ĐKXĐ: \(x\ge-\dfrac{1}{2}\)

\(\Leftrightarrow\left(x^2-8x+16\right)+\left(2x+1-6\sqrt{2x+1}+9\right)=0\)

\(\Leftrightarrow\left(x-4\right)^2+\left(\sqrt{2x+1}-3\right)^2=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-4=0\\\sqrt{2x+1}-3=0\end{matrix}\right.\) \(\Leftrightarrow x=4\)

\(|x-6|=-5x+9\)

Xét \(x\ge6\)thì \(pt< =>x-6=-5x+9\)

\(< =>x-6+5x-9=0\)

\(< =>6x-15=0\)

\(< =>x=\frac{15}{6}\)(ktm)

Xét \(x< 6\)thì \(pt< =>x-6=5x-9\)

\(< =>4x-9+6=0\)

\(< =>4x-3=0< =>x=\frac{3}{4}\)(tm)

Vậy ...

\(|x+1|=x^2+x\)

Xét \(x\ge-1\)thì \(pt< =>x+1=x^2+x\)

\(< =>x^2+x-x-1=0\)

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

\(< =>\orbr{\begin{cases}x=1\\x=-1\end{cases}\left(tm\right)}\)

Xét \(x< -1\)thì \(pt< =>-x-1=x^2+x\)

\(< =>x^2+2x+1=0\)

\(< =>\left(x+1\right)^2=0\)

\(< =>x=-1\left(ktm\right)\)

Vậy ...

\(\Leftrightarrow x^2-6x+8=6\sqrt{2x+1}-18\left(Đk:x\ge-\dfrac{1}{2}\right)\)

\(\Leftrightarrow\left(x-2\right)\left(x-4\right)=\dfrac{12\left(x-4\right)}{\sqrt{2x+1}+3}\left(\sqrt{2x+1}+3>0\right)\)

+) \(x=4\left(TM\right)\)

+) \(x\ne4\Rightarrow x-2=\dfrac{12}{\sqrt{2x+1}+3}\)

            \(\Leftrightarrow x-4=\dfrac{12-2\left(\sqrt{2x+1}+3\right)}{\sqrt{2x+1}+3}\)

             \(\Leftrightarrow x-4+\dfrac{2\left(x-4\right)}{\left(\sqrt{2x+1}+3\right)^2}=0\)

             \(\Leftrightarrow1+\dfrac{2}{\left(\sqrt{2x+1}+3\right)^2}=0\left(x\ne4\right)\)

    Vì \(\dfrac{2}{\left(\sqrt{2x+1}+3\right)^2}>0\forall x\) => VT>0

=> phương trình vô nghiệm

Vậy \(S=\left\{4\right\}\)

3x² + 2x - 1 = 0

=> 3x² + 3x - x - 1 = 0

=> 3x(x + 1) - (x + 1) = 0

=> (3x - 1)(x + 1) = 0

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

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

x² - 5x + 6 = 0

=> x² - 2x - 3x + 6 = 0

=> x(x - 2) - 3(x - 2) = 0

=> (x - 3)(x - 2) = 0

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

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

3x² + 7x + 2 = 0

=> 3x² + 6x + x  + 2 = 0

=> 3x(x + 2) + (x + 2) = 0

=> (3x + 1)(x + 2) = 0

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

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

1, \(3x^2+2x-1=0\Leftrightarrow3x^2+3x-x-1=0\)

\(\Leftrightarrow3x\left(x+1\right)-\left(x+1\right)=0\)

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

\(\Leftrightarrow\orbr{\begin{cases}x+1=0\\3x-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-1\\x=\frac{1}{3}\end{cases}}}\)

2, \(x^2-5x+6=0\Leftrightarrow x^2-2x-3x+6=0\)

\(\Leftrightarrow x\left(x-2\right)-3\left(x-2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x-3\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\x-3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=2\\x=3\end{cases}}}\)

3, \(3x^2+7x+2=0\Leftrightarrow3x^2+6x+x+2=0\)

\(\Leftrightarrow3x\left(x+2\right)+\left(x+2\right)=0\)

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

\(\Leftrightarrow\orbr{\begin{cases}x+2=0\\3x+1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-2\\x=-\frac{1}{3}\end{cases}}}\)

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

\(\Leftrightarrow\left(x^2-4x+4\right)=3\)

\(\Leftrightarrow\left(x-2\right)^2=3\)

\(\Leftrightarrow x=\sqrt{3}+2;x=2-\sqrt{3}\)

\(2x^2-6x+1=0\)

\(\Leftrightarrow4x^2-12x+2=0\)

\(\Leftrightarrow\left(2x-3\right)^2=7\)

\(\Leftrightarrow x=\frac{\sqrt{7}+3}{2};x=\frac{3-\sqrt{7}}{2}\)

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

\(\Leftrightarrow3x^2-2x+6x-4=0\)

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

\(\Leftrightarrow x=-2;x=\frac{2}{3}\)

a) căn(2x+5) - căn(3-x) = x² -5x + 8 
Điều kiện : \(-\frac{5}{2}\Leftarrow x\Leftarrow3\)
căn(2x+5) - căn(3-x) = x^2-5x+8 
\(\Leftrightarrow\)[căn(2x+5)-3]-[căn(3-x)-1]=x² -5x+6 
nhân liên hợp 
\(\Leftrightarrow\)(2x+5-9) / [căn(2x+5)+3] -(3-x-1) / [căn (3-x)+1]=(x-2)(x-3) 
\(\Leftrightarrow\)(2x-4) / [căn (2x+5)+3] -(2-x) /  [ căn (3-x)+1]-(x-2)(x-3)=0 
\(\Leftrightarrow\)(x-2).M=0 
\(\Leftrightarrow\)x=2 hoặc M=0 
M=2 / [căn(2x+5)+3]+1 / [căn(3-x)+1]-x+3 

