\(A=\left(x+2\right)^2+\left(x-2\right)^2\\ =x^2+4x+4+x^2-4x+4\\ =2x^2+8\)

Thay `x=-1/2` vào A ta có:

\(A=2\cdot\left(-\dfrac{1}{2}\right)^2+8=2\cdot\dfrac{1}{4}+8=\dfrac{1}{2}+8=\dfrac{17}{2}\)

\(B=\left(3x-2\right)^2-\left(3x+5\right)^2\\ =\left(3x-2-3x-5\right)\left(3x-2+3x+5\right)\\ =-7\left(6x+3\right)\)

Thay `x=-4` vào B ta có:

\(B=-7\cdot\left(6\cdot-4+3\right)=-7\cdot-21=147\)

\(C=\left(2x+5y\right)^2-5y\left(4x+5y\right)\\ =4x^2+20xy+25y^2-20xy-25y^2\\ =4x^2\)

Thay `x=-1/2;y=-756` vào C ta có:

\(C=4\cdot\left(-\dfrac{1}{2}\right)^2=4\cdot\dfrac{1}{4}=1\)

\(D=\left(x+5\right)^2-\left(x-3\right)^2\\ =\left(x+5-x+3\right)\left(x+5+x-3\right)\\ =8\left(2x+2\right)\)

Thay `x=-3/4` vào D ta có:

\(D=8\cdot\left(2\cdot\dfrac{-3}{4}+2\right)=8\cdot\left(-\dfrac{3}{2}+2\right)=8\cdot\dfrac{1}{2}=4\)

1. How many people are there in Linh's family?
2. My friend doesn't live with his family in Ha Noi.
3. Hoa gets up at six o'clock and brushes her teeth.
4. Our classroom is on the first floor.
5. There are six rooms in Minh's house.
6. I am doing my homework.
7. They usually go to the cinema.
8. These glasses are my grandpa's.
9. My sister has long black hair.
10. We live in a large house in the country.
11. Jane's puppy is sleeping under the table.
12. English is Ha's favourite subject.
13. We are taking our English test today.
14. My father works in a big factory in Vinh Long.
15. There are four members in her family.

1: there/ family/ many/ in/ are/ How/ Linh's/ people/ ?

How many people are there in Linh's family? 
2; his/ friend/ in/ family/ My/ Ha Noi/ doesn't/ with/ live.

My friend doesn't live with his family in Ha Noi
3: brushes/ six/ gets/ her/ o'clock/ Hoa/ at/ up/ teeth/ and.

Hoa gets up and brushed her teeth at six o'clock
4: on/ floor/ classroom/ the/ is/ Our/ first.

Our classroom is on the first floor
5: Minh's/ six/ There/ in/ rooms/ house/ are.

There are six room in Minh's house

a: Ta có: AD+DB=AB

AE+EC=AC

mà DB=EC và AB=AC

nên AD=AE

Xét ΔABC có \(\dfrac{AD}{AB}=\dfrac{AE}{AC}\)

nên DE//BC

b: Xét ΔAEB và ΔADC có

AE=AD

\(\widehat{BAE}\) chung

AB=AC

Do đó: ΔAEB=ΔADC

c: ΔAEB=ΔADC

=>\(\widehat{AEB}=\widehat{ADC}\)

mà \(\widehat{AEB}+\widehat{CEB}=180^0\)(hai góc kề bù) và \(\widehat{ADC}+\widehat{CDB}=180^0\)(hai góc kề bù)

nên \(\widehat{CEB}=\widehat{CDB}\)

ΔAEB=ΔADC

=>\(\widehat{ABE}=\widehat{ACD}\)

Xét ΔIDB và ΔIEC có

\(\widehat{IDB}=\widehat{IEC}\)

DB=EC

\(\widehat{IBD}=\widehat{ICE}\)

Do đó: ΔIDB=ΔIEC

d: Ta có: ΔIDB=ΔIEC

=>IB=IC

Xét ΔABI và ΔACI có

AB=AC

BI=CI

AI chung

Do đó: ΔAIB=ΔAIC

=>\(\widehat{BAI}=\widehat{CAI}\)

=>AI là phân giác của góc BAC

e: Ta có: IB=IC

=>I nằm trên đường trung trực của BC(1)

Ta có: AB=AC

=>A nằm trên đường trung trực của BC(2)

