6. Egypt is believed to be the driest country in the world.

7. The Taj is said to have been built with blind people who couldn't see how beautiful it is.

8. He is alleged to have kicked a policeman.

9. The train was supposed to arrive at 11.30.

10. The two injured men are thought to have been repairing overhead cables.

Bài 9:

a: ĐKXĐ: \(\begin{cases}5-2x\ge0\\ x-1\ge0\end{cases}\Rightarrow1\le x\le\frac52\)

\(\sqrt{5-2x}=\sqrt{x-1}\)

=>5-2x=x-1

=>-2x-x=-1-5

=>-3x=-6

=>x=2(nhận)

b: ĐKXĐ: \(\begin{cases}x^2-3x+2\ge0\\ x-1\ge0\end{cases}\Rightarrow\begin{cases}\left(x-2\right)\left(x-1\right)>=0\\ x-1\ge0\end{cases}\)

=>(x>=2 hoặc x<=1) hoặc x>=1

=>(x>=2 hoặc x=1)

\(\sqrt{x^2-3x+2}=\sqrt{x-1}\)

=>\(x^2-3x+2=x-1\)

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

=>(x-1)(x-3)=0

=>\(\left[\begin{array}{l}x-1=0\\ x-3=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=1\left(nhận\right)\\ x=3\left(nhận\right)\end{array}\right.\)

c: ĐKXĐ: \(\begin{cases}x^2-3x+2\ge0\\ 5x+2\ge0\end{cases}\)

=>\(\begin{cases}\left(x-1\right)\left(x-2\right)\ge0\\ 5x+2\ge0\end{cases}\)

=>(x>=2 hoặc x<=1) và x>=-2/5

=>x>=2 hoặc -2/5<=x<=1

\(\sqrt{x^2-3x+2}-\sqrt{5x+2}=0\)

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

=>\(x^2-3x+2=5x+2\)

=>\(x^2-8x=0\)

=>x(x-8)=0

=>x=0(nhận) hoặc x=8(nhận)

d: ĐKXĐ: \(\begin{cases}3x+7\ge0\\ x+1\ge0\end{cases}=>x\ge-1\)

\(\sqrt{3x+7}-\sqrt{x+1}=0\)

=>\(\sqrt{3x+7}=\sqrt{x+1}\)

=>3x+7=x+1

=>2x=-6

=>x=-3(loại)

Bài 8:

a: \(\left|x^2-5x+4\right|=x+4\)

=>\(\begin{cases}x+4\ge0\\ x^2-5x+4=\left(x+4\right)^2\end{cases}\)

=>\(\begin{cases}x\ge-4\\ x^2-5x+4-x^2-8x-16=0\end{cases}\Rightarrow\begin{cases}x\ge-4\\ -13x-12=0\end{cases}\)

=>\(x=-\frac{12}{13}\)

b: \(\left|x^2-7x+12\right|=15-5x\)

=>\(\begin{cases}15-5x\ge0\\ \left(15-5x\right)^2=x^2-7x+12\end{cases}\)

=>\(\begin{cases}5x\le15\\ 25\left(x-3\right)^2=\left(x-3\right)\left(x-4\right)\end{cases}\Rightarrow\begin{cases}x\le3\\ \left(x-3\right)\left(x-4\right)-25\left(x-3\right)^2=0\end{cases}\)

=>\(\begin{cases}x\le3\\ \left(x-3\right)\left(x-4-25x+75\right)=0\end{cases}\Rightarrow\begin{cases}x\le3\\ \left(x-3\right)\left(-24x+71\right)=0\end{cases}\)

=>x=3(nhận) hoặc x=71/24(nhận)

c: \(\left|x^2-6x+5\right|+1=x\)

=>\(\left|x^2-6x+5\right|=x-1\)

=>\(\begin{cases}x-1\ge0\\ x^2-6x+5=\left(x-1\right)^2\end{cases}\Rightarrow\begin{cases}x\ge1\\ x^2-6x+5=x^2-2x+1\end{cases}\)

=>\(\begin{cases}x\ge1\\ -6x+5=-2x+1\end{cases}\Rightarrow\begin{cases}x\ge1\\ -4x=-4\end{cases}\Rightarrow x=1\)

d: \(3x^2+5\left|x-3\right|+7=0\) (1)

TH1: x>=3

(1) sẽ trở thành: \(3x^2+5\left(x-3\right)+7=0\)

=>\(3x^2+5x-15+7=0\)

=>\(3x^2+5x-8=0\)

=>\(3x^2+8x-3x-8=0\)

=>(3x+8)(x-1)=0

=>x=-8/3(loại) hoặc x=1(loại)

TH2: x<3

(1) sẽ trở thành: \(3x^2+5\left(3-x\right)+7=0\)

=>\(3x^2+15-5x+7=0\)

=>\(3x^2-5x+22=0\)

\(\Delta=\left(-5\right)^2-4\cdot3\cdot22=25-12\cdot22<0\)

=>Phương trình vô nghiệm

e: ĐKXĐ: x<>2

\(\frac{x^2-1}{\left|x-2\right|}=x\)

=>\(x^2-1=x\cdot\left|x-2\right|\) (1)

TH1: x>2

(1) sẽ trở thành:

\(x\left(x-2\right)=x^2-1\)

=>\(x^2-2x=x^2-1\)

=>-2x=-1

=>x=1/2(loại)

TH2: x<2

(1) sẽ trở thành: \(x\left(x-2\right)=1-x^2\)

