1)

<=> \(x^2-3x=0\)

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

x= 0 

x = 3

2) <=> \(x\left(x-3\right)=4\)

=> \(x=\dfrac{4}{x}+3\)

\(2,x^2-3x=4\)

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

\(\Delta=b^2-4ac=\left(-3\right)^2-4\left(-4\right)=25>0\)

\(\Rightarrow\)Pt có 2 nghiệm pb

\(\left\{{}\begin{matrix}x_1=\dfrac{-b+\sqrt{\Delta}}{2a}=\dfrac{3+5}{2}=4\\x_2=\dfrac{-b-\sqrt{\Delta}}{2a}=\dfrac{-3-5}{2}=-1\end{matrix}\right.\)

Vậy \(S=\left\{4;-1\right\}\)

\(3,x^4-5x^2+6=0\)

Đặt \(t=x^2\left(t\ge0\right)\)

Pt trở thành

\(t^2-5t+6=0\)

\(\Delta=b^2-4ac=\left(-5\right)^2-4.6=1>0\)

\(\Rightarrow\)Pt ó 2 nghiệm pb

\(\left\{{}\begin{matrix}x_1=\dfrac{-b+\sqrt{\Delta}}{2a}=\dfrac{5+1}{2}=3\\x_2=\dfrac{-b-\sqrt{\Delta}}{2a}=\dfrac{-5-1}{2}-3\end{matrix}\right.\)

\(\Rightarrow t=x^2\Leftrightarrow t=\pm\sqrt{3}\)

Vậy \(S=\left\{\pm\sqrt{3}\right\}\)

\(4,x^3=9x\)

\(\Leftrightarrow x^3-9x=0\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x^2=9\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\pm3\end{matrix}\right.\)

Vậy \(S=\left\{0;\pm3\right\}\)

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

\(\Leftrightarrow x^2-3x+2x-6-x^2+4=0\)

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

\(\Leftrightarrow-x=2\)

\(\Leftrightarrow x=-2\)

Vậy \(S=\left\{-2\right\}\)

Hung nguyen, Trần Thanh Phương, Sky SơnTùng, @tth_new, @Nguyễn Việt Lâm, @Akai Haruma, @No choice teen

help me, pleaseee

Cần gấp lắm ạ!

a, (3x - 1)(5x + 3) = (2x + 3)(3x - 1)

⇔ 5x + 3 = 2x + 3

⇔ 3x = 0

⇔ x = 0

Vậy phương trình có nghiệm là x = 0

Mình làm lại rồi nhé!

a, (3x - 1)(5x + 3) = (2x + 3)(3x - 1)

⇔ 5x + 3 = 2x + 3

⇔ 3x = 0

⇔ x = 0

Vậy phương trình có nghiệm là x = 3.

b, 9x² - 1 = (3x + 1)(2x - 1)

⇔ (3x + 1)(3x - 1) = (3x + 1)(2x - 1)

⇔ 3x - 1 = 2x - 1

⇔ x = 0

Vậy phương trình có nghiệm là x = 0

\(a,-5x\left(x-3\right)\left(2x+4\right)-\left(x+3\right)\left(x-3\right)+\left(5x-2\right)\left(3x+4\right)\)

\(=-5x\left(2x^2-x-12\right)-\left(x^2-9\right)+15x^2+20x-6x-8\)

\(=-10x^3+5x^2+60x-x^2+9+15x^2+20x-6x-8\)

\(=-10x^3+19x^2+74x+1\)

\(b,\left(4x-1\right)x\left(3x+1\right)-5x^2.x\left(x-3\right)-\left(x-4\right)x\left(x-5\right)\)\(-7\left(x^3-2x^2+x-1\right)\)

\(=\left(4x^2-x\right)\left(3x+1\right)-5x^4-15x^3-\left(x^2-4x\right)\left(x-5\right)\)\(-7x^3+14x^2-7x+7\)

\(=12x^3+x^2-x-5x^4-15x^3-x^3+9x^2+20x\)\(-7x^3+14x^2-7x+7\)

