\(21x^3-15x^2-6x=0\)
Giải pt
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a) X^3-x^2-21x+45=0
x^3-3x^2+2x^2-6x-15x+45=0
x^2(x-3)+2x(x-3)-15(x-3)=0
(x-3)(x^2+2x-15)=0
(x-3)(x^2-3x+5x-15)=0
(x-3)[x(x-3)+5(x-3)]=0
(x-3)^2(x+5)=0
<=> x=3 hoặc x=-5
Câu 2 đề ko rõ lắm bn sửa lại đề để mk giải hộ nha
Bích Ngọc bạn xem lời giải dưới đây nhé :
X^3-x^2-21x+45=0\(\Leftrightarrow\)(x+5)(x^2-6x+9)=0
\(\Leftrightarrow\)(x+5)(x-3)^2=0
Rồi đó tới đây bạn tự tìm x nhé!
g: =>(x-1)(x-2)=0
=>x=1 hoặc x=2
i: \(\Leftrightarrow x^4-x^3+x^3-x^2+2x^2-2x+8x-8=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^3+x^2+2x+8\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+2\right)\left(x^2-x+4\right)=0\)
=>x=1 hoặc x=-2
g: \(x^2-3x+2=0\)
=>(x-1)(x-2)=0
=>x=1 hoặc x=2
i: \(x^4+x^2+6x-8=0\)
\(\Leftrightarrow x^4-x^3+x^3-x^2+2x^2-2x+8x-8=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^3+x^2+2x+8\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left[\left(x+2\right)\left(x^2-2x+4\right)+x\left(x+2\right)\right]=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+2\right)\left(x^2-x+4\right)=0\)
=>x=1 hoặc x=-2
\(x^6-6x^5+15x^4-20x^3+15x^2-6x+1=0\)
\(\Leftrightarrow x^6-x^5-5x^5+5x^4+10x^4-10x^3-10x^3+10x^2+5x^2-5x-x+1=0\)
\(\Leftrightarrow x^5\left(x-1\right)-5x^4\left(x-1\right)+10x^3\left(x-1\right)-10x^2\left(x-1\right)+5x\left(x-1\right)-\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^5-5x^4+10x^3-10x^2+5x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left[x^5-x^4-4x^4+4x^3+6x^3-6x^2-4x^2+4x+x-1\right]=0\)
\(\Leftrightarrow\left(x-1\right)\left[x^4\left(x-1\right)-4x^3\left(x-1\right)+6x^2\left(x-1\right)-4x\left(x-1\right)+x-1\right]=0\)
\(\Leftrightarrow\left(x-1\right)^2\left[x^4-4x^3+6x^2-4x+1\right]=0\)
\(\Leftrightarrow\left(x-1\right)^2\left[x^4-x^3-3x^3+3x^2+3x^2-3x-x+1\right]=0\)
\(\Leftrightarrow\left(x-1\right)^3\left[x^3-3x^2+3x-1\right]=0\)
\(\Leftrightarrow\left(x-1\right)^3\left[x^3-x^2-2x^2+2x+x-1\right]=0\)
\(\Leftrightarrow\left(x-1\right)^4\left[x^2-2x+1\right]=0\Leftrightarrow\left(x-1\right)^6=0\Leftrightarrow x=1\)
a) \(21x+7=15x+35\)
\(\Leftrightarrow21x-15x=35-7\)
\(\Leftrightarrow6x=28\)
\(\Leftrightarrow x=\frac{28}{6}\)
b)\(|5x+3|-2x=x-17\)
\(\Leftrightarrow|5x+3|=3x-17\)
\(\Leftrightarrow\orbr{\begin{cases}5x+3=3x-17\\5x+3=17-3x\end{cases}\Leftrightarrow\orbr{\begin{cases}2x=-20\\8x=14\end{cases}}}\Leftrightarrow\orbr{\begin{cases}x=-10\\x=\frac{4}{7}\end{cases}}\)
c) \(\left(5x+2\right)\left(3x-4\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}5x+2=0\\3x-4=0\end{cases}\Leftrightarrow\orbr{\begin{cases}5x=-2\\3x=4\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=-\frac{2}{5}\\x=\frac{4}{3}\end{cases}}}\)
a) \(21x+7=15x+35\)
\(\Leftrightarrow\)\(6x=28\)
\(\Leftrightarrow\)\(x=\frac{14}{3}\)
Vậy...
b) \(\left|5x+3\right|-2x=x-17\)
\(\Leftrightarrow\)\(\left|5x+3\right|=3x-17\)
Nếu \(x\ge-\frac{3}{5}\)thì pt trở thành:
\(5x+3=3x-17\)
\(\Leftrightarrow\)\(2x=-20\)
\(\Leftrightarrow\)\(x=-10\)(loại)
Nếu \(x< -\frac{3}{5}\)thì pt trở thành:
\(-5x-3=3x-17\)
\(\Leftrightarrow\)\(8x=14\)
\(\Leftrightarrow\) \(x=\frac{7}{4}\) (loại)
Vậy pt vô nghiệm
c) \(\left(5x+2\right)\left(3x-4\right)=0\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}5x+2=0\\3x-4=0\end{cases}}\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x=-\frac{2}{5}\\x=\frac{4}{3}\end{cases}}\)
Vậy...
\(\dfrac{15x^3y^5-10x^4y^4-21x^5y^3z}{-6x^3y^2}=-\dfrac{5}{2}y^3+\dfrac{5}{3}xy^2+\dfrac{7}{2}x^2yz\)
x1=1
x2=\(\dfrac{-2}{7}\)
\(21x^3-15x^2-6x=0\\ \Leftrightarrow x\left(21x^2-15x-6\right)=0\\ \Leftrightarrow x\left[\left(21x^2-21x\right)+\left(6x-6\right)\right]=0\\ \Leftrightarrow x\left[21x\left(x-1\right)+6\left(x-1\right)\right]=0\\ \Leftrightarrow x\left(x-1\right)\left(21x+6\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x=\dfrac{-2}{7}\end{matrix}\right.\)