\(A=\dfrac{2024}{x^2-2x+8}\)

Ta có :

\(x^2-2x+8=x^2-2x+1+7=\left(x-1\right)^2+7\ge7,\forall x\inℝ\)

\(\Rightarrow\dfrac{1}{x^2-2x+8}\le\dfrac{1}{7}\)

\(\Rightarrow A=\dfrac{2024}{x^2-2x+8}\le\dfrac{2024}{7}\)

\(\Rightarrow Max\left(A\right)=\dfrac{2024}{7}\left(tạix=1\right)\)

a) \(\dfrac{x}{2}=\dfrac{y}{3}\Rightarrow\dfrac{x^2}{4}=\dfrac{y^2}{9}=\dfrac{x^2-y^2}{4-9}=\dfrac{-16}{-5}=\dfrac{16}{5}\)

\(\Rightarrow\left\{{}\begin{matrix}x^2=4.\dfrac{16}{5}\\y^2=9.\dfrac{16}{5}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=\pm\left(2.\dfrac{4}{\sqrt[]{5}}\right)=\pm\dfrac{8\sqrt[]{5}}{5}\\y=\pm\left(3.\dfrac{4}{\sqrt[]{5}}\right)=\pm\dfrac{12\sqrt[]{5}}{5}\end{matrix}\right.\)

\(\dfrac{y}{4}=\dfrac{z}{5}\Rightarrow z=\dfrac{5}{4}y=\dfrac{5}{4}.\left(\pm\dfrac{12\sqrt[]{5}}{5}\right)=\pm3\sqrt[]{5}\)

b) \(\left|2x+3\right|=x+2\)

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

\(\Rightarrow\left[{}\begin{matrix}x=-1\\3x=-5\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-1\\3x=-\dfrac{5}{3}\end{matrix}\right.\)

Đính chính

Dòng cuối \(3x=-\dfrac{5}{3}\rightarrow x=-\dfrac{5}{3}\)

\(y^2+xy+x+2y+1\)

\(=y^2+2y+1+xy+x\)

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

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

(y + 1)(y + 1 + x)

giup voi

ua

A B C E K H D M

a/

Ta có

\(\widehat{B}=\widehat{C}\) (góc ở đáy tg cân ABC)

EK//AB \(\Rightarrow\widehat{EKC}=\widehat{B}\) (góc đồng vị)

\(\Rightarrow\widehat{EKC}=\widehat{C}\) => tg EKC cân tại E => CE=EK

Mà AD=CE 

=> AD=EK (1)

Ta có

EK//AB => EK//AD (2)

Từ (1) và (2) => ADKE là hình bình hành (Tứ giác có 1 cặp cạnh đối // và bằng nhau là hbh)

=> MA=MK; MD=ME (Trong hbh 2 đường chéo cắt nhau tại trung điểm mỗi đường)

b/

Ta có \(H\in\left(M;MK\right)\) => MH=MK

Mà MK=MA (cmt) 

=> MH=MK=MA

=> tg MHK cân tại M \(\Rightarrow\widehat{MHK}=\widehat{MKH}\)

\(\widehat{HMK}+\widehat{MHK}+\widehat{MKH}=\widehat{HMK}+2\widehat{MHK}=180^o\)  (tổng các góc trong của 1 tg = 180 độ)

MH=MK=MA (cmt) => tg MAH cân tại M

\(\Rightarrow\widehat{MAH}=\widehat{MHA}\)

\(\widehat{HMK}=\widehat{MAH}+\widehat{MHA}\) (trong tg góc ngoài bằng tổng 2 góc trong không kề với nó)

\(\Rightarrow\widehat{HMK}=2\widehat{MHA}\)

Từ \(\widehat{HMK}+2\widehat{MHK}=180^o\Rightarrow2\widehat{MHA}+2\widehat{MHK}=180^o\)

\(\Rightarrow\widehat{MHA}+\widehat{MHK}=\widehat{AHK}=90^o\Rightarrow AH\perp BC\)

Xét tg vuông ABH và tg vuông ACH có

AH chung

AB=AC (cạnh bên tg cân ABC)

=> tg AHB = tg AHC (Hai tg vuông có cạnh huyền và cạnh góc vuông bằng nhau)

