a: \(BM=\dfrac{1}{4}BC\)

\(BN=\dfrac{1}{2}BC\)(N là trung điểm của BC)

Do đó: BN=2BM

=>M là trung điểm của BN

=>MB=MN

Xét ΔMBE và ΔMNA có

MB=MN

\(\widehat{BME}=\widehat{NMA}\)(hai góc đối đỉnh)

ME=MA

Do đó: ΔMBE=ΔMNA

=>\(\widehat{MBE}=\widehat{MNA}\)

=>BE//NA

Xét ΔMAB và ΔMEN có

MA=ME

\(\widehat{AMB}=\widehat{EMN}\)(hai góc đối đỉnh)

MB=MN

Do đó: ΔMAB=ΔMEN

=>AB=EN

1

\(A=\left(\dfrac{x+1}{x^3-1}-\dfrac{1}{x-1}\right)\left(\dfrac{x+2}{x-1}-\dfrac{1}{x}\right)\left(x\ne1;0\right)\\ =\left[\dfrac{x+1}{\left(x-1\right)\left(x^2+x+1\right)}-\dfrac{x^2+x+1}{\left(x-1\right)\left(x^2+x+1\right)}\right]\left[\dfrac{x\left(x+2\right)}{x\left(x-1\right)}-\dfrac{x-1}{x\left(x-1\right)}\right]\\ =\dfrac{x+1-x^2-x-1}{\left(x-1\right)\left(x^2+x+1\right)}\cdot\dfrac{x^2+2x-x+1}{x\left(x-1\right)}\\ =\dfrac{-x^2}{\left(x-1\right)\left(x^2+x+1\right)}\cdot\dfrac{x^2+x+1}{x\left(x-1\right)}\\ =\dfrac{-x}{\left(x-1\right)^2}\\ =\dfrac{-x}{x^2-2x+1}\)

ĐKXĐ: \(x\notin\left\{1;0\right\}\)

\(A=\left(\dfrac{x+1}{x^3-1}-\dfrac{1}{x-1}\right)\left(\dfrac{x+2}{x-1}-\dfrac{1}{x}\right)\)

\(=\left(\dfrac{x+1}{\left(x-1\right)\left(x^2+x+1\right)}-\dfrac{1}{x-1}\right)\cdot\left(\dfrac{x\left(x+2\right)-x+1}{x\left(x-1\right)}\right)\)

\(=\dfrac{x+1-x^2-x-1}{\left(x-1\right)\left(x^2+x+1\right)}\cdot\dfrac{x^2+x+1}{x\left(x-1\right)}\)

\(=\dfrac{-x^2}{\left(x-1\right)\cdot x\left(x-1\right)}=\dfrac{-x}{\left(x-1\right)^2}\)

$2^{4-x}=128$

$\Rightarrow 2^{4-x}=2^7$

$\Rightarrow 4-x=7$

$\Rightarrow x=4-7$

$\Rightarrow x=-3$

\(2^{4-x}=128\)

\(2^{4-x}=2^7\)

\(4-x=7\)

      \(x=4-7\)

      \(x=-3\)

a: Xét (O) có

CM,CA là các tiếp tuyến

Do đó: CM=CA và OC là phân giác của góc MOA

Xét (O) có

DM,DB là các tiếp tuyến

Do đó: DM=DB và OD là phân giác của góc MOB

AC+BD

=CM+MD

=CD
b: \(\widehat{COD}=\widehat{COM}+\widehat{DOM}=\dfrac{1}{2}\cdot\widehat{MOA}+\dfrac{1}{2}\cdot\widehat{MOB}\)

\(=\dfrac{1}{2}\left(\widehat{MOA}+\widehat{MOB}\right)=\dfrac{1}{2}\cdot\widehat{AOB}=90^0\)

=>ΔCOD vuông tại O

c: Xét ΔCOD vuông tại O có OM là đường cao

nên \(OM^2=MC\cdot MD\)

giúp tôi ý d với bạn ơi

\(M=\dfrac{1}{1000}+\dfrac{1}{1002}+\dfrac{1}{1004}+...+\dfrac{1}{2000}\)

