M = (x - 3)³  + (-x - 1)³ 

 = (x - 3)³ - (x + 1)³ 

 = (x - 3 - x - 1)[(x - 3)² + (x - 3)(x + 1) + (x + 1)²]

= -4(x² - 6x + 9 + x² - 2x - 3 + x² + 2x + 1) 

= -4(3x² - 6x + 7) 

= - 4(3x² - 6x + 3 + 4) 

 = -4.3.(x² - 2x + 1) -16 

= -12(x - 1)² - 16 \(\le\)-16

=> Max M = -16

Dáu "=" xảy ra <=> x - 1 = 0 

<=> x = 1

Vậy Max M = -16 <=> x = 1

\(a^{100}+a^{100000000000000000000000000}\)

\(=a^{100+100000000000000000000000000}\)

\(=a^{100000000000000000000000100}\)

\(\frac{xy-x+y-1}{x^2y+xy-x-1}\)

\(=\frac{x\left(y-1\right)+\left(y-1\right)}{xy\left(x+1\right)-\left(x+1\right)}\)

\(=\frac{\left(y-1\right)\left(x+1\right)}{\left(xy-1\right)\left(x+1\right)}\)

\(=\frac{y-1}{xy-1}\)

\(\frac{2xy^2\left(x^2-y^2\right)}{xy\left(x+y\right)^2}=\frac{2xy^2\left(x-y\right)\left(x+y\right)}{xy\left(x+y\right)^2}=\frac{2y\left(x-y\right)}{x+y}\);

\(\frac{2x^2-2x}{3x^3}=\frac{2x\left(x-1\right)}{3x^3}=\frac{2\left(x-1\right)}{3x^2}\);

\(\frac{x^3+x^2-4x-4}{x^3+8x^2+17x+10}=\frac{x^2\left(x+1\right)-4\left(x+1\right)}{x^3+x^2+7x^2+7x+10x+10}=\frac{\left(x+1\right)\left(x^2-4\right)}{x^2\left(x+1\right)+7x\left(x+1\right)+10\left(x+1\right)}\)

                                             \(=\frac{\left(x+1\right)\left(x-2\right)\left(x+2\right)}{\left(x+1\right)\left(x^2+7x+10\right)}=\frac{\left(x+1\right)\left(x-2\right)\left(x+2\right)}{\left(x+1\right)\left(x^2+7x+10\right)}=\frac{\left(x+1\right)\left(x-2\right)\left(x+2\right)}{\left(x+1\right)\left(x^2+2x+5x+10\right)}\)

\(=\frac{\left(x+1\right)\left(x-2\right)\left(x+2\right)}{\left(x+1\right)\left(x^2+2x+5x+10\right)}=\frac{\left(x+1\right)\left(x-2\right)\left(x+2\right)}{\left(x+1\right)\left[x\left(x+2\right)+5\left(x+2\right)\right]}=\frac{\left(x+1\right)\left(x-2\right)\left(x+2\right)}{\left(x+1\right)\left(x+2\right)\left(x+5\right)}=\frac{x-2}{x+5}\)

a,ˆD=ˆC=700(t/c.hthang.cân)AB//CD⇒ˆA+ˆD=1800(2.góc.trong.cùng.phía)⇒ˆA=1100ˆA=ˆB=1100(t/c.hthang.cân)b,⎧⎪ ⎪ ⎪⎨⎪ ⎪ ⎪⎩AD=BC(t/c.hthang.cân)ˆAHD=ˆBKC(=900)ˆD=ˆC(cm.trên)⇒ΔAHD=ΔBKC(ch−gn)⇒DH=CK

a) Ta có tam giác ABC cân tại A

=> góc B= góc C

=> 1/2 góc C= 1/2 góc B

=> ABE=ACF

Xét tam giác ABE và tam giác AFC có:

AB=AC(gt)

A(chung)

ABE=ACF(cmt)

=> tam giac ABE= tam giác ACF(g.c.g)

=> AF=AE

=> tam giác AEF cân tại A

b)Ta có góc B= góc C

=> 1/2 góc B=1/2 góc C=>EBC=FCB

Theo câu a, ta có tam giác ABE= tam giác ACF(g.c.g)

=> BE=CF

Xét tam giác BFC vá tam giác CEB có

BE=CF(tam giác ABE= tam giác ACF)

FCB=ECB(cmt)

BC(chung)

=> tam giác BFC= tam giác CEB(c.g.c0

c) Tam giác AFE cân tại A

=>góc AFE=(180*-A)/2

Tam giác ABC cân tại B=>ABC=(180*-A)/2

=> ABC=AFE

=> FE//BC(1)

Ta có: FB=AB-AF

          EC=AC-AE

          AB=AC

        AF=AE

=> FB=EC(2)

Từ (1)(2)=> tứ giác BFEC là hình thang cân