\(x^2+\left(x+3\right)\left(x-9\right)=-27\\ \Rightarrow x^2+x^2+3x-9x-27=-27\\ \Rightarrow2x^2-6x=0\\ \Rightarrow2x\left(x-3\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\)

\(x^2+\left(x+3\right)\left(x-9\right)=-27\)

\(\Rightarrow2x^2-6x=0\)

\(\Rightarrow2x\left(x-3\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\)

\(\Rightarrow x^2+x^2-6x-27=-27\\ \Rightarrow2x^2-6x=0\\ \Rightarrow2x\left(x-3\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\)

B

b

B đó bạn

a: Xét ΔADC có 

E là tđiểm của AD

I là tđiểm của AC

Do đó: EI là đường trung bình

=>EI//DC

Xét ΔBCA có 

F là tđiểm của BC

I là tđiểm của AC

Do đó: IF là đường trung bình

=>IF//AB

a) Ta có: \(6x^4-9x^3\)

\(=3x^3\cdot2x-3x^3\cdot3\)

\(=3x^3\left(2x-3\right)\)

b) Ta có: \(x^2y^2z+xy^2z^2+x^2yz^2\)

\(=xyz\cdot\left(xy+yz+xz\right)\)

c) Ta có: \(2x\left(x+3\right)+2\left(x+3\right)\)

\(=2\cdot\left(x+3\right)\cdot x+2\cdot\left(x+3\right)\cdot1\)

\(=2\left(x+3\right)\left(x+1\right)\)

d) Ta có: \(\left(x+5\right)^2-3\left(x+5\right)\)

\(=\left(x+5\right)\left(x+5-3\right)\)

\(=\left(x+5\right)\left(x+2\right)\)

e) Ta có: \(2x\left(x-3\right)-\left(x-3\right)^2\)

\(=\left(x-3\right)\left(2x-x+3\right)\)

\(=\left(x-3\right)\left(x+3\right)\)

a, 6x⁴ - 9x³ = 3x³ (2x-3x) = 3x³ (-x) = -3x⁴

b, x²y²z + xy²z² + x²yz² = xyz (xy+yz+xz)

c, 2x (x+3) + 2 (x+3) = (x+3) (2x+2) = (x+3) 2 (x+1)

d, (x+5)² - 3 (x+5) = (x+5) (x+5-3) = (x+5) (x+2)

e, 2x (x-3) - (x-3)² = (x-3) [2x-(x-3)] = (x-3) (2x-x+3) = (x-3) (x+3) = x² - 9

Tự làm á! Đúng sai thì chịu

a: Xét tứ giác ABNM có 

AM//BN

AM=BN

Do đó: ABNM là hình bình hành

mà \(\widehat{MAB}=90^0\)

nên ABNM là hình chữ nhật

câu a, \(\dfrac{x}{x+1}\); \(\dfrac{x^2}{1-x}\); \(\dfrac{1}{x^2-1}\)  (đk \(x\)≠ -1; 1)

          \(x^2\) - 1 = ( \(x\) - 1).(\(x\) + 1)

          \(\dfrac{x}{x+1}\) = \(\dfrac{x.\left(x-1\right)}{\left(x+1\right).\left(x-1\right)}\);

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

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

b, \(\dfrac{10}{x+2}\); \(\dfrac{5}{2x-4}\); \(\dfrac{1}{6-3x}\) (đk \(x\) ≠ -2; 2)

    2\(x-4\) = 2.(\(x\) - 2); 6 - 3\(x\) = - 3.(\(x\)  - 2)

   \(\dfrac{10}{x+2}\) = \(\dfrac{10.2.3\left(x-2\right)}{2.3\left(x+2\right)\left(x-2\right)}\) = \(\dfrac{60\left(x-2\right)}{6\left(x-2\right)\left(x+2\right)}\)

    \(\dfrac{5}{2x-4}\) = \(\dfrac{5.3\left(x+2\right)}{2.3\left(x-2\right).\left(x+2\right)}\) = \(\dfrac{15.\left(x+2\right)}{6.\left(x-2\right)\left(x+2\right)}\)

    \(\dfrac{1}{6-3x}\) = \(\dfrac{-1}{3.\left(x-2\right)}\) = \(\dfrac{-1.\left(x+2\right)}{3.2.\left(x-2\right)\left(x+2\right)}\) = \(\dfrac{-2.\left(x+2\right)}{6.\left(x-2\right).\left(x+2\right)}\)

c, \(\dfrac{x}{2x-4}\); \(\dfrac{1}{2x+4}\) và \(\dfrac{3}{4-x^2}\)  đk \(x\) ≠ 2; -2

\(\dfrac{x}{2x-4}\)  =   \(\dfrac{x}{2.\left(x-2\right)}\) = \(\dfrac{x.\left(x+2\right)}{2.\left(x-2\right).\left(x+2\right)}\) 

  \(\dfrac{1}{2x+4}\) = \(\dfrac{1}{2.\left(x+2\right)}\) = \(\dfrac{\left(x-2\right)}{2.\left(x+2\right).\left(x-2\right)}\)

\(\dfrac{3}{4-x^2}\) = \(\dfrac{-3}{\left(x-2\right)\left(x+2\right)}\)  = \(\dfrac{-6}{2.\left(x-2\right)\left(x+2\right)}\)

\(\dfrac{4x^2-3x+5}{x^3-1}\) =  \(\dfrac{4x^2-3x+5}{\left(x-1\right)\left(x^2+x+1\right)}\) Đk \(x\) ≠ 1
\(\dfrac{6}{x-1}\) = \(\dfrac{6.\left(x^2+x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)

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

9:

a: XétΔABC vuông tại A và ΔHBA vuông tại H có

góc B chung

=>ΔABC đồng dạng với ΔHBA

=>BA/BH=BC/BA
=>BA^2=BH*BC

b: BC=25cm; AB=căn 9*25=15cm; AC=căn 16*25=20cm

S ABC=1/2*15*20=150cm²

C ABC=25+15+20=60cm