}+\dfrac{-5\left(x+3\right)}{x^2-9}=\dfrac{2\left(x-3\right)}{x^2-9}\)

<=> \(-3+\left(-5\right)\left(x+3\right)=2\left(x-3\right)\)

<=> -3 + (-5x) + (-15) = 2x - 6

<=> -5x -2x = 15 - 6 + 3

<=> -7x = 12

<=> x = \(\dfrac{-12}{7}\)

Vậy ........

b. \(\left|x+5\right|=2x-1\)

Nếu x \(\ge\) -5 => \(\left|x+5\right|\) = x + 5

Nếu x < -5 => \(\left|x+5\right|\) = -(x + 5)

TH1: Nếu x \(\ge\) -5

<=> x + 5 = 2x - 1

<=> x - 2x = -1 - 5

<=> -x = -6 

<=> x = 6

TH2: Nếu x < -5 

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

<=> -x - 5 = 2x - 1

<=> -5 + 1 = 2x + x

<=> -4 = 3x

<=> x = \(\dfrac{-4}{3}\)

Vậy .........

c. Bạn tự giải câu này nhé (có thể tách các hạng tử rồi tính)

a. 3(x-2)-10=5(2x + 1)

<=> 3x - 6 - 10 = 10x + 5

<=> 3x - 10x = 5 + 6 + 10

<=> -7x = 21

<=> x = -3

b. 3x + 2=8 -2(x-7)

<=> 3x + 2 = 8 - 2x + 14

<=> 3x + 2x = 8 + 14 - 2

<=> 5x = 20

<=> x = 4

c. 2x-(2+5x)= 4(x + 3)

<=> 2x - 2 - 5x = 4x + 12

<=> 2x - 5x - 4x = 12 + 2

<=> -7x = 14

<=> x = -2

d. 5-(x +8)=3x + 3(x-9)

<=> 5 - x - 8 = 3x + 3x - 27

<=> -x - 3x - 3x = -27 + 8 - 5

<=> -7x = -24

<=> x = 24/7

e. 3x - 18 + x= 12-(5x + 3)

<=> 3x - 18 + x = 12 - 5x - 3

<=> 3x + x - 5x = 12 - 3 + 18

<=> -x = 27

<=> x = - 27

a. 3(x-2)-10=5(2x + 1)

<=> 3x - 6 - 10 = 10x + 5

<=> 3x - 10x = 5 + 6 + 10

<=> -7x = 21

<=> x = -3

b. 3x + 2=8 -2(x-7)

<=> 3x + 2 = 8 - 2x + 14

<=> 3x + 2x = 8 + 14 - 2

<=> 5x = 20

<=> x = 4

c. 2x-(2+5x)= 4(x + 3)

<=> 2x - 2 - 5x = 4x + 12

<=> 2x - 5x - 4x = 12 + 2

<=> -7x = 14

<=> x = -2

d. 5-(x +8)=3x + 3(x-9)

<=> 5 - x - 8 = 3x + 3x - 27

<=> -x - 3x - 3x = -27 + 8 - 5

<=> -7x = -24

<=> x = 24/7

e. 3x - 18 + x= 12-(5x + 3)

<=> 3x - 18 + x = 12 - 5x - 3

<=> 3x + x - 5x = 12 - 3 + 18

<=> -x = 27

<=> x = - 27

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