`15)(-5sqrtx+4)/(3sqrtx-2)+(6sqrtx+4)/(2sqrtx+3)+(29sqrtx-28)/(3(6x+5sqrtx-6))`
`=(3(-5sqrtx+4)(2sqrtx+3)+3(6sqrtx+4)(3sqrtx-2)+29sqrtx-28)/(3(3sqrtx-2)(2sqrtx+3))`
`=(-30x-21sqrtx+36+54x-24+29sqrtx-28)/(3(3sqrtx-2)(2sqrtx+3))`
`=(24x+8sqrtx-16)/(3(3sqrtx-2)(2sqrtx+3))`

`=(8(3sqrtx-2)(sqrtx+1))/(3(3sqrtx-2)(2sqrtx+3))`

`=(8(sqrtx+1))/(3(2sqrtx+3))`

thank

8: Ta có: \(\dfrac{\sqrt{x}-3}{\sqrt{x}-2}-\dfrac{2\sqrt{x}-1}{\sqrt{x}-1}+\dfrac{x-2}{x-3\sqrt{x}+2}\)

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

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

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

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

16. \(\dfrac{2\sqrt{x}-4}{3\sqrt{x}-4}-\dfrac{4+2\sqrt{x}}{\sqrt{x}-2}+\dfrac{x+13\sqrt{x}-20}{3x-10\sqrt{x}+8}\)

=\(\dfrac{\left(2\sqrt{x}-4\right)\left(\sqrt{x}-2\right)-\left(4+2\sqrt{x}\right)\left(3\sqrt{x}-4\right)+x+13\sqrt{x}-20}{\left(3\sqrt{x}-4\right)\left(\sqrt{x}-2\right)}\)

=\(\dfrac{2x-8\sqrt{x}+8-\left(4\sqrt{x}+6x-16\right)+x+13\sqrt{x}-20}{\left(3\sqrt{x}-4\right)\left(\sqrt{x}-2\right)}\)

=​\(\dfrac{2x-8\sqrt{x}+8-4\sqrt{x}-6x+16+x+13\sqrt{x}-20}{\left(3\sqrt{x}-4\right)\left(\sqrt{x}-2\right)}\)​​​

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

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

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

14.

=\(\dfrac{-\left(7\sqrt{x}+7\right)\left(\sqrt{x}+2\right)}{\left(5\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)+\(\dfrac{\left(2\sqrt{x}-2\right)\left(5\sqrt{x}-1\right)}{\left(5\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)+\(\dfrac{39\sqrt{x}+12}{\left(5\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)

=\(\dfrac{-7x-21\sqrt{x}-14}{\left(5\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)+\(\dfrac{10x-12\sqrt{x}+2}{\left(5\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)+​​​​​​\(\dfrac{39\sqrt{x}+12}{\left(5\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)​​​​​​​​

=\(\dfrac{-7x-21\sqrt{x}-14+10x-12\sqrt{x}+2+39\sqrt{x}+12}{\left(5\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)

=\(\dfrac{3x-6\sqrt{x}}{\left(5\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)=\(\dfrac{3\sqrt{x}\left(\sqrt{x}-2\right)}{\left(5\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)

17.

\(\dfrac{3}{1-\sqrt{2}}+\dfrac{\sqrt{2}-1}{\sqrt{2}+1}=\dfrac{3\left(\sqrt{2}+1\right)-\left(\sqrt{2}-1\right)^2}{-1}=-\left(3\sqrt{2}+3-3+2\sqrt{2}\right)=-5\sqrt{2}\)

\(\dfrac{\sqrt{5}-1}{\sqrt{5}+1}+\dfrac{6}{1-\sqrt{5}}=\dfrac{\left(\sqrt{5}-1\right).\left(1-\sqrt{5}\right)+6.\left(\sqrt{5}+1\right)}{-4}=\dfrac{6-2\sqrt{5}-6\sqrt{5}-6}{4}=\dfrac{-8\sqrt{5}}{4}=-2\sqrt{5}\)

\(\dfrac{\sqrt{2}-\sqrt{3}}{2-\sqrt{6}}+\dfrac{\sqrt{3}-\sqrt{2}}{\sqrt{6}+2}=\dfrac{\left(\sqrt{2}-\sqrt{3}\right).\left(\sqrt{6}+2\right)+\left(\sqrt{3}-\sqrt{2}\right).\left(2-\sqrt{6}\right)}{-2}=\dfrac{2\left(\sqrt{12}-\sqrt{18}\right)}{-2}=\sqrt{18}-\sqrt{12}\)

