a) Ta có: \(A=\sqrt{4x^2+4x+2}=\sqrt{\left(4x^2+4x+1\right)+1}\)

\(=\sqrt{\left(2x+1\right)^2+1}\ge\sqrt{1}=1\left(\forall x\right)\)

Dấu "=" xảy ra khi: \(\left(2x+1\right)^2=0\Rightarrow x=-\frac{1}{2}\)

Vậy Min(A) = 1 khi x = -1/2

b) Ta có: \(B=\sqrt{2x^2-4x+5}=\sqrt{\left(2x^2-4x+2\right)+3}\)

\(=\sqrt{2\left(x-1\right)^2+3}\ge\sqrt{3}\left(\forall x\right)\)

Dấu "=" xảy ra khi: \(\left(x-1\right)^2=0\Rightarrow x=1\)

Vậy Min(B) = \(\sqrt{3}\) khi x = 1

1.A lot of plays has been written by my brother

2. His second tragedy has been finished

3.My homework haven't been done.

4.The moon has been seen by you

5.That book has been read by him when he was a boy

6.The thief has been caught by the police

7.This film has been seen by her a several times

8.

1 We often eat ....some.............ham for breakfast

2 Diane keeps .......some.................photos in her bag

3.There's hardly  any....................honey in the jar

4. I can not rarely take ..any.................photos because I haven't got a camera  (sai đề mình sửa  nha)

5 Susan has ...some............coffee and cake in the afternoons

6.Did you buy ...any..............butter yesterday?

7. They have......some..........surprises for us

study well

Biết rồi, khổ lắm, nói mãi

Answers:

1/ We often eat ........some.........ham for breakfast

2/ Diane keeps ..........some..............photos in her bag

3/ There's hardly ........any............honey in the jar

4/ I can rarely take .........some..........photos because I haven't got a camera 

5/ Susan has .......some........coffee and cake in the afternoons

6/ Did you buy ..........any.......butter yesterday?

7/ They have.......some.........surprises for us

\(ĐKXĐ:x\ge-\frac{2}{3}\)

Ta có : \(4x^2+6x+1=4\sqrt{6x+4}\)

\(\Leftrightarrow4x^2+6x+1+6x+4+4=6x+4+4\sqrt{6x+4}+4\)

\(\Leftrightarrow4x^2+12x+9=\left(\sqrt{6x+4}\right)^2+2.\sqrt{6x+4}.2+2^2\)

\(\Leftrightarrow\left(2x+3\right)^2=\left(\sqrt{6x+4}+2\right)^2\)

\(\Leftrightarrow\orbr{\begin{cases}2x+3=\sqrt{6x+4}+2\left(1\right)\\2x+3=-\sqrt{6x+4}-2\left(2\right)\end{cases}}\)

+) Pt (1) \(\Leftrightarrow\sqrt{6x+4}=2x+1\)

\(\Leftrightarrow\hept{\begin{cases}5x+4=4x^2+4x+1\\x\ge-\frac{1}{2}\end{cases}}\) \(\Leftrightarrow\hept{\begin{cases}\left(x-1\right)\left(4x+3\right)=0\\x\ge-\frac{1}{2}\end{cases}}\) \(\Leftrightarrow x=1\) ( Thỏa mãn )

+) Pt (2) \(\Leftrightarrow\sqrt{6x+4}=-2x-5\)

\(\Leftrightarrow\hept{\begin{cases}6x+4=\left(-2x-5\right)^2\\x\le-\frac{5}{2}\end{cases}}\) \(\Leftrightarrow\hept{\begin{cases}6x+4=4x^2+25+20x\\x\le-\frac{5}{2}\end{cases}}\) ( Vô nghiệm )

Vậy phương trình đã cho có nghiệm duy nhất \(x=1\)

a) \(2x-\frac{3}{4}=-\left(\frac{2}{3}\right)\)

\(\Leftrightarrow2x=\frac{1}{12}\)

\(\Rightarrow x=\frac{1}{24}\)

b) \(x^5\div x^3=\frac{1}{16}\)

\(\Leftrightarrow x^2=\left(\frac{1}{4}\right)^2\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{1}{4}\\x=-\frac{1}{4}\end{cases}}\)

c) \(\frac{2}{9}x\left(x-3\frac{7}{8}\right)=0\)

\(\Leftrightarrow x\left(x-\frac{31}{8}\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x-\frac{31}{8}=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=\frac{31}{8}\end{cases}}\)

Answers:

1/ Grandpa gave .........some............chocolate to the kids

2/ Mike hasn't got ........any.............mugs

3/ My friends never paint ..........any..............pictures

4/ There's........some........sand in my shoes

5/ Steve did his homework without ........any.......help

6/ Paul is thinking of ........some..........good memories

Trần Công Mạnh cảm ơn

e)

ta có :ab+ba=110

10a + b+10b+a=110

11a+11b=110

11(a+b)=110

(a+b)=110 : 11

a+b=10

a+b+a-b=10+8

suy ra

a=9

b=1

g)

tự làm như trên

h)

tương tự hết

A\(\in\){14; 21;28}

a) \(\left(\frac{11}{12}:\frac{44}{16}\right)\cdot\left(-\frac{1}{3}+\frac{1}{2}\right)=\left(\frac{11}{12}\cdot\frac{16}{44}\right)\cdot\frac{1}{6}=\left(\frac{1}{3}\cdot\frac{4}{4}\right)\cdot\frac{1}{6}=\frac{1}{3}\cdot\frac{1}{6}=\frac{1}{18}\)

b) \(\frac{\left(-5\right)^2\cdot\left(-5\right)^3\cdot16}{5^4\cdot\left(-2\right)^4}=\frac{5^2\cdot\left(-5\right)^3\cdot2^4}{5^4\cdot2^4}=\frac{5^2\cdot\left(-5\right)^3\cdot2^4}{5^2\cdot5^2\cdot2^4}=\frac{\left(-5\right)^3}{5^2}=-5\)

c) \(7,5:\left(-\frac{5}{3}\right)+2\frac{1}{2}:\left(-\frac{5}{3}\right)=7,5:\left(-\frac{5}{3}\right)+2,5:\left(-\frac{5}{3}\right)=\left(7,5+2,5\right):\left(-\frac{5}{3}\right)\)

\(=10:\left(-\frac{5}{3}\right)=10\cdot\left(-\frac{3}{5}\right)=-6\)

d) \(\left|-\frac{3}{7}\right|:\left(-3\right)^2-\sqrt{\frac{4}{49}}=\frac{3}{7}:9-\frac{2}{7}=\frac{3}{7}\cdot\frac{1}{9}-\frac{2}{7}=\frac{1}{7}\cdot\frac{1}{3}-\frac{2}{7}=\frac{1}{21}-\frac{2}{7}=-\frac{5}{21}\)