\(2M=\dfrac{1}{500}+\dfrac{1}{501}+\dfrac{1}{502}+...+\dfrac{1}{1000}\)

\(2M< \dfrac{1}{500}+\dfrac{1}{500}+\dfrac{1}{500}+...+\dfrac{1}{500}=\dfrac{500}{500}=1\)

\(M< \dfrac{1}{2}\)

\(\dfrac{3^{10}.15^5}{25^3.9^7}\)

\(=\dfrac{3^{10}.3^55^5}{\left(5^2\right)^3.\left(3^2\right)^7}\)

\(=\dfrac{3^{15}.5^5}{5^6.3^{14}}\)

\(=\dfrac{3.1}{5.1}\)

\(=\dfrac{3}{5}\)

a)

\(P=\left(\dfrac{2+x}{2-x}-\dfrac{4x^2}{x^2-4}-\dfrac{2-x}{2+x}\right):\dfrac{x^2-3x}{2x^2-x^3}\\ =\left(\dfrac{2+x}{2-x}+\dfrac{4x^2}{4-x^2}-\dfrac{2-x}{2+x}\right):\dfrac{x\left(x-3\right)}{x^2\left(2-x\right)}\\ =\left[\dfrac{\left(2+x\right)^2}{\left(2+x\right)\left(2-x\right)}+\dfrac{4x^2}{\left(2+x\right)\left(2-x\right)}-\dfrac{\left(2-x\right)^2}{\left(2+x\right)\left(2-x\right)}\right]:\dfrac{x-3}{x\left(2-x\right)}\\ =\dfrac{\left(2+x\right)^2+4x^2-\left(2-x\right)^2}{\left(2+x\right)\left(2-x\right)}:\dfrac{x-3}{x\left(2-x\right)}\\ =\dfrac{4+4x+x^2+4x^2-4+4x-x^2}{\left(2+x\right)\left(2-x\right)}\cdot\dfrac{x\left(2-x\right)}{x-3}\\ =\dfrac{4x^2+8x}{\left(2+x\right)\left(2-x\right)}\cdot\dfrac{x\left(2-x\right)}{x-3}\\ =\dfrac{4x\left(x+2\right)}{\left(2+x\right)\left(2-x\right)}\cdot\dfrac{x\left(2-x\right)}{x-3}\\ =\dfrac{4x^2}{x-3}\)

b) \(\left|x-5\right|=2\Leftrightarrow\left[{}\begin{matrix}x-5=2\left(x\ge5\right)\\x-5=-2\left(x< 5\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=7\left(tm\right)\\x=3\left(ktm\right)\end{matrix}\right.\)

\(\Rightarrow P=\dfrac{4\cdot7^2}{7-3}=\dfrac{4\cdot7^2}{4}=7^2=49\)

c) \(P>0\Rightarrow\dfrac{4x^2}{x-3}>0\)

Mà: \(4x^2\ge0\forall x\) 

=> \(x-3>0\Leftrightarrow x>3\) 

d) 

\(P=\dfrac{4x^2}{x-3}=\dfrac{4\left(x^2-6x+9\right)+24x-36}{x-3}\\ =4\left(x-3\right)+\dfrac{24x-36}{x-3}=4\left(x-3\right)+\dfrac{24\left(x-3\right)+36}{x-3}\\ =4\left(x-3\right)+\dfrac{36}{x-3}+24\)
Mà: x > 3 \(\Rightarrow\left\{{}\begin{matrix}4\left(x-3\right)>0\\\dfrac{36}{x-3}>0\end{matrix}\right.\)

Áp dụng bđt cô-si ta có:

\(P\ge2\sqrt{4\left(x-3\right)\cdot\dfrac{36}{x-3}}+24=2\sqrt{4\cdot36}+24=24+24=48\)

Dấu "=" xảy ra khi: 

\(4\left(x-3\right)=\dfrac{36}{x-3}\\ \Leftrightarrow\left(x-3\right)^2=9\\ \Leftrightarrow\left[{}\begin{matrix}x=6\left(tm\right)\\x=0\left(ktm\right)\end{matrix}\right.\)  

