\(C=2\left(x^2+x-5\right)^2-5\left(x^2+x\right)+28\)

Đặt t=\(x^2+x\)

\(\Rightarrow C=2\left(t-5\right)^2-5t+28=2t^2-20t+50-5t+28=2t^2-25t+78=2\left(t-\dfrac{13}{2}\right)\left(t-6\right)\)

Thay t: \(C=2\left(t-\dfrac{13}{2}\right)\left(t-6\right)=2\left(x^2+x-\dfrac{13}{2}\right)\left(x^2+x-6\right)=2\left(x-2\right)\left(x+3\right)\left(x^2+x-\dfrac{13}{2}\right)\)

 1.Their children often play football at weekends

--> What do their children often do at weekends?

Viêt lại câu sao  cho nghĩa ko đổi

   1.If you don't work hard, you will be sacked

-->Unless if you work harder, you will be sacked.

   2. Mr. Hung owns this house

--> Mr. Hung is the owner of this house.

   3. Her bag isn't the same as mine

-->Her bag is different from mine.

a) \(A=3\left(x-1\right)^2-\left(x+1\right)^2+2\left(x-3\right)\left(x+3\right)-\left(2x+3\right)^2-\left(5-20x\right)\)

         \(=\left(3x^2-6x+3\right)-\left(x^2+2x+1\right)+2\left(x^2-9\right)-\left(4x^2+12x+9\right)-5+20x\)

\(=-30\)

b) \(B=-x\left(x+2\right)^2+\left(2x+1\right)^2+\left(x+3\right)\left(x^2-3x+9\right)-1\)

\(=-x\left(x^2+4x+4\right)+\left(4x^2+4x+1\right)+\left(x^3-3x^2+9x+3x^2-9x+27\right)-1\)

\(=27\)

\(2x^2-5x+7=2\left(x^2-\dfrac{5}{2}x+\dfrac{25}{16}\right)-\dfrac{25}{8}+7=2\left(x-\dfrac{5}{4}\right)^2-\dfrac{25}{8}+7\ge\dfrac{31}{8}\)

ĐTXR⇔\(x=\dfrac{5}{4}\)

Vậy minA =\(\dfrac{31}{8}\)khi x=\(\dfrac{5}{4}\)

Đáp án: \(B:\dfrac{31}{8}\)

b) \(3x^2+2x-5=3\left(x-1\right)\left(x+\dfrac{5}{3}\right)\)

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

d) \(x^2+7x+12=\left(x+3\right)\left(x+4\right)\)

e) \(x^2-x-12=\left(x-4\right)\left(x+3\right)\)

\(3x\left(x-2\right)-x^2+2x=0\Rightarrow3x\left(x-2\right)-x\left(x-2\right)=0\Rightarrow\left(x-2\right)\left(3x-x\right)=0\Rightarrow\left[{}\begin{matrix}x-2=0\\2x=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2\\x=0\end{matrix}\right.\)

Ta có: H đối xứng với M qua BC => HM là đường trung trực của BC

\(\Rightarrow\left\{{}\begin{matrix}BH=BM\\HC=MC\end{matrix}\right.\)

Xét tam giác BHC và tam giác BMC có

BH=BM

HC=MC

BC chung

=> Tam giác BHC = tam giác BMC( c.c.c)

b) Gọi CE, BD lần lượt là 2 đường cao của tam giác ABC

Xét tam giác ACE vuông tại E có:

\(\widehat{BAC}+\widehat{ACE}=90^0\Rightarrow\widehat{ACE}=90^0-60^0=30^0\)

Xét tam giác HDC vuông tại D có:

\(\widehat{DHC}+\widehat{HCD}=90^0\Rightarrow\widehat{DHC}=90^0-30^0=60^0\Rightarrow\widehat{BMC}=\widehat{BHC}=180^0-\widehat{DHC}=180^0-60^0=120^0\)

\(2a^2\left(1-4a\right)+\left(2a-1\right)^3=2a^2-8a^3+8a^2-12a^2+6a-1=-10a^2+6a-1=-10\left(a^2-2.\dfrac{3}{10}.a+\dfrac{9}{100}\right)+\dfrac{9}{10}-1=-10\left(a-\dfrac{3}{10}\right)^2-\dfrac{1}{10}\le-10.0-\dfrac{1}{10}=-\dfrac{1}{10}\)Dấu "=" xảy ra \(\Leftrightarrow a=\dfrac{3}{10}\)

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