d) ĐKXĐ: \(x\ne5,x\ne-2\)

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

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

e) ĐKXĐ: \(x\ne1,x\ne2\)

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

\(\Leftrightarrow\dfrac{\left(x+1\right)\left(x-2\right)+\left(x-1\right)-3x-4}{\left(x-1\right)\left(x-2\right)}=0\Leftrightarrow x^2-x-2+x-1-3x-4=0\Leftrightarrow x^2-3x-7=0\)

\(\Delta=\left(-3\right)^2-4.\left(-7\right)=37>0\)

PT có 2 nghiệm: \(\left\{{}\begin{matrix}x_1=\dfrac{3-\sqrt{37}}{2}\left(tm\right)\\x_2=\dfrac{3+\sqrt{37}}{2}\left(tm\right)\end{matrix}\right.\)

Phương trình có 2 nghiệm khi:

\(\Delta=m^2-12\left(2m+1\right)\ge0\Leftrightarrow m^2-24m-12\ge0\)

Khi đó theo hệ thức Viet: \(\left\{{}\begin{matrix}x_1+x_2=\dfrac{m}{3}\\x_1x_2=\dfrac{2m+1}{3}\end{matrix}\right.\)

Tích 2 nghiệm bằng -3 khi:

\(\dfrac{2m+1}{3}=-3\Rightarrow2m+1=-9\)

\(\Rightarrow m=-5\)

Khi đó tổng 2 nghiệm là: \(x_1+x_2=\dfrac{m}{3}=-\dfrac{5}{3}\)

a) \(\Leftrightarrow x^2+10x+25-x^2+8x-15=-8\\ \Leftrightarrow18x=-18\\ \Leftrightarrow x=-1\)

b) \(\Leftrightarrow\left(2x+1\right)^2-3\left(2x+1\right)=0\\ \Leftrightarrow\left(2x+1\right)\left(2x+1-3\right)=0\\ \Leftrightarrow\left(2x+1\right)\left(2x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=1\end{matrix}\right.\)

3x(2-x)-5=1-(3x²+2)

<=>6x-3x²-5=-3x²-2

<=>6x=3

<=>x=1/2

\(\Leftrightarrow2x^2+10x-x^2+6x-9=x^2+6\)

=>16x-9=6

=>16x=15

hay x=15/16

\(PT\Leftrightarrow2x^2+10x-x^2+6x-9-x^2-6=0.\)

\(\Leftrightarrow16x-15=0.\\ \Leftrightarrow x=\dfrac{15}{16}.\)

ĐK: \(x\ge0\)

Dễ thấy \(1-\sqrt{2\left(x^2-x+1\right)}\le1-\sqrt{2}< 0\)

Khi đó bất phương trình tương đương:

\(x-\sqrt{x}\le1-\sqrt{2\left(x^2-x+1\right)}\)

\(\Leftrightarrow\sqrt{x}-\dfrac{1}{\sqrt{x}}-1+\sqrt{2\left(x+\dfrac{1}{x}-1\right)}\le0\)

\(\Leftrightarrow\sqrt{x}-\dfrac{1}{\sqrt{x}}-1+\sqrt{2\left(\sqrt{x}-\dfrac{1}{\sqrt{x}}\right)^2+2}\le0\)

\(\Leftrightarrow t-1+\sqrt{2t^2+2}\le0\)

Ta có:

     (2 - 3x)(x + 8) = (3x - 2)(3 - 5x)

⇔ (2 - 3x)(x + 8) - (3x - 2)(3 - 5x) = 0

⇔ (2 - 3x)(x + 8) + (2 - 3x)(3 - 5x) = 0

⇔ (2 - 3x)(x + 8 + 3 - 5x) = 0

⇔ (2 - 3x)(11 - 4x) = 0

⇔ 2 - 3x = 0 hay 11 - 4x = 0

⇔ 2 = 3x hay 11 = 4x

⇔ x = \(\dfrac{2}{3}\) hay x = \(\dfrac{11}{4}\)

Vậy tập nghiệm của pt S = \(\left\{\dfrac{2}{3};\dfrac{11}{4}\right\}\)

khum bt mới lớp 5 hoi à(k10)

<=> (2-3x ) (x+8) + (2-3x ) (3-5x)=0
<=> (2-3x ) ( x+8 +  3-5x ) =0 
<=> (2-3x ) ( 11 - 4x ) = 0
 => 2-3x  =0 hoặc 11-4x =0  
       3x = 2            4x =11
         x = 2/3         x    = 11/4

18) ĐKXĐ: \(x\ne-1,x\ne4\)

\(pt\Leftrightarrow\dfrac{x^2-x+8-2x\left(x-4\right)}{\left(x+1\right)\left(x-4\right)}=0\Leftrightarrow-x^2+7x+8=0\)

\(\Leftrightarrow\left(x-8\right)\left(x+1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=8\left(tm\right)\\x=-1\left(ktm\right)\end{matrix}\right.\)

19) ĐKXĐ: \(x\ne\pm3\)

\(pt\Leftrightarrow\dfrac{\left(x+3\right)^2-48-\left(x-3\right)^2}{\left(x-3\right)\left(x+3\right)}=0\Leftrightarrow12x-48=0\Leftrightarrow x=4\left(tm\right)\)

20) ĐKXĐ: \(x\ne\pm1\)

\(pt\Leftrightarrow\dfrac{\left(x+1\right)^2-\left(x-1\right)^2-16}{\left(x-1\right)\left(x+1\right)}=0\Leftrightarrow4x-16=0\Leftrightarrow x=4\left(tm\right)\)

đó là kt cơ bản nhất luôn , mấy bài này k bt lm thì lm b j nx :<

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

\(\Leftrightarrow2x^2+x^2-2x+1+4=0\)

\(\Leftrightarrow2x^2+\left(x-1\right)^2=-4\)

Phương trình vô nghiệm.

Sai đề phải là 3x²-2x-5=0

Ta có:3x²-2x-5=0

  <=> 3x²+3x-5x-5=0

  <=> 3x(x+1)-5(x+1)=0

  <=> (x+1)(3x-5)=0

  \(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=\dfrac{5}{3}\end{matrix}\right.\)

Ta có: \(3x^2-2x+5=0\)

\(\text{Δ}=\left(-2\right)^2-4\cdot3\cdot5\)

\(=4-60\)

\(=-56< 0\)

vậy: Phương trình vô nghiệm