\(\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{x+6\sqrt{x}+9}{9-x}-\dfrac{\sqrt{x}}{\sqrt{x}+3}\left(dkxd:x\ge0,x\ne9\right)\)

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

\(=\dfrac{2\sqrt{x}\left(\sqrt{x}-3\right)-\left(x+6\sqrt{x}+9\right)-\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)

\(=\dfrac{2x-6\sqrt{x}-x-6\sqrt{x}-9-x+3\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)

\(=\dfrac{-9\sqrt{x}-9}{x-9}\) với \(x\ge0,x\ne9\)

Ta có: \(A=\sqrt{x}+1-\dfrac{17}{1-\sqrt{x}}\)

\(=\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\sqrt{x}-1}+\dfrac{17}{\sqrt{x}-1}\)

\(=\dfrac{x-1+17}{\sqrt{x}-1}\)

\(=\dfrac{x+16}{\sqrt{x}-1}\)

Ta có: \(B=\dfrac{x-7}{x-4\sqrt{x}+3}+\dfrac{1}{\sqrt{x}-1}-\dfrac{1}{\sqrt{x}-3}\)

\(=\dfrac{x-7}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-3\right)}+\dfrac{\sqrt{x}-3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-3\right)}-\dfrac{\sqrt{x}-1}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-1\right)}\)

\(=\dfrac{x-7+\sqrt{x}-3-\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-3\right)}\)

\(=\dfrac{x-9}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-3\right)}\)

\(=\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-3\right)}\)

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

Ta có: P=A:B

\(\Leftrightarrow P=\dfrac{x+16}{\sqrt{x}-1}:\dfrac{\sqrt{x}+3}{\sqrt{x}-1}\)

\(\Leftrightarrow P=\dfrac{x+16}{\sqrt{x}-1}\cdot\dfrac{\sqrt{x}-1}{\sqrt{x}+3}\)

\(\Leftrightarrow P=\dfrac{x+16}{\sqrt{x}+3}\)

Vừa nhầm X+16 nha không phải x-16

phần tìm GTN trong phần Bình luận bài của bạn Nguyễn Lê Phước Thịnh

a: \(P=\dfrac{x\sqrt{x}-3-2x+12\sqrt{x}-18-x-4\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+1\right)}\)

\(=\dfrac{x\sqrt{x}-3x+8\sqrt{x}-24}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+1\right)}\)

\(=\dfrac{x+8}{\sqrt{x}+1}\)

b: Khi x=3-2 căn 2 thì \(P=\dfrac{3-2\sqrt{2}+8}{\sqrt{2}-1+1}=\dfrac{11-2\sqrt{2}}{\sqrt{2}}\)

