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\(=x\left(2x^2+3x-2\right)=x\left(2x^2+4x-x-2\right)=x\left[2x\left(x+2\right)-\left(x+2\right)\right]=x\left(2x-1\right)\left(x+2\right)\)
2x3 + 3x2 - 2x
= x ( 2x2 + 3x - 2 )
= x ( 2\(x^2\) + 4\(x-x-2\) )
= x [ ( 2\(x^2\) + 4x ) - ( x + 2 )]
= x [ 2x ( x + 2 ) - ( x + 2 )]
= x ( 2x - 1 ) ( x + 2 )
Bài 1:
a: \(8x^3-2x=2x\left(4x^2-1\right)=2x\left(2x-1\right)\left(2x+1\right)\)
c: \(-5m^3\left(m+1\right)+m+1=\left(m+1\right)\left(-5m^3+1\right)\)
Bài 1:
a: \(=6x^3-10x^2+6x\)
b: \(=-2x^3-10x^2-6x\)
Bài 4:
a: =>3x+10-2x=0
=>x=-10
c: =>3x2-3x2+6x=36
=>6x=36
hay x=6
Bài 1:
\(a,=6x^3-10x^2+6x\\ b,=-2x^3-10x^2-6x\)
Bài 4:
\(a,\Leftrightarrow3x+10-2x=0\Leftrightarrow x=-10\\ b,\Leftrightarrow x\left(2x^2+9x-5\right)-\left(2x^3+9x^2+x+4,5\right)=3,5\\ \Leftrightarrow2x^3+9x^2-5x-2x^3-9x^2-x-4,5=3,5\\ \Leftrightarrow-6x=8\Leftrightarrow x=-\dfrac{4}{3}\\ c,\Leftrightarrow3x^2-3x^2+6x=36\Leftrightarrow x=6\)
Bài 1:
\(a,=7xy\left(2x-3y+4xy\right)\\ b,=x\left(x+y\right)-5\left(x+y\right)=\left(x-5\right)\left(x+y\right)\\ c,=\left(x-y\right)\left(10x+8\right)=2\left(5x+4\right)\left(x-y\right)\\ d,=\left(3x+1-x-1\right)\left(3x+1+x+1\right)\\ =2x\left(4x+2\right)=4x\left(2x+1\right)\\ e,=5\left[\left(x-y\right)^2-4z^2\right]=5\left(x-y-2z\right)\left(x-y+2z\right)\\ f,=x^2+8x-x-8=\left(x+8\right)\left(x-1\right)\\ g,\left(x+y\right)^3-\left(x+y\right)=\left(x+y\right)\left[\left(x+y\right)^2-1\right]\\ =\left(x+y\right)\left(x+y-1\right)\left(x+y+1\right)\\ h,=x^2+3x+x+3=\left(x+3\right)\left(x+1\right)\)
a: Ta có: \(x^2-4y^2-2x-4y\)
\(=\left(x-2y\right)\left(x+2y\right)-2\left(x+2y\right)\)
\(=\left(x+2y\right)\left(x-2y-2\right)\)
c: Ta có: \(x^3+2x^2y-x-2y\)
\(=x^2\left(x+2y\right)-\left(x+2y\right)\)
\(=\left(x+2y\right)\left(x-1\right)\left(x+1\right)\)
d: Ta có: \(3x^2-3y^2-2\cdot\left(x-y\right)^2\)
\(=3\left(x-y\right)\left(x+y\right)-2\cdot\left(x-y\right)^2\)
\(=\left(x-y\right)\left(3x+3y-2x+2y\right)\)
\(=\left(x-y\right)\left(x+5y\right)\)
e: Ta có: \(x^3-4x^2-9x+36\)
\(=x^2\left(x-4\right)-9\left(x-4\right)\)
\(=\left(x-4\right)\left(x-3\right)\left(x+3\right)\)
f: Ta có: \(x^2-y^2-2x-2y\)
\(=\left(x-y\right)\left(x+y\right)-2\left(x+y\right)\)
\(=\left(x+y\right)\left(x-y-2\right)\)
Bài 3:
b: \(x^2+2x+1=\left(x+1\right)^2\)
c: \(x^2-16=\left(x-4\right)\left(x+4\right)\)
