tk:

a) ĐKXĐ: \(\left\{{}\begin{matrix}x\le-1\\x\ge2\end{matrix}\right.\)

\(\sqrt{x^2-x-2}-\sqrt{x-2}=0\\ \Leftrightarrow\sqrt{x^2-x-2}=\sqrt{x-2}\\ \Leftrightarrow x^2-x-2=x-2\\ \Leftrightarrow x^2-2x=0\\ \Leftrightarrow x\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\left(ktm\right)\\x=2\left(tm\right)\end{matrix}\right.\)

\(a,ĐK:x\ge2\\ PT\Leftrightarrow x^2-x-2=x-2\\ \Leftrightarrow x^2-2x=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2\left(tm\right)\\x=0\left(ktm\right)\end{matrix}\right.\Leftrightarrow x=2\\ b,ĐK:\left[{}\begin{matrix}x\le-1\\x\ge1\end{matrix}\right.\\ PT\Leftrightarrow\sqrt{x^2-1}=x^2-1\\ \Leftrightarrow x^2-1=\left(x^2-1\right)^2\\ \Leftrightarrow\left(x^2-1\right)\left(x^2-1-1\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x+1\right)\left(x-\sqrt{2}\right)\left(x+\sqrt{2}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\left(tm\right)\\x=-1\left(tm\right)\\x=\sqrt{2}\left(tm\right)\\x=-\sqrt{2}\left(tm\right)\end{matrix}\right.\)

\(c,ĐK:\left[{}\begin{matrix}x\le-2\\x\ge1\end{matrix}\right.\\ PT\Leftrightarrow\sqrt{x^2-x}=-\sqrt{x^2+x-2}\\ \Leftrightarrow x^2-x=x^2+x-2\\ \Leftrightarrow2x=2\\ \Leftrightarrow x=1\left(tm\right)\)

\(d,ĐK:x\ge1\\ PT\Leftrightarrow\sqrt{\left(x^2-1\right)^2}=x-1\\ \Leftrightarrow\left|x^2-1\right|=x-1\\ \Leftrightarrow\left[{}\begin{matrix}x^2-1=x-1\left(x\le-1;x\ge1\right)\\x^2-1=1-x\left(-1< x< 1\right)\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x=0\left(tm\right)\\x=1\left(tm\right)\end{matrix}\right.\\\left[{}\begin{matrix}x=1\left(ktm\right)\\x=-2\left(ktm\right)\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)

\(e,PT\Leftrightarrow\left|x+2\right|+\left|x-4\right|=0\Leftrightarrow\left\{{}\begin{matrix}x+2=0\\x-4=0\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=-2\\x=4\end{matrix}\right.\Leftrightarrow x\in\varnothing\)

\(g,\Leftrightarrow x\in\varnothing\left(\sqrt{x-2}+\sqrt{x-3}\ge0>-5\right)\\ f,\Leftrightarrow\left|x-1\right|+\left|x-3\right|=1\\ \Leftrightarrow\left[{}\begin{matrix}1-x+3-x=1\left(x< 1\right)\\x-1+3-x=1\left(1\le x< 3\right)\\x-1+x-3=1\left(x\ge3\right)\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\left(ktm\right)\\0x=-1\left(ktm\right)\\x=\dfrac{5}{2}\left(ktm\right)\end{matrix}\right.\\ \Leftrightarrow x\in\varnothing\)

3:

a:Các tia trên hình là Ax,Ay,Bx,By,Cx,Cy

=>Có 6 tia

b: AB<AC

=>B nằm giữa A và C

=>AB+BC=AC

=>BC=4cm

c: AI=3/2=1,5cm

CI=7-1,5=5,5cm

Bài 3:

\(\dfrac{a}{b}>\dfrac{c}{d}\)

\(\Leftrightarrow\dfrac{a}{b}.b>\dfrac{c}{d}.b\)

\(\Leftrightarrow a>\dfrac{bc}{d}\)

\(\Leftrightarrow ad>\dfrac{bc}{d}.d\)

\(\Leftrightarrow ad>bc\) (điều này đúng do giả thiết và \(b,d>0\))

Bài 4: 

a: \(17A=\dfrac{17^{201}+85}{17^{201}+5}=1+\dfrac{80}{17^{201}+5}\)

\(17B=\dfrac{17^{202}+85}{17^{202}+5}=1+\dfrac{80}{17^{202}+5}\)

mà \(17^{201}+5< 17^{202}+5\)

nên 17A>17B

hay A>B

8.31:

a: Xét ΔABD có AM/AB=AQ/AD

nên MQ//BD và MQ=BD/2

Xét ΔCBD có CN/CB=CP/CD

nên NP//BD và NP=BD/2

=>MQ//NP và MQ=NP

XétΔBAC có BM/BA=BN/BC

nên MN//AC

=>MN vuông góc BD

=>MN vuông góc MQ

Xét tứ giác MNPQ có

MQ//NP

MQ=NP

góc NMQ=90 độ

=>MNPQ là hình chữ nhật

=>M,N,P,Q cùng nằm trên 1 đường tròn

Giusp mình với mọi người ơi!!!

1, Nam isn't so short as Hưng

2,Lan drives to work

3,How many fans are there in the classroom ?

4,It took 15 minutes 

5,Unless she work hard ,giữ nguyên

\(n_{FeS_2}=\dfrac{240}{120}.80\%=1,6\left(mol\right)\)

PTHH: 4FeS₂ + 11O₂ --t^o--> 2Fe₂O₃ + 8SO₂

               1,6 -------------------------------> 3,2

2SO₂ + O₂ --t^o--> 2SO₃

3,2 --------------------> 3,2

SO₃ + H₂O ---> H₂SO₄

3,2 ----------------> 3,2

\(m_{ddH_2SO_4}=\dfrac{3,2.98}{49\%}=640\left(g\right)\)

4FeS2+11O2-to>2Fe2O3+8SO2

1,6----------------------------------3,2

2SO2+O2-to,V2O5->2SO3

3,2----------------------------3,2

SO3+H2o->H2SO4

3,2-----------------3,2

n FeS2=2 mol

H=80%

=>n FeS2=2.80%=1,6 mol

=>m H2SO4=3,2.98=313,6g

=>mdd=640g

=>B