\(x^2-7x+2=x^2-2\cdot x\cdot\dfrac{7}{2}+\dfrac{49}{4}-\dfrac{41}{4}\)

\(=\left(x-\dfrac{7}{2}\right)^2-\dfrac{41}{4}>=-\dfrac{41}{4}\forall x\)

Dấu '=' xảy ra khi \(x-\dfrac{7}{2}=0\)

=>\(x=\dfrac{7}{2}\)

\(x^2-12x+5\)

\(=x^2-12x+36-31=\left(x-6\right)^2-31>=-31\forall x\)

Dấu '=' xảy ra khi x-6=0

=>x=6

`x^2 - 7x + 2`

`= x^2 - 2.x . 7/2 + (7/2)^2 - 41/4`

`= (x - 7/2)^2 - 41/4`

Do `(x - 7/2)^2 >= 0=>  (x - 7/2)^2 - 41/4 >= - 41/4`

Dấu = xảy ra khi: 

`x - 7/2 = 0`

`<=> x = 7/2`

Vậy ...

-----------------------

`x^2 - 12x + 5`

`= x^2 - 2.x.6 + 6^2 - 31`

`= (x-6)^2 - 31`

Do `(x-6)^2 >= 0 =>  (x-6)^2 - 31>= -31`

Dấu = có khi: 

`x - 6 = 0`

`<=> x = 6`

Vậy .... (không có max )

\(1)A=x^2-7x+2\\ =\left(x^2-2\cdot x\cdot\dfrac{7}{2}+\dfrac{49}{4}\right)-\dfrac{41}{4}\\ =\left(x-\dfrac{7}{2}\right)^2-\dfrac{41}{4}\)

Ta có: `(x-7/2)^2>=0` với mọi x

`=>A=(x-7/2)^2-41/4>=-41/4` với mọi x

Dấu "=" xảy ra: `x-7/2=0<=>x=7/2` 

\(2)B=9x^2-12x+5\\ =\left(9x^2-12x+4\right)+1\\ =\left[\left(3x\right)^2-2\cdot3x\cdot2+2^2\right]+1\\ =\left(3x-2\right)^2+1\)

Ta có: `(3x-2)^2>=0` với mọi x

`=>B=(3x-2)^2+1>=1` với mọi x

Dấu "=" xảy ra: `3x-2=0<=>x=2/3` 

\(1,x^2+9-16y^2+6x\\ =\left(x^2+6x+9\right)-\left(4y\right)^2\\ =\left(x+3\right)^2-\left(4y\right)^2\\ =\left(x-4y+3\right)\left(x+4y+3\right)\\ 2,x^2-9+y^2+2xy\\ =\left(x^2+2xy+y^2\right)-9\\ =\left(x+y\right)^2-3^2\\ =\left(x+y-3\right)\left(x+y+3\right)\\ 3,x^2-4x+4-9y^2\\ =\left(x-2\right)^2-\left(3y\right)^2\\ =\left(x-3y-2\right)\left(x+3y-2\right)\\ 4,x^2-4xy+4y^2-81\\ =\left(x-2y\right)^2-9^2\\ =\left(x-2y-9\right)\left(x-2y+9\right)\\ 5,6x^2+6y^2-24+12xy\\ =\left(6x^2+12xy+6y^2\right)-24\\ =6\left[\left(x^2+2xy+y^2\right)-4\right]\\ =6\left[\left(x+y\right)^2-2^2\right]\\ =6\left(x+y-2\right)\left(x+y+2\right)\\ 6,9x^2-6x+1-25\\ =\left(3x-1\right)^2-5^2\\ =\left(3x-1-5\right)\left(3x-1+5\right)\\ =\left(3x-6\right)\left(3x+4\right)\)

7: \(x^2+4x+4-49y^2\)

\(=\left(x^2+4x+4\right)-49y^2\)

\(=\left(x+2\right)^2-49y^2\)

=(x+2+7y)(x+2-7y)

8: \(a^3+9a-ab^2-6a^2\)

\(=a\left(a^2-6a+9-b^2\right)\)

\(=a\left[\left(a-3\right)^2-b^2\right]\)

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

9: \(8x^2-16x+8-32y^2\)

\(=8\left(x^2-2x+1-4y^2\right)\)

\(=8\left[\left(x-1\right)^2-\left(2y\right)^2\right]\)

=8(x-1-2y)(x-1+2y)

10: \(4x^2-4x+1-81a^2\)

\(=\left(4x^2-4x+1\right)-81a^2\)

\(=\left(2x-1\right)^2-\left(9a\right)^2\)

=(2x-1-9a)(2x-1+9a)

11: \(x^2-6xy+9y^2-121\)

\(=\left(x^2-6xy+9y^2\right)-121\)

\(=\left(x-3y\right)^2-11^2=\left(x-3y-11\right)\left(x-3y+11\right)\)

12: \(12x^2-24x+12-3y^2\)

\(=3\left(4x^2-8x+4-y^2\right)\)