d: \(\left(2x-1\right)^2-\left(x+3\right)^2\)
\(=\left(2x-1-x-3\right)\left(2x-1+x+3\right)\)
\(=\left(x-4\right)\left(3x+2\right)\)
1/(x+2)2 -(3x-1)2=(x+2+3x-1)(x+2-3x+1)=4x(-2x+3)=-8x2+12x
2/(x4+x2)(-2x3-2x)=x2(x2+1)-2x(x2+1)=(x2+1)(x2-2x)
Bài 1:
b: \(3x-6=x^2-16\)
\(\Leftrightarrow x^2-3x-10=0\)
\(\Leftrightarrow\left(x-5\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-2\end{matrix}\right.\)
thành nhân tử ta được:
)(9x2-x+
) 
)(9x2+x+
Câu 6:Thực hiện phép nhân -2x(x2 + 3x - 4) ta được:
A.-2x3 - 6x2 – 8x B. 2x3 -6x2 – 8x C. -2x3 - 6x2 + 8x D. -2x3 + 3x2 -4
Câu 7 : Phân tích đa thức x2 + 2xy + y2 – 9z2 thành nhân tử ta được:
A. (x+y+3z)(x+y–3z)
B. (x-y+3z)(x+y–3z)
C.(x - y +3z)(x - y – 3z)
D. (x + y +3z)(x -y – 3z)
Câu 9: Phân tích đa thức x2 + 7x + 12 thành nhân tử ta được:
A. (x - 3)( x + 4 ) B. (x + 3)( x + 4 ) C.(x + 5)( x + 2 ) D. (x -5)( x + 2 )
Câu 10: Giá trị của biểu thức (x2 + 4x + 4) tại x = - 2 là:
A. 4 B. -2 C. 0 D. -8
Mấy câu còn lại bị lỗi r nhé
x⁴ - 2x³ + 2x - 1
= (x⁴ - 1) - (2x³ - 2x)
= (x² - 1)(x² + 1) - 2x(x² - 1)
= (x² - 1)(x² + 1 - 2x)
= (x - 1)(x + 1)(x² - 2x + 1)
= (x - 1)(x + 1)(x - 1)²
= (x - 1)³(x + 1)
a)\(2x^3+3x^2+2x+3=0\)
\(\Leftrightarrow2x^3+2x+3x^2+3=0\)
\(\Leftrightarrow2x\left(x^2+1\right)+3\left(x^2+1\right)=0\)
\(\Leftrightarrow\left(2x+3\right)\left(x^2+1\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}2x+3=0\\x^2+1=0\end{array}\right.\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}2x=-3\\x^2+1>0\left(loai\right)\end{array}\right.\)
\(\Leftrightarrow x=-\frac{3}{2}\)
b)\(x\left(2x-1\right)\left(1-2x\right)=0\)
\(\Leftrightarrow-x\left(2x-1\right)\left(2x-1\right)=0\)
\(\Leftrightarrow-x\left(2x-1\right)^2=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}-x=0\\\left(2x-1\right)^2=0\end{array}\right.\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\2x=1\end{array}\right.\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\x=\frac{1}{2}\end{array}\right.\)
\(2x^3+3x^2+2x+3=0\)
\(2x\left(x^2+1\right)+3\left(x^2+1\right)=0\)
\(\left(2x+3\right)\left(x^2+1\right)=0\)
\(2x+3=0\left(x^2+1\ge1>0\right)\)
\(2x=-3\)
\(x=-\frac{3}{2}\)
\(x\left(2x-1\right)\left(1-2x\right)=0\)
\(\left[\begin{array}{nghiempt}x=0\\2x-1=0\\1-2x=0\end{array}\right.\)
\(\left[\begin{array}{nghiempt}x=0\\2x=1\\2x=1\end{array}\right.\)
\(\left[\begin{array}{nghiempt}x=0\\x=\frac{1}{2}\end{array}\right.\)