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)

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

b) Ta có: 5x-8=2x+7

\(\Leftrightarrow5x-2x=7+8\)

\(\Leftrightarrow3x=15\)

hay x=5

c) Ta có: \(4x^2+2x-5=4x^2-2x+3\)

\(\Leftrightarrow2x-5+2x-3=0\)

\(\Leftrightarrow4x=8\)

hay x=2

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

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