Vậy: ... 

e) 

\(P=-8\Rightarrow\dfrac{4x^2}{x-3}=-8\\ \Leftrightarrow4x^2+8\left(x-3\right)=0\\ \Leftrightarrow4x^2+8x-24=0\\ \Leftrightarrow\left(4x^2+8x+4\right)-28=0\\ \Leftrightarrow4\left(x+1\right)^2=28\\ \Leftrightarrow\left(x+1\right)^2=7\\ \Leftrightarrow\left[{}\begin{matrix}x+1=\sqrt{7}\\x+1=-\sqrt{7}\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\sqrt{7}-1\\x=-\sqrt{7}-1\end{matrix}\right.\)

a)

b) |x - 5| = 2

*) Với x 5, ta có:

|x - 5| = 2

x - 5 = 2

x = 2 + 5

x = 7 (nhận)

*) Với x < 5, ta có:

|x - 5| = 2

5 - x = 2

x = 5 - 2

x = 3 (nhận)

+) Với x = 7, ta có:

 +) Với x = 3, ta có mẫu thức là 3 - 3 = 0 nên P không xác định tại x = 3

a) \(mx-y=2m-1\Leftrightarrow y=mx-2m+1\left(a=m;b=-2m+1\right)\)
Để (d) đi qua góc tọa độ thì: \(b=0\Rightarrow-2m+1=0\Leftrightarrow m=\dfrac{1}{2}\)

b) 

\(mx-y=2m-1\\ \Leftrightarrow mx-2m-y+1=0\\ \Leftrightarrow m\left(x-2\right)-\left(y-1\right)=0\\ \Leftrightarrow m\left(x_0-2\right)-\left(y_0-1\right)=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x_0=2\\y_0=1\end{matrix}\right.\)

=> (d) luôn đi qua điểm (2;1) cố định

c) (d) cắt trục tung tại: \(\left(0;-2m+1\right)=>AO=\left|-2m+1\right|\) 

(d) cắt trục tung tại: \(\left(\dfrac{2m-1}{m};0\right)=>BO=\left|\dfrac{2m-1}{m}\right|\)

\(S_{AOB}=\dfrac{1}{2}AO\cdot OB=\dfrac{1}{2}\left|-2m+1\right|\cdot\left|\dfrac{2m-1}{m}\right|\) 

\(=\left|\dfrac{1}{2}\left(-2m+1\right)\dfrac{2m-1}{m}\right|\\ =\left|\dfrac{1}{2}\cdot\dfrac{-\left(2m-1\right)^2}{m}\right|\\ =\left|\dfrac{-\left(2m-1\right)^2}{2m}\right|=\dfrac{\left(2m-1\right)^2}{\left|2m\right|}\)  

Để \(S_{AOB}=4=>\dfrac{\left(2m-1\right)^2}{\left|2m\right|}=4\Leftrightarrow4m^2-4m+1=8\left|m\right|\)

TH1: m≥0 \(\Rightarrow4m^2-12m+1=0\Leftrightarrow\left[{}\begin{matrix}m=\dfrac{3+2\sqrt{2}}{2}\\m=\dfrac{3-2\sqrt{2}}{2}\end{matrix}\right.\left(tm\right)\)

TH2: m<0 \(\Rightarrow4m^2+4m+1=\Leftrightarrow m=-\dfrac{1}{2}\left(tm\right)\)

a) 