2/[can(2x+5)+3]+1/[can(3-x)+1]>0 voi moi x 
voi -5/2<=x<=3 <->3-x thuoc[0;11/2] 
nen M>0 
vay x=2 
b/ 2+ căn(3-8x) = 6x + căn(4x-1) 
dk[1/4;8/3] 
6x-2+căn(4x-1)-căn(3-8x)=0 
<->2(3x-1)+(4x-1-3+8x)/[căn(4x-1)+căn(... 
<->2(3x-1)+(12x-4)/[căn(4x-1)+căn(3-8x... 
<->2(3x-1)+4(3x-1)/[căn(4x-1)+căn(3-8x... 
<->(3x-1){2+4/[căn(4x-1)+căn(3-8x)]}=0 
2+4/[căn(4x-1)+căn(3-8x)>0 
nen 3x-1=0 
x=1/3

 a)  căn(2x+5) - căn(3-x) = x^2-5x+8 
dkxd -5/2<=x<=3 
căn(2x+5) - căn(3-x) = x^2-5x+8 
<->[can(2x+5)-3]-[can(3-x)-1]=x^2-5x+6 
nhan lien hop 
<->(2x+5-9)/[can(2x+5)+3] -(3-x-1)/[can(3-x)+1]=(x-2)(x-3) 
<->(2x-4)/[can(2x+5)+3] -(2-x)/[can(3-x)+1]-(x-2)(x-3)=0 
<->(x-2).M=0 
<->x=2 hoac M=0 
M=2/[can(2x+5)+3]+1/[can(3-x)+1]-x+3 

2/[can(2x+5)+3]+1/[can(3-x)+1]>0 voi moi x 
voi -5/2<=x<=3 <->3-x thuoc[0;11/2] 
nen M>0 
vay x=2 
b/ 2+ căn(3-8x) = 6x + căn(4x-1) 
dk[1/4;8/3] 
6x-2+căn(4x-1)-căn(3-8x)=0 
<->2(3x-1)+(4x-1-3+8x)/[căn(4x-1)+căn(... 
<->2(3x-1)+(12x-4)/[căn(4x-1)+căn(3-8x... 
<->2(3x-1)+4(3x-1)/[căn(4x-1)+căn(3-8x... 
<->(3x-1){2+4/[căn(4x-1)+căn(3-8x)]}=0 
2+4/[căn(4x-1)+căn(3-8x)>0 
nen 3x-1=0 
x=1/3

a, \(\left(x-3\right)\left(x^2+x-20\right)\ge0\)

\(\Leftrightarrow\) \(\left(x-3\right)\left(x-4\right)\left(x+5\right)\ge0\)

+) \(x-3=0\Leftrightarrow x=3\); \(x-4=0\Leftrightarrow x=4\); \(x+5=0\Leftrightarrow x=-5\)

+) Lập trục xét dấu f(x) (Bạn tự kẻ trục nha)

\(\Rightarrow\) Bpt có tập nghiệm S = \(\left[-5;3\right]\cup\) [4; \(+\infty\))

b, \(\dfrac{x^2-4x-5}{2x+4}\ge0\)

\(\Leftrightarrow\) \(\dfrac{\left(x-5\right)\left(x+1\right)}{2x+4}\ge0\)

+) \(x-5=0\Leftrightarrow x=5\); \(x+1=0\Leftrightarrow x=-1\); \(2x+4=0\Leftrightarrow x=-2\)

+) Lập trục xét dấu f(x) 

\(\Rightarrow\) Bpt có tập nghiệm S = (-2; -1] \(\cup\) [5; \(+\infty\))

c, \(\dfrac{-1}{x^2-6x+8}\le1\)

\(\Leftrightarrow\) \(\dfrac{\left(x-3\right)^2}{\left(x-4\right)\left(x-2\right)}\ge0\)

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

+) Lập trục xét dấu f(x)

\(\Rightarrow\) Bpt có tập nghiệm S = (\(-\infty\); 2) \(\cup\) (4; \(+\infty\))

Chúc bn học tốt!

\(\dfrac{x-4}{x+2}+\dfrac{x+3}{x+4}=\dfrac{2x+1}{x^2+6x+8}\\ =\dfrac{x-4.2}{x+4}+\dfrac{x+3}{x+4}=\dfrac{2x+1}{x^2+6x+8}\\ =\dfrac{x-8+x+3}{x+4}=\dfrac{2x+1}{x^2+6x+8}\\ =\dfrac{2x-5}{x+4}=\dfrac{2x+1}{x^2+6x+8}\)

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

\(\Delta'=\left(-3\right)^2+3.2=15>0\)

⇒ Phương trình có hai nghiệm phân biệt với mọi m.

Theo hệ thức viét ta có : \(\left\{{}\begin{matrix}x_1+x_2=3\\x_1.x_2=-\dfrac{3}{2}\end{matrix}\right.\)

Ta có : \(B=3x_1x_2-x_1^2-x_2^2=-\left(x_1+x_2\right)^2+5x_1x_2=-9+5.\left(-\dfrac{3}{2}\right)=\dfrac{135}{2}\)

Vậy \(B=-\dfrac{135}{2}\) với hai nghiệm phân biệt thỏa mãn.