Giải:

Ta có: x = 1

=> \(\frac{7}{3a-1}=1\)

=> \(3a-1=7\)

=> 3a = 8

=> a = 8/3

b) Ta có: x = 7

=> \(\frac{7}{3a-1}=7\)

=> 3a - 1 = 7 : 7

=> 3a - 1 = 1

=> 3a = 2

=> a = 2/3

#)Giải :

a) \(x=\frac{7}{3a-1}\)

Theo đề : \(-1=\frac{7}{3a-1}\)

Từ đây giải ra a = - 2 

b) \(x=\frac{7}{3a-1}\)

theo đề : \(7=\frac{7}{3a-1}\)

Từ đây ra a = \(\frac{2}{3}\)

\(=>A=\frac{2}{3}\)

~Hok tốt~

a) \(\dfrac{5}{3a-1}=1\)

\(\Rightarrow3a-1=5\)

\(\Rightarrow3a=6\)

\(\Rightarrow a=\dfrac{6}{3}=2\)

b) \(\dfrac{5}{3a-1}=-5\)

\(\Rightarrow3a-1=5:\left(-5\right)=-1\)

\(\Rightarrow3a=-1+1=0\)

\(\Rightarrow a=0:3=0\)

a) x = 1

⇒ 3a - 1 = 5

⇒ 3a = 6

⇒ a = 2

b) x = 5

⇒ 3a - 1 = 1

⇒ 3a = 2

⇒ a = 2/3

2: \(\left(\dfrac{7}{a+7}+\dfrac{a^2+49}{a^2-49}-\dfrac{7}{a-7}\right):\dfrac{a+1}{2}\)

\(=\dfrac{7a-49+a^2+49-7a-49}{\left(a-7\right)\left(a+7\right)}\cdot\dfrac{2}{a+1}\)

\(=\dfrac{a^2-49}{\left(a-7\right)\left(a+7\right)}\cdot\dfrac{2}{a+1}=\dfrac{2}{a+1}\)

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

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

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

\(=\dfrac{x^4}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{x^3-4x+2x-8-3x^2+12x}{x\left(x-2\right)\left(x-4\right)}\)

\(=\dfrac{x^4}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{x^3-3x^2+10x-8}{x\left(x-2\right)\left(x-4\right)}\)

\(=\dfrac{x^4}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{x^3-x^2-2x^2+2x+8x-8}{x\left(x-2\right)\left(x-4\right)}\)

\(=\dfrac{x^3\left(x-1\right)\left(x^2-2x+8\right)}{\left(x-2\right)^2\cdot\left(x+2\right)\left(x-4\right)}\)

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

\(P=\dfrac{A}{B}=\sqrt{x}+1\)

P<7/4

=>căn x<3/4

=>0<x<9/16

đk x > 0 

\(\dfrac{A}{B}=\dfrac{\dfrac{x+2\sqrt{x}}{x}}{\dfrac{\sqrt{x}+2}{\sqrt{x}+1}}=\dfrac{\dfrac{\sqrt{x}+2}{\sqrt{x}}}{\dfrac{\sqrt{x}+2}{\sqrt{x}+1}}=\dfrac{\sqrt{x}+1}{\sqrt{x}}-\dfrac{7}{4}< 0\)

\(\Leftrightarrow\dfrac{4\sqrt{x}+4-7\sqrt{x}}{4\sqrt{x}}< 0\Leftrightarrow\dfrac{-3\sqrt{x}+4}{4\sqrt{x}}< 0\)

\(\Leftrightarrow\left\{{}\begin{matrix}-3\sqrt{x}+4\ne0\\-3\sqrt{x}+4< 0\\4\sqrt{x}\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne\dfrac{16}{9}\\x< \dfrac{16}{9}\\x\ne0\end{matrix}\right.\)

a) \(x=-1\Leftrightarrow\frac{7}{3a-1}=-1\)

\(\Leftrightarrow3a-1=-7\Leftrightarrow a=-2\)

b) \(x=7\Leftrightarrow\frac{7}{3a-1}=7\)

\(\Leftrightarrow3a-1=1\Leftrightarrow a=\frac{2}{3}\)

a) x = -1

7/3a - 1 = -1

7 = -3a + 1

7 - 1 = -3a

6 = -3a

6 : (-3) = a

-2 = a

=> a = -2

b) x = 7

7/3a - 1 = 7

7 = 7(3a - 1)

7 : 7 = 3a - 1

1 = 3a - 1

1 + 1 = 3a

2 = 3a

2/3 = a

=> a = 2/3

                                                      Bài giả

Ta có :  \(x=\frac{7}{3a-1}\)

\(a,\text{ }x=-1\) \(\Rightarrow\text{ }7\text{ : }\left(3a-1\right)=-1\)

\(3a-1=7\text{ : }\left(-1\right)\)

\(3a-1=-7\)

\(3a=-7+1\)

\(3a=-6\)

\(a=-6\text{ : }3\)

\(a=-2\)

\(b,\text{ }x=7\) \(\Rightarrow\text{ }7\text{ : }\left(3a-1\right)=7\)

\(3a-1=7\text{ : }7\)

\(3a-1=1\)

\(3a=1+1\)

\(3a=2\)

\(a=\frac{2}{3}\)