\(\dfrac{-31+8\sqrt{x}-x}{x-8\sqrt{x}+15}-\dfrac{\sqrt{x}+5}{\sqrt{x}-3}-\dfrac{3\sqrt{x}-1}{5-\sqrt{x}}\)

\(=\dfrac{-31+8\sqrt{x}-x}{\left(\sqrt{x}-5\right)\left(\sqrt{x}-3\right)}-\dfrac{\sqrt{x}+5}{\sqrt{x}-3}+\dfrac{3\sqrt{x}-1}{\sqrt{x}-5}\)

\(=\dfrac{-31+8\sqrt{x}-x-x+25+3x-9\sqrt{x}-\sqrt{x}+3}{\left(\sqrt{x}-5\right)\left(\sqrt{x}-3\right)}=\dfrac{x-2\sqrt{x}-3}{\left(\sqrt{x}-5\right)\left(\sqrt{x}-3\right)}=\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-5\right)\left(\sqrt{x}-3\right)}=\dfrac{\sqrt{x}+1}{\sqrt{x}-5}\)

14, \(\frac{-7\sqrt{x}+7}{5\sqrt{x}-1}+\frac{2\sqrt{x}-2}{\sqrt{x}+2}+\frac{39\sqrt{x}+12}{5x+9\sqrt{x}-2}\)

\(=\frac{-7\sqrt{x}+7}{5\sqrt{x}-1}+\frac{2\sqrt{x}-2}{\sqrt{x}+2}+\frac{39\sqrt{x}+12}{\left(\sqrt{x}+2\right)\left(5\sqrt{x}-1\right)}\)

\(=\frac{\left(-7\sqrt{x}+7\right)\left(\sqrt{x}+2\right)+\left(2\sqrt{x}-2\right)\left(5\sqrt{x}-1\right)+39\sqrt{x}+12}{\left(5\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)

\(=\frac{-7x-14\sqrt{x}+7\sqrt{x}+14+10x-2\sqrt{x}-10\sqrt{x}+2+39\sqrt{x}+12}{\left(5\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)

\(=\frac{3x+20\sqrt{x}+28}{\left(5\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)

\(=\frac{\left(3\sqrt{x}+14\right)\left(\sqrt{x}+2\right)}{\left(5\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)

\(=\frac{3\sqrt{x}+14}{5\sqrt{x}-1}\)

thank

ĐKXĐ : \(\left\{{}\begin{matrix}x\ge0\\x\ne49\end{matrix}\right.\)

Ta có : \(\dfrac{7\sqrt{x}-1}{\sqrt{x}-7}-\dfrac{6\sqrt{x}+1}{\sqrt{x}+1}+\dfrac{1-55\sqrt{x}}{\left(\sqrt{x}-7\right)\left(\sqrt{x}+1\right)}\)

\(=\dfrac{\left(\sqrt{x}+1\right)\left(7\sqrt{x}-1\right)-\left(\sqrt{x}-7\right)\left(6\sqrt{x}+1\right)+1-55\sqrt{x}}{\left(\sqrt{x}-7\right)\left(\sqrt{x}+1\right)}\)

\(=\dfrac{7x+7\sqrt{x}-\sqrt{x}-1-\left(6x-42\sqrt{x}+\sqrt{x}-7\right)+1-55\sqrt{x}}{\left(\sqrt{x}-7\right)\left(\sqrt{x}+1\right)}\)

\(=\dfrac{7x+7\sqrt{x}-\sqrt{x}-1-6x+42\sqrt{x}-\sqrt{x}+7+1-55\sqrt{x}}{\left(\sqrt{x}-7\right)\left(\sqrt{x}+1\right)}\)

\(=\dfrac{x-8\sqrt{x}+7}{\left(\sqrt{x}-7\right)\left(\sqrt{x}+1\right)}=\dfrac{\left(\sqrt{x}-7\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-7\right)\left(\sqrt{x}+1\right)}=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\)

19) Ta có: \(\dfrac{7\sqrt{x}-1}{\sqrt{x}-7}-\dfrac{6\sqrt{x}+1}{\sqrt{x}+1}+\dfrac{1-55\sqrt{x}}{x-6\sqrt{x}-7}\)

\(=\dfrac{\left(7\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-7\right)\left(\sqrt{x}+1\right)}-\dfrac{\left(6\sqrt{x}+1\right)\left(\sqrt{x}-7\right)}{\left(\sqrt{x}-7\right)\left(\sqrt{x}+1\right)}+\dfrac{1-55\sqrt{x}}{\left(\sqrt{x}-7\right)\left(\sqrt{x}+1\right)}\)

