a) \(\Leftrightarrow x^2-4x-x^2+6x-9=0\\ \Leftrightarrow2x=9\\ \Leftrightarrow x=4,5\)

b) \(\Leftrightarrow x^2-3x-10=0\\ \Leftrightarrow\left(x^2+2x\right)-\left(5x+10\right)=0\\ \Leftrightarrow x\left(x+2\right)-5\left(x+2\right)=0\\ \left(x-5\right)\left(x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=5\\x=-2\end{matrix}\right.\)

c) \(\Leftrightarrow\left(2x-3-7\right)\left(2x-3+7\right)=0\\ \Leftrightarrow\left(2x-10\right)\left(2x+4\right)=0\\ \Leftrightarrow\left(x-5\right)\left(x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=5\\x=-2\end{matrix}\right.\)

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

a) (x-2)³+6(x+1)²-x³+12=0

\(\Rightarrow\)x³-6x²+12x-8+6(x²+2x+1)-x³+12=0

\(\Rightarrow\)x³-6x²+12x-8+6x²+12x+6-x³+12=0

\(\Rightarrow\)24x+10=0

\(\Rightarrow\)24x=-10

\(\Rightarrow\)x=\(\dfrac{-10}{24}=\dfrac{-5}{12}\)

b)(x-5)(x+5)-(x+3)²+3(x-2)²=(x+1)²-(x-4)(x+4)+3x²

\(\Rightarrow\)x²-25-(x²+6x+9)+3(x²-4x+4)=x²+2x+1-(x²-16)+3x²

\(\Rightarrow\)x²​-25-x²-6x-9+3x²-12x+12=x²+2x+1-x²+16+3x²

\(\Rightarrow\)3x²-18x-22=3x²+2x+17

\(\Rightarrow\)3x²-18x-22-3x²-2x-17=0

\(\Rightarrow\)-20x-39=0

\(\Rightarrow\)-20x=39

\(\Rightarrow\)x=\(-\dfrac{39}{20}\)

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

⇒4x²+12x+9+x²-1=5(x²+4x+4)-(x²+x-5x-5)+x+4

⇒5x²+12x+8=5x²+20x+20-x²-x+5x+5+x+4

⇒5x²+12x+8-5x²-20x-20+x²+x-5x-5-x-4=0

⇒x²-13x-21=0

Bài 1.

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

     x²-16-x²-2x = 0

     -16 - 2x = 0

             2x = -16 

               x = -16/2

               x = -8

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

     (3x-1)(x-2)=0

=> x ∈ {1/3 ; 2 }

c) 6x - 12x² = 0

    6x(1-2x) = 0

=> x ∈ {0; 1/2 }

d) mình thấy có vẻ hơi sai đề nên mình ko giải được, bạn thông cảm nha

d/ 4x (3 - ¹/₄ x) + (x -2) ( x+ 2)

câu d bị  sai đề 

a

\(x^2\left(2x+15\right)+4\left(2x+15\right)=0\\ \Leftrightarrow\left(2x+15\right)\left(x^2+4\right)=0\\ \Leftrightarrow2x+15=0\left(x^2+4>0\forall x\right)\\ \Leftrightarrow2x=-15\\ \Leftrightarrow x=-\dfrac{15}{2}\)

b

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

c

\(2\left(x+3\right)-x^2-3x=0\\ \Leftrightarrow2\left(x+3\right)-\left(x^2+3x\right)=0\\ \Leftrightarrow2\left(x+3\right)-x\left(x+3\right)=0\\ \Leftrightarrow\left(x+3\right)\left(2-x\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x+3=0\\2-x=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0-3=-3\\x=2-0=2\end{matrix}\right.\)

a: =>(2x+15)(x^2+4)=0

=>2x+15=0

=>2x=-15

=>x=-15/2

b; =>(x-2)(5x-3)=0

=>x=2 hoặc x=3/5

c: =>(x+3)(2-x)=0

=>x=2 hoặc x=-3

a: \(\Leftrightarrow\left(x+2\right)\left(12-x\right)=0\)

\(\Leftrightarrow x\in\left\{-2;12\right\}\)

b: \(\Leftrightarrow\left(2x+5\right)\left(x-1\right)=0\)

\(\Leftrightarrow x\in\left\{-\dfrac{5}{2};1\right\}\)

\(a,=3x-9-4x+12=-x+3=0\)

\(\Leftrightarrow x=3\)

Vậy ..

\(b,=\left(x+2\right)\left(x+2-x+2\right)=4\left(x+2\right)=0\)

\(\Leftrightarrow x+2=0\)

\(\Leftrightarrow x=-2\)

Vậy ..

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

\(\Leftrightarrow x=1\)

Vậy ..

\(d,\Leftrightarrow x\left(x-2\right)\left(x+2\right)=0\)

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

Vậy ..

