\(=9x^2+2.3x.5+25=9x^2+30x+25\)

a, Xét tam giac amb va tam giac DMC có

MB = MC ( M la trung điểm BC )

góc aMB = góc CMD ( đối đỉnh)

Ma = MD ( gt)

=> tam giac amb = tam giac DMC (c-g-c)

=> aB = CD; am = MD

góc BaM = góc MDC ; góc aBM = góc MCD 

ma hai góc vị trí so le trong

=> aB // CD

b, Xét tam giac aHM va tam giac DKM

góc aHM = góc DKM = 90 độ

góc aMH = góc DMK ( đối đỉnh )

am = MD ( cmt)

=> tam giac aHM = tam giac DKM ( ch-gn)

=> MH = MK

c, Xét tam giac BHa va tam giac CKD có

ab = CD ( cmt)

góc BHa = góc CKD = 90 độ

góc aBH = góc DCK ( cmt)

=> tam giac BHa = tam giac CKD(ch-gn)

\(\text{a},=3x^2+15x-\left(3x^2+18x-3x-18\right)=3x^2+15x-3x^2-18x+3x+18=18\\ b,=\left(x^3+1\right)-\left(x^3-1\right)=x^3+1-x^3+1=2\)

\(=\left(3x+2-2\right)\left(3x+2+2\right)=\left(3x\right)\left(3x+4\right)=9x^2+12x\)

\(\text{a},=7-\left(-24\right)+16=31+16=47\\ b,=2+12-\left(-3\right)=14-\left(-3\right)=17\\ c,=-14-24+1=-38+1=-37\\ d,=98.\left(-3\right)+32=-294+32=-262\\ e,=8+8-16=0\)

C

C

C

\(\text{a},=\left(x+2\right)^2\\ b,=\left(x-4\right)^2\\ c,=\left(x^2-25\right)\\ d,=\left(x+4\right)^3\\ e,=\left(x-2\right)^3\\ f,=\left(x^3+8\right)\\ g,=\left(x^3-27\right)\\ h,=\left(x+1\right)^2\\ i,=\left(x-1\right)\left(x+1\right)\\ k,=\left(x+3\right)^2\\ l,=\left(2x-3\right)\left(2x+3\right)\\ m,=\left(4x-1\right)^2\\ n,=\left(3x+1\right)^2\\ o,=\left(6x+3\right)^2\\ p,=\left(x+3\right)\left(x^2-3x+9\right)\)

Với  x >0, x ≠ 1 \(\text{a},Q=\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{2}{\sqrt{x}+1}-\dfrac{2}{x-1}\\ =\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)-2\left(\sqrt{x}-1\right)-2}{x-1}=\dfrac{x+\sqrt{x}-2\sqrt{x}+2-2}{x-1}=\dfrac{x-\sqrt{x}}{x-1}=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{x-1}=\dfrac{\sqrt{x}}{\sqrt{x}+1}\)

b, Với x0, x ≠ 1

Với x = 9 ( t/m đk)

Thay x = 9 vao Q

\(\dfrac{\sqrt{9}}{\sqrt{9}+1}=\dfrac{3}{3+1}=\dfrac{3}{4}\)

Vay....