a) \(Q=\dfrac{\left(x+2\right)^2}{x}\cdot\left(1-\dfrac{x^2}{x+2}\right)-\dfrac{x^2+10x+4}{x}\left(x\ne0;x\ne-2\right)\)

\(Q=\dfrac{\left(x+2\right)^2}{x}\cdot\dfrac{\left(x+2\right)-x^2}{x+2}-\dfrac{x^2+10x+4}{x}\)

\(Q=\dfrac{\left(x+2\right)^2}{x}\cdot\dfrac{-x^2+x+2}{x+2}-\dfrac{x^2+10x+4}{x}\)

\(Q=\dfrac{\left(x+2\right)\left(-x^2+x+2\right)}{x}-\dfrac{x^2+10x+4}{x}\)

\(Q=\dfrac{-x^3+x^2+2x-2x^2+2x+4-x^2-10x-4}{x}\)

\(Q=\dfrac{-x^3-2x^2-6x}{x}\)

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

\(Q=-x^2-2x-6\)

b) Ta có:

\(Q=-x^2-2x-6\)

\(Q=-\left(x^2+2x+6\right)\)

\(Q=-\left[\left(x^2+2x+1\right)+5\right]\)

\(Q=-\left(x+1\right)^2-5\)

Mà: \(-\left(x+1\right)^2\le0\forall x\)

\(\Rightarrow Q=-\left(x+1\right)^2-5\le-5\forall x\)

Dấu "=" xảy ra khi:

\(x+1=0\Rightarrow x=-1\)

Vậy: \(Q_{max}=-5\Leftrightarrow x=-1\)

a, \(-\dfrac{315}{540}\) = \(\dfrac{-315:45}{540:45}\) = \(\dfrac{-7}{12}\)        b,   \(\dfrac{25.13}{26.35}\) = \(\dfrac{25.13:5:13}{26.35:13:5}\) = \(\dfrac{5}{14}\)

c,     \(\dfrac{6.9-2.17}{63.3-119}\) = \(\dfrac{2.3.9-2.17}{7.9.3-7.17}\) = \(\dfrac{2.(27-17)}{7.(7-17)}\) = \(\dfrac{2}{7}\)

d, \(\dfrac{3.13-13.18}{15.40-80}\) = \(\dfrac{3.13(1-6)}{40.(15-2)}\) = \(\dfrac{-3.13.5}{40.13}\) = \(\dfrac{-15}{40}\) = \(\dfrac{-15:5}{40:5}\) = \(-\dfrac{3}{8}\)

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

Vì x>2 suy ra 2-x<0

Suy ra |2-x|=x-2

Vì x>2 suy ra x+1>3

Suy ra |x+1|=x+1

B=-x-1+x-2=-3

Vì x > 2 => 2 - x < 0

=> 2 - x = x - 2

Vì x > 2  => x + 1 > 3

=> x+1= x+1

B = -x - 1 + x - 2 =  -3

hok tốt

*YOUTUBER*

-2/3

7/13

3/5

TK
-2/3

7/13

3/5

tham khảo

-2/3

7/13

3/5

\(\dfrac{2sin8a-sin16a}{2sin8a+sin16a}=\dfrac{2sin8a-2sin8a.cos8a}{2sin8a+2sin8a.cos8a}=\dfrac{2sin8a\left(1-cos8a\right)}{2sin8a\left(1+cos8a\right)}=\dfrac{1-cos8a}{1+cos8a}=\dfrac{1-\left(1-2sin^24a\right)}{1+\left(1-2sin^24a\right)}=\dfrac{2sin^24a}{2-2sin^24a}=\dfrac{sin^24a}{1-sin^24a}=\dfrac{sin^24a}{cot^24a}=tan^24a\)

\(=\dfrac{2sin8a-2sin8a.cos8a}{2sin8a+2sin8a.cos8a}=\dfrac{2sin8a\left(1-cos8a\right)}{2sin8a\left(1+cos8a\right)}=\dfrac{1-cos8a}{1+cos8a}\)

\(=\dfrac{1-\left(1-2sin^24a\right)}{1+\left(2cos^24a-1\right)}=\dfrac{2sin^24a}{2cos^24a}=tan^24a\)

