1:

a: BC=8-3=5cm

b: MN=MC+CN=1/2(CA+CB)

=1/2*AB=4cm

2: 

a: Có 2 tia là OA và OB

b: AB=OB+OA=11cm

c: AC=BC=11/2=5,5cm

vì be là đường cao của tam giác abc :

=)góc bec = góc bea =90

vì ch là đường cao của tam giác abc (gt)

=) góc bch = góc cha =90

xét tam giác ahc và tam giác aeb ta có 

góc ahc =góc aeb( =90)

góc a chung

=) tam giác ahc ~TAM GIÁC EAB(g.g)

b, gọi i là giao điểm của be và ch 

xét tam giác hib và tam giác eic ta có 

h2 =e2 =90

i1=i2 ( 2gocs đối đỉnh)

=)hib ~eic (g.g)

do mk đang bận nên chỉ giải đc đến đây thôi ạ

\(P=\left(\dfrac{x\sqrt{x}-1}{x-\sqrt{x}}-\dfrac{x\sqrt{x}+1}{x+\sqrt{x}}\right):\left(\dfrac{2\left(x-2\sqrt{x}+1\right)}{x-1}\right)\left(x>0,x\ne1\right)\)

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

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

\(=\dfrac{2\sqrt{x}}{\sqrt{x}}:\dfrac{2\left(\sqrt{x}-1\right)}{\sqrt{x}+1}=2.\dfrac{\sqrt{x}+1}{2\left(\sqrt{x}-1\right)}=\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)

b) \(P=\dfrac{\sqrt{x}+1}{\sqrt{x}-1}=1+\dfrac{2}{\sqrt{x}-1}\)

Để \(P\in Z\Rightarrow2⋮\sqrt{x}-1\Rightarrow\sqrt{x}-1\in\left\{1;2;-1;-2\right\}\)

\(\Rightarrow\sqrt{x}\in\left\{2;3;0\right\}\Rightarrow x\in\left\{4;9;0\right\}\)

a) Với  x>0; x≠1 ta có:

\(P=\left(\dfrac{x\sqrt{x}-1}{x-\sqrt{x}}-\dfrac{x\sqrt{x}+1}{x+\sqrt{x}}\right).\left(\dfrac{2\left(x-2\sqrt{x}+1\right)}{x-1}\right)\)

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

\(=\left(\dfrac{x+\sqrt{x}+1}{\sqrt{x}}-\dfrac{x-\sqrt{x}+1}{\sqrt{x}}\right).\left(\dfrac{2\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)}\right)\)

\(=\left(\dfrac{2\sqrt{x}}{\sqrt{x}}\right).\left(\dfrac{2\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)}\right)=\dfrac{4x-4\sqrt{x}}{\sqrt{x}+1}\)

Vậy....

a) Ta có: \(P=\left(\dfrac{x\sqrt{x}-1}{x-\sqrt{x}}-\dfrac{x\sqrt{x}+1}{x+\sqrt{x}}\right):\left(\dfrac{2\left(x-2\sqrt{x}+1\right)}{x-1}\right)\)

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

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

\(=\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)

a. \(\dfrac{2^{11}.9^2}{3^5.16^2}\)

= \(\dfrac{1^{11}.3^2}{1^5.8^2}\)

= \(\dfrac{9}{64}\)

\(\dfrac{2^{11}.9^2}{3^5.16^2}=\dfrac{2^{11}.\left(3^2\right)^2}{3^5.\left(2^4\right)^2}=\dfrac{8}{3}\)

\(\dfrac{9^3.27^2}{6^2.3^{10}}=\dfrac{\left(3^2\right)^3.\left(3^3\right)^2}{\left(2.3\right)^2.3^{10}}=\dfrac{1}{2}\)

\(\dfrac{27^{15}.5^3.8^4}{25^2.81^{11}.2^{11}}=\dfrac{\left(3^3\right)^{15}.5^3.\left(2^3\right)^4}{\left(5^2\right)^2.\left(3^4\right)^{11}.2^{11}}=\dfrac{3.2}{5}=\dfrac{6}{5}\)

a:\(\dfrac{2^{11}\cdot9^2}{3^5\cdot16^2}=\dfrac{2^{11}\cdot3^4}{3^5\cdot2^8}=\dfrac{2^3}{3}=\dfrac{8}{3}\)

b: \(\dfrac{9^3\cdot27^2}{6^2\cdot3^{10}}=\dfrac{3^6\cdot3^6}{3^2\cdot3^{10}\cdot2^2}=\dfrac{1}{4}\)

c: \(\dfrac{27^{15}\cdot5^3\cdot8^4}{25^2\cdot81^{11}\cdot2^{11}}=\dfrac{3^{45}\cdot5^3\cdot2^{12}}{5^4\cdot3^{44}\cdot2^{11}}=\dfrac{1}{5}\cdot3\cdot2=\dfrac{6}{5}\)

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

Giusp mình với mọi người ơi!!!