\(B=\left(\dfrac{9-3x}{x^2+4x-5}-\dfrac{x+5}{1-x}-\dfrac{x+1}{x+5}\right):\dfrac{7x-14}{x^3-1}\\ =\left[\dfrac{9-3x}{\left(x-1\right)\left(x+5\right)}+\dfrac{x+5}{x-1}-\dfrac{x+1}{x+5}\right]:\dfrac{7\left(x-2\right)}{\left(x-1\right)\left(x^2+x+1\right)}\\ =\left[\dfrac{9-3x}{\left(x-1\right)\left(x+5\right)}+\dfrac{\left(x+5\right)^2}{\left(x-1\right)\left(x+5\right)}-\dfrac{\left(x+1\right)\left(x-1\right)}{\left(x-1\right)\left(x+5\right)}\right]:\dfrac{7\left(x-2\right)}{\left(x-1\right)\left(x^2+x+1\right)}\\ =\dfrac{9-3x+\left(x^2+10x+25\right)-\left(x^2-1\right)}{\left(x-1\right)\left(x+5\right)}\cdot\dfrac{\left(x-1\right)\left(x^2+x+1\right)}{7\left(x-2\right)}\\ =\dfrac{9-3x+x^2+10x+25-x^2+1}{\left(x-1\right)\left(x+5\right)}\cdot\dfrac{\left(x-1\right)\left(x^2+x+1\right)}{7\left(x-2\right)}\\ =\dfrac{7x+35}{x+5}\cdot\dfrac{x^2+x+1}{7\left(x-2\right)}\\ =\dfrac{7\left(x+5\right)}{x+5}\cdot\dfrac{x^2+x+1}{7\left(x-2\right)}\\ =\dfrac{x^2+x+1}{x-2}\)

b) 

\(\left(x+5\right)^2-9x-45=0\\ \Leftrightarrow\left(x+5\right)^2-9\left(x+5\right)=0\\ \Leftrightarrow\left(x+5\right)\left(x+5-9\right)=0\\ \Leftrightarrow\left(x+5\right)\left(x-4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-5\left(ktm\right)\\x=4\left(tm\right)\end{matrix}\right.\)
Thay `x=4` vào B ta có: 

\(B=\dfrac{4^2+4+1}{4-2}=\dfrac{16+4+1}{2}=\dfrac{21}{2}\) 

c) 

\(B=\dfrac{x^2+x+1}{x-2}=\dfrac{x^2-4x+4+5x-3}{x-2}=x-2+\dfrac{5x-3}{x-2}\\ =x-2+\dfrac{5\left(x-2\right)+7}{x-2}=x+3+\dfrac{7}{x-2}\) 

Để B nguyên thì 7 ⋮ x - 2

=> x - 2 ∈ Ư(7) = {1;-1;7;-7}

=> x ∈ {3; 1; 8; -5} 

kết hợp với đk: => x ∈ {3;8}

d) 

\(B=-\dfrac{3}{4}=>\dfrac{x^2+x+1}{x-2}=-\dfrac{3}{4}\\ \Leftrightarrow-4\left(x^2+x+1\right)=3\left(x-2\right)\\ \Leftrightarrow-4x^2-4x-4-3x+6=0\\ \Leftrightarrow-4x^2-7x+2=0\\ \Leftrightarrow4x^2+7x-2=0\\ \Leftrightarrow\left(4x^2-x\right)+\left(8x-2\right)=0\\ \Leftrightarrow x\left(4x-1\right)+2\left(4x-1\right)=0\\ \Leftrightarrow\left(x+2\right)\left(4x-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{1}{4}\end{matrix}\right.\left(tm\right)\) 

e) \(B< 0\Rightarrow\dfrac{x^2+x+1}{x-2}< 0\)

Mà: \(x^2+x+1=\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0\forall x\) 

\(\Rightarrow x-2< 0\Leftrightarrow x< 2\)

Kết hợp với đk: \(x\ne-5;x< 2\)

f) 

\(M=\dfrac{2}{x-2}:B=\dfrac{2}{x-2}:\dfrac{x^2+x+1}{x-2}\\ =\dfrac{2}{x-2}\cdot\dfrac{x-2}{x^2+x+1}\\ =\dfrac{2}{x^2+x+1}\) 

g) 

Ta có: \(B=x+3+\dfrac{7}{x-2}=\left(x-2\right)+\dfrac{7}{x-2}+6\)

Vì x > 2 \(\Rightarrow\left\{{}\begin{matrix}x-2>0\\\dfrac{7}{x-2}>0\end{matrix}\right.\)

Áp dụng bất đẳng thức cô-si ta có: 