\(=\dfrac{7x+7\sqrt{x}-\sqrt{x}-1-6x+42\sqrt{x}-\sqrt{x}+7+1-55\sqrt{x}}{\left(\sqrt{x}-7\right)\left(\sqrt{x}+1\right)}\)

\(=\dfrac{x-8\sqrt{x}+7}{\left(\sqrt{x}-7\right)\left(\sqrt{x}+1\right)}\)

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

`19)(7sqrtx-1)/(sqrtx-7)-(6sqrtx+1)/(sqrtx+1)+(1-55sqrtx)/(x-6sqrtx-7)`

`=((7sqrtx-1)(sqrtx+1)-(6sqrtx+1)(sqrtx-7)+1-55sqrtx)/((sqrtx+1)(sqrtx-7))`

`=(7x+6sqrtx-1-6x+41sqrtx+7+1-55sqrtx)/((sqrtx+1)(sqrtx-7))`

`=(x-8sqrtx+7)/((sqrtx+1)(sqrtx-7))`

`=((sqrtx-1)(sqrtx-7))/((sqrtx+1)(sqrtx-7))`

`=(sqrtx-1)/(sqrtx+1)`

a) 7.111-777=777-777=0

b) -2+(-2)+...+(-2)+22=(-2).5+22=-10+22=12

c) Số các số hạng là:(108-10):2+1=50

G.trị biểu thức trên là: (108+10).(50:2) = 118.25 = 2950

d) -1+(-1)+(-1)+...+(-1)+10+11+77

=(-1).5+98=(-5)+98+93

Nhớ tk cho mk nha!

a, ta có 7+7+7+7+...........+7(111 số 7)=7x 111=777

ta có 777-777=0

b,ta sẽ có;2-4=-2;6-8=-2;.........;18-20=-2+22

=(-2)+(-2) +(-2)+(-2)+(-2 )+22=(-10)+22

                                          =12

c,10+12+14+16+....+108=(108+10)x[(108-10):+1):2=2950

d,95

a,0

b,12

c,2950

d,71

các bạn nhơ cho mình nhé

Chụp dọc đi bn oi!Chụp ngang thế này gãy cổ mất thôi!

Số số hạng là: (19-11)*1+1=9(số hạng) 

Có số cặp số là:9:2=4 dư 1

Vậy có 4 cặp và dư 1 số

Ta có : 11+(12+19)+(13+18)+(14+17)+(15+16)

=11+21+21+21+21

=11+(21.4)

=11+84

=95

11+12+13+14+15+16+17+18+19=(11+19)+(12+18)+(13+17)+(16+14)+15    

                                                     =30+30+30+30+15

                                                      =30*4+15                                                

                                                      =120+15                                              

                                                      =135 

11 + 12 + 13 + 14 + 15 + 16 + 17 + 18 + 19

= ( 11 + 19 ) + ( 12 + 18 ) + ( 13 + 17 ) + ( 14 + 16 ) + 15

= 30 + 30 + 30 + 30 + 15

= 135

`11)1/(3+sqrt5)+1/(sqrt5-3)=(3-sqrt5)/(9-5)+(sqrt5+3)/(5-9)=(3-sqrt5-3-sqrt5)/4=-sqrt5/2` $\\$ `12)1/(sqrt2-sqrt6)-1/(sqrt6-sqrt2)=(sqrt2+sqrt6)/(2-6)-(sqrt6-sqrt2)/(6-2)=(-sqrt2-sqrt6-sqrt6+sqrt2)/4=-sqrt6/2` $\\$ `13)1/(sqrt2-sqrt3)-3/(sqrt{18}+2sqrt3)=(sqrt2+sqrt3)/(2-3)-(3(sqrt{18}-2sqrt3))/(18-12)=-(sqrt2+sqrt3)-(sqrt{18}-3sqrt2)/2=(-2sqrt2-2sqrt3-3sqrt2+2sqrt3)/2=-(5sqrt2)/2` $\\$ `14)3/(1-sqrt2)+(sqrt2-1)/(sqrt2+1)=(3(1+sqrt2))/(1-2)+(sqrt2-1)^2/(2-1)=-3(1+sqrt2)+3-2sqrt2=-5sqrt2`

Mình đọc không kĩ xin lỗi bạn.

`10)(sqrt5+sqrt6)/(sqrt5-sqrt6)+(sqrt6-sqrt5)/(sqrt6+sqrt5)`

`=(sqrt5+sqrt6)^2/(5-6)+(sqrt6-sqrt5)^2/(6-5)`

`=((sqrt6-sqrt5)^2-(sqrt6+sqrt5)^2)/1`

`=11-2sqrt{30}-11-2sqrt{30}=-4sqrt{30}`

biết làm bài 10 ko bạn