c(x-1)^2=4

x^2-2x+1=4

x^2-2x+1-4=0

x^2-2x-3=0

x^2-3x+x-3=0

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

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

\(\Rightarrow\hept{\begin{cases}x-3=0\\x+1=0\end{cases}\Rightarrow\hept{\begin{cases}x=3\\x=-1\end{cases}}}\)

d, x^3+2x^2-x-2=0

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

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

\(\Rightarrow\hept{\begin{cases}x=-2\\x=+-1\end{cases}}\)

e, (3x+2)^2-(2x-1)^2=0

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

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

x+3=0=>x=-3

5x-1=0=>5x=1=>x=1/5

bn kiểm tra giúp mk đề 2 câu cuối , mk làm ko ra

Khó quá,hay quá !!!

Tìm x:

1. \(25x^2-20x+4=0\)

⇔ \(\left(5x-2\right)^2=0\)

⇔ \(5x-2=0\)

⇔ \(5x=2\)

⇔ \(x=\dfrac{2}{5}\)

⇒ S = \(\left\{\dfrac{2}{5}\right\}\)

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

⇔ \(4x^2-12x+9-\left(4x^2-1\right)=0\)

⇔ \(4x^2-12x+9-4x^2+1=0\)

⇔ \(-12x+10=0\)

⇔ \(-12x=-10\)

⇔ \(x=\dfrac{5}{6}\)

⇒ S \(=\left\{\dfrac{5}{6}\right\}\)

3. \(\left(\dfrac{1}{2}x-1\right)\left(\dfrac{1}{2}x+1\right)-\left(\dfrac{1}{2}x-1\right)^2=0\)

⇔ \(\dfrac{1}{4}x^2-1-\left(\dfrac{1}{4}x^2-x+1\right)=0\)

⇔ \(\dfrac{1}{4}x^2-1-\dfrac{1}{4}x^2+x-1=0\)

⇔ \(-2+x=0\)

⇔ \(x=2\)

⇒ S \(=\left\{2\right\}\)

4. \(\left(2x-3\right)^2+\left(2x+5\right)^2=8\left(x+1\right)^2\)

⇔ \(4x^2-12x+9+4x^2+20x+25=8\left(x^2+2x+1\right)\)

⇔ \(8x^2+8x+34=8x^2+16x+8\)

⇔ \(8x+34=16x+8\)

⇔ \(8x-16x=8-34\)

⇔ \(-8x=-26\)

⇔ \(x=\dfrac{13}{4}\)

⇒ S \(=\left\{\dfrac{13}{4}\right\}\)

5.\(4x^2+12x-7=0\)

⇔ \(4x^2+14x-2x-7=0\)

⇔ \(2x\left(2x+7\right)-\left(2x+7\right)=0\)

⇔ \(\left(2x+7\right)\left(2x-1\right)=0\)

⇔ \(\left[{}\begin{matrix}2x+7=0\\2x-1=0\end{matrix}\right.\)

⇔ \(\left[{}\begin{matrix}x=\dfrac{-7}{2}\\x=\dfrac{1}{2}\end{matrix}\right.\)

⇒ S \(=\left\{\dfrac{-7}{2};\dfrac{1}{2}\right\}\)

6. \(\dfrac{1}{4}x^2+\dfrac{2}{3}x-\dfrac{5}{9}=0\)

⇔ \(9x^2+24x-20=0\)

⇔ \(9x^2+30x-6x-20=0\)

⇔ \(3x\left(3x+10\right)-2\left(3x+10\right)=0\)

⇔ \(\left(3x+10\right)\left(3x-2\right)=0\)

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

⇔ \(\left[{}\begin{matrix}x=\dfrac{-10}{3}\\x=\dfrac{2}{3}\end{matrix}\right.\)

⇒ S \(=\left\{\dfrac{-10}{3};\dfrac{2}{3}\right\}\)

7. \(24\dfrac{8}{9}-\dfrac{1}{4}x^2-\dfrac{1}{3}x=0\)

⇔ \(\dfrac{224}{9}-\dfrac{1}{4}x^2-\dfrac{1}{3}x=0\)

⇔ \(896-9x^2-12x=0\)

⇔ \(-896+9x^2+12x=0\)

⇔ \(9x^2+12x-896=0\)

⇔ \(9x^2-84x+96x-896=0\)

⇔ \(3x\left(3x-28\right)+32\left(3x-28\right)=0\)

⇔ \(\left(3x-28\right)\left(3x+32\right)=0\)

⇔ \(\left[{}\begin{matrix}3x-28=0\\3x+32=0\end{matrix}\right.\)

⇔ \(\left[{}\begin{matrix}x=\dfrac{28}{3}\\x=\dfrac{-32}{3}\end{matrix}\right.\)

⇒ S \(=\left\{\dfrac{-32}{3};\dfrac{28}{3}\right\}\)

* a mũ 2 hay 4 hay 6 ,... ( những số tự nhiên chẵn khác 0 ) đều lớn hơn hoặc bằng 0 với mọi a

Áp dụng :

a) (2x-8)^4 + (3y+45)^2 = 0

Vì : (2x-8)^4 >=0 , (3y+45)^2 >=0 với mọi x,y

=> (2x-8)^4 + (3y+45)^2 >=0

Dấu "=" xảy ra khi : 2x-8=3y+45=0

->(x;y)=(4;-15)

Những câu sau làm tương tự, ta được :

b) ...