Từ (1),(2) suy ra AI là đường trung trực của BC

=>AI\(\perp\)BC

f: BD=DE

=>ΔDEB cân tại D

=>\(\widehat{DEB}=\widehat{DBE}\)

mà \(\widehat{DEB}=\widehat{EBC}\)(DE//BC)

nên \(\widehat{ABE}=\widehat{CBE}\)

=>BE là phân giác của góc ABC

=>E là chân đường phân giác kẻ từ B xuống AC

Ta có: DE=EC

=>ΔEDC cân tại E

=>\(\widehat{ECD}=\widehat{EDC}\)

mà \(\widehat{EDC}=\widehat{DCB}\)(ED//BC)

nên \(\widehat{ACD}=\widehat{BCD}\)

=>CD là phân giác của góc ACB

=>D là chân đường phân giác kẻ từ C xuống AB

\(51^2=\left(50+1\right)^2=50^2+2\cdot50\cdot1+1^2\\ =2500+100+1=2601\\ 502^2=\left(500+2\right)^2=500^2+2\cdot500\cdot2+2^2\\ =250000+2000+4=252004\\ 98^2=\left(100-2\right)^2=100^2-2\cdot100\cdot2+2^2\\ =10000-400+4 =9604\\ 199^2=\left(200-1\right)^2=200^2-2\cdot200\cdot1+1^2\\ =40000-400+1=39601\)

\(25x^2+10x+4\\ =\left(25x^2+10x+1\right)+3\\ =\left[\left(5x\right)^2+2\cdot5x\cdot1+1^2\right]+3\\ =\left(5x+1\right)^2+3\)

Ta có: `(5x+1)^2>=0` với mọi x 

`=>(5x+1)^2+3>=3>0` với mọi x 

`=>` Đpcm 

\(a.4+12x+9x^2\\ =2^2+2\cdot2\cdot3x+\left(3x\right)^2\\ =\left(2+3x\right)^2\\ b.25+4a^2+10a\\ =5^2+2\cdot5\cdot2a+\left(2a\right)^2\\ =\left(5+2a\right)^2\\ c.14y+y^2+49\\ =y^2+2\cdot y\cdot7+7^2\\ =\left(y+7\right)^2\\ d.9x^2+y^2-6xy\\ =\left(3x\right)^2-2\cdot3x\cdot y+y^2\\ =\left(3x-y\right)^2\)

\(\widehat{A}=180^o-\widehat{B}-\widehat{C}=180^o-40^o-40^o=100^o\)

=> \(\widehat{A_{ngoai}}=180^o-100^o=80^o\) 

=> \(\widehat{DAB}=\dfrac{1}{2}\widehat{A_{ngoai}}=\dfrac{1}{2}\cdot80^o=40^o\)

Ta có: \(\widehat{DAB}=\widehat{ABC}\left(=40^o\right)\)

Mà 2 góc này ở vị trí so le trong

=> AD//BC 

cậu giúp mik nhiều ghê, cám ơn nha

\(a.\left(x^2-2\right)^2=\left(x^2\right)^2-2\cdot x^2\cdot2+2^2=x^4-4x^2+4\\ b.\left(x+2y\right)^2=x^2+2\cdot x\cdot2y+\left(2y\right)^2=x^2+4xy+4y^2\\ c.\left(\dfrac{1}{2}-x\right)^2=\left(\dfrac{1}{2}\right)^2-2\cdot x\cdot\dfrac{1}{2}+x^2=\dfrac{1}{4}-x+x^2\\ d.\left(2a-1\right)^2=\left(2a\right)^2-2\cdot2a\cdot1+1^2=4a^2-4a+1\\ e.\left(\dfrac{1}{3}a+2\right)^2=\left(\dfrac{1}{3}a\right)^2+2\cdot\dfrac{1}{3}a\cdot2+2^2=\dfrac{1}{9}a^2+\dfrac{4}{3}a+4\\ f.\left(2a-\dfrac{2}{3}\right)^2=\left(2a\right)^2-2\cdot2a\cdot\dfrac{2}{3}+\left(\dfrac{2}{3}\right)^2=4a^2-\dfrac{8}{3}a+\dfrac{4}{9}\)

Xét ΔABC vuông tại A có \(\widehat{ABC}+\widehat{ACB}=90^0\)

=>\(2\cdot\left(\widehat{OBC}+\widehat{OCB}\right)=90^0\)

=>\(\widehat{OBC}+\widehat{OCB}=45^0\)

Xét ΔOBC có \(\widehat{BOC}+\widehat{OBC}+\widehat{OCB}=180^0\)

=>\(\widehat{BOC}+45^0=180^0\)

=>\(\widehat{BOC}=135^0\)