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

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 $x-\frac{1}{2}=0\Leftrightarrow x=\frac{1}{2}$

`bb5)`

`x/3 - ( 2x + 1 )/6 = x/6 - x`

`<=> ( 2x )/6 - ( 2x+1 )/6 = x/6 - ( 6x )/6`

`=> 2x-( 2x+1) = x - 6x`

`<=> 2x-2x-1=-5x`

`<=> -1=-5x`

`<=> 1=5x`

`<=> x = 1/5``(tm)`

Vậy `S={1/5}` 

`5)x/3-[2x+1]/6=x/6-x`

`<=>[2x-2x-1]/6=[x-6x]/6`

`<=>2x-2x-1=x-6x`

`<=>2x-2x-x+6x=1`

`<=>5x=1`

`<=>x=1/5`

Vậy `S={1/5}`

____________________________________

`6)[x+1]/2004+[x+2]/2003=[x+3]/2002+[x+4]/2001`

`<=>[x+1]/2004+1+[x+2]/2003+1=[x+3]/2002+1+[x+4]/2001+1`

`<=>[x+1+2004]/2004+[x+2+2003]/2003-[x+3+2002]/2002-[x+4+2001]/2001=0`

`<=>[x+2005]/2004+[x+2005]/2003-[x+2005]/2002-[x+2005]/2001=0`

`<=>(x+2005)(1/2004+1/2003-1/2002-1/2001)=0`

     Mà `1/2004+1/2003-1/2002-1/2001 \ne 0`

  `=>x+2005=0`

`<=>x=-2005`

Vậy `S={-2005}`

\(\dfrac{x}{3}-\dfrac{2x+1}{6}=\dfrac{x}{6}-x\)

\(\Leftrightarrow\dfrac{2x-\left(2x+1\right)}{6}=\dfrac{x-6x}{6}\)

\(\Leftrightarrow2x-2x-1=x-6x\)

\(\Leftrightarrow-5x=-1\)

\(\Leftrightarrow x=\dfrac{1}{5}\)

Vậy \(S=\left\{\dfrac{1}{5}\right\}\)

08:43 :vvvv

Vì \(\widehat{MIA}=90^0\left(\text{góc nt chắn nửa đường tròn}\right)\) nên \(MI\perp IA\)

Xét \(\Delta MBP\) có \(\left\{{}\begin{matrix}PK\perp MB\left(PK\perp MN\right)\\MI\perp PB\left(MI\perp IA\right)\\\left\{H\right\}=PK\cap MI\end{matrix}\right.\) nên H là trực tâm 

Do đó \(HB\perp PM\)

Mà \(AM\perp PM\Rightarrow HB\text{//}AM\)

Vì \(HB\text{//}OA\Rightarrow\dfrac{PB}{PA}=\dfrac{HB}{OA}\)

Ta có \(\sin MPB=\sin MPA=\dfrac{MA}{PA}=\dfrac{2OA}{PA}\)

\(\Rightarrow\dfrac{1}{2}BP\cdot\sin MPB=\dfrac{PB\cdot\dfrac{2OA}{PA}}{2}=\dfrac{PB\cdot2OA}{2PA}=\dfrac{PB}{PA}\cdot OA=\dfrac{HB}{OA}\cdot OA=HB\left(đpcm\right)\)

mn ơi mình cần gấp ạ

1 you are => are you

2 fiveteen => fifteen

3 are => is

4 am => are

5 is => bỏ

6 thanks => thank

7 year => years

8 are => is

9 phong is => is phong

10 is => are

Câu 3 bài đầu là IS THIS PHONG ?

Bài 3 câu 3 là My name is Linh

Chúc bn học tốt nè

\(a,ĐK:x\ge2\\ PT\Leftrightarrow\sqrt{x-2}=x-2\\ \Leftrightarrow x-2=x^2-4x+4\\ \Leftrightarrow x^2-5x+6=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2\left(tm\right)\\x=3\left(tm\right)\end{matrix}\right.\\ b,ĐK:x\ge5\\ PT\Leftrightarrow x^2-2x+3=x^2-10x+25\\ \Leftrightarrow8x=22\Leftrightarrow x=\dfrac{11}{4}\left(ktm\right)\\ \Leftrightarrow x\in\varnothing\\ c,ĐK:x\ge\dfrac{1}{2}\\ PT\Leftrightarrow9\left(2x-1\right)=\left(2x-5\right)^2\\ \Leftrightarrow18x-9=4x^2-20x+25\\ \Leftrightarrow4x^2-38x+24=0\\ \Leftrightarrow2x^2-19x+12=0\\ \Delta=361-96=265\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{19-\sqrt{265}}{4}\left(tm\right)\\x=\dfrac{19+\sqrt{265}}{4}\left(tm\right)\end{matrix}\right.\)

\(d,ĐK:x\ge2\\ PT\Leftrightarrow\sqrt{\left(\sqrt{x-2}+1\right)^2}=\sqrt{x-2}+1\\ \Leftrightarrow\sqrt{x-2}+1-\sqrt{x-2}-1=0\\ \Leftrightarrow0=0\left(luôn.đúng\right)\Leftrightarrow x\in R;x\ge2\)

\(e,ĐK:x\ge3\\ PT\Leftrightarrow\sqrt{x-3}\left(\sqrt{x+3}-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-3=0\\x+3=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\left(tm\right)\\x=2\left(ktm\right)\end{matrix}\right.\Leftrightarrow x=3\\ f,ĐK:x\ge1\\ PT\Leftrightarrow3\sqrt{x-1}-2\sqrt{x-1}+4\sqrt{x-1}-3\sqrt{x-1}=16\\ \Leftrightarrow\sqrt{x-1}=8\\ \Leftrightarrow x-1=64\Leftrightarrow x=65\left(tm\right)\)

a,

c, Gọi \(\left(D_3\right):y=ax+b\) là đt cần tìm

\(\Leftrightarrow\left\{{}\begin{matrix}a=-2;b\ne0\\3x+3=ax+b,\forall x=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=-2\\-a+b=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=-2\\b=-2\end{matrix}\right.\)

Vậy \(\left(D_3\right):y=-2x-2\)