\(Q=\frac{x+2}{x+3}-\frac{5}{x^2+x-6}+\frac{1}{2-x}\)

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

\(\Rightarrow\) \(Q=\left(x-2\right)\left(x+2\right)+5-1\)

\(\Leftrightarrow\) \(Q=x^2-4+5-1\)

\(\Leftrightarrow\) \(Q=x^2\)

Thay \(Q=\frac{-3}{4}\) ta được:

\(x^2=\frac{-3}{4}\)

Vì \(\frac{-3}{4}>0\forall x\)

\(\Rightarrow\) Pt vô nghiệm

Vậy không có giả trị nào của x thỏa mãn \(Q=\frac{-3}{4}\)

Chúc bn học tốt!!

cảm ơn nhiều nha

e) \(2\left(x+5\right)-x^2-5x=0\)

\(=>2\left(x+5\right)-x\left(x+5\right)=0\)

\(=>\left(x+5\right)\left(2-x\right)=0\)

\(=>\hept{\begin{cases}x+5=0\\2-x=0\end{cases}}\)

\(=>\hept{\begin{cases}x=-5\\x=2\end{cases}}\)

f) \(x^2-2x-3=0\)

\(=>x^2-3x+x-3=0\)

\(=>x\left(x-3\right)+\left(x-3\right)=0\)

\(=>\left(x+1\right)\left(x-3\right)=0\)

\(=>\hept{\begin{cases}x+1=0\\x-3=0\end{cases}}\)

\(=>\hept{\begin{cases}x=-1\\x=3\end{cases}}\)

g) \(2x^2+5x-3=0\)

\(=>2x^2-6x+x-3=0\)

\(=>2x\left(x-3\right)+\left(x-3\right)=0\)

\(=>\left(2x+1\right)\left(x-3\right)=0\)

\(=>\hept{\begin{cases}2x+1=0\\x-3=0\end{cases}}\)

\(=>\hept{\begin{cases}x=\frac{-1}{2}\\x=3\end{cases}}\)

h) \(x^2+x-6=0\)

\(=>x^2-2x+3x-6=0\)

\(=>x\left(x-2\right)+3\left(x-2\right)=0\)

\(=>\left(x+3\right)\left(x-2\right)=0\)

\(=>\hept{\begin{cases}x+3=0\\x-2=0\end{cases}}\)

\(=>\hept{\begin{cases}x=-3\\x=2\end{cases}}\)

r3t4yjytuky

ai luot wa xinn co tam tra loi ho

a) ĐKXĐ: x≠-5

Ta có: \(\frac{2x-5}{x+5}=3\)

\(\Leftrightarrow\frac{2x-5}{x+5}-3=0\)

\(\Leftrightarrow\frac{2x-5}{x+5}-\frac{3\left(x+5\right)}{x+5}=0\)

\(\Leftrightarrow2x-5-3\left(x+5\right)=0\)

\(\Leftrightarrow2x-5-3x-15=0\)

\(\Leftrightarrow-x-20=0\)

\(\Leftrightarrow-x=20\)

\(\Leftrightarrow x=-20\)(tmđk)

Vậy: x=-20

b) ĐKXĐ: x≠1;x≠-1

Ta có: \(\frac{2}{x-1}=\frac{6}{x+1}\)

\(\Leftrightarrow\frac{2}{x-1}-\frac{6}{x+1}=0\)

\(\Leftrightarrow\frac{2\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}-\frac{6\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}=0\)

\(\Leftrightarrow2\left(x+1\right)-6\left(x-1\right)=0\)

\(\Leftrightarrow2x+2-6x+6=0\)

\(\Leftrightarrow-4x+8=0\)

\(\Leftrightarrow-4x=-8\)

\(\Leftrightarrow x=2\)(tmđk)

Vậy: x=2

c) ĐKXĐ: x≠1;x≠-1

Ta có: \(\frac{2x+1}{x-1}=\frac{5\left(x-1\right)}{x+1}\)

\(\Leftrightarrow\frac{2x+1}{x-1}-\frac{5\left(x-1\right)}{x+1}=0\)

