`(6/11 +5/11) xx 3/7`

`= 11/11xx 3/7`

`=1xx3/7`

`=3/7`

__

`3/5 xx 7/9 - 3/5 xx 2/9`

`= 3/5 xx (7/9-2/9)`

`= 3/5 xx 5/9`

`= 15/45`

`= 1/3`

__

`(6/7 -4/7):2/5`

`= 2/7 : 2/5`

`= 2/7 xx 5/2`

`= 10/14`

`= 5/7`

\(\left(\dfrac{6}{11}+\dfrac{5}{11}\right)\times\dfrac{3}{7}\)

\(=1\times\dfrac{3}{7}=\dfrac{3}{7}\)

___________

\(\dfrac{3}{5}\times\dfrac{7}{9}-\dfrac{3}{5}\times\dfrac{2}{9}\)

\(=\dfrac{3}{5}\times\left(\dfrac{7}{9}-\dfrac{2}{9}\right)\)

\(=\dfrac{3}{5}\times\dfrac{5}{9}\)

\(=\dfrac{1}{3}\)

__________

\(\left(\dfrac{6}{7}-\dfrac{4}{7}\right):\dfrac{2}{5}\)

\(=\dfrac{2}{7}\times\dfrac{5}{2}\)

\(=\dfrac{5}{7}\)

\(#WendyDang\)

\(\left(\dfrac{6}{11}+\dfrac{5}{11}\right)\times\dfrac{3}{7}\)

\(=\dfrac{11}{11}\times\dfrac{3}{7}\\ =1\times\dfrac{3}{7}=\dfrac{3}{7}\)

_____

\(\dfrac{3}{5}\times\dfrac{7}{9}-\dfrac{3}{5}\times\dfrac{2}{9}\\ =\dfrac{3}{5}\times\left(\dfrac{7}{9}-\dfrac{2}{9}\right)\\ =\dfrac{3}{5}\times\dfrac{5}{9}\\ =\dfrac{3}{9}\\ =\dfrac{1}{3}\)

_________

\(\left(\dfrac{6}{7}-\dfrac{4}{7}\right):\dfrac{2}{5}\\ =\dfrac{2}{7}\times\dfrac{5}{2}\\ =\dfrac{10}{14}\\ =\dfrac{5}{7}\)

1. \(x=\frac{1}{9}\) thỏa mãn đk: \(x\ge0;x\ne9\)

Thay \(x=\frac{1}{9}\) vào A ta có:

\(A=\frac{\sqrt{\frac{1}{9}}+1}{\sqrt{\frac{1}{9}}-3}=-\frac{1}{2}\)

2. \(B=...\)

    \(B=\frac{3\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}+\frac{\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}-\frac{4x+6}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)

    \(B=\frac{3x-9\sqrt{x}+x+3\sqrt{x}-4x-6}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)

     \(B=\frac{-6\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)

3. \(P=A:B=\frac{\sqrt{x}+1}{\sqrt{x}-3}:\frac{-6\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)

\(P=\frac{\sqrt{x}+3}{-6}\)

Vì \(\sqrt{x}+3\ge3\forall x\)\(\Rightarrow\frac{\sqrt{x}+3}{-6}\le\frac{3}{-6}=-\frac{1}{2}\)

hay \(P\le-\frac{1}{2}\)

Dấu "=" xảy ra <=> x=0

toán lớp 9 khó zậy em đọc k hỉu 1 phân số

\(=\dfrac{2}{3}+\dfrac{1}{3}.\dfrac{7}{18}.\dfrac{12}{7}\)

\(=\dfrac{2}{3}+\dfrac{7.3.2.2}{3.7.3.2.3}\)

\(=\dfrac{2}{3}+\dfrac{2}{9}=\dfrac{8}{9}\)

TICK CHO MÌNH NHÉ

Giải:

\(\dfrac{2}{3}\) + \(\dfrac{1}{3}\) . (\(-\dfrac{4}{9}\) + \(\dfrac{5}{6}\) ) : \(\dfrac{7}{12}\)

 = \(\dfrac{2}{3}\)  + \(^{\dfrac{1}{3}}\) . \(\dfrac{7}{18}\) : \(\dfrac{7}{12}\)

 = \(\dfrac{2}{3}\) + \(\dfrac{7}{54}\) : \(\dfrac{7}{12}\)

 = \(\dfrac{2}{3}\) + \(\dfrac{2}{9}\)

 = \(\dfrac{8}{9}\)

2 /3 + 1/ 3 . ( − 4 9 + 5 6 ) : 7 /12

= 2/ 3 + 1 /3 . 7 /18 . 12/ 7

= 2/ 3 + 7 /48 . 12 /7

= 2/ 3 + 1/ 4

= 11/ 12

\(=\dfrac{2}{3}+\dfrac{1}{3}.\left(\dfrac{7}{18}\right):\dfrac{7}{12}\)

\(=\dfrac{2}{3}+\dfrac{7}{54}:\dfrac{7}{12}\)

\(=\dfrac{2}{3}+\dfrac{2}{9}\)

\(=\dfrac{8}{9}\)

