b.

\(\dfrac{x+5}{x-1}-\dfrac{x+1}{x-3}=\dfrac{-8}{\left(x-1\right)\left(x-3\right)}\\ \Leftrightarrow\dfrac{x^2+2x-15}{\left(x-1\right)\left(x-3\right)}-\dfrac{x^2-1}{\left(x-1\right)\left(x-3\right)}=\dfrac{-8}{\left(x-1\right)\left(x-3\right)}\\ \Rightarrow x^2+2x-15-x^2+1=0\\ \Leftrightarrow2x-14=0\\ \Leftrightarrow x=7\)

Vậy x = 7

b.

\(\dfrac{x+5}{x-1}-\dfrac{x+1}{x-3}=\dfrac{-8}{\left(x-1\right)\left(x-3\right)}\\ \Leftrightarrow\dfrac{x^2+2x-15}{\left(x-1\right)\left(x-3\right)}-\dfrac{x^2-1}{\left(x-1\right)\left(x-3\right)}=\dfrac{-8}{\left(x-1\right)\left(x-3\right)}\\ \Rightarrow x^2+2x-15-x^2+1=-8\\ \Leftrightarrow2x-14=-8\\ \Leftrightarrow2x-6=0\\ \Leftrightarrow x=3\)

Bài 1: 

c) |2x - 1| = x + 2

<=> 2x - 1 = +(x + 2) hoặc -(x + 2)

* 2x - 1 = x + 2      

<=> 2x - x = 2 + 1

<=> x = 3

* 2x - 1 = -(x + 2)

<=> 2x - 1 = x - 2

<=> 2x - x = -2 + 1

<=> x = -1

Vậy.....

a-b=5 và (a,b)/[a,b]. Tim a:b

\(a,ĐK:1\le x\le3\)

Đặt \(\left\{{}\begin{matrix}\sqrt{x-1}=a\\\sqrt{3-x}=b\end{matrix}\right.\left(a,b\ge0\right)\)

\(PT\Leftrightarrow a+b-ab=1\Leftrightarrow a+b-ab-1=0\\ \Leftrightarrow\left(a-1\right)\left(1-b\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}a=1\\b=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x-1=1\\3-x=1\end{matrix}\right.\Leftrightarrow x=2\left(tm\right)\)

\(b,ĐK:0\le x\le9\\ PT\Leftrightarrow9+2\sqrt{x\left(9-x\right)}=-x^2+9x+9\\ \Leftrightarrow2\sqrt{-x^2+9x}-\left(-x^2+9x\right)=0\\ \Leftrightarrow\sqrt{-x^2+9x}\left(2-\sqrt{-x^2+9x}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}-x^2+9x=0\\\sqrt{-x^2+9x}=2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=9\\x^2-9x+4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(n\right)\\x=9\left(n\right)\\x=\dfrac{9+\sqrt{65}}{2}\left(n\right)\\x=\dfrac{9-\sqrt{65}}{2}\left(n\right)\end{matrix}\right.\)

a) Ta có: \(\sqrt{49\left(x^2-2x+1\right)}-35=0\)

\(\Leftrightarrow7\left|x-1\right|=35\)

\(\Leftrightarrow\left|x-1\right|=5\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=5\\x-1=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-4\end{matrix}\right.\)

b)

ĐKXĐ: \(\left[{}\begin{matrix}x\ge3\\x\le-3\end{matrix}\right.\)

Ta có: \(\sqrt{x^2-9}-5\sqrt{x+3}=0\)

\(\Leftrightarrow\sqrt{x+3}\left(\sqrt{x-3}-5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x+3}=0\\\sqrt{x-3}=5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x-3=25\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\left(nhận\right)\\x=28\left(nhận\right)\end{matrix}\right.\)

c) ĐKXĐ: \(x\ge0\)

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

\(\Leftrightarrow x-1=x+\sqrt{x}-6\)

\(\Leftrightarrow\sqrt{x}-6=-1\)

\(\Leftrightarrow\sqrt{x}=5\)

hay x=25(nhận)

`1/9(x-3)^2-1/25(x+5)^2=0`

`<=>(1/3x-1)^2-(1/5x+1)^2=0`

`<=>(1/3x-1-1/5x-1)(1/3x-1+1/5x+1)=0`

`<=>(2/15x-2). 8/15x=0`

`<=>2/15x-2=0` hoặc `8/15x=0`

`<=>x=15`         hoặc `x=0`

Vậy `S=`{`15;0`}

3:

a: =>x=0 hoặc x+5=0

=>x=0 hoặc x=-5

b: =>x^2=4

=>x=2 hoặc x=-2

c: =>(x-5)(2x+1+x+6)=0

=>(x-5)(3x+7)=0

=>x=5 hoặc x=-7/3

1.

a. 2x - 6 > 0 

\(\Leftrightarrow\)  2x  > 6

\(\Leftrightarrow\)    x  > 3

S = \(\left\{x\uparrow x>3\right\}\) 

b. -3x + 9 > 0

\(\Leftrightarrow\)  - 3x   > - 9 

\(\Leftrightarrow\)      x < 3

S = \(\left\{x\uparrow x< 3\right\}\) 

c. 3(x - 1) + 5 > (x - 1) + 3

\(\Leftrightarrow\) 3x - 3 + 5 > x - 1 + 3

\(\Leftrightarrow\) 3x - 3 + 5 - x + 1 - 3 > 0

\(\Leftrightarrow\) 2x > 0 

\(\Leftrightarrow\)   x > 0

S = \(\left\{x\uparrow x>0\right\}\) 

d. \(\dfrac{x}{3}-\dfrac{1}{2}>\dfrac{x}{6}\) 

\(\Leftrightarrow\dfrac{2x}{6}-\dfrac{3}{6}>\dfrac{x}{6}\)

\(\Leftrightarrow2x-3>x\)

\(\Leftrightarrow2x-3-x>0\)

\(\Leftrightarrow x-3>0\)

\(\Leftrightarrow x>3\)

\(S=\left\{x\uparrow x>3\right\}\)

2.

a. 

Ta có: a > b

3a > 3b (nhân cả 2 vế cho 3)

3a + 7 > 3b + 7 (cộng cả 2 vế cho 7)

b. Ta có: a > b

a > b (nhân cả 2 vế cho 1)

a + 3 > b + 3 (cộng cả 2 vế cho 3) (1)

Ta có; 3 > 1

b + 3 > b + 1 (nhân cả 2 vế cho 1b) (2)

Từ (1) và (2) \(\Rightarrow\) a + 3 > b + 1 

c.

5a - 1 + 1 > 5b - 1 + 1 (cộng cả 2 vế cho 1)

5a . \(\dfrac{1}{5}\) > 5b . \(\dfrac{1}{5}\) (nhân cả 2 vế cho \(\dfrac{1}{5}\) )

a > b

3.

a. 2x(x + 5) = 0

\(\Leftrightarrow\left[{}\begin{matrix}2x=0\\x+5=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\) 

\(S=\left\{0,-5\right\}\)

b. x² - 4 = 0 

\(\Leftrightarrow x\left(x-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\)

\(S=\left\{0,4\right\}\)

d. (x - 5)(2x + 1) + (x - 5)(x + 6) = 0

\(\Leftrightarrow\left(x-5\right)\left(2x+1+x+6\right)=0\)

\(\Leftrightarrow\left(x-5\right)\left(3x+7\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-5=0\\3x+7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=\dfrac{-7}{3}\end{matrix}\right.\)

\(S=\left\{5,\dfrac{-7}{3}\right\}\)