\(\Leftrightarrow\frac{\left(2x+1\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}-\frac{5\left(x-1\right)\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}=0\)

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

\(\Leftrightarrow2x^2+2x+x+1-5x^2+10x-5=0\)

\(\Leftrightarrow-3x^2+13x-4=0\)

\(\Leftrightarrow-3x^2+x+12x-4=0\)

\(\Leftrightarrow x\left(-3x+1\right)+4\left(3x-1\right)=0\)

\(\Leftrightarrow x\left(1-3x\right)-4\left(1-3x\right)=0\)

\(\Leftrightarrow\left(1-3x\right)\left(x-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}1-3x=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=1\\x=4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{3}\\x=4\end{matrix}\right.\)(thỏa mãn điều kiện)

Vậy: \(x\in\left\{\frac{1}{3};4\right\}\)

d) ĐKXĐ: x≠1;x≠-1

Ta có: \(\frac{x}{x-1}-\frac{2x}{x^2-1}=0\)

\(\Leftrightarrow\frac{x\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}-\frac{2x}{\left(x-1\right)\left(x+1\right)}=0\)

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

\(\Leftrightarrow x^2+x-2x=0\)

\(\Leftrightarrow x^2-x=0\)

\(\Leftrightarrow x\left(x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(tm\right)\\x=1\left(ktm\right)\end{matrix}\right.\)

Vậy: x=0

e) ĐKXĐ: x≠2

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

⇔\(\frac{1}{x-2}+3-\frac{x-3}{2-x}=0\)

⇔\(\frac{1}{x-2}+3+\frac{x-3}{x-2}=0\)

⇔\(\frac{1}{x-2}+\frac{3\left(x-2\right)}{x-2}+\frac{x-3}{x-2}=0\)

\(\Leftrightarrow1+3\left(x-2\right)+x-3=0\)

\(\Leftrightarrow1+3x-6+x-3=0\)

\(\Leftrightarrow4x-8=0\)

\(\Leftrightarrow4x=8\)

\(\Leftrightarrow x=2\)(không thỏa mãn)

Vậy: x∈∅

f) ĐKXĐ: \(x\ne\pm2\)

Ta có: \(\frac{x+1}{x-2}+\frac{x-1}{x+2}=\frac{2\left(x^2+2\right)}{x^2-4}\)

⇔\(\frac{x+1}{x-2}+\frac{x-1}{x+2}-\frac{2\left(x^2+2\right)}{x^2-4}=0\)

⇔\(\frac{\left(x+1\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{\left(x-1\right)\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}-\frac{2x^2+4}{\left(x+2\right)\left(x-2\right)}=0\)

\(\Leftrightarrow\left(x+1\right)\left(x+2\right)+\left(x-1\right)\left(x-2\right)-2x^2-4=0\)

\(\Leftrightarrow x^2+2x+x+2+x^2-2x-x+2-2x^2-4=0\)

\(\Leftrightarrow0=0\)

Vậy: x∈R

g) ĐKXĐ: \(x\ne\pm2\)

Ta có: \(\frac{x+2}{x-2}+\frac{1}{x+2}=\frac{x\left(x-5\right)}{x^2-4}\)

⇔\(\frac{x+2}{x-2}+\frac{1}{x+2}-\frac{x\left(x-5\right)}{\left(x-2\right)\left(x+2\right)}=0\)

⇔\(\frac{\left(x+2\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{x-2}{\left(x+2\right)\left(x-2\right)}-\frac{x^2-5x}{\left(x-2\right)\left(x+2\right)}=0\)

⇔\(\left(x+2\right)^2+x-2-x^2+5x=0\)

\(\Leftrightarrow x^2+4x+4+x-2-x^2+5x=0\)

\(\Leftrightarrow10x-2=0\)

\(\Leftrightarrow10x=2\)

\(\Leftrightarrow x=\frac{2}{10}=\frac{1}{5}\)(thỏa mãn)

