a,2x(x-3)-3(x-3)=0 b,x²(x-1)+4(1-x)=0 c,2x(x-5)+(x-5)² d,(2x-1)=(4-3x)² e,2x(3-4X)-5x²(4x-3)=0

0 bình luận về “a,2x(x-3)-3(x-3)=0 b,x²(x-1)+4(1-x)=0 c,2x(x-5)+(x-5)² d,(2x-1)=(4-3x)² e,2x(3-4X)-5x²(4x-3)=0”

1. `\text{~~Holi~~}`

   `a. 2x(x-3)-3(x-3)=0`

   `-> (x-3)(2x-3)=0`

   `->`\(\left[ \begin{array}{l}x-3=0\\2x-3=0\end{array} \right.\) 

   `->`\(\left[ \begin{array}{l}x=3\\x=\dfrac{3}{2}\end{array} \right.\) 

   Vậy `S={3;3/2}`

   `b. x^2 (x-1)+4(1-x)=0`

   `-> x^2 (x-1)-4(x-1)=0`

   `-> (x-1)(x^2-4)=0`

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

   `->`\(\left[ \begin{array}{l}x-1=0\\x+2=0\\x-2=0\end{array} \right.\) 

   `->`\(\left[ \begin{array}{l}x=1\\x=-2\\x=2\end{array} \right.\) 

   Vậy `S={1;±2}`

   `c. 2x(x-5)+(x-5)^2=0` (bổ sung pt`=0`)

   `-> (x-5)(2x+x-5)=0`

   `-> (x-5)(3x-5)=0`

   `->`\(\left[ \begin{array}{l}x-5=0\\3x-5=0\end{array} \right.\) 

   `->`\(\left[ \begin{array}{l}x=5\\x=\dfrac{5}{3}\end{array} \right.\) 

   Vậy `S={5;5/3}`

   `d. (2x-1)=(4-3x)^2`

   `-> 2x-1=16-24x+9x^2`

   `-> 2x-1-16+24x-9x^2=0`

   `-> 26x-17-9x^2=0`

   `-> -26x+17+9x^2=0`

   `-> 9x^2-26x+17=0`

   `-> 9x^2-9x-17x+17x=0`

   `-> (x-1)(9x-7)=0`

   `->`\(\left[ \begin{array}{l}x-1=0\\9x-17=0\end{array} \right.\) 

   `->`\(\left[ \begin{array}{l}x=1\\x=\dfrac{17}{9}\end{array} \right.\) 

   Vậy `S={1;17/9}`

   `e. 2x(3-4x)-5x^2 (4x-3)=0`

   `-> 2x(3-4x)+5x^2 (3-4x)=0`

   `-> (3-4x)(2x+5x^2)=0`

   `-> (3-4x)x(2+5x)=0`

   `->`\(\left[ \begin{array}{l}3-4x=0\\x=0\\2+5x=0\end{array} \right.\) 

   `->`\(\left[ \begin{array}{l}x=\dfrac{3}{4}\\x=0\\x=-\dfrac{2}{5}\end{array} \right.\) 

   Vậy `S={3/4;0;-2/5}`

   Trả lời

2. Đáp án:

    

   Giải thích các bước giải:

    a,$2x(x-3)-3(x-3)=0$

   $(x-3)(2x-3)=0$

   \(\left[ \begin{array}{l}x-3=0\\2x-3=0\end{array} \right.\)

   \(\left[ \begin{array}{l}x=3\\x=\frac{3}{2}\end{array} \right.\)

   Vậy \(\left[ \begin{array}{l}x=3\\x=\frac{3}{2}\end{array} \right.\)

   b,$x^2(x-1)+4(1-x)=0$

   $x^2(x-1)-4(x-1)=0$

   $(x^2-4)(x-1)=0$

   \(\left[ \begin{array}{l}x-1=0\\x^2-4=0\end{array} \right.\)

   \(\left[ \begin{array}{l}x=1\\x=±2\end{array} \right.\)

   Vậy \(\left[ \begin{array}{l}x=1\\x=±2\end{array} \right.\)

   c,$2x(x-5)+(x-5)^2=0$

   $2x^2-10x+x^2-10x+25=0$

   $3x^2+25-20x=0$

   $(x-5)(3x-5)=0$

   \(\left[ \begin{array}{l}x-5=0\\3x-5=0\end{array} \right.\)

   \(\left[ \begin{array}{l}x=5\\x=\frac{5}{3}\end{array} \right.\)

   Vậy \(\left[ \begin{array}{l}x=5\\x=\frac{5}{3}\end{array} \right.\)

   d,$(2x-1)=(4-3x)^2$

   $2x-1=9x^2-24x+16$

   $2x-1-9x^2+24x-16=0$

   $9x^2+26x-17=0$

   $(x-1)(9x-7)=0$

   \(\left[ \begin{array}{l}x-1=0\\9x-7=0\end{array} \right.\)

   \(\left[ \begin{array}{l}x=1\\x=\frac{17}{9}\end{array} \right.\)

   Vậy \(\left[ \begin{array}{l}x=1\\x=\frac{17}{9}\end{array} \right.\)

   e,$2x(3-4x)-5x^2(4x-3)=0$

   $2x(3-4x)+5x^2(3-4x)=0$

   $x(2+5x^2)(3-4x)=0$

   \(\left[ \begin{array}{l}x=0\\3-4x=0;2-5x=0\end{array} \right.\)

   \(\left[ \begin{array}{l}x=0;x=\frac{2}{5}\\x=\frac{3}{4}\end{array} \right.\)

   Vậy \(\left[ \begin{array}{l}x=0;x=\frac{2}{5}\\x=\frac{3}{4}\end{array} \right.\)

   Trả lời