     2\(\sqrt{\dfrac{16}{3}}\)  - 3\(\sqrt{\dfrac{1}{27}}\) - \(\dfrac{3}{2\sqrt{3}}\)

= \(\dfrac{8}{\sqrt{3}}\) - \(\dfrac{3}{3\sqrt{3}}\)  - \(\dfrac{3}{2\sqrt{3}}\)

= \(\dfrac{8}{\sqrt{3}}\) - \(\dfrac{1}{\sqrt{3}}\) - \(\dfrac{3}{2\sqrt{3}}\)

= \(\dfrac{16}{2\sqrt{3}}\) - \(\dfrac{2}{2\sqrt{3}}\) - \(\dfrac{3}{2\sqrt{3}}\)

= \(\dfrac{11}{2\sqrt{3}}\)

= \(\dfrac{11\sqrt{3}}{6}\)

f, 2\(\sqrt{\dfrac{1}{2}}\)- \(\dfrac{2}{\sqrt{2}}\) + \(\dfrac{5}{2\sqrt{2}}\)

= \(\dfrac{2}{\sqrt{2}}\) - \(\dfrac{2}{\sqrt{2}}\) + \(\dfrac{5}{2\sqrt{2}}\)

= \(\dfrac{5}{2\sqrt{2}}\)

= \(\dfrac{5\sqrt{2}}{4}\)

(1 + \(\dfrac{3-\sqrt{3}}{\sqrt{3}-1}\)).(1- \(\dfrac{3+\sqrt{3}}{\sqrt{3}+1}\)) 

= \(\dfrac{\sqrt{3}-1+3-\sqrt{3}}{\sqrt{3}-1}\).\(\dfrac{\sqrt{3}+1-3+\sqrt{3}}{\sqrt{3}+1}\)

= \(\dfrac{2}{\sqrt{3}-1}\).\(\dfrac{-2}{\sqrt{3}+1}\)

= \(\dfrac{-4}{3-1}\)

= \(\dfrac{-4}{2}\)

= -2

   \(\dfrac{2}{\sqrt{6}-2}+\dfrac{2}{\sqrt{6}+2}+\dfrac{5}{\sqrt{6}}\)

= \(\dfrac{2.\left(\sqrt{6}+2\right)+2\left(\sqrt{6}-4\right)}{\left(\sqrt{6}-2\right)}\) + \(\dfrac{5}{\sqrt{6}}\)

= \(\dfrac{2\sqrt{6}+4+2\sqrt{6}-4}{6-4}\) + \(\dfrac{5\sqrt{6}}{6}\)

= \(\dfrac{4\sqrt{6}}{2}\) + \(\dfrac{5\sqrt{6}}{6}\)

= \(\dfrac{12\sqrt{6}+5\sqrt{6}}{6}\)

= \(\dfrac{17\sqrt{6}}{6}\)

\(1,\sqrt{4\left(a-4\right)^2}\left(dkxd:a\ge4\right)\)

\(=\sqrt{4}.\sqrt{\left(a-4\right)^2}\)

\(=\sqrt{2^2}.\left|a-4\right|\)

\(=2\left(a-4\right)\)

\(=2a-8\)

\(2,\sqrt{9\left(b-5\right)^2}\left(dkxd:b< 5\right)\)

\(=\sqrt{9}.\sqrt{\left(b-5\right)^2}\)

\(=\sqrt{3^2}.\left|b-5\right|\)

\(=3\left(-b+5\right)\)

\(=-3b+15\)

a) điều kiện xác định: x≠3 và x≠2

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

Tại x=13 ta có \(\dfrac{13+2}{13-3}\)=\(\dfrac{3}{2}\)

\(\left(x+2\right)^2-\left(x+2\right)\left(x-2\right)\\ =\left(x+2\right)\left(x+2-x+2\right)\\ =4\left(x+2\right)=4x+8\)

Ta có:(x+2)²-(x+2)(x-2)=(x+2)(x+2-x+2)=4(x+2)

`(x+2)^2-(x+2)(x-2)`

`=x^2+4x+4-(x^2-4)`

`=x^2+4x+4-x^2+4`

`=4x+8`