\(=-5x^4-11x^3+24x^2+12x+7\)

\(c,\left(5x-7\right)\left(x-9\right)-\left(3-x\right)\left(2-5x\right)-2x\left(x-4\right)\)

\(=5x^2-52x+63-6+17x-5x^2-2x^2+8x\)

\(=-2x^2-27x+57\)

\(d,\left(5x-4\right)\left(x+5\right)-\left(x+1\right)\left(x^2-6\right)-5x+19\)

\(=5x^2+21x-20-x^3-x^2+6x+6-5x+19\)

\(=-x^3+4x^2+22x+5\)

\(e,\left(9x^2-5\right)\left(x-3\right)-3x^2\left(3x+9\right)-\left(x-5\right)\left(x+4\right)-9x^3\)

\(=9x^3-27x^2-5x+15-9x^3-27x^2-x^2+x+20-9x^3\)

\(=-9x^3-55x^2+4x+35\)

\(g,\left(x-1\right)^2-\left(x+2\right)^2\)

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

\(=-6x-3\)

a)

\(\left(5x+3\right)\cdot\left(x^2+4\right)\cdot\left(x-4\right)=0\\ \Rightarrow\left[{}\begin{matrix}5x+3=0\\x-4=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-\frac{3}{5}\\x=4\end{matrix}\right.\)

b)

\(\left(4x-1\right)\cdot\left(x-3\right)-\left(x-2\right)\cdot\left(5x+2\right)=0\\ \Leftrightarrow4x^2-12x-x+3-5x^2-2x+10x+4=0\\ \Leftrightarrow-x^2-5x+7=0\\ \Rightarrow x=\left[{}\begin{matrix}-\frac{5+\sqrt{53}}{2}\\-\frac{5-\sqrt{53}}{2}\end{matrix}\right.\)

c)

\(\left(x+3\right)\cdot\left(x-5\right)+\left(x+3\right)\cdot\left(3x-4\right)=0\\ \Leftrightarrow\left(x+3\right)\cdot\left(x-5+3x-4\right)=0\\ \Leftrightarrow\left(x+3\right)\cdot\left(4x-9\right)=0\\ \Rightarrow\left[{}\begin{matrix}x+3=0\\4x-9=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-3\\x=\frac{9}{4}\end{matrix}\right.\)

d)

\(\left(x+6\right)\cdot\left(3x-1\right)+x^2-36=0\\ \Leftrightarrow\left(x+6\right)\cdot\left(3x-1\right)+\left(x^2-36\right)=0\\ \Leftrightarrow\left(x+6\right)\cdot\left(3x-1\right)+\left(x+6\right)\cdot\left(x-6\right)=0\\ \Leftrightarrow\left(x+6\right)\cdot\left(3x-1+x-6\right)=0\\ \Leftrightarrow\left(x+6\right)\cdot\left(4x-7\right)=0\\ \Rightarrow\left[{}\begin{matrix}x+6=0\\4x-7=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-6\\x=\frac{7}{4}\end{matrix}\right.\)

e)

\(0.75x\cdot\left(x+5\right)=\left(x+5\right)\cdot\left(3-1.25x\right)\\ \Leftrightarrow0.75x\cdot\left(x+5\right)-\left(x+5\right)\cdot\left(3-1.25x\right)=0\\ \Leftrightarrow\left(x+5\right)\cdot\left(0.75x-3+1.25x\right)=0\\ \Leftrightarrow\left(x+5\right)\cdot\left(2x-3\right)=0\\ \Rightarrow\left[{}\begin{matrix}x+5=0\\2x-3=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-5\\x=\frac{3}{2}\end{matrix}\right.\)

g: Ta có: \(3\left(2x-1\right)\left(3x-1\right)-\left(2x-3\right)\left(9x-1\right)=0\)

\(\Leftrightarrow3\left(6x^2-5x+1\right)-\left(18x^2-29x+3\right)=0\)

\(\Leftrightarrow18x^2-15x+3-18x^2+29x-3=0\)

\(\Leftrightarrow14x=0\)

hay x=0

w