1) \(\frac{x-1}{x+3}-\frac{x}{x-3}=\frac{4x+15}{9-x^2}\)

ĐKXĐ : \(x\ne\pm3\)

\(\Leftrightarrow\frac{x-1}{x+3}-\frac{x}{x-3}=\frac{-4x-15}{x^2-9}\)

\(\Leftrightarrow\frac{\left(x-1\right)\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}-\frac{x\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}=\frac{-4x-15}{\left(x-3\right)\left(x+3\right)}\)

\(\Leftrightarrow\frac{x^2-4x+3}{\left(x-3\right)\left(x+3\right)}-\frac{x^2+3x}{\left(x-3\right)\left(x+3\right)}=\frac{-4x-15}{\left(x-3\right)\left(x+3\right)}\)

\(\Leftrightarrow\frac{x^2-4x+3-x^2-3x}{\left(x-3\right)\left(x+3\right)}=\frac{-4x-15}{\left(x-3\right)\left(x+3\right)}\)

\(\Leftrightarrow-7x+3=-4x-15\)

\(\Leftrightarrow-7x+4x=-15-3\)

\(\Leftrightarrow-3x=-18\)

\(\Leftrightarrow x=6\)( tmđk )

Vậy x = 6 là nghiệm của phương trình

2) 2x + 3 < 6 - ( 3 - 4x )

<=> 2x + 3 < 6 - 3 + 4x

<=> 2x - 4x < 6 - 3 - 3

<=> -2x < 0

<=> x > 0

Vậy nghiệm của bất phương trình là x > 0

x² - 4x + y² - 6y + 13

= ( x² - 4x + 4 ) + ( y² - 6y + 9 )

= ( x - 2 )² + ( y - 3 )² 

          Bài làm :

Nếu bạn muốn viết PT thành tổng 2 bình phương thì mình làm như sau

Ta có :

x² - 4x + y² - 6y + 13

= ( x² - 4x + 4 ) + ( y² - 6y + 9 )

= ( x - 2 )² + ( y - 3 )² 

A) Với \(x>y>0\),ta có: \(x^2+y^2< x^2+y^2+2xy=\left(x+y\right)^2\Rightarrow\frac{1}{x^2+y^2}>\frac{1}{\left(x+y\right)^2}\)

Xét: \(\frac{x^2-y^2}{x^2+y^2}>\frac{x^2-y^2}{\left(x+y\right)^2}=\frac{\left(x-y\right)\left(x+y\right)}{\left(x+y\right)^2}=\frac{x-y}{x+y}\)--->ĐPCM

B) \(3^{16}+1=\left(3^{16}-1\right)+2=\left(3^8+1\right)\left(3^8-1\right)+2\)

\(=\left(3^8+1\right)\left(3^4+1\right)\left(3^4-1\right)+2\)

\(=\left(3^8+1\right)\left(3^4+1\right)\left(3^2+1\right)\left(3^2-1\right)+2\)

\(=\left(3^8+1\right)\left(3^4+1\right)\left(3^2+1\right)\left(3+1\right)\left(3-1\right)+2\)

\(>\left(3^8+1\right)\left(3^4+1\right)\left(3^2+1\right)\left(3+1\right)\)--->ĐPCM

Bài làm :

\(a,\left(8-5x\right)\left(x+2\right)+4\left(x-2\right)\left(x+1\right)+2\left(x-2\right)\left(x+2\right)+10\)

\(=8x+16-5x^2-10x+\left(4x-8\right)\left(x+1\right)+2\left(x^2-2^2\right)+10\)

\(=8x+16-5x^2-10x+4x^2+4x-8x-8+2x^2-8+10\)

\(=\left(8x-10x+4x-8x\right)+\left(-5x^2+4x^2+2x^2\right)+\left(16-8-8+10\right)\)

\(=-6x+x^2+10\)

a)\(\left(8-5x\right)\left(x+2\right)+4\left(x-2\right)\left(x+1\right)+2\left(x-2\right)\left(x+2\right)+10\)\(=8x+16-5x^2-2+4x-8x-8+2x-4x-4+10\)\(=\left(8x+4x-8x+2x-4x\right)+\left(16-2-8-4+10\right)+5x^2\)