Dấu "=" xảy ra khi : 2x-10=0 và x+y-7=0

->x=5 và 5+y-7=0

->(x;y)=(5;2)

c) 5x-15=0 và 2x-y+4=0

->x=3 và 6-y+4=0

->(x;y)=(3;10)

d) Trùng câu a

a)x=4,y=-15

b)x=5,y=2

còn câu c) mik chịu 

\(a.2x-3=4x+6\) 

\(\Leftrightarrow2x-3-4x-6=0\) 

\(\Leftrightarrow-2x-9=0\)

\(\Leftrightarrow x=\dfrac{9}{2}\)

\(S=\left\{\dfrac{9}{2}\right\}\) 

\(b.x\left(x-1\right)+x\left(x+3\right)=0\) 

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

\(\Leftrightarrow2x^2+2=0\)

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

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

\(S=\left\{0,-1\right\}\) 

Mấy câu khác bn gửi lại đc ko tại mik chx hiểu lắm

a: =>-2x=9

=>x=-9/2

c: =>x(x-1+x+3)=0

=>x(2x+2)=0

=>x=0 hoặc x=-1

\(a,2x-3=4x+6\)

\(\Leftrightarrow2x-4x=6+3\)

\(\Leftrightarrow-2x=9\)

\(\Leftrightarrow x=-\dfrac{9}{2}\)

\(b,\) Ghi vậy mình không làm được.

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

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)

\(d,\dfrac{x}{2x-6}-\dfrac{x}{2x+2}=\dfrac{2x}{\left(x+1\right)\left(x-3\right)}\)

\(\Leftrightarrow\dfrac{x}{2\left(x-3\right)}-\dfrac{x}{2\left(x+1\right)}-\dfrac{2}{\left(x+1\right)\left(x-3\right)}=0\left(dkxd:x\ne-1;x\ne3\right)\)

\(\Leftrightarrow\dfrac{x\left(x+1\right)-x\left(x-3\right)-2.2}{2\left(x+1\right)\left(x-3\right)}=0\)

\(\Leftrightarrow x^2+x-x^2+3x-4=0\)
\(\Leftrightarrow4x-4=0\)

\(\Leftrightarrow x=1\left(tmdk\right)\)

Vậy \(S=\left\{1\right\}\)

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

\(\Leftrightarrow x^2-2x-x^3+4x^2-3x=0\)

\(\Leftrightarrow-x^3+5x^2-5x=0\)

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

\(\Leftrightarrow\orbr{\begin{cases}-x=0\\x^2-5x+5=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\\left(x-\frac{5}{2}\right)^2-\frac{5}{4}=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\\left(x-\frac{5}{2}\right)^2=\frac{5}{4}\end{cases}\Leftrightarrow}\hept{\begin{cases}x=0\\x-\frac{5}{2}=\frac{\sqrt{5}}{2}\\x-\frac{5}{2}=-\frac{\sqrt{5}}{2}\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x=0\\x=\frac{5+\sqrt{5}}{2}\\x=\frac{5-\sqrt{5}}{2}\end{cases}}\)

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

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

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

\(\Leftrightarrow x\left(x-\frac{5+\sqrt{5}}{2}\right)\left(x-\frac{5-\sqrt{5}}{2}\right)=0\)

=> \(x\in\left\{0;\frac{5+\sqrt{5}}{2};\frac{5-\sqrt{5}}{2}\right\}\)

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

\(\Leftrightarrow2x^2+x-15-2x^2-x+3=0\)

\(\Leftrightarrow-12=0\left(vn\right)\)

c) \(\left(x-2\right)\left(x^2+2x+8\right)-x^3-2x+1=0\)

\(\Leftrightarrow x^3+4x-16-x^3-2x+1=0\)

\(\Leftrightarrow2x=15\)

\(\Rightarrow x=\frac{15}{2}\)

a) x( x - 2 ) - x( x - 1 )( x - 3 ) = 0

<=> x² - 2x - x( x² - 4x + 3 ) = 0

<=> x² - 2x - x³ + 4x² - 3x = 0

<=> -x³ + 5x² - 5x = 0

<=> -x( x² - 5x + 5 ) = 0

<=> \(\orbr{\begin{cases}-x=0\\x^2-5x+5=0\end{cases}}\)

+) -x = 0 => x = 0

+) x² - 5x + 5 = 0 (*)

\(\Delta=b^2-4ac=\left(-5\right)^2-4\cdot1\cdot5==25-20=5\)

\(\Delta>0\)nên (*) có hai nghiệm phân biệt

\(\hept{\begin{cases}x_1=\frac{-b+\sqrt{\Delta}}{2a}=\frac{5+\sqrt{5}}{2}\\x_2=\frac{-b-\sqrt{\Delta}}{2a}=\frac{5-\sqrt{5}}{2}\end{cases}}\)

Vậy S = { \(0;\frac{5\pm\sqrt{5}}{2}\)}

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

<=> 2x² + x - 15 - ( 2x² + x - 3 ) = 0

<=> 2x² + x - 15 - 2x² - x + 3 = 0

<=> -12 = 0 ( vô lí )

Vậy phương trình vô nghiệm

c) ( x - 2 )( x² + 2x + 8 ) - x³ - 2x + 1 = 0

<=> x³ + 4x - 16 - x³ - 2x + 1 = 0

<=> 2x - 15 = 0

<=> 2x = 15

<=> x = 15/2