a) 1/2

b) 3/8

~~~~~~~~~hok tốt~~~~~~~~~~~~

A/ \(\frac{1}{2}\)

B/ \(\frac{3}{8}\)  

Bài 2:

a: 2/6x5/3=10/18=5/9

b: 11/9x5/10=55/90=11/18

c: 3/9x6/8=1/3x3/4=1/4

d: 4/9x12/16=48/144=1/3

e: 25/15x6/7=5/3x6/7=30/21=10/7

f: 6/10x15/20=90/200=9/20

Bài 1

4/5 x 6/7= 24/35

2/9 x 1/2= 2/18= 1/9

1/2 x 8/3= 8/6= 4/3

7/9 x 6/5= 42/45= 14/15

8/7 x 5/9= 40/63

10/11 x 22/15= 220/165= 4/3

Bài 2

2/6 x 5/3= 1/3 x 5/3=5/9

11/9 x 5/10= 11/9 x 1/2= 11/18

3/9 x 6/8= 1/3 x 3/4 =3/12= 1/4

4/9 x 12/16= 4/9 x 3/4= 12/36= 1/3

25/15 x 6/7= 5/3 x 6/7= 30/21= 10/7

6/10 x 15/20= 3/5 x 3/4= 9/20

bài 1

4/5 x 6/7 = 24/35

2/9 x 1/2 = 1/9

1/2 x 8/3 = 4/3

7/9 x 6/5 = 14/15

8/7 x 5/9 = 14/15

8/7 x 5/9 = 40/63

10/11 x 22/15 = 4/3

\(a)\frac{\sqrt{7}+\sqrt{5}}{\sqrt{7}-\sqrt{5}}+\frac{\sqrt{7}-\sqrt{5}}{\sqrt{7}+\sqrt{5}}\)

\(=\frac{\left(\sqrt{7}+\sqrt{5}\right)^2+\left(\sqrt{7}+\sqrt{5}\right)^2}{\left(\sqrt{7}-\sqrt{5}\right)\left(\sqrt{7}+\sqrt{5}\right)}\)

\(=\frac{7+2\sqrt{35}+5+7-2\sqrt{35}+5}{7-5}\)

\(=\frac{24}{2}\)

\(=12\)

\(b)\frac{4+\sqrt{2}-\sqrt{3}-\sqrt{6}+\sqrt{8}}{2+\sqrt{2}-\sqrt{3}}\)

\(=\frac{\left(2+\sqrt{2}-\sqrt{3}\right)+\left(2+\sqrt{8}-\sqrt{6}\right)}{2+\sqrt{2}-\sqrt{3}}\)

\(=\frac{\left(2+\sqrt{2}-\sqrt{3}\right)+\sqrt{2}\left(\sqrt{2}+2-\sqrt{3}\right)}{2+\sqrt{2}-\sqrt{3}}\)

\(=\frac{\left(2+\sqrt{2}-\sqrt{3}\right)\left(1+\sqrt{2}\right)}{2+\sqrt{2}-\sqrt{3}}\)

\(=1+\sqrt{2}\)

\(c)A=\left(\sqrt{3}+1\right)\sqrt{\frac{14-6\sqrt{3}}{5+\sqrt{3}}}\)

\(A=\left(\sqrt{3}+1\right)\sqrt{\frac{\left(14-6\sqrt{3}\right)\left(5-\sqrt{3}\right)}{\left(5+\sqrt{3}\right)\left(5-\sqrt{3}\right)}}\)

\(A=\left(\sqrt{3}+1\right)\sqrt{\frac{70-14\sqrt{3}-30\sqrt{3}+18}{25-3}}\)

\(A=\left(\sqrt{3}+1\right)\sqrt{\frac{88-44\sqrt{3}}{22}}\)

\(A=\left(\sqrt{3}+1\right)\sqrt{\frac{44\left(2-\sqrt{3}\right)}{22}}\)

\(A=\left(\sqrt{3}+1\right)\sqrt{2\left(2-\sqrt{3}\right)}\)

\(A=\left(\sqrt{3}+1\right)\sqrt{4+2\sqrt{3}}\)

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

\(A=\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)\)

\(A=3-1=2\)

P/s: nếu đề là vậy thì t ra kết quả như vậy ạ, nhưng lần sau khi đăng câu hỏi bạn nên viết rõ hơn ra nhé

(1+1+1)! = 3! = 6

2+2+2=6

3x3-3=6

\(\sqrt{4}+\sqrt{4}+\sqrt{4}=6\)

5:5+5=6

6+(6-6)=6

7-7:7=6

\(8-\sqrt{\sqrt{8+8}}=6\)

\(\sqrt{9}.\sqrt{9}-\sqrt{9}=6\)

Ta có: M=A+B

\(=\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}+1}{\sqrt{x}-3}+\dfrac{11\sqrt{x}-3}{x-9}\)

\(=\dfrac{2x-6\sqrt{x}+x+4\sqrt{x}+3+11\sqrt{x}-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)

\(=\dfrac{3x+9\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}=\dfrac{3\sqrt{x}}{\sqrt{x}-3}\)

`c)root{3}{4}.root{3}{1-sqrt3}.root{6}{(sqrt3+1)^2}`

`=root{3}{4(1-sqrt3)}.root{3}{1+sqrt3}`

`=root{3}{4(1-sqrt3)(1+sqrt3)}`

`=root{3}{4(1-3)}=-2`

`d)2/(root{3}{3}-1)-4/(root{9}-root{3}{3}+1)`

`=(2(root{3}{9}+root{3}{3}+1))/(3-1)-(4(root{3}{3}+1))/(3+1)`

`=root{3}{9}+root{3}{3}+1-root{3}{3}-1`

`=root{3}{9}`

`a)root{3}{8sqrt5-16}.root{3}{8sqrt5+16}`

`=root{3}{(8sqrt5-16)(8sqrt5+16)}`

`=root{3}{320-256}`

`=root{3}{64}=4`

`b)root{3}{7-5sqrt2}-root{6}{8}`

`=root{3}{1-3.sqrt{2}+3.2.1-2sqrt2}-root{6}{(2)^3}`

`=root{3}{(1-sqrt2)^3}-sqrt2`

`=1-sqrt2-sqrt2=1-2sqrt2`