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

b)\(4\left(x-1\right)\left(x+5\right)-\left(x+2\right)\left(x+5\right)-3\left(x-1\right)\left(x+2\right)\)

\(=4x-4x-20-\left[x^2+5x+2x+10\right]-3\left[x^2+2x-1x-2\right]\)

\(=4x-4x-20-x^2-5x-2x-10-3x^2-6x+3x+6\)

\(=\left(4x-4x-5x-2x-6x+3x\right)+\left(-20-10+6\right)+\left(-x^2-3x^2\right)\)

\(=-10x-24-4x^2\)

c)\(\left(x^{2n}+x^ny^n+y^{2n}\right)\left(x^n-y^n\right)\left(x^{3n}+y^{3n}\right)\)

Xét tích \(\left(x^{2n}+x^ny^n+y^{2n}\right)\left(x^n-y^n\right)\Leftrightarrow\left(x^n\right)^3-\left(y^n\right)^3=x^{3n}-y^{3n}\)

Thay vào bt đã cho ta có \(\left(x^{3n}-y^{3n}\right)\left(x^{3n}+y^{3n}\right)\)

\(\Leftrightarrow\left(x^{3n}\right)^2-\left(y^{3n}\right)^2=x^{6n}-y^{6n}\)

\(9876543.9876545-9876544^2\)

\(=9876543.9876545-9876544.9876544\)

\(=9876543.\left(9876544+1\right)-9876544.9876544\)

\(=9876543.9876544+9876544.9876544+9876544.1\)

\(=9876544.\left(9876543+1\right)+9876544.9876544\)

\(=\left(9876544+9876544\right)^2\)

\(=19753088^2\)

9 876 543 . 9 876 545 - ( 9 876 544 )²

= ( 9 876 544 - 1 )( 9 876 544 + 1 ) - ( 9 876 544 )².1

= [ ( 9 876 544 )² - 1² ] - ( 9 876 544 )²

= ( 9 876 544 )² - 1 - ( 9 876 544 )²

= -1

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

=> \(2x^2+3\left(x^2-1\right)=5x^2+5x\)

=> \(2x^2+3x^2-3-5x^2-5x=0\)

=> \(-3-5x=0\)

=> \(5x=-3\Rightarrow x=-\frac{3}{5}\)

\(x\left(2x-1\right)\left(x+5\right)-\left(2x^2+1\right)\left(x+\frac{9}{2}\right)=\frac{7}{2}\)

=> \(x\left[2x\left(x+5\right)-1\left(x+5\right)\right]-2x^2\left(x+\frac{9}{2}\right)-1\left(x+\frac{9}{2}\right)=\frac{7}{2}\)

=> \(x\left(2x^2+10x-x-5\right)-2x^3-9x^2-x-\frac{9}{2}=\frac{7}{2}\)

=> \(2x^3+10x^2-x^2-5x-2x^3-9x^2-x-\frac{9}{2}=\frac{7}{2}\)

=> \(\left(2x^3-2x^3\right)+\left(10x^2-x^2-9x^2\right)+\left(-5x-x\right)-\frac{9}{2}=\frac{7}{2}\)

=> \(-6x-\frac{9}{2}=\frac{7}{2}\)

=> \(-6x=8\Rightarrow x=-\frac{8}{6}=-\frac{4}{3}\)

\(\left(12x-5\right)\left(4x-1\right)+\left(3x-7\right)\left(1-16x\right)=81\)

=> 12x(4x - 1) - 5(4x - 1) + 3x(1 - 16x) - 7(1 - 16x) = 81

=> 48x² - 12x - 20x + 5 + 3x - 48x² - 7 + 112x = 81

=> -12x - 20x + 3x + 112x + 5 - 7 = 81

=> 83x + 5 - 7 = 81

=> 83x = 81 + 7 - 5

=> 83x = 83

=> x = 1

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

\(\Leftrightarrow2x^2+3x^2-3-5x^2-5x=0\)

