Bài làm

a) 2x²y - 4xy² + 6xy

= 2xy( x - 2y + 3 )

b) 4x³y² - 8x²y³ + 2x⁴y

= 2x²y( 2xy - 4y² + x² )

c) 9x²y³ - 3x⁴y² - 6x³y² + 18y⁴ 

= 3y²( 3x²y - x⁴ - 2x³ + 6y² )

d) 7x²y² - 21xy²z + 7xyz - 14xy

= 7xy( xy - 3yz + z - 2 )

# Học tốt #

A B C O D E F

a

Ta có:

\(OA=AD-OD=\frac{2S_{ABC}}{BC}-\frac{2S_{BOC}}{BC}=\frac{2\left(S_{ABC}-S_{BOC}\right)}{BC}\)

\(OD=2S_{BOC}\Rightarrow\frac{OA}{OD}=\frac{S_{ABC}-S_{BOC}}{S_{BOC}}=\frac{S_{ABC}}{S_{BOC}}-1\Rightarrow\frac{OA}{OD}+1=\frac{S_{ABC}}{S_{BOC}}\)

Tương tự 

\(\frac{OB}{OE}+1=\frac{S_{ABC}}{S_{COA}};\frac{OC}{OD}+1=\frac{S_{ABC}}{S_{AOB}}\)

Cộng vế theo vế ta có:

\(\frac{OA}{OD}+\frac{OB}{OE}+\frac{OC}{OF}+3=S_{ABC}\left(\frac{1}{S_{AOB}+S_{BOC}+S_{COA}}\right)\left(1\right)\)

Áp dụng BĐT s-vác ta có:

\(\frac{OA}{OD}+\frac{OB}{OE}+\frac{OC}{OF}+3\ge S_{ABC}\cdot\frac{9}{S_{AOB}+S_{BOC}+S_{COA}}=\frac{9S_{ABC}}{S_{ABC}}=9\)

\(\Rightarrowđpcm\)

Dấu "=" xảy ra tại \(S_{OAB}=S_{OBC}=S_{COA}\Leftrightarrow O\) là trọng tâm của tam giác.

b

Em nghĩ đề là \(\frac{OA}{OD}\cdot\frac{OB}{OE}\cdot\frac{OC}{OF}\ge8\)

Nếu vậy thì e lm như sau:

Ta có:\(\frac{OA}{OD}=\frac{S_{ABC}-S_{BOC}}{S_{BOC}}=\frac{S_{AOC}+S_{AOB}}{S_{BOC}}\)

Tương tự ta có:\(\frac{OB}{OE}=\frac{S_{BOA}+S_{BOC}}{S_{COA}};\frac{OC}{OF}=\frac{S_{COA}+S_{COB}}{S_{BOA}}\)

Đặt \(\left(S_{COA};S_{BOA};S_{BOC}\right)\rightarrow\left(S_1;S_2;S_3\right)\)

Ta có:

\(\frac{OA}{OD}\cdot\frac{OB}{OE}\cdot\frac{OC}{OF}=\frac{\left(S_1+S_2\right)\left(S_2+S_3\right)\left(S_3+S_1\right)}{S_1\cdot S_2\cdot S_3}\)

Áp dụng BĐT Cô si ta có:

\(S_1+S_2\ge2\sqrt{S_1\cdot S_2};S_2+S_3\ge2\sqrt{S_2\cdot S_3};S_3+S_1\ge2\sqrt{S_3\cdot S_1}\)

\(\Rightarrow\frac{OA}{OD}\cdot\frac{OB}{OE}\cdot\frac{OC}{OF}\ge\frac{8\cdot S_1\cdot S_2\cdot S_3}{S_1\cdot S_2\cdot S_3}=8\)

Dấu "=" xảy ra tại \(S_1=S_2=S_3\Leftrightarrow O\) là trọng tâm tam giác ABC.

Câu a. Dòng đầu tiên là nhầm rồi Huy.  AD đâu phải đường cao đâu thế tại sao: \(AD=\frac{2S_{\Delta ABC}}{BC}\)???

Bài này có thể giải:

a. 

Có: \(\frac{OA}{OD}=\frac{AD-OD}{OD}=\frac{AD}{OD}-1=\frac{S_{\Delta ABC}}{S_{\Delta OBC}}-1\)

Tương tự: \(\frac{OB}{OE}=\frac{S_{BAC}}{S_{OAC}}-1\); \(\frac{OC}{OF}=\frac{S_{CAB}}{S_{OAB}}-1\)

=> \(\frac{OA}{OD}+\frac{OB}{OE}+\frac{OC}{OF}=\frac{S_{ABC}}{S_{OBC}}+\frac{S_{ABC}}{S_{OAC}}+\frac{S_{ABC}}{S_{OAB}}-3\)

\(=S_{ABC}\left(\frac{1}{S_{OBC}}+\frac{1}{S_{OAC}}+\frac{1}{S_{OAB}}\right)-3\ge S_{ABC}.\frac{\left(1+1+1\right)^2}{S_{OBC}+S_{OAC}+S_{OAB}}-3=\frac{S_{ABC}.9}{S_{ABC}}-3=6\)

"="  xảy ra <=> O là trọng tâm

b. Làm đúng rồi.

