\(\frac{x^2+3xy+2y^2}{x^3+2x^2y-xy^2-2y^3}\)

\(=\frac{\left(x^2+2xy+y^2\right)+xy+y^2}{\left(x^3+x^2y+xy^2+y^3\right)+x^2y-2xy^2-3y^3}\)

\(=\frac{\left(x+y\right)^2+y\left(x+y\right)}{\left(x+y\right)^3+y.\left(x^2-2xy-2y^2\right)}\)

Bài 1:

\(a+b=15\)

\(\Rightarrow\left(a+b\right)^2=225\)

\(\Leftrightarrow a^2+2ab+b^2=225\)

\(\Leftrightarrow a^2+4+b^2=225\)

\(\Leftrightarrow a^2+b^2=221\)

Ta có: \(\left(a-b\right)^2=a^2-2ab+b^2\)

                               \(=221-4\)

                                \(217\)

Bài 2:

Vì \(x:7\)dư 6

\(\Rightarrow x\equiv-1\left(mod7\right)\)

\(\Rightarrow x^2\equiv1\left(mod7\right)\)

Vậy \(x^2:7\)dư 1

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

\(\Rightarrow K=\frac{\left(x-1\right)^2}{x-2}=\frac{\left(x-2\right)^2+2\left(x-2\right)+1}{x-2}\)

\(\Rightarrow K=x-2+2+\frac{1}{x-2}\ge2+2\sqrt{\left(x-2\right).\frac{1}{x-2}=4}\)

Vậy GTTN là :x=3

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

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

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

\(=\left(x+1\right)^4-1\)

\(\ge-1\)

Dấu "=" xảy ra tại \(x=-1\)

a) \(27x^3-0,001\)

\(=\left(3x\right)^3-\left(\frac{1}{10}\right)^3\)

\(=\left(3x-\frac{1}{10}\right)\left(9x^2+\frac{3}{10}x+\frac{1}{100}\right)\)

b) \(a^4-2a^2+1\)

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

\(=\left(a^2-1\right)^2\)

c)\(\left(a^2+4\right)^2-16a^2\) 

\(=\left(a^2+4\right)^2-\left(4a\right)^2\)

\(=\left(a^2+4-4a\right)\left(a^2+4+4a\right)\)

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

a

Xét \(\Delta AEB\) có:\(\widehat{ABE}=90^0;\widehat{BAE}=60^0\Rightarrow\widehat{AEB}=30^0\)

Ta có:\(\widehat{ABC}=\widehat{ABO}+\widehat{OBE}+\widehat{EBC}\Rightarrow\widehat{OBE}=180^0-90^0-60^0=30^0\)

Khi đó \(\widehat{AEB}=\widehat{OBE}=30^0\) suy ra \(\Delta EOB\) cân tại O

b

Ta có:\(\widehat{AOE}=\widehat{AOB}+\widehat{BOC}+\widehat{COE}\Rightarrow\widehat{BOC}=180^0-90^0-60^0=30^0\)

Khi đó:\(\widehat{BOI}=\widehat{IBO}=30^0\Rightarrow\Delta IOB\) cân tại I

\(\Rightarrow IO=IB\)

Xét \(\Delta OIE\) và \(\Delta BIC\) có:

\(OI=BI;\widehat{EOI}=\widehat{CBI}=90^0;\widehat{OIE}=\widehat{BIC}\left(đ.đ\right)\Rightarrow\Delta OIE=\Delta BIC\left(cgv.gn\right)\)

\(\Rightarrow OE=BC\Rightarrow OE+OA=BC+AB\Rightarrow AE=AC\)

\(\Rightarrow\Delta AEC\) cân tại A có \(\widehat{A}=60^0\) nên nó là tam giác đều.

c

Xét \(\Delta OCE\) và \(\Delta BEC\) có:\(OE=BC;\widehat{EBC}=\widehat{COE}=60^0;\widehat{EOC}=\widehat{EBC}=90^0\)

\(\Rightarrow\Delta OCE=\Delta BEC\left(cgv.gn\right)\Rightarrow OC=BE\) ( 1 )

Mặt khác:\(\widehat{ABO}=\widehat{BCE}=60^0\Rightarrow OB//CE\Rightarrow OBCE\) là hình thang.

Kết hợp với ( 1 ) ta có được tứ giác OBCE là hình thang cân.

Có viết sai gì ko đấy

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

48x²-12x-20x+5-(48x²-112x)=81

48x²-12x-20x+5-48x²+112x=81

(48x²-48x²)+(112x-12x-20x)+5=81

0+100x-20x+5=81

80x+5=81

80x=81-5

80x=76

x=76:80

x=19/20

Vậy x=19/20

Hok tốt

đề bài hỏi j

chứng minh với mọi a,b thuộc Z thì ....

A B C M N D E

Xét \(\Delta DAM\) và \(\Delta CBM\) có:

\(BM=AM\left(gt\right);\widehat{DMA}=\widehat{CMB}\left(đ.đ\right);DM=MC\left(đ.đ\right)\Rightarrow\Delta DAM=\Delta CBM\left(c.g.c\right)\)

\(\Rightarrow\widehat{DAM}=\widehat{CBM}\)  ( 1 ) 

Tương tự \(\Delta AEN=\Delta CBN\left(c.g.c\right)\Rightarrow\widehat{EAN}=\widehat{BCN}\)( 2 )

Từ ( 1 ) và ( 2 ) suy ra:\(\widehat{DAM}+\widehat{EAN}=\widehat{CBM}+\widehat{BCN}\)

\(\Rightarrow\widehat{DAM}+\widehat{EAN}+\widehat{BAC}=\widehat{CBM}+\widehat{BCN}+\widehat{BAC}\Rightarrow\widehat{DAE}=180^0\)

=> D,A,E thẳng hàng.

Mặt khác \(DA=BC;EA=BC\Rightarrow DA+EA=2BC\Rightarrow DE=2BC\Rightarrow DA=EA\Rightarrowđpcm\)