a) x³+4x²+x-6=0

<=> x³+x²-2x+3x²+3x-6=0

<=>x(x²+x-2)+3(x²+x-2)=0

<=>(x+3)(x²+x-2)=0

<=>(x+3)(x²+2x-x-2)=0

<=>(x+3)[x(x+2)-(x+2)]=0

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

=> x+3=0 hay

x-1=0 hay

x+2=0

<=> x=-3 hay x=1 hay x=-2

b)x³-3x²+4=0

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

\(\Leftrightarrow x\left(x^2-4x+4\right)+\left(x^2-4x+4\right)=0\)

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

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

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

a) 3x² - 7x + 2

= 3x² - 6x - x + 2

= (3x² - 6x) - (x - 2)

= 3x (x - 2) - (x - 2)

= (3x - 1) (x - 2)

Ví dụ cho bạn một bài, còn lại tương tự.

a)Ta có: \(3x^4-5x^3+8x^2-5x+3\)

\(=3x^2\left(x-\frac{5}{6}\right)^2+\frac{71}{12}\left(x-\frac{30}{71}\right)^2+\frac{138}{71}>0\)

Vậy phương trình vô nghiệm.

tth_new bạn làm hết ra đc ko. mình đọc không hiểu đc

tự làm đi dễ vầy

Tui cũng tự làm chứ không làm đâu mà nói nhưng có vài câu chia thấy sai nên mới hỏi để xem đáp án của mình có đúng hay không .

\(5X\left(X-2020\right)+X=2020\)

\(\Leftrightarrow5X^2-10100X+X=2020\)

\(\Leftrightarrow5X^2-10099X=2020\)

\(\Leftrightarrow5X^2-10099X-2020=0\)

\(\Leftrightarrow5X^2-10100X+x-2020=0\)

\(\Leftrightarrow5X\left(X-2020\right)+X-2020=0\)

\(\Leftrightarrow\left(X-2020\right)\left(5X+1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=2020\\x=-\frac{1}{5}\end{cases}}\)

\(4\left(x-5\right)^2-\left(2x+1\right)^2=0\)

\(\Leftrightarrow\left[2\left(x-5\right)\right]^2-\left(2x+1\right)^2=0\)

\(\Leftrightarrow\left[2\left(x-5\right)-2x-1\right]\left[2\left(x-5\right)+2x+1\right]=0\)

\(\Leftrightarrow\left(2x-10-2x-1\right)\left(2x-10+2x+1\right)=0\)

\(\Leftrightarrow-11\left(4x-9\right)=0\)

\(\Leftrightarrow x=\frac{9}{4}\)

a. \(\left(8x^2-4x\right):\left(-4x\right)-\left(x+2\right)=8\)

\(\Leftrightarrow-2x+1-x-2=8\)

\(\Leftrightarrow-2x-x=8+2-1\)

\(\Leftrightarrow-3x=9\)

\(\Leftrightarrow x=-3\)

Vậy...............

b. \(\left(2x^4-3x^3+x^2\right):\left(-\dfrac{1}{2}x^2\right)+4\left(x-1\right)^2=0\)

\(\Leftrightarrow-4x^2+6x-2+4\left(x^2-2x+1\right)=0\)

\(\Leftrightarrow-4x^2+6x-2+4x^2-8x+4=0\)

\(\Leftrightarrow-4x^2+4x^2+6x-8x=-4+2\)

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

\(\Leftrightarrow x=1\)

Vậy...............

a, \(x^6-x^4+2x^3+2x^2=x^2\left(x^4-x^2+2x+2\right)=x^2\left[x^2\left(x^2-1\right)+2\left(x+1\right)\right]\)

\(=x^2\left[x^2\left(x-1\right)\left(x+1\right)+2\left(x+1\right)\right]=x^2\left[\left(x+1\right)\left(x^3+x^2+2\right)\right]\)

\(=x^2\left(x+1\right)\left[\left(x^3+1\right)-\left(x^2-1\right)\right]=x^2\left(x+1\right)\left(x^2-2x+2\right)\)

b, \(\left(x+y\right)^3-\left(x-y\right)^3=\left(x+y-x+y\right)\left[\left(x+y\right)^2+\left(x+y\right)\left(x-y\right)+\left(x-y\right)^2\right]\)

\(=2y\left[\left(x+y\right)\left(x+y+x-y\right)+\left(x-y\right)^2\right]\)\(=2y\left[2x\left(x+y\right)+\left(x-y\right)^2\right]\)

\(=2y\left(2x^2+2xy+x^2-2xy+y^2\right)\)\(=2y\left(3x^2+y^2\right)\)

c, \(x^3-3x^2+3x-1-y^3=\left(x-1\right)^3-y^3=\left(x-1-y\right)\left[\left(x-1\right)^2+\left(x-1\right)y+y^2\right]\)

\(=\left(x-1-y\right)\left[x^2-2x+1+xy-y+y^2\right]\)

d, \(x^5+x^4+1=x^5+x^4+x^3-x^3+1\)

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

e, \(4x^4+81=\left(2x^2\right)^2+36x^2+9^2-36x^2=\left(2x^2+9\right)^2-\left(6x\right)^2\)

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

d, \(64x^4+y^4=\left(8x^2\right)^2+16x^2y^2+\left(y^2\right)^2-16x^2y^2=\left(8x^2+y^2\right)^2-\left(4xy\right)^2=\left(8x^2+y^2+4xy\right)\left(8x^2+y^2-4xy\right)\)