a: \(\dfrac{x-1}{4}-\dfrac{3x}{2}=1+\dfrac{2x-1}{3}\)

=>\(\dfrac{x-1-6x}{4}=\dfrac{3+2x-1}{3}\)

=>\(\dfrac{-5x-1}{4}=\dfrac{2x+2}{3}\)

=>\(4\left(2x+2\right)=3\left(-5x-1\right)\)

=>-15x-3=8x+8

=>-23x=11

=>\(x=-\dfrac{11}{23}\)

b: \(\left(4x-1\right)\left(x+5\right)=x^2-25\)

=>\(\left(4x-1\right)\left(x+5\right)-\left(x-5\right)\left(x+5\right)=0\)

=>\(\left(x+5\right)\left(4x-1-x+5\right)=0\)

=>(x+5)(3x+4)=0

=>\(\left[{}\begin{matrix}x=-5\\x=-\dfrac{4}{3}\end{matrix}\right.\)

c: \(x\left(x-1\right)\left(x^2-x+1\right)-6=0\)

=>\(\left(x^2-x\right)\left(x^2-x+1\right)-6=0\)

=>\(\left(x^2-x\right)^2+\left(x^2-x\right)-6=0\)

=>\(\left(x^2-x+3\right)\left(x^2-x-2\right)=0\)

mà \(x^2-x+3=\left(x-\dfrac{1}{2}\right)^2+\dfrac{11}{4}>0\forall x\)

nên \(x^2-x-2=0\)

=>(x-2)(x+1)=0

=>\(\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)

a: Thay x=5 vào A, ta được:

\(A=\dfrac{5+3}{5^2}=\dfrac{8}{25}\)

b: \(B=\dfrac{x-6}{x^2-4}+\dfrac{3}{x-2}+\dfrac{x}{x+2}\)

\(=\dfrac{x-6}{\left(x-2\right)\left(x+2\right)}+\dfrac{3}{x-2}+\dfrac{x}{x+2}\)

\(=\dfrac{x-6+3\left(x+2\right)+x\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}\)

\(=\dfrac{x-6+3x+6+x^2-2x}{\left(x-2\right)\left(x+2\right)}\)

\(=\dfrac{x^2+2x}{\left(x-2\right)\left(x+2\right)}=\dfrac{x}{x-2}\)

c: \(P=A\cdot B=\dfrac{x}{x-2}\cdot\dfrac{x+3}{x^2}=\dfrac{x+3}{x\left(x-2\right)}\)

\(P=\dfrac{1}{x+2}\)

=>\(\dfrac{x+3}{x\left(x-2\right)}=\dfrac{1}{x+2}\)

=>\(x\left(x-2\right)=\left(x+3\right)\left(x+2\right)\)

=>\(x^2+5x+6=x^2-2x\)

=>7x=-6

=>\(x=-\dfrac{6}{7}\left(nhận\right)\)

a: \(3x\left(x-7\right)+2xy-14y\)

\(=3x\left(x-7\right)+2y\left(x-7\right)\)

=(x-7)(3x+2y)

b: \(\left(4x-y\right)\left(a+b\right)-\left(y-4x\right)\left(b-1\right)\)

\(=\left(4x-y\right)\left(a+b\right)+\left(4x-y\right)\left(b-1\right)\)

\(=\left(4x-y\right)\left(a+b+b-1\right)\)

\(=\left(4x-y\right)\left(a+2b-1\right)\)

c: \(2x^2-x-6xy+3y\)

\(=\left(2x^2-x\right)-\left(6xy-3y\right)\)

\(=x\left(2x-1\right)-3y\left(2x-1\right)\)

\(=\left(2x-1\right)\left(x-3y\right)\)

d: \(x^2-z^2+4xy+4y^2\)

\(=\left(x^2+4xy+4y^2\right)-z^2\)

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

\(=\left(x+2y+z\right)\left(x+2y-z\right)\)

a: \(\left(2-3x\right)^2-5x\left(x-4\right)+4\left(x-1\right)\)

\(=9x^2-12x+4-5x^2+20x+4x-4\)

\(=4x^2+12x\)

b: \(\left(x-1\right)^2-2\left(x-1\right)\left(x+2\right)+\left(x+2\right)^2+5\left(2x-3\right)\)

\(=x^2-2x+1-2\left(x^2+x-2\right)+x^2+4x+4+10x-15\)

\(=2x^2+12x-14-2x^2-2x+4\)

=10x-10

c: \(\left(x-2\right)^3-6\left(x+4\right)\left(x-4\right)-\left(x-2\right)\left(x^2+2x+4\right)\)

\(=x^3-6x^2+12x-8-\left(x^3-8\right)-6\left(x^2-16\right)\)

\(=x^3-6x^2+12x-8-x^3+8-6x^2+96\)

\(=-12x^2+12x+96\)

1. The man asked where the next bus station was?

2. Mary said that she wanted to be a writer when she grew up.

3. Marry said life in the city is better than life in the country.

4. Tom said that he wanted to buy a gift for mother.

5. Hoa said that she went to school every day.

6. Minh said that he had a sport car.

7. Linda said that she wanted to go to the bookshop with her sister.

8. Tam said that she liked listening to music withe her father.

1 The man asked where the next bus station was

2 Mary said that she wanted to be a writer when she grew up

3 Mary said that life in the city was better than life in the country

4 Tom told his sister that he wanted to buy a gift for mother

1. C

Cu không thể phản ứng với dung dịch HCl và H₂SO₄ loãng.

2. B

Các base không tan như Zn(OH)₂ , Mg(OH)₂, Ba(OH)₂ và Fe(OH)₃ không làm phenol phthalein hóa đỏ

Tam said that she enjoyed listening to music with her father.

Tam said that she liked listening to music withe her father.

hoa said she goes to school everyday

Marry said life in the city is better than life in the country.

a: ΔABC cân tại A

mà AH là đường trung tuyến

nên AH⊥BC tại H và AH là phân giác của góc BAC

Xét ΔAHE vuông tại E và ΔBCK vuông tại K có

\(\hat{AHE}=\hat{BCK}\left(=90^0-\hat{EHC}\right)\)

Do đó: ΔAHE~ΔBCK

b: Ta có: HE⊥AC

BK⊥AC

Do đó: HE//BK

Xét ΔCKB có

H là trung điểm của CB

HE//BK

Do đó: E là trung điểm của CK

ΔAHE~ΔBCK

=>\(\frac{HE}{CK}=\frac{AE}{BK}\)

=>\(\frac{AE}{BK}=\frac{2\cdot EO}{2\cdot EK}=\frac{EO}{EK}\)

=>\(AE\cdot EK=BK\cdot EO\)

c: Gọi M là trung điểm của EC

Xét ΔEHC có

O,M lần lượt là trung điểm của EH,EC

=>OM là đường trung bình của ΔEHC

=>OM//HC

mà HC⊥HA

nên MO⊥HA

Xét ΔAHM có

MO,HE là các đường cao

MO cắt HE tại O

Do đó: O là trực tâm của ΔAHM

=>AO⊥HM

Xét ΔBEC có

H,M lần lượt là trung điểm của CB,CE

=>HM là đường trung bình của ΔBEC

=>HM//BE

mà AO⊥HM

nên AO⊥BE