a: Ta có: \(A=\dfrac{1}{2}\)

\(\Leftrightarrow x+2=2x-6\)

\(\Leftrightarrow-x=-8\)

hay x=8

Thay x=8 vào B,ta được:

\(B=-\dfrac{2}{8+2}=-\dfrac{2}{10}=-\dfrac{1}{5}\)

`a, 2/3 +3/4 = (8+9)/12=17/12.`

`1 1/3+4/5 = 4/3 + 4/5 = (20+12)/15=32/15`.

`=> x=2.`

`b, 5/6-1/4=(20-6)/24=7/12`.

`2 1/3-2/5= 7/3-2/5 = (35-6)/15=29/15`.

`=> x=1`.

a) \(\dfrac{2}{3}+\dfrac{3}{4}=\dfrac{8+9}{12}=\dfrac{17}{12}\)

-> 1 1/3 + 4/5 = 4/3 + 4/5 =  20+12/15 = 32/15

vậy x có thể = 14/14 = 1 (x thuộc N)

\(a,P=\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{2}{x-\sqrt{x}}\right):\left(\dfrac{1}{\sqrt{x}+1}-\dfrac{2}{1-x}\right)\left(dkxd:x\ge0,x\ne1\right)\)

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

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

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

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

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

\(=\dfrac{2\sqrt{3}+4-2}{\sqrt{\sqrt{3}^2+2\sqrt{3}+1}}\)

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

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

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

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

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

b: Khi x=4+2căn 3 thì \(P=\dfrac{2+2\sqrt{3}}{\sqrt{3}+1}=2\)

\(\Rightarrow x+\dfrac{1}{6}=\dfrac{3}{4}\\ \Rightarrow x=\dfrac{7}{12}\)

X+

\(x+\dfrac{1}{6}=\dfrac{3}{4}\)

nên x=7/12

\(3,=\left(\dfrac{13}{25}-\dfrac{38}{25}\right)+\left(\dfrac{14}{9}-\dfrac{5}{9}\right)=-1+1=0\\ 4,=\left(\dfrac{4}{9}\right)^5\cdot\left(\dfrac{9}{49}\right)^5=\left(\dfrac{4}{9}\cdot\dfrac{9}{49}\right)^5=\left(\dfrac{4}{49}\right)^5\\ 5,\Rightarrow\dfrac{x}{5}=\dfrac{y}{3}=\dfrac{x-y}{5-3}=\dfrac{x+y}{5+3}=\dfrac{2}{2}=\dfrac{x+y}{8}\Rightarrow x+y=8\\ 6,\Rightarrow\left[{}\begin{matrix}x=\sqrt{2}\\x=-\sqrt{2}\end{matrix}\right.\Rightarrow2\text{ giá trị}\\ 7,=\dfrac{3^{10}\cdot2^{30}}{2^9\cdot3^9\cdot2^{20}}=2\cdot3=6\)

Câu 7:

=6

a) \(x=\dfrac{-2}{7}+\dfrac{9}{7}=1\) 

b) \(\dfrac{x}{3}=\dfrac{2}{5}+\dfrac{-4}{3}\) 

     \(\dfrac{x}{3}=\dfrac{-14}{15}\) 

\(\Rightarrow x=\dfrac{3.-14}{15}=\dfrac{-14}{5}\)

\(x=\dfrac{-2}{7}+\dfrac{9}{7}\) 

\(x=1\)

x=1

a: Ta có: \(M=\dfrac{A}{B}\)

\(=\dfrac{x-3}{x+2}:\dfrac{-2}{x+2}\)

\(=\dfrac{x-3}{-2}\)

Để |M|=-M thì \(M\le0\)

\(\Leftrightarrow x\ge3\)

Đề bài yêu cầu gì?

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

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

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

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

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

\(=\dfrac{3\sqrt{x}-6}{\sqrt{x}-1}:\dfrac{1}{\sqrt{x}-1}\)

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

\(=3\sqrt{x}-6\)

b) \(P=\dfrac{4\sqrt{x}-1}{\sqrt{x}}\)

\(\Leftrightarrow3\sqrt{x}-6=\dfrac{4\sqrt{x}-1}{\sqrt{x}}\)   (1)

ĐKXĐ: \(x>0\)

\(\left(1\right)\Leftrightarrow3x-6\sqrt{x}=4\sqrt{x}-1\)

\(\Leftrightarrow3x-6\sqrt{x}-4\sqrt{x}+1=0\)

\(\Leftrightarrow3x-10\sqrt{x}+1=0\)   (2)

Đặt \(t=\sqrt{x}\ge0\)

\(\left(2\right)\Leftrightarrow3t^2-10t+1=0\)

\(\Delta'=25-4=22\)

Phương trình có hai nghiệm phân biệt:

\(t_1=\dfrac{5+\sqrt{22}}{3}\) (nhận)

\(t_2=\dfrac{5-\sqrt{22}}{3}\) (nhận)

Với \(t=\dfrac{5+\sqrt{22}}{3}\) \(\Leftrightarrow\sqrt{x}=\dfrac{5+\sqrt{22}}{3}\Leftrightarrow x=\dfrac{47+10\sqrt{22}}{9}\) (nhận)

Với \(t=\dfrac{5-\sqrt{22}}{3}\Leftrightarrow\sqrt{x}=\dfrac{5-\sqrt{22}}{3}\Leftrightarrow x=\dfrac{47-10\sqrt{22}}{9}\) (nhận)

Vậy \(x=\dfrac{47+10\sqrt{22}}{9};x=\dfrac{47-10\sqrt{22}}{9}\) thì \(P=\dfrac{4\sqrt{x}-1}{\sqrt{x}}\)

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

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

b: P=(4căn x-1)/căn x

=>3x-6căn x-4căn x+1=0

=>3x-10căn x+1=0

=>x=(47+10căn 22)/9 hoặc x=(47-10căn 22)/9