Vậy: \(x=\frac{1}{5}\)

cảm ơn bạn nha

1. \(B=\left(\frac{21}{\left(x-3\right)\left(x+3\right)}+\frac{\left(x-4\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}-\frac{\left(x-3\right)\left(x-1\right)}{\left(x-3\right)\left(x+3\right)}\right):\frac{x+3-1}{x+3}\)\(=\frac{3x+6}{\left(x-3\right)\left(x+3\right)}.\frac{x+3}{x+2}=\frac{3\left(x+2\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)\left(x+2\right)}=\frac{3}{x-3}\)
2. Điều kiện \(x\ne3\) \(\Rightarrow\frac{-3}{5}=\frac{3}{x-3}\Leftrightarrow x-3=-5\Leftrightarrow x=-2\)
3. \(B=\frac{3}{x-3}< 0\Leftrightarrow x-3< 0\Leftrightarrow x< 3\)

a) B=(\(\frac{21}{x^2-9}\)-\(\frac{x-4}{3-x}\)-\(\frac{x-1}{3+x}\)) : (1-\(\frac{1}{x+3}\)) (ĐK: x khác +-3)

=(\(\frac{21}{\left(x-3\right).\left(x+3\right)}\)+\(\frac{x-4}{x-3}\)-\(\frac{x-1}{x+3}\)) : (1-\(\frac{1}{x+3}\))

=(\(\frac{21+\left(x+4\right).\left(x+3\right)-\left(x-1\right).\left(x-3\right)}{\left(x-3\right).\left(x+3\right)}\):(\(\frac{x+3-1}{x+3}\))

=(\(\frac{3x+6}{\left(x-3\right).\left(x+3\right)}\)) . (\(\frac{x+3}{x+2}\))

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

=\(\frac{3}{x-3}\)

b) B=\(\frac{3}{x-3}\)=\(\frac{-3}{5}\)

(=) \(\frac{3.5}{x-3}\)=-3

(=) -3.(x-3) = 15

(=) -3x=6

(=) x=-2

vậy x=2 thì B=\(\frac{-3}{5}\)

c) B=\(\frac{3}{x-3}\)<0

(=) 3 < x - 3

(=) -x < - 3 - 3

(=) x > 6

Vậy với x > 6 thì B < 0

a) x²+8x+16-x²+1=16

8x+17=16

8x=-1

x=-1/8

b) 4x²-4x+1+x²+6x+9-5x²+245=0

2x+255=0

x=-255/2

a) \(B=\left[\frac{21}{\left(x+3\right)\left(x-3\right)}+\frac{x-4}{x-3}-\frac{\left(x-1\right)}{x+3}\right]:\left(\frac{x+3-1}{x+3}\right)\)

ĐK: \(\hept{\begin{cases}x\ne3\\x\ne-3\end{cases}}\)

\(=\left[\frac{21+x-4-\left(x-1\right)\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}\right]:\left(\frac{x+2}{x+3}\right)\)

\(=\left[\frac{21+x-4-x^2+3x+x-3}{\left(x+3\right)\left(x-3\right)}\right]\times\left(\frac{x+3}{x+2}\right)\)

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

\(=\frac{-\left(x^2+2x-7x-14\right)}{\left(x-3\right)\left(x+2\right)}\)

\(=\frac{-\left(x+2\right)\left(x-7\right)}{\left(x-3\right)\left(x+2\right)}\)

\(=\frac{7-x}{x-3}\)

b) \(\Rightarrow\orbr{\begin{cases}2x+1=5\\2x+1=-5\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=2\\x=-3\end{cases}}\)

Mà \(x\ne-3\)

\(\Rightarrow x=2\)

Thế \(x=2\)vào B ta được:

\(B=\frac{7-2}{2-3}=-5\)

c) \(B=\frac{7-x}{x-3}=\frac{-3}{5}\)

\(\Leftrightarrow5\left(7-x\right)=-3\left(x-3\right)\)

\(\Leftrightarrow35-5x+3x-9=0\)

\(\Leftrightarrow-2x=-26\)

\(\Leftrightarrow x=13\)

