a) 2x(x+1)−x2(x+2)+x3−x+4=02x(x+1)−x2(x+2)+x3−x+4=0

⇔2x2+2x−x3−2x2+x3−x+4=0⇔2x2+2x−x3−2x2+x3−x+4=0

⇔x+4=0⇔x+4=0

⇔x=−4⇔x=−4

Vậy ...

b) 4x(3x+2)−6x(2x+5)+21(x−1)=04x(3x+2)−6x(2x+5)+21(x−1)=0

⇔12x2+8x−12x2−30x+21x−21=0⇔12x2+8x−12x2−30x+21x−21=0

⇔−x−21=0⇔−x−21=0

⇔x=−21⇔x=−21

Vậy ...

c) 5x(12x+7)−3x(2x−5)=−1005x(12x+7)−3x(2x−5)=−100

⇔60x2+35x−6x2+15x+100=0⇔60x2+35x−6x2+15x+100=0

⇔54x2+50x+100=0⇔54x2+50x+100=0

⇔54(x2+2527x+6252916)+2909752916=0⇔54(x2+2527x+6252916)+2909752916=0

⇔54(x+2554)2+2909752916=0(ktm)⇔54(x+2554)2+2909752916=0(ktm)

Vậy phương trình vô nghiệm.

d) x(x−1)−x2+2x=5x(x−1)−x2+2x=5

⇔x2−x−x2+2x−5=0⇔x2−x−x2+2x−5=0

⇔x−5=0⇔x−5=0

⇔x=5⇔x=5

Vậy ...

e) 2x3(2x−3)−x2(4x2−6x+2)=02x3(2x−3)−x2(4x2−6x+2)=0

⇔4x4−6x3−4x3+6x3−2x2=0⇔4x4−6x3−4x3+6x3−2x2=0

⇔−2x2=0⇔−2x2=0

⇔x=0⇔x=0

Vậy x = 0

mình trả lời sai hả

A = ( x - 5 )( x² + 5x + 25 ) - x³ + 2 ( đã sửa )

= x³ - 5³ - x³ + 2

= x³ - 125 - x³ + 2

= -123 ( không phụ thuộc vào biến )

=> đpcm

B = ( 2x + 3 )( 4x² - 6x + 9 ) - 8x( x² + 2 ) + 16x + 5

= ( 2x )³ + 3³ - 8x³ - 16x + 16x + 5

= 8x³ + 27 - 8x³ - 16x + 16x + 5

= 27 + 5 = 32 ( không phụ thuộc vào biến )

=> đpcm

mình đang cần gấp

a)Xet hinh binh hanh ABCD co:

AB = DC va AB song song voi DC (t/c hinh binh hanh)

ma M la trung diem  AB, N la trung diem DC(gt)

=>AM=DN va AM song song voi DN

=>AMND la hinh binh hanh (t/g co 1 cap canh doi song song va bang nhau)

Ta co: AB=2AD(gt)

ma M la trung diem AD(gt)

=>AM=AD

=>AMND la hinh thoi (hinh binh hanh co 2 canh ke bang nhau)

Vì ABCD là hình thang ( AB // CD )

⇒\hept{ˆA+ˆB=180oˆC+ˆD=180o⇒\hept{A^+B^=180oC^+D^=180o

⇒\hept{ˆB=(180+50):2=165oˆC=165−50=95o⇒\hept{B^=(180+50):2=165oC^=165−50=95o

+) ˆA=13ˆDA^=13D^

⇒ˆD=3ˆA

a) \(\frac{x+1}{24}+\frac{x+2}{23}+\frac{x+3}{22}+\frac{x+4}{21}+4=0\)

\(\Leftrightarrow\frac{x+1}{24}+1+\frac{x+2}{23}+1+\frac{x+3}{22}+1+\frac{x+4}{21}+1=0\)

\(\Leftrightarrow\frac{x+25}{24}+\frac{x+25}{23}+\frac{x+25}{22}+\frac{x+25}{21}=0\)

\(\Leftrightarrow\left(x+25\right)\left(\frac{1}{24}+\frac{1}{23}+\frac{1}{22}+\frac{1}{21}\right)=0\)

\(\Leftrightarrow x+25=0\)

\(\Leftrightarrow x=-25\)

b) \(\frac{x+2}{498}+\frac{x+3}{497}=\frac{x+4}{496}+\frac{x+5}{495}\)

\(\Leftrightarrow\frac{x+2}{498}+1+\frac{x+3}{497}+1=\frac{x+4}{496}+1+\frac{x+5}{495}+1\)

\(\Leftrightarrow\frac{x+500}{498}+\frac{x+500}{497}=\frac{x+500}{496}+\frac{x+500}{495}\)

\(\Leftrightarrow x+500=0\)

\(\Leftrightarrow x=-500\)

(x+3)² - (x-3)(x+5)

= x² + 6x + 9 - (x² + 5x - 3x -15)

= x² + 6x +9 -x² - 2x +15

=(x² - x²) + (6x -2x) + (9+15)

= 4x + 24