ĐKXĐ: \(2\le x\le\frac{10}{3}\)

Ta có: \(\sqrt{x-2}+\sqrt{10-3x}=5-x\)

=>\(\sqrt{x-2}-1+\sqrt{10-3x}-1=5-x-2\)

=>\(\frac{x-2-1}{\sqrt{x-2}+1}+\frac{10-3x-1}{\sqrt{10-3x}+1}=3-x\)

=>\(\frac{x-3}{\sqrt{x-2}+1}+\frac{9-3x}{\sqrt{10-3x}+1}=3-x\)

=>\(\left(x-3\right)\left(\frac{1}{\sqrt{x-2}+1}-\frac{3}{\sqrt{10-3x}+1}+1\right)=0\)

=>x-3=0

=>x=3(nhận)

Teenagers become more independent and responsible to their parents.

Many teenagers in the 21st century are less selfish and ready to help other people in need

Teenagers are also more business-orientated and have to find their own source of opportunity

Most high school students believe they should be treated like adults

The primary role of young people is to get a good education to become better citizens of tomorrow.

Câu b

5x-8=2x+7

<=> 3x=7+8=15

<=>x=5

Câu c:

<=>4x=3+5

<=>4x=8

<=>x=2

Câu d

<=>6=4x

<=> x=3/2

Câu e

<=> 2x+8-6x+15=3

<=>4x=20

<=>x=5

b)5x-8=2x+7

⇔3x=15

⇔ x=5

c)4x²+2x-5=4x²-2x+3

⇔ 4x=8

⇔ x=2

d)2x³-3x+6=2x³+x

⇔ 4x=6

⇔ \(x=\dfrac{3}{2}\)

e)2(x+4)-3(2x-5)=3

⇔ 2x+4-6x+15=3

⇔ -4x+19=3

⇔ 4x=16

⇔ x=4

\(\Delta'=4-\left(m-1\right)=5-m\)

để pt có nghiệm kép khi \(5-m=0\Leftrightarrow m=5\)

chọn B 

Phương trình có nghiệm kép khi:

\(\Delta'=4-\left(m-1\right)=0\Leftrightarrow5-m=0\)

\(\Rightarrow m=5\)

Ex6

1 was

2 had been

3 found

4 had talked

5 had spoken

6 snowed

7 were

8 painted

9 had

10 had known

Ex7

2 had got

3 had hurt

4 were

5 were

6 was

7 wanted

8 ?

9 stopped

10 had eaten

\(pt\text{⇔}\left(x^2+3x+2\right)\left(x+5\right)-x^3-8x^2-27=0\text{⇔}x^3+5x^2+3x^2+15x+2x+10-x^3-8x^2-27=0\\ \text{⇔}17x=17\text{⇔}x=1\)

Vậy nghiệm của phương trình : \(S=\left\{1\right\}\)

Ta có: \(\left(x+1\right)\left(x+2\right)\left(x+5\right)-x^3-8x^2=27\)

\(\Leftrightarrow\left(x^2+3x+2\right)\left(x+5\right)-x^3-8x^2=27\)

\(\Leftrightarrow x^3+5x^2+3x^2+15x+2x+10-x^3-8x^2=27\)

\(\Leftrightarrow17x=17\)

hay x=1

h: \(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{9\cdot10}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{9}-\dfrac{1}{10}\)

\(=1-\dfrac{1}{10}=\dfrac{9}{10}\)

m: \(\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}\)

\(=\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+\dfrac{1}{5\cdot6}+\dfrac{1}{6\cdot7}+\dfrac{1}{7\cdot8}\)

\(=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}\)

\(=\dfrac{1}{2}-\dfrac{1}{8}=\dfrac{3}{8}\)

Câu 1:

$(x^2-1)(4x-x^2)=0$

$\Leftrightarrow (x-1)(x+1)x(4-x)=0$

$\Rightarrow x=\pm 1$ hoặc $x=0$ hoặc $x=4$

Vì $x\in\mathbb{N}$ nên $x\in\left\{0;4;1\right\}$

Đáp án B

Câu 2: C

Câu 3: D

Câu 4:

ĐKXĐ: $x^2-7x+12\neq 0$

$\Leftrightarrow (x-3)(x-4)\neq 0$

$\Leftrightarrow x-3\neq 0$ và $x-4\neq 0$

$\Leftrightarrow x\neq 3$ và $x\neq 4$

$\Leftrightarrow x\in\mathbb{R}\setminus\left\{3;4\right\}$

Đáp án D

Câu 5:

ĐKXĐ: \(\left\{\begin{matrix} 2-x\geq 0\\ x+7\geq 0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\leq 2\\ x\geq -7\end{matrix}\right.\Leftrightarrow x\in [-7;2]\)

Đáp án C.

Câu 6: 

ĐKXĐ: \(\left\{\begin{matrix} 5-2x\geq 0\\ x-1\geq 0\\ (x-2)\sqrt{x-1}\neq 0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\leq \frac{5}{2}\\ x>1\\ x\ne 2\end{matrix}\right.\)