\(e,=\left(2x-3-5\right)\left(2x-3+5\right)=\left(2x-8\right)\left(2x+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{8}{2}=4\\x=-\dfrac{2}{2}=-1\end{matrix}\right.\)

Vậy ...

a) Ta có: 3(x-3)-4x+12=0

\(\Leftrightarrow3\left(x-3\right)-4\left(x-3\right)=0\)

\(\Leftrightarrow x-3=0\)

hay x=3

Vậy: S={3}

b) Ta có: \(\left(x+2\right)^2-\left(x+2\right)\left(x-2\right)=0\)

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

\(\Leftrightarrow4x=-8\)

hay x=-2

Vậy: S={-2}

c) Ta có: \(x^3+3x=3x^2+1\)

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

\(\Leftrightarrow x-1=0\)

hay x=1

Vậy: S={1}

d) Ta có: \(\dfrac{2}{3}x\left(x^2-4\right)=0\)

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

Vậy: S={0;2;-2}

a) 3.(x-3)-4x+12=0

=> 3x - 9 - 4x + 12 = 0

=> -x + 3 = 0

=> x = 3

b) (x+2)^2-(x+2).(x-2) =0

\(\Rightarrow\left(x+2\right)^2-x^2+4=0\)

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

=> 4x + 8 = 0

=> x = -2

c) x^3+3x=3x^2+1

\(\Rightarrow x^3+3x-3x^2-1=0\)

\(\Rightarrow\left(x-1\right)^3=0\)

=> x = 1

d) \(\dfrac{2}{3}x\left(x^2-4\right)=0\)

\(\Rightarrow\dfrac{2}{3}x\left(x-2\right)\left(x+2\right)=0\)

=> x = 0 hoặc x = 2 hoặc x = -2

e) \(\left(2x-3\right)^2-5^2=0\)

\(\Rightarrow\left(2x-8\right)\left(2x+2\right)=0\)

=> x = 4 hoăc x = -1

a) Ta có: \(7x^2-28=0\)

\(\Leftrightarrow7\left(x^2-4\right)=0\)

\(\Leftrightarrow7\left(x-2\right)\left(x+2\right)=0\)

mà 7>0

nên (x-2)(x+2)=0

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

Vậy: \(x\in\left\{2;-2\right\}\)

b) Ta có: \(\dfrac{2}{3}x\left(x^2-4\right)=0\)

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

mà \(\dfrac{2}{3}>0\)

nên x(x-2)(x+2)=0

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

Vậy: \(x\in\left\{0;-2;2\right\}\)

c) Ta có: \(2x\left(3x-5\right)-\left(5-3x\right)=0\)

\(\Leftrightarrow2x\left(3x-5\right)+\left(3x-5\right)=0\)

\(\Leftrightarrow\left(3x-5\right)\left(2x+1\right)=0\)

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

Vậy: \(x\in\left\{\dfrac{5}{3};-\dfrac{1}{2}\right\}\)

d) Ta có: \(\left(2x-1\right)^2-25=0\)

\(\Leftrightarrow\left(2x-1-5\right)\left(2x-1+5\right)=0\)

\(\Leftrightarrow\left(2x-6\right)\left(2x+4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-6=0\\2x+4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=6\\2x=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)

Vậy: \(x\in\left\{3;-2\right\}\)

a,7x² - 28 = 0

=> 7x² = 28 => x² = 4 => x = 2

b,2/3x(x² - 4) = 0

=>2/3x(x - 2)(x + 2) = 0

=> x ∈ {0 ; 2 ; -2}

c,2x(3x - 5) - (5 - 3x) = 0

= 2x(3x - 5) + (3x - 5)

= (3x - 5)(2x + 1) = 0

=> x ∈ { 5/3 ; -1/2}

d, (2x - 1)² - 25 = 0

=> (2x - 4)(2x - 6) = 0

=> x ∈ {2 ;3}

a,7x² - 28 = 0

=> 7x² = 28 => x² = 4 => x = 2

b,2/3x(x² - 4) = 0

=>2/3x(x - 2)(x + 2) = 0

=> x ∈ {0 ; 2 ; -2}

c,2x(3x - 5) - (5 - 3x) = 0

= 2x(3x - 5) + (3x - 5)

= (3x - 5)(2x + 1) = 0

=> x ∈ { 5/3 ; -1/2}

d, (2x - 1)² - 25 = 0

=> (2x - 4)(2x - 6) = 0

=> x ∈ {2 ;3}

trình bày đầy đủ các bước

a: Ta có: \(\left(x+1\right)^2-3\left(x+1\right)=0\)

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

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

b: Ta có: \(2\left(3x-2\right)^2=9x^2-4\)

\(\Leftrightarrow2\left(3x-2\right)^2-\left(3x-2\right)\left(3x+2\right)=0\)

\(\Leftrightarrow\left(3x-2\right)\left(6x-4-3x-2\right)=0\)

\(\Leftrightarrow\left(3x-2\right)\left(3x-6\right)=0\)

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

a)(2x-3)(x+5)=0

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

Vậy x=3/2 hoặc x=-5

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

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

Vậy phương trình có tập nghiệm là: \(S=\left\{\dfrac{3}{2};-5\right\}\)

b) \(3x\left(x-2\right)-7\left(x-2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(3x-7\right)=0\)

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

Vậy phương trình có tập nghiệm là: \(S=\left\{2;\dfrac{7}{2}\right\}\)

c) \(5x\left(2x-3\right)-6x+9=0\)

\(\Leftrightarrow5x\left(2x-3\right)-3\left(2x-3\right)=0\)

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

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

Vậy phương trình có tập nghiệm là: \(S=\left\{\dfrac{3}{2};\dfrac{3}{5}\right\}\)

a: Ta có: \(\left(2x-3\right)\left(x+5\right)=0\)

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

b: Ta có: \(3x\left(x-2\right)-7\left(x-2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(3x-7\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{7}{3}\end{matrix}\right.\)

c: Ta có: \(5x\left(2x-3\right)-6x+9=0\)

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

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