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

\(\dfrac{x+5}{2006}+\dfrac{x+6}{2005}+\dfrac{x+7}{2004}=-3\)

=>\(\left(\dfrac{x+5}{2006}+1\right)+\left(\dfrac{x+6}{2005}+1\right)+\left(\dfrac{x+7}{2004}+1\right)=-3+3=0\)

=>\(\dfrac{x+2011}{2006}+\dfrac{x+2011}{2005}+\dfrac{x+2011}{2004}=0\)

=>\(\left(x+2011\right)\left(\dfrac{1}{2006}+\dfrac{1}{2005}+\dfrac{1}{2004}\right)=0\)

=>x+2011=0

=>x=-2011

1: \(\left(x-y\right)^2-\left(x+y\right)^2\)

\(=x^2-2xy+y^2-x^2-2xy-y^2\)

=-4xy

2: \(\left(7n-2\right)^2-\left(2n-7\right)^2\)

\(=\left(7n-2+2n-7\right)\left(7n-2-2n+7\right)\)

\(=\left(9n-9\right)\left(5n+5\right)\)

\(=9\left(n-1\right)\left(5n+5\right)⋮9\)

3: \(P=-x^2+6x+1\)

\(=-\left(x^2-6x-1\right)\)

\(=-\left(x^2-6x+9-10\right)\)

\(=-\left(x-3\right)^2+10< =10\forall x\)

Dấu '=' xảy ra khi x-3=0

=>x=3

4: \(\left(a^2+b^2\right)\left(x^2+y^2\right)=\left(ax+by\right)^2\)

=>\(a^2x^2+a^2y^2+b^2x^2+b^2y^2=a^2x^2+b^2y^2+2abxy\)

=>\(a^2y^2-2abxy+b^2x^2=0\)

=>\(\left(ay-bx\right)^2=0\)

=>ay-bx=0

a: Xét ΔAFH vuông tại F và ΔADB vuông tại D có

\(\widehat{FAH}\) chung

DO đó: ΔAFH~ΔADB

b: ΔAFH~ΔADB

=>\(\dfrac{AF}{AD}=\dfrac{AH}{AB}\)

=>\(\dfrac{AF}{AH}=\dfrac{AD}{AB}\)

Xét ΔAFD và ΔAHB có

\(\dfrac{AF}{AH}=\dfrac{AD}{AB}\)

\(\widehat{FAD}\) chung

Do đó: ΔAFD~ΔAHB

c: ΔAFD~ΔAHB

=>\(\widehat{ADF}=\widehat{ABH}\)

=>\(\widehat{ADF}=\widehat{ACH}\)

Xét ΔAEH vuông tại E và ΔADC vuông tại D có

\(\widehat{EAH}\) chung

DO đó: ΔAEH~ΔADC

=>\(\dfrac{AE}{AD}=\dfrac{AH}{AC}\)

=>\(\dfrac{AE}{AH}=\dfrac{AD}{AC}\)

Xét ΔAED và ΔAHC có

\(\dfrac{AE}{AH}=\dfrac{AD}{AC}\)

\(\widehat{EAD}\) chung

Do đó: ΔAED~ΔAHC

=>\(\widehat{ADE}=\widehat{ACH}\)

=>\(\widehat{FDA}=\widehat{EDA}\)

=>DA là phân giác của góc FDE

4x+3y-xy=1

=>\(4x-xy+3y=1\)

=>\(x\left(4-y\right)+3y-12=-11\)

=>-x(y-4)+3(y-4)=-11

=>(-x+3)(y-4)=-11

=>(x-3)(y-4)=11

=>\(\left(x-3;y-4\right)\in\left\{\left(1;11\right);\left(11;1\right);\left(-1;-11\right);\left(-11;-1\right)\right\}\)

=>\(\left(x;y\right)\in\left\{\left(4;15\right);\left(14;5\right);\left(2;-7\right);\left(-8;3\right)\right\}\)

a: Xét ΔDBE vuông tại D và ΔCDE vuông tại C có

\(\widehat{DEB}\) chung

Do đó: ΔDBE~ΔCDE

b:

Ta có: CH\(\perp\)DE

DB\(\perp\)DE

Do đó: CH//DB

Xét ΔHCD vuông tại H và ΔCDB vuông tại C có

\(\widehat{HCD}=\widehat{CDB}\)(hai góc so le trong, CH//DB)

Do đó: ΔHCD~ΔCDB

=>\(\dfrac{HC}{CD}=\dfrac{CD}{DB}\)

=>\(HC\cdot DB=CD^2\)

c: ABCD là hình chữ nhật

=>AC cắt BD tại trung điểm của mỗi đường

=>O là trung điểm của BD

=>OB=OD(1)

Xét ΔEOD có HK//OD

nên \(\dfrac{HK}{OD}=\dfrac{EK}{EO}\left(2\right)\)

Xét ΔEOB có KC//OB

nên \(\dfrac{KC}{OB}=\dfrac{EK}{EO}\left(3\right)\)

Từ (1),(2),(3) suy ra HK=KC

=>K là trung điểm của HC

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