\(1,P=\left(\dfrac{\sqrt{x}}{3+\sqrt{x}}+\dfrac{2x}{9-x}\right):\left(\dfrac{\sqrt{x}-1}{x-3\sqrt{x}}-\dfrac{2}{\sqrt{x}}\right)\left(dkxd:x\ge0,x\ne9\right)\)

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

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

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

\(=\dfrac{-x-3\sqrt{x}}{\left(\sqrt{x}+3\right)}.\dfrac{\sqrt{x}}{-\sqrt{x}+5}\)

\(=\dfrac{-\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)}.\dfrac{\sqrt{x}}{5-\sqrt{x}}\)

\(=-\dfrac{x}{5-\sqrt{x}}\)

\(2,x=\sqrt{7+4\sqrt{3}}+\sqrt{7-4\sqrt{3}}\)

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

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

\(=2+\sqrt{3}+2-\sqrt{3}=4\)

\(x=4\Rightarrow P=-\dfrac{4}{5-\sqrt{4}}=\dfrac{-4}{5-2}=-\dfrac{4}{3}\)

Bài `1`

\(\sqrt{4-2\sqrt{3}}-\dfrac{2}{\sqrt{3}+1}+\dfrac{\sqrt{3}-3}{\sqrt{3}-1}\\ =\sqrt{3-2\sqrt{3}+1}-\dfrac{2\left(\sqrt{3}-1\right)}{3-1}-\dfrac{\sqrt{3}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}\\ =\sqrt{\left(\sqrt{3}\right)^2-2\cdot\sqrt{3}\cdot1+1^2}-\dfrac{2\left(\sqrt{3}-1\right)}{2}-\sqrt{3}\\ =\sqrt{\left(\sqrt{3}-1\right)^2}-\sqrt{3}+1-\sqrt{3}\\ =\sqrt{3}-1-\sqrt{3}+1-\sqrt{3}\\ =-\sqrt{3}\)

2:

a: \(B=\dfrac{\sqrt{x}}{\sqrt{x}-3}+\dfrac{2\sqrt{x}-24}{x-9}\)

\(=\dfrac{\sqrt{x}}{\sqrt{x}-3}+\dfrac{2\sqrt{x}-24}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)

\(=\dfrac{\sqrt{x}\left(\sqrt{x}+3\right)+2\sqrt{x}-24}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)

\(=\dfrac{x+5\sqrt{x}-24}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\dfrac{\left(\sqrt{x}+8\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)

\(=\dfrac{\sqrt{x}+8}{\sqrt{x}+3}\)

b: B=5

=>\(5\left(\sqrt{x}+3\right)=\sqrt{x}+8\)

=>\(5\sqrt{x}+15=\sqrt{x}+8\)

=>\(4\sqrt{x}=-7\)(loại)

Vậy: \(x\in\varnothing\)

7/5.3/4:4/5

=21/20:4/5

=21/16

x-3/9=8/7

x=8/7+3/9

x=31/21

\(\dfrac{7}{5}\times\dfrac{3}{4}:\dfrac{4}{5}=\dfrac{21}{20}:\dfrac{4}{5}=\dfrac{21}{20}\times\dfrac{5}{4}=\dfrac{105}{80}\)

\(x-\dfrac{3}{9}=\dfrac{8}{7}\)

\(x=\dfrac{8}{7}+\dfrac{3}{9}\)

\(x=\dfrac{72}{63}+\dfrac{21}{63}\)

\(x=\dfrac{93}{63}\)

a,ĐK:  \(\hept{\begin{cases}x\ne0\\x\ne\pm3\end{cases}}\)

b, \(A=\left(\frac{9}{x\left(x-3\right)\left(x+3\right)}+\frac{1}{x+3}\right):\left(\frac{x-3}{x\left(x+3\right)}-\frac{x}{3\left(x+3\right)}\right)\)

\(=\frac{9+x\left(x-3\right)}{x\left(x-3\right)\left(x+3\right)}:\frac{3\left(x-3\right)-x^2}{3x\left(x+3\right)}\)

\(=\frac{x^2-3x+9}{x\left(x-3\right)\left(x+3\right)}.\frac{3x\left(x+3\right)}{-x^2+3x-9}=\frac{-3}{x-3}\)

c, Với x = 4 thỏa mãn ĐKXĐ thì

\(A=\frac{-3}{4-3}=-3\)

d, \(A\in Z\Rightarrow-3⋮\left(x-3\right)\)

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

Mà \(x\ne0\Rightarrow x\in\left\{2;4;6\right\}\)

\(\dfrac{9}{5}-\dfrac{3}{4}:\dfrac{5}{6}=\dfrac{9}{5}-\dfrac{9}{10}=\dfrac{9}{10}\)

\(x=\dfrac{7}{9}:\dfrac{5}{7}=\dfrac{49}{45}\)

\(=\dfrac{9}{5}-\dfrac{3}{4}x\dfrac{6}{5}=\dfrac{9}{5}-\dfrac{9}{10}=\dfrac{18}{10}-\dfrac{9}{10}=\dfrac{9}{10}\)

\(x=\dfrac{7}{9}:\dfrac{5}{7}\)

\(x=\dfrac{49}{45}\)

9/5-3/4:5/6=9/5-9/10=9/10

7/9:x=5/7

x=7/9:5/7

x=49/45

\(A=\dfrac{1}{3}x+x-\dfrac{4}{3}x=0\)