=>\(1-x^2=x^2-2x\)

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

=>\(2x^2-2x-1=0\)

=>\(x^2-x-\frac12=0\)

=>\(x^2-x+\frac14-\frac34=0\)

=>\(\left(x-\frac12\right)^2=\frac34\)

=>\(\left[\begin{array}{l}x-\frac12=\frac{\sqrt3}{2}\\ x-\frac12=-\frac{\sqrt3}{2}\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac{\sqrt3+1}{2}\left(nhận\right)\\ x=\frac{1-\sqrt3}{2}\left(nhận\right)\end{array}\right.\)

f: \(\frac{\left|x-1\right|}{x^2-x-6}=1\)

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

=>\(\begin{cases}x^2-x-6\ge0\\ \left(x^2-x-6\right)^2=\left(x-1\right)^2\end{cases}\Rightarrow\begin{cases}\left(x-3\right)\left(x+2\right)\ge0\\ \left(x^2-x-6-x+1\right)\left(x^2-x-6+x-1\right)=0\end{cases}\)

=>(x>=3 hoặc x<=-2) và \(\left(x^2-2x-5\right)\left(x^2-7\right)=0\)

=>(x>=3 hoặc x<=-2) và \(\left[\begin{array}{l}x^2-2x+1-6=0\\ x^2-7=0\end{array}\right.\)

=>(x>=3 hoặc x<=-2) và \(\left[\begin{array}{l}\left(x-1\right)^2=6\\ x^2=7\end{array}\right.\)

=>(x>=3 hoặc x<=-2) và \(\left[\begin{array}{l}x-1=\sqrt6\\ x-1=-\sqrt6\\ x=\pm\sqrt7\end{array}\right.\)

=>(x>=3 hoặc x<=-2) và x\(\in\left\lbrace\sqrt6+1;-\sqrt6+1;\sqrt7;-\sqrt7\right\rbrace\)

=>\(x\in\left\lbrace\sqrt6+1;-\sqrt7\right\rbrace\)

Bài 6:

a: ĐKXĐ: \(x^2-4<>0\)

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

=>x∉{2;-2}

\(\frac{x^2-3x+5}{x^2-4}=-1\)

=>\(x^2-3x+5=-x^2+4\)

=>\(2x^2-3x+1=0\)

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

=>\(\left[\begin{array}{l}x-1=0\\ 2x-1=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=1\left(nhận\right)\\ x=\frac12\left(nhận\right)\end{array}\right.\)

b: ĐKXĐ: x∉{2;-2/3}

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

=>(3x+2)(x+1)=(2x+1)(x-2)

=>\(3x^2+3x+2x+2=2x^2-4x+x-2\)

=>\(3x^2+5x+2-2x^2+3x+2=0\)

=>\(x^2+8x+4=0\)

=>\(x^2+8x+16-12=0\)

=>\(\left(x+4\right)^2=12\)

=>\(\left[\begin{array}{l}x+4=2\sqrt3\\ x+4=-2\sqrt3\end{array}\right.\Rightarrow\left[\begin{array}{l}x=2\sqrt3-4\left(nhận\right)\\ x=-2\sqrt3-4\left(nhận\right)\end{array}\right.\)

c: ĐKXĐ: x∉{2;-3}

\(1+\frac{2}{x-2}=\frac{10}{x+3}-\frac{50}{\left(2-x\right)\left(x+3\right)}\)

=>\(1+\frac{2}{x-2}=\frac{10}{x+3}+\frac{50}{\left(x-2\right)\left(x+3\right)}\)

=>

 thôi ngay trò spam nếu ko muốn bay acc

a, 25/12

b,  25/72

a) = 3/4 + 4/3 = 25/12

b) = 3/8 - 16/9 = -101/72 * hình như hơi sai sai:"> *

Gọi số ở chỗ trống đó là x

\(\dfrac{6}{8}< \dfrac{x}{18}< \dfrac{7}{8}\)

\(\dfrac{6\times9}{8\times9}< \dfrac{4\times x}{18\times4}< \dfrac{7\times9}{8\times9}\)

\(\dfrac{54}{72}< \dfrac{4\times x}{72}< \dfrac{63}{72}\)

\(54< 4\times x< 63\)

\(52< 4\times x< 60\)

\(\dfrac{52}{4}< x< \dfrac{60}{4}\)

\(13< x< 15\)

\(x=14\)

Vậy số cần điền vào ... là 14 

đáp án là khoảng từ 13.5 đến 15.75

4

Đợi tí

cảm ơn nhanh giùm minh nha

\(b,N=\left(2x-1\right)^2-4\ge-4\\ N_{min}=-4\Leftrightarrow x=\dfrac{1}{2}\\ c,P=\left(2x-5\right)^2+6\left(2x-5\right)+9-4\\ P=\left(2x-5+3\right)^2-4=\left(2x-2\right)^2-4\ge-4\\ P_{min}=-4\Leftrightarrow x=1\\ d,Q=\left(x^2-2x+1\right)+\left(y^2+4y+4\right)+1\\ Q=\left(x-1\right)^2+\left(y+2\right)^2+1\ge1\\ Q_{min}=1\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\)

6a.

$M=x^2-x+1=(x^2-x+\frac{1}{4})+\frac{3}{4}$

$=(x-\frac{1}{2})^2+\frac{3}{4}\geq \frac{3}{4}$

Vậy $M_{\min}=\frac{3}{4}$ khi $x-\frac{1}{2}=0\Leftrightarrow x=\frac{1}{2}$