=> HB=HC

Em cảm ơn ạ

VD5. Đặt a = x - y , b = y - z , c = z - x

=> a + b + c = 0

nên P = (x  - y)³ + (y - z)³ + (z - x)³ 

= a³ + b³ + c³ 

= (a + b)³ - 3ab(a + b) + c³ 

= (-c)³ - 3ab(-c) + c³ 

= 3abc = 3(x - y)(y - z)(z - x)

VD7 : Đặt x + y - z = a ; x - y + z = b ; -x + y + z = c

ta thấy : a + b + c = x + y + z 

Nên ta được Q = (x + y + z)³ - (x + y - z)³ - (x - y + z)³ - (-x + y + z)³

= (a + b + c)³ - a³ - b³ - c³ 

= (a + b)³ + 3(a + b)²c + 3(a + b)c² + c³ - a³ - b³ - c³

= a³ + b³ + 3ab(a + b) + 3(a + b)c(a + b + c) + c³ - a³ - b³ - c³ 

= 3(a + b)[ab + c.(a + b + c)] 

 = 3(a + b)(b + c)(c + a)

= 24xyz

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

\(=\left(2x+y\right).3y\)

b) \(\left(x+1\right)^3+\left(x-1\right)^3\)

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

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

c) \(9x^2-3x+2y-4y^2\)

\(=9x^2-4y^2-3x+2y\)

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

\(=\left(3x-2y\right)\left[3x+2y-1\right]\)

d) \(4x^2-4xy+2x-y+y^2\)

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

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

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

e) \(x^3+3x^2+3x+1-y^3\)

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

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

g) \(x^3-2x^2y+xy^2-4x\)

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

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

\(=x\left[\left(x-y\right)^2-4\right]\)

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

a) (x + 2y)² - (x - y)²

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

= 3y(2x + y)

b) (x + 1)³ + (x - 1)³

= (x + 1 + x - 1)[(x + 1)² - (x + 1)(x - 1) + (x - 1)²]

= 2x(x² + 2x + 1 - x² + 1 + x² - 2x + 1)

= 2x(x² + 3)

c) 9x² - 3x + 2y - 4y²

= (9x² - 4y²) - (3x - 2y)

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

= (3x - 2y)(3x + 2y - 1)

d) 4x² - 4xy + 2x - y + y²

= (4x² - 4xy + y²) + (2x - y)

= (2x - y)² + (2x - y)

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

e) x³ + 3x² + 3x + 1 - y³

= (x³ + 3x² + 3x + 1) - y³

= (x + 1)³ - y³

= (x + 1 - y)[(x + 1)² + (x + 1)y + y²]

= (x - y + 1)(x² + 2x + 1 + xy + y + y²)

g) x³ - 2x²y + xy² - 4x

= x(x² - 2xy + y² - 4)

= x[(x² - 2xy + y²) - 4]

= x[(x - y)² - 2²]

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

2.26:

a. $x^2-6x+9-y^2=(x^2-6x+9)-y^2=(x-3)^2-y^2$
$=(x-3-y)(x-3+y)$

b. $4x^2-y^2+4y-4=4x^2-(y^2-4y+4)$

$=(2x)^2-(y-2)^2=(2x-y+2)(2x+y-2)$
c. $xy+z^2+xz+yz=(xy+xz)+(z^2+yz)=x(y+z)+z(z+y)$
$=(y+z)(x+z)$
c.

$x^2-4xy+4y^2+xz-2yz$
$=(x^2-4xy+4y^2)+(xz-2yz)$
$=(x-2y)^2+z(x-2y)=(x-2y)(x-2y+z)$

2.27:

a. $x^3+y^3+x+y=(x^3+y^3)+(x+y)$
$=(x+y)(x^2-xy+y^2)+(x+y)=(x+y(x^2-xy+y^2+1)$
b. $x^3-y^3+x-y=(x^3-y^3)+(x-y)=(x-y)(x^2+xy+y^2)+(x-y)$

$=(x-y)(x^2+xy+y^2+1)$

c.

$(x-y)^3+(x+y)^3=(x^3-3x^2y+3xy^2-y^3)+(x^3+3x^2y+3xy^2+y^3)$
$=2x^3+6xy^2=2x(x^2+3y^2)$

d.

$x^3-3x^2y+3xy^2-y^3+y^2-x^2$
$=(x^3-3x^2y+3xy^2-y^3)-(x^2-y^2)$
$=(x-y)^3-(x-y)(x+y)=(x-y)[(x-y)^2-(x+y)]$

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