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

3x(2-x)-5=1-(3x²+2)

<=>6x-3x²-5=-3x²-2

<=>6x=3

<=>x=1/2

\(\Leftrightarrow\left\{{}\begin{matrix}\left(x+y+1\right)\left(x+y-6\right)=0\\y-x-3=0\left(3\right)\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x=-\left(y+1\right)\left(1\right)\\x=6-y\left(2\right)\end{matrix}\right.\\y-x-3=0\left(3\right)\end{matrix}\right.\)

\(thế\left(1\right)\left(2\right)vào\left(3\right)\Rightarrow\left(x;y\right)\)

Bài đâu bạn?

\(1,4x\left(1-x\right)-8=1-\left(4x^2+3\right)\\ \Leftrightarrow4x-4x^2-8=1-4x^2-3\\ \Leftrightarrow4x-4x^2-8-1+4x^2+3=0\\ \Leftrightarrow4x-6=0\\ \Leftrightarrow x=\dfrac{3}{2}\)

\(2,\left(2-3x\right)\left(x+11\right)=\left(3x-2\right)\left(2-5x\right)\\ \Leftrightarrow\left(2-3x\right)\left(x+11\right)-\left(2-3x\right)\left(5x-2\right)=0\\ \Leftrightarrow\left(2-3x\right)\left(x+11-5x+2\right)=0\\ \Leftrightarrow\left(2-3x\right)\left(-4x+13\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=\dfrac{13}{4}\end{matrix}\right.\)

bạn nhấn vào  đúng 0 sẽ ra đáp án

em mới lớp 6 thui anh ơi 

\(x^2-x+1-m=0\left(1\right)\\ \text{PT có 2 nghiệm }x_1,x_2\\ \Leftrightarrow\Delta=1-4\left(1-m\right)\ge0\\ \Leftrightarrow4m-3\ge0\Leftrightarrow m\ge\dfrac{3}{4}\\ \text{Vi-ét: }\left\{{}\begin{matrix}x_1+x_2=1\\x_1x_2=1-m\end{matrix}\right.\\ \text{Ta có }5\left(\dfrac{1}{x_1}+\dfrac{1}{x_2}\right)-x_1x_2+4=0\\ \Leftrightarrow5\cdot\dfrac{x_1+x_2}{x_1x_2}-x_1x_2+4=0\\ \Leftrightarrow\dfrac{5}{1-m}+m-1+4=0\\ \Leftrightarrow\dfrac{5}{1-m}+m+3=0\\ \Leftrightarrow5+\left(1-m\right)\left(m+3\right)=0\\ \Leftrightarrow m^2+2m-8=0\\ \Leftrightarrow m^2-2m+4m-8=0\\ \Leftrightarrow\left(m-2\right)\left(m+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}m=2\left(n\right)\\m=-4\left(l\right)\end{matrix}\right.\)

Vậy $m=2$

Lời giải:
Để pt có 2 nghiê pb thì:

$\Delta'=1-(m-3)>0\Leftrightarrow m< 4$

Áp dụng định lý Viet: \(\left\{\begin{matrix} x_1+x_2=2\\ x_1x_2=m-3\end{matrix}\right.\)

Khi đó:
\(x_1^2-2x_2+x_1x_2=-12\)

\(\Leftrightarrow x_1^2-2(2-x_1)+x_1(2-x_1)=-12\)

\(\Leftrightarrow x_1=-2\Leftrightarrow x_2=2-x_1=4\)

$m-3=x_1x_2=(-2).4=-8$

$\Leftrightarrow m=-5$ (tm)

\(x^2\left(x+4,5\right)=13,5\)

<=>\(x^3+4,5x^2-13,5=0\)

<=> \(x^3+3x^2+1,5x^2+4,5x-4,5x-13,5=0\)

<=>\(x^2\left(x+3\right)+1,5x\left(x+3\right)-4,5\left(x+3\right)=0\)

<=>\(\left(x+3\right)\left(x^2+1,5x-4,5\right)=0\)

<=>\(\left(x+3\right)\left[x^2+3x-1,5-4,5\right]=0\)

<=>\(\left(x+3\right)\left[x\left(x+3\right)-1,5\left(x+3\right)\right]=0\)

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

<=> \(\left[{}\begin{matrix}\left(x+3\right)^2=0\\x-1,5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=1,5\end{matrix}\right.\)

Vậy...

Ta có: \(x^2\left(x+4.5\right)=13.5\)

\(\Leftrightarrow x^3+\dfrac{9}{2}x^2-\dfrac{27}{2}=0\)

\(\Leftrightarrow2x^3+9x^2-27=0\)

\(\Leftrightarrow2x^3-3x^2+12x^2-18x+18x-27=0\)

\(\Leftrightarrow x^2\left(2x-3\right)+12x\left(2x-3\right)+9\left(2x-3\right)=0\)

\(\Leftrightarrow\left(2x-3\right)\left(x^2+12x+9\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-3=0\\x^2+12x+9=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=3\\\left(x+6\right)^2=27\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x+6=3\sqrt{3}\\x+6=-3\sqrt{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=3\sqrt{3}-6\\x=-3\sqrt{3}-6\end{matrix}\right.\)

Vậy: \(S=\left\{\dfrac{3}{2};3\sqrt{3}-6;-3\sqrt{3}-6\right\}\)