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a) \(81-\left(3x+2\right)^2=9^2-\left(3x+2\right)^2=\left(9-3x-2\right)\left(9+3x+2\right)=\left(7-3x\right)\left(11+3x\right)\)

b) \(\left(7x-4\right)^2-\left(2x+1\right)^2=\left(7x-4-2x-1\right)\left(7x-4+2x+1\right)\)

\(=\left(5x-5\right)\left(9x-3\right)=15\left(x-1\right)\left(3x-1\right)\)

c) \(9\left(x-5y\right)^2-16\left(x+y\right)^2=\left[3\left(x-5y\right)-4\left(x+y\right)\right]\left[3\left(x-5y\right)+4\left(x+y\right)\right]\)

\(=\left(-x-19y\right)\left(7x-11y\right)\)

`(2x-5)^2-64=(2x-5)^2-8^2=(2x-5+8)(2x-5-8)=(2x+3)(2x-13)`

k cho mik

Kẻ BH ⊥ CD

Ta có: AD ⊥ CD ( Vì ABCD là hình thang vuông có  ∠ A =  ∠ D = 90 0  )

Suy ra: BH // AD

Hình thang ABHD có hai cạnh bên song song nên HD = AB và BH = AD

AB = AD = 2cm (gt)

⇒ BH = HD = 2cm

CH = CD – HD = 4 – 2 = 2 (cm)

Suy ra: ∆ BHC vuông cân tại H

Hình đây ạ !!:

H ở đâu ra v ạ

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k cho mik

kb nữa nha

thanks

hok tốt

Ta có (ax³ + bx² - 11x + 30) : (x² - 3x - 10) = ax + 3a + b (dư (19a  +3b - 11)x + 10(b + 3a  +3)]

Để  (ax³ + bx² - 11x + 30) \(⋮\) (x² - 3x - 10) khi \(\hept{\begin{cases}19a+3b-11=0\\b+3a+3=0\end{cases}}\Leftrightarrow\hept{\begin{cases}a=2\\b=-9\end{cases}}\)

Vậy a = 2 ; b = -9

\(P=x^4+3y^4-xy^3-x^3y+2021\)

   \(=x^3\left(x-y\right)-y^3\left(x-y\right)+4y^4+2021\)

   \(=\left(x-y\right)\left(x^3-y^3\right)+4y^4+2021\)

  \(=\left(x-y\right)^2\left(x^2+xy+y^2\right)+4y^2+2021\)

\(\Rightarrow P\ge2021\)\(\Rightarrow Pmin=2021\Leftrightarrow x=y=0\)

AI ny mk k? mk 2k10

ế quá rồi khùng=)))

Ta có:

\(\left(a+1\right)\left(a-2\right)\le0;\left(b+1\right)\left(b-2\right)\le0;\left(c+1\right)\left(c-2\right)\le0\)

\(\Leftrightarrow a^2\le2+a;b^2\le2+b;c^2\le2+c\)

\(\Rightarrow a^2+b^2+c^2\le6+a+b+c=6\)