3)

\(x^3-7x+6=0\)

\(\Leftrightarrow x^3+3x^2-3x^2-9x+2x+6=0\)

\(\Leftrightarrow\left(x^3+3x^2\right)-\left(3x^2+9x\right)+\left(2x+6\right)=0\)

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

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

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

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

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

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

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

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

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

Vậy ................

2.

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

\(\Leftrightarrow2x^3+6x^2-x^2-3x=0\)

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

\(\Leftrightarrow2x^2\left(x+3\right)-x\left(x+3\right)=0\)

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

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

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

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

x²+10x+25-4x(x+5)=0

⇔(x+5)²-4x(x+5)=0

⇔(x+5)(x+5-4x)=0

⇔(x+5)(5-3x)=0

⇔\(\left\{{}\begin{matrix}x+5=0\\5-3x=0\end{matrix}\right.\Leftrightarrow\left\{{} }\left\{{}\begin{matrix}x=-5\\x=\dfrac{5}{3}\end{matrix}\right.\)

Bạn tham khảo câu trả lời tại đây:

Câu hỏi của Nguyễn Kim Chi - Toán lớp 8 - Học toán với OnlineMath

a: =>2x-3x^2-x<15-3x^2-6x

=>x<-6x+15

=>7x<15

=>x<15/7

b: =>4x^2-24x+36-4x^2+4x-1>=12x

=>-20x+35>=12x

=>-32x>=-35

=>x<=35/32

\(a,2x-x\left(3x+1\right)< 15-3x\left(x+2\right)\\ \Leftrightarrow2x-3x^2-x< 15-3x^2-6x\\ \Leftrightarrow3x^2-3x^2+2x+6x-x< 15\\ \Leftrightarrow7x< 15\\ \Leftrightarrow x< \dfrac{15}{7}\)

Vậy S={-∞; 15/7}

\(b,4\left(x-3\right)^2-\left(2x-1\right)^2\ge12x\\ \Leftrightarrow4\left(x^2-6x+9\right)-\left(4x^2-4x+1\right)-12x\ge0\\ \Leftrightarrow4x^2-4x^2-24x+4x-12x\ge-36+1\\ \Leftrightarrow-32x\ge-35\\ \Leftrightarrow x\le\dfrac{35}{32}\)

Vậy S={-∞; 35/32]

b: \(\Leftrightarrow2x\left(x^2+4x+4\right)-8x^2=2\left(x^3-8\right)\)

\(\Leftrightarrow2x^3+8x^2+8x-8x^2-2x^3+16=0\)

=>8x+16=0

=>x=-2

d: \(\Leftrightarrow x^3-6x^2+12x-8+9x^2-1-x^3-3x^2-3x-1=0\)

\(\Leftrightarrow9x-10=0\)

=>x=10/9