\(\Leftrightarrow5x=-3\)

\(\Rightarrow x=-\frac{3}{5}\)

2) \(x\left(2x-1\right)\left(x+5\right)-\left(2x^2+1\right)\left(x+\frac{9}{2}\right)=\frac{7}{2}\)

\(\Leftrightarrow2x^3+9x^2-5x-2x^3-9x^2-x-\frac{9}{2}=\frac{7}{2}\)

\(\Leftrightarrow-6x=8\)

\(\Rightarrow x=-\frac{4}{3}\)

3) \(\left(12x-5\right)\left(4x-1\right)+\left(3x-7\right)\left(1-16x\right)=81\)

\(\Leftrightarrow48x^2-32x+5-48x^2+115x-7=81\)

\(\Leftrightarrow83x=83\)

\(\Rightarrow x=1\)

Khai triển: \(\left(a+b+c\right)^2+\left(a+b-c\right)^2\)

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

\(=2a^2+2b^2+2c^2+4ab\)

( a + b + c )² + ( a + b - c )²

= [ ( a + b ) + c ]² + [ ( a + b ) - c ]²

= [ ( a + b )² + 2( a + b )c + c² ] + [ ( a + b )² - 2( a + b )c + c² ]

= a² + b² + c² + 2ab + 2bc + 2ca + a² + b² + c² + 2ab - 2bc - ca

= 2a² + 2b² + 2c² + 4ab

= 2( a² + b² + 2ab + c² )

= 2[ ( a + b )² + c² ]

a)Xét tam giác AKC và tam giác AHB có

Góc A chung

AB=AC(ABC cân)

góc AKC=góc AHB(=90 độ)

Suy ra tam giác AKC=tam giác AHB(g.c.g)

Suy ra AK=AH(hai góc tương ứng)

Vậy AKH là tam giác cân

Ta có góc AKH=(180 độ -góc A)/2

lại có góc ABC=(180 độ -góc A)/2

vậy góc AKH=góc ABC

MÀ hai góc này nằm ở vị trí đồng vị nên KH//BC

Vậy tứ giácBCHK là hình thang

Ta lại có góc B = góc C(ABC cân)

Suy ra tứ giác BCHK là hình thang cân

                                                    Bài giải

a, Xét \(\Delta KBC\) và \(\Delta HCB\)có :

\(\widehat{BKC}=\widehat{CHB}=90^o\text{ }\left(gt\right)\)

BC : cạnh chung

\(\widehat{KBC}=\widehat{HCB}\text{ }\left(gt\right)\)

\(\Rightarrow\text{ }\Delta KBC=\Delta HCB\text{ }\left(ch\text{ - }gn\right)\)

\(\Rightarrow\text{ }BK=HC\)

Ta có :

\(AB=AK+BK\)

\(AC=AH+HC\)

Mà : \(AB=BC\text{ }\left(gt\right)\text{ ; }BK=HC\text{ }\left(gt\right)\)

\(\Rightarrow\text{ }AK=AH\)

\(\Rightarrow\text{ }\Delta AKH\) cân tại A \(\Rightarrow\text{ }\widehat{AKH}=\frac{180^o-\widehat{A}}{2}\text{ }\left(1\right)\)

\(\Rightarrow\text{ }\Delta ABC\) cân tại A \(\Rightarrow\text{ }\widehat{ABC}=\frac{180^o-\widehat{A}}{2}\text{ }\left(2\right)\)

Từ ( 1 ) ( 2 ) \(\Rightarrow\text{ }\widehat{AKB}=\widehat{ABC}\) Mà hai góc này ở vị trí đồng vị \(\Rightarrow\text{ }KH\text{ }//\text{ }BC\)

Mà \(\widehat{B}=\widehat{C}\text{ }\left(gt\right)\) \(\Rightarrow\text{ }BCHK\)là hình thang cân

b, Dễ mà !