\(\left(x^2-5\right)\left(x^2-10\right)\left(x^2-15\right)\left(x^2-20\right)< 0\)

\(\Rightarrow\)Phải có 1 hoặc 3 số âm (còn lại dương)

Mà \(x^2-20< x^2-15< x^2-10< x^2-5\)

TH1: 3 số âm, 1 số dương

\(\Rightarrow\hept{\begin{cases}x^2-10< 0\\x^2-5>0\end{cases}}\Leftrightarrow5< x^2< 10\Leftrightarrow x=\pm3\)

TH2: 1 số âm, 3 số dương

\(\Rightarrow\hept{\begin{cases}x^2-20< 0\\x^2-15>0\end{cases}}\Leftrightarrow15< x^2< 20\left(L\right)\)

Vậy x = 3 hoặc x = -3

        (x²-5)(x²-10)(x²-15)(x²-20) <0

<=> (x²-5)(1.2.3.4) < 0

<=> (x²-5) . 36 < 0

<=> 36x² -180 < 0

<=> 36x² < 180

<=> x² < 5

<=> x < √5

Đó là cách mình làm

Bài làm

a) 4x² - 6x 

= 2x( 2x - 3 )

b) 9x⁴y³ + 3x²y⁴ 

= 3x²y³( 3x² + y )

c) x³ - 2x² + 5x

= x( x² - 2x + 5 )

d) 3x( x - 1 ) + 5( x - 1 )

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

e) 2x²( x + 1 ) + 4( x + 1 )

= ( x + 1 )( 2x² + 4 )

= ( x + 1 )2( x² + 2 )

= 2( x + 1 )( x² + 2 )

f) -3x - 6xy + 9xz

= -( 3x + 6xy - 9xz )

= -3x( 1 + 2y - 3z )

# Học tốt #

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a) \(2x^2-5x-12\)

\(=2x^2-8x+3x-12\)

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

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

b) \(x^3+5x^2+8x+4\)

\(=\left(x^3+3x^2+2x\right)+\left(2x^2+6x+4\right)\)

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

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

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

\(=\left[x\left(x+1\right)+2\left(x+1\right)\right]\left(x+2\right)\)

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

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

c) \(x^4+x^2+1\)

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

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

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

a. 2x²-5x-12

= 2x²-8x+3x-12

=2x(x-4)+3(x-4)

=(x-4)(2x+3)

c.x⁴+x²+1

=x⁴-x³+x²+x³+1

=x²(x²-x+1)+(x+1)(x²-x+1)

=(x²-x+1)(x²+x+1)

\(A=2x^2+y^2+2xy+60+8x+8y\)

    \(=\left(x^2+y^2+2xy\right)+8x+8y+16+y^2+44\)

    \(=\left(x+y\right)^2+2\left(x+y\right).4+16+y^2+44\)

    \(=\left(x+y+4\right)^2+y^2+44\)

Vì \(\hept{\begin{cases}\left(x+y+4\right)^2\ge0\forall x\\y^2\ge0\forall y\end{cases}}\)

\(\Rightarrow A\ge44\)

Dấu = xảy ra \(\Leftrightarrow\hept{\begin{cases}x+y+4=0\\y=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-4\\y=0\end{cases}}\)

Vậy \(minA=44\Leftrightarrow\hept{\begin{cases}x=-4\\y=0\end{cases}}\)

bạn cũng dc đó

\(a\left(b^3-c^3\right)+b\left(c^3-a^3\right)+c\left(a^3-b^3\right)\)

\(=a\left(b^3-c^3\right)+b\left[\left(c^3-b^3\right)-\left(a^3-b^3\right)\right]+c\left(a^3-b^3\right)\)

\(=a\left(b^3-c^3\right)-b\left(b^3-c^3\right)-b\left(a^3-b^3\right)+c\left(a^3-b^3\right)\)

\(=\left(b^3-c^3\right)\left(a-b\right)-\left(a^3-b^3\right)\left(b-c\right)\)

\(=\left(b-c\right)\left(b^2+ac+c^2\right)\left(a-b\right)-\left(a-b\right)\left(a^2+ab+b^2\right)\left(b-c\right)\)

\(=\left(b-c\right)\left(a-b\right)\left(b^2+ac+c^2-a^2-ab-b^2\right)\)