\(B\ge2\sqrt{\left(x-2\right)\cdot\dfrac{7}{\left(x-2\right)}}+6=2\sqrt{7}+6\)

Dấu "=" xảy ra: \(x-2=\dfrac{7}{x-2}\Leftrightarrow\left(x-2\right)^2=7\Leftrightarrow\left[{}\begin{matrix}x=\sqrt[]{7}+2\left(tm\right)\\x=2-\sqrt{7}\left(ktm\right)\end{matrix}\right.\)

ĐKXĐ: \(x\notin\left\{1;-5;2\right\}\)

a: \(B=\left(\dfrac{9-3x}{x^2+4x-5}-\dfrac{x+5}{1-x}-\dfrac{x+1}{x+5}\right):\dfrac{7x-14}{x^3-1}\)

\(=\left(\dfrac{9-3x}{\left(x+5\right)\left(x-1\right)}+\dfrac{\left(x+5\right)^2-\left(x+1\right)\left(x-1\right)}{\left(x+5\right)\left(x-1\right)}\right)\cdot\dfrac{\left(x-1\right)\left(x^2+x+1\right)}{7x-14}\)

\(=\dfrac{9-3x+x^2+10x+25-x^2+1}{\left(x+5\right)\left(x-1\right)}\cdot\dfrac{\left(x-1\right)\left(x^2+x+1\right)}{7x-14}\)

\(=\dfrac{7x+35}{\left(x+5\right)}\cdot\dfrac{x^2+x+1}{7x-14}=\dfrac{7\left(x^2+x+1\right)}{7x-14}=\dfrac{x^2+x+1}{x-2}\)

b: \(\left(x+5\right)^2-9x-45=0\)

=>\(\left(x+5\right)^2-9\left(x+5\right)=0\)

=>(x+5)(x-4)=0

=>\(\left[{}\begin{matrix}x=-5\left(loại\right)\\x=4\left(nhận\right)\end{matrix}\right.\)

Thay x=4 vào B, ta được:

\(B=\dfrac{4^2+4+1}{4-2}=\dfrac{21}{2}\)

c: Để B nguyên thì \(x^2+x+1⋮x-2\)

=>\(x^2-2x+3x-6+7⋮x-2\)

=>\(7⋮x-2\)

=>\(x-2\in\left\{1;-1;7;-7\right\}\)

=>\(x\in\left\{3;1;9;-5\right\}\)

Kết hợp ĐKXĐ, ta được: \(x\in\left\{3;9\right\}\)

d: \(B=\dfrac{-3}{4}\)

=>\(\dfrac{x^2+x+1}{x-2}=\dfrac{-3}{4}\)

=>\(4\left(x^2+x+1\right)=-3\left(x-2\right)\)

=>\(4x^2+4x+4+3x-6=0\)

=>\(4x^2+7x-2=0\)

=>(x+2)(4x-1)=0

=>\(\left[{}\begin{matrix}x=-2\left(nhận\right)\\x=\dfrac{1}{4}\left(nhận\right)\end{matrix}\right.\)

e: Để B<0 thì \(\dfrac{x^2+x+1}{x-2}< 0\)

mà \(x^2+x+1=\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0\forall x\) thỏa mãn ĐKXĐ

nên x-2<0

=>x<2

Kết hợp ĐKXĐ, ta được: \(\left\{{}\begin{matrix}x< 2\\x\notin\left\{-5;1\right\}\end{matrix}\right.\)

f: \(M=\dfrac{2}{x-2}:B=\dfrac{2}{x-2}:\dfrac{x^2+x+1}{x-2}=\dfrac{2}{x^2+x+1}\)

\(=\dfrac{2}{\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}}\)

\(\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}>=\dfrac{3}{4}\forall x\) thỏa mãn ĐKXĐ

=>\(M=\dfrac{2}{\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}}< =2:\dfrac{3}{4}=\dfrac{8}{3}\forall x\) thỏa mãn ĐKXĐ

Dấu '=' xảy ra khi x+1/2=0

=>x=-1/2(nhận)