TL

trong các hình sau hình nào không có trục đối xứng 

A. hình bình hành                                                 B. hình thoi

C.hình chữ nhật                                                      D hình thang cân

là hình bình hành nha

Answer:

\(7x^3+3x^2-3x+1=0\)

\(\Rightarrow7x^3+7x^2-4x^2-4x+x+1=0\)

\(\Rightarrow7x^2.\left(x+1\right)-4x.\left(x+1\right)+\left(x+1\right)=0\)

\(\Rightarrow\left(x+1\right).\left(7x^2-4x+1\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x+1=0\\7x^2-4x+1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=-1\\7x^2-4x+1=0\text{(Vô lý)}\end{cases}}\)

\(S=\frac{1}{x\left(x+1\right)\left(x+2\right)}+\frac{1}{\left(x+1\right)\left(x+2\right)\left(x+3\right)}+...+\frac{1}{\left(x+8\right)\left(x+9\right)\left(x+10\right)}\)

\(=\frac{1}{2}\left[\frac{2}{x\left(x+1\right)\left(x+2\right)}+\frac{2}{\left(x+1\right)\left(x+2\right)\left(x+3\right)}+...+\frac{2}{\left(x+8\right)\left(x+9\right)\left(x+10\right)}\right]\)

\(=\frac{1}{2}\left[\frac{x+2-x}{x\left(x+1\right)\left(x+2\right)}+\frac{\left(x+3\right)-\left(x+1\right)}{\left(x+1\right)\left(x+2\right)\left(x+3\right)}+...+\frac{\left(x+10\right)-\left(x+8\right)}{\left(x+8\right)\left(x+9\right)\left(x+10\right)}\right]\)

\(=\frac{1}{2}\left[\frac{1}{x\left(x+1\right)}-\frac{1}{\left(x+1\right)\left(x+2\right)}+...+\frac{1}{\left(x+8\right)\left(x+9\right)}-\frac{1}{\left(x+9\right)\left(x+10\right)}\right]\)

\(=\frac{1}{2}\left[\frac{1}{x\left(x+1\right)}-\frac{1}{\left(x+9\right)\left(x+10\right)}\right]\)

(x - 4)(x² + 4x + 16) - x(x² - 6) = 2

x³ - 64 - x³ + 6x = 2

6x = 2 + 64

6x = 66

x = 66 : 6

x = 11

x³ - 27 + 3x(x - 3)

= (x - 3)(x² + 3x + 9) + 3x(x - 3)

= (x - 3)(x² + 3x + 9 + 3x)

= (x - 3)(x² + 6x + 9)

= (x - 3)(x + 3)²

5x³ - 7x² + 10x - 14

= 5x(x² + 2) - 7(x² + 2)

= (x² + 2)(5x - 7)

\(x^3+2x^2-x-2\)

\(=x^3+3x^2+2x-1x^2-3x-2\)

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

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

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

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

\(x^3+3x+2\)

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

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

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

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

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

Answer:

\(5xy^2-10xyz+5xz^2\)

\(=5x.\left(y^2-2yz+z^2\right)\)

\(=5x.\left(y-z\right)^2\)

a/ ˆDCE+ˆECF=180oDCE^+ECF^=180o

=> ˆECF=90oECF^=90o

Xét t/g DEC và t/g BFC có

EC = FC (GT)

ˆDCE=ˆBCF=90oDCE^=BCF^=90o

DC = BC (do ABCD là hình vuông)

=> t/g DEC = t/g BFC (c.g.c)

=> DE = BF (2 cạnh t/ứ(

b/ Xét t/g BEH và t/g DEC có

ˆBEH=ˆDECBEH^=DEC^ (đối đỉnh)

ˆEBF=ˆEDCEBF^=EDC^ (do t/g BFC = t/g DEC)

 ⇒ΔBEH∼ΔDEC⇒ΔBEH∼ΔDEC (g.g)

=> ˆBHE=ˆDCB=90oBHE^=DCB^=90o

=> DE⊥BFDE⊥BF

Xét t/g BDF có

DE ⊥ BF

BC ⊥ DF

DE cắt BC tại E

=> E là trực tâm t/g BDF

=> .... đpcm

c/ Xét t/g CEF có CE = CF ; M là trung điểm EF

=> CM ⊥ EF

=> ˆKMC=90oKMC^=90o

Tự cm OKMC làhcn

=> OC = KM => AO = KM

Mà AO // KM (cùng vuông góc vs BD)

=> AOMK là hbh

=> OM // AK

😱😱😱😱😱 oh mai gót!

  16x - 5x² - 3

= 5x² - 16x - 3

= 5x² - 15x - x - 3

= 5x(x - 3) - (x - 3)

= (x - 3)(5x - 1)