Yêu cầu chứng minh của đề chưa rõ bạn nhé!

ΔABC vuông cân tại A

mà AM là đường trung tuyến

nên AM=MB=MC và AM\(\perp\)BC

Ta có: \(\widehat{AME}+\widehat{AMF}=\widehat{EMF}=90^0\)

\(\widehat{CMF}+\widehat{AMF}=90^0\)

Do đó: \(\widehat{AME}=\widehat{CMF}\)

Xét ΔAME và ΔCMF có

\(\widehat{MAE}=\widehat{MCF}\left(=45^0\right)\)

AM=CM

\(\widehat{AME}=\widehat{CMF}\)

Do đó: ΔAME=ΔCMF

=>AE=CF

a: \(\dfrac{4}{15}-\left(2,9-\dfrac{11}{15}\right)\)

\(=\dfrac{4}{15}-2,9+\dfrac{11}{15}\)

=1-2,9=-1,9

b: \(\left(-36,75\right)+\left(\dfrac{37}{10}-63,25\right)-\left(-6,3\right)\)

\(=-36,75-63,25+\dfrac{37}{10}+6,3\)

=-100+10

=-90

c: \(6,5-\left(-\dfrac{10}{71}\right)-\left(-\dfrac{7}{2}\right)-\dfrac{7}{17}\)

\(=6,5+\dfrac{10}{71}+3,5-\dfrac{7}{17}\)

\(=10+\dfrac{10}{71}-\dfrac{7}{17}=\dfrac{11743}{1207}\)

d: \(\left(-39,1\right)\cdot\dfrac{13}{25}-60,9\cdot\dfrac{13}{25}\)

\(=\dfrac{13}{25}\left(-39,1-60,9\right)\)

\(=\dfrac{13}{25}\cdot\left(-100\right)=-52\)

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

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

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

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

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

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

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) 

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

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)

a: \(A=\dfrac{x^2+2x}{x^2-4x+4}:\left(\dfrac{x+2}{x}-\dfrac{1}{2-x}+\dfrac{6-x^2}{x^2-2x}\right)\)

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

\(=\dfrac{x\left(x+2\right)}{\left(x-2\right)^2}:\dfrac{\left(x+2\right)\left(x-2\right)+x+6-x^2}{x\left(x-2\right)}\)

\(=\dfrac{x\left(x+2\right)}{\left(x-2\right)^2}\cdot\dfrac{x\left(x-2\right)}{x^2-4+x+6-x^2}\)

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

b: |2x+1|=3

=>\(\left[{}\begin{matrix}2x+1=3\\2x+1=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\left(nhận\right)\\x=-2\left(loại\right)\end{matrix}\right.\)

Khi x=1 thì \(A=\dfrac{1^2}{1-2}=-1\)

c: A<0

=>\(\dfrac{x^2}{x-2}< 0\)

mà \(x^2>0\forall x\) thỏa mãn ĐKXĐ

nên x-2<0

=>x<2

Kết hợp ĐKXĐ, ta được: \(\left\{{}\begin{matrix}x< 2\\x\notin\left\{0;-2\right\}\end{matrix}\right.\)

d: Để A là số nguyên thì \(x^2⋮x-2\)

=>\(x^2-4+4⋮x-2\)

=>\(4⋮x-2\)

=>\(x-2\in\left\{1;-1;2;-2;4;-4\right\}\)

=>\(x\in\left\{3;1;4;0;6;-2\right\}\)

Kết hợp ĐKXĐ, ta được: \(x\in\left\{3;1;4;6\right\}\)

e) 

\(A=\dfrac{x^2}{x-2}=\dfrac{x^2-4x+4+4x-4}{x-2}\\ =\dfrac{\left(x-2\right)^2}{x-2}+\dfrac{4x-4}{x-2}\\ =x-2+\dfrac{4x-8+4}{x-2}\\ =x-2+4+\dfrac{4}{x-2}\\ =x+2+\dfrac{4}{x-2}\\ =x-2+\dfrac{4}{x-2}+4\) 

Vì \(x>2\Rightarrow\left\{{}\begin{matrix}x-2>0\\\dfrac{4}{x-2}>0\end{matrix}\right.\)

Áp dụng bất đẳng thức cô-si ta có:

\(A\ge2\sqrt{\left(x-2\right)\cdot\dfrac{4}{x-2}}+4=2\cdot2+4=8\)

Dấu "=" xảy ra khi: \(x-2=\dfrac{4}{x-2}\Leftrightarrow\left(x-2\right)^2=4\Leftrightarrow\left[{}\begin{matrix}x=6\left(tm\right)\\x=0\left(ktm\right)\end{matrix}\right.\)

Vậy: ...

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Vd:\(3^{10}\) 

\(\dfrac{3^{10}\cdot15^5}{25^3\cdot9^7}=\dfrac{3^{10}\cdot\left(3\cdot5\right)^5}{\left(5^2\right)^3\cdot\left(3^2\right)^7}=\dfrac{3^{10}\cdot3^5\cdot5^5}{5^6\cdot3^{14}}\)

\(=\dfrac{3^{15}}{5\cdot3^{14}}=\dfrac{3}{5}\)

a/

$\frac{1}{x}+\frac{1}{y}=\frac{1}{5}$

$\Rightarrow \frac{x+y}{xy}=\frac{1}{5}$

$\Rightarrow 5(x+y)=xy$

$\Rightarrow 5x+5y-xy=0$

$\Rightarrow x(5-y)+5y=0$

$\Rightarrow x(5-y)-5(5-y)=-25$

$\Rightarrow (x-5)(5-y)=-25$

$\Rightarrow (x-5)(y-5)=25$

Do $x,y$ nguyên nên $x-5,y-5$ nguyên. Mà tích $(x-5)(y-5)=25$ nên xảy ra các TH sau đây:

TH1: $x-5=1, y-5=25\Rightarrow x=6; y=30$

TH2: $x-5=-1, y-5=-25\Rightarrow x=4; y=-20$

TH3: $x-5=25, y-5=1\Rightarrow x=30; y=6$

TH4: $x-5=-25, y-5=-1\Rightarrow x=-20; y=4$

TH5: $x-5=5, y-5=5\Rightarrow x=10; y=10$

TH6: $x-5=-5, y-5=-5\Rightarrow x=0; y=0$

b/

$\frac{2}{x}+\frac{1}{y}=3$

$\Rightarrow \frac{x+2y}{xy}=3$

$\Rightarrow x+2y=3xy$

$\Rightarrow 3xy-x-2y=0$

$\Rightarrow x(3y-1)-2y=0$

$\Rightarrow 3x(3y-1)-6y=0$

$\Rightarrow 3x(3y-1)-2(3y-1)=2$

$\Rightarrow (3x-2)(3y-1)=2$

Do $x,y$ nguyên nên $3x-2, 3y-1$ cũng là số nguyên. Mà tích của chúng bằng 2 nên ta xét các TH sau:

TH1: $3x-2=1, 3y-1=2\Rightarrow x=y=1$

TH2: $3x-2=2, 3y-1=1\Rightarrow x=\frac{4}{3}$ (loại) 

TH3: $3x-2=-1, 3y-1=-2\Rightarrow x=\frac{1}{3}$ (loại)

TH4: $3x-2=-2, 3y-1=-1\Rightarrow x=y=0$ (loại do $x,y\neq 0$)

Vậy $x=y=1$