Vậy để \(B=\frac{-3}{5}\)thì \(x=13\)

d) B<0\(\Rightarrow\frac{7-x}{x-3}< 0\)

TH1: \(\hept{\begin{cases}7-x< 0\\x-3>0\end{cases}\Rightarrow\hept{\begin{cases}x>7\\x>3\end{cases}\Rightarrow}x>7}\)

TH2: \(\hept{\begin{cases}7-x>0\\x-3< 0\end{cases}\Rightarrow\hept{\begin{cases}x< 7\\x< 3\end{cases}\Rightarrow}x< 3}\)

Để B<0 thì x>7 hoặc x<3

a) \(B=\left(\frac{21}{x^2-9}-\frac{x-4}{3-x}-\frac{x-1}{3+x}\right):\left(1-\frac{1}{x+3}\right)\)         ĐKXĐ: x khác =-3; x khác -2

\(B=\frac{21+x^2-x-12-x^2+4x-3}{\left(x+3\right)\left(x-3\right)}:\frac{x+2}{x+3}\)

\(B=\frac{3x+6}{\left(x+3\right)\left(x-3\right)}:\frac{x+2}{x+3}\)

\(B=\frac{3\left(x+2\right)}{\left(x+3\right)\left(x-3\right)}\cdot\frac{x+3}{x+2}\)

\(B=\frac{3}{x-3}\)

b) bước đầu tiên ta phải tìm x:

 \(\left|2x+1\right|=5\)

TH1: 2x+1=5                      TH2: 2x+1=-5

            2x=4                                 2x=-6

          x=2 (nhận)                             x=-3 (loại)

thay x=2 vào biểu thức B, ta được:

\(B=\frac{3}{2-3}=\frac{3}{-1}=-3\)

vậy B=-3 tại x=2

c) Để \(B=-\frac{3}{5}\)thì \(\frac{3}{x-3}=-\frac{3}{5}\)

\(\Leftrightarrow-3\left(x-3\right)=15\)

\(\Leftrightarrow x-3=-5\)

\(\Leftrightarrow x=-2\)

vậy \(x=-2\)thì \(B=-\frac{3}{5}\)

d) để B<0 thì \(\frac{3}{x-3}< 0\Leftrightarrow x-3< 0\Leftrightarrow x< 3\)

vậy để B<0 thì x phải < 3 và x khác -3

a) \(\left(4x+3\right)\left(4x-3\right)-\left(4x-5\right)^2=46\)

\(\Leftrightarrow16x^2-9-16x^2+40x-25=46\)

\(\Leftrightarrow40x=46+9+25=80\)

\(\Leftrightarrow x=2\)

b) \(\left(x+1\right)^3+2x-\left(x-1\right)^3-3\left[\left(x+1\right)^2+\left(x-1\right)^2\right]+5=0\)

\(=x^3+3x^2+3x+1+2x-x^3+3x^2-3x+1-3\left(x^2+2x+1+x^2-2x+1\right)+5=0\)

\(=6x^2+2x+2-3\left(2x^2+2\right)+5=0\)

\(\Leftrightarrow6x^2+2x+2-6x^2-6+5=0\)

\(\Leftrightarrow2x=-2+6-5=-1\)

\(\Leftrightarrow x=\frac{1}{2}\)

Bài 1: 

a: \(P=\left(\dfrac{x-2}{\left(x-1\right)\left(x+1\right)}-\dfrac{x+2}{\left(x+1\right)^2}\right)\cdot\dfrac{\left(x-1\right)^2\cdot\left(x+1\right)^2}{4}\)

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

\(=\dfrac{-2x}{1}\cdot\dfrac{x-1}{4}=-\dfrac{x\left(x-1\right)}{2}\)

b: Để \(\dfrac{P-4}{5}=x\) thì P-4=5x

=>P=5x+4

\(\Leftrightarrow-\dfrac{x\left(x-1\right)}{2}=5x+4\)

=>-x²+x=10x+8

=>x²-x=-10x-8

=>x²+9x+8=0

=>x=-8(nhận) hoặc x=-1(loại)