a: \(=25x^4-10x^3+5x^2\)

c: \(=2x^3-3x-5x^3-x^2+x^2=-3x^3-3x\)

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

a) \(\left(2x-5\right)^2-\left(2x+3\right)\left(2x-3\right)=10\Leftrightarrow\left(4x^2-20x+25\right)-\left(4x^2-9\right)-10=0\)

\(\Leftrightarrow-20x+24=0\Leftrightarrow x=\frac{6}{5}\)

b) \(\left(4x-1\right)\left(x+2\right)-\left(2x+3\right)^2-5\left(x-1\right)=9\Leftrightarrow-10x-15=0\)

\(\Leftrightarrow x=\frac{-3}{2}\)

c) \(\left(x+1\right)^3-\left(x-1\right)^3-2=6\Leftrightarrow\left(x^3+3x^2+3x+1\right)-\left(x^3-3x^2+3x-1\right)-8=0\)

\(\Leftrightarrow6x^2-6=0\Leftrightarrow x=\pm1\)

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

\(\Leftrightarrow\left(x^3+8\right)-\left(x^3+1\right)+3x+6=5\Leftrightarrow3x+8=0\Leftrightarrow x=\frac{-8}{3}\)

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

1) x² + x²y - y - 1

= x²( 1 + y ) - ( 1 + y )

= ( 1 + y )( x² - 1 )

= ( 1 + y )( x - 1 )( x + 1 )

2) x² + y² - 2xy - 25

= ( x² - 2xy + y² ) - 25

= ( x - y )² - 5²

= ( x - y - 5 )( x - y + 5 )

3) ( 2x - 1 )( x² + 2x - 1 ) - ( 1 - 2x )( x - 3 )

= ( 2x - 1 )( x² + 2x - 1 ) + ( 2x - 1 )( x - 3 )

= ( 2x - 1 )( x² + 2x - 1 + x - 3 )

= ( 2x - 1 )( x² + 3x - 4 )

= ( 2x - 1 )( x² - x + 4x - 4 )

= ( 2x - 1 )[ x( x - 1 ) + 4( x - 1 ) ]

= ( 2x - 1 )( x - 1 )( x + 4 )

4) a² + x² - 16 + 2ax

= ( a² + 2ax + x² ) - 16

= ( a + x )² - 4²

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

\(\text{a)}\Rightarrow x-1-x-1-x+2=5\)

\(\Rightarrow-x=5\)

\(\Rightarrow x=-5\)

     \(\text{Vậy x=-5}\)

\(\text{b)}\left(2x-1\right)^2-\left(2x+3\right)^2=7\)

\(\Rightarrow\left(4x^2-4x+1\right)-\left(4x^2+12x+9\right)=7\)

\(\Rightarrow4x^2-4x+1-4x^2-12x-9=7\)

\(\Rightarrow-16x-8=7\)

\(\Rightarrow-16x=15\)

\(\Rightarrow x=\frac{-15}{16}\)

      \(\text{Vậy }x=\frac{-15}{16}\)

\(\text{c)}\Rightarrow16x^2-9-\left(16x^2-8x+1\right)=8\)

\(\Rightarrow-9+8x-1=8\)

\(\Rightarrow8x=18\)

\(\Rightarrow x=\frac{18}{8}=\frac{9}{4}\)

      \(\text{Vậy }x=\frac{9}{4}\)

\(\text{Phần d số rất lẻ, có thể bạn chép sai đề nên mình ko chữa nha~}\)

a) (x + 5)² - x² = 45

<=> (x + 5 - x)(x + 5 + x) = 45

<=> 5(2x + 5) = 45

<=>> 2x + 5 = 9

<=> 2x = 4

<=> x = 2

Vậy S = {2}

b) (2x + 1)(1 - 2x) + (2x - 1)² = 22

<=> (2x - 1)(-2x - 1 + 2x - 1) = 22

<=> -2(2x - 1) = 22

<=> 2x - 1 = -11

<=> 2x = -10

<=> x = -5

c) (3x - 5)² - ((3x + 1)² = -3

<=> (3x - 5 - 3x - 1)(3x - 5 + 3x + 1) = -3

<=> -6(6x - 4) = -3

<=> 6x - 4 = 1/2

<=> 6x = 9/2

<=> x = 3/4

a,(x+5)²-x²=45

x²+10x+25-x²=45

10x+25=45

10x=20=> x=2

b,(2x+1)(1-2x)+(2x-1)²=22

(1+2x)(1-2x)+4x²-4x+1=22

1-4x²+4x²-4x+1=22

2-4x=22

4x=-20

x=-5