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\(G=2x^2-3x+1=2x^2-2x-x+1\)
\(=2x\left(x-1\right)-\left(x-1\right)=\left(2x-1\right)\left(x-1\right)\)
\(H=-x^2+5x-4=-x^2+4x+x-4\)
\(=-x\left(x-4\right)+\left(x-4\right)=\left(1-x\right)\left(x-4\right)\)
\(I=x^2+4x+3=x^2+3x+x+3\)
\(=x\left(x+3\right)+\left(x+3\right)=\left(x+1\right)\left(x+3\right)\)
\(K=2x^2+7x+5=2x^2+2x+5x+5\)
\(=2x\left(x+1\right)+5\left(x+1\right)=\left(2x+5\right)\left(x+1\right)\)
\(L=-3x^2-5x-2=-3x^2-3x-2x-2\)
\(=-3x\left(x+1\right)-2\left(x+1\right)=\left(-3x-2\right)\left(x+1\right)\)
G = 2x2 - 3x +1 = 2x2 -2x -x +1 =(x-1).(2x-1)
H = -x2 + 5x - 4 = -x2 + 4x +x-4 = (x-4).(1-x)
I = x2 + 4x + 3 = x2 + 3x + x + 3 =(x+3).(x+1)
K = 2x2 + 7x + 5 = 2x2 + 2x + 5x + 5 = (x+1).(2x+5)
L = -3x2 -5x -2 = -3x2 - 3x - 2x - 2 = -3.x(x+1) - 2.(x+1) = (x+1).(-3x-2)
\(A=3x^2-14x^2+4x+3\)
Giả sử:
\(A=\left(3x+a\right)\left(x^2+bx+c\right)\)
\(=3x^3+3bx^2+3cx+ax^{2\:}+abx+ac\)
\(=3x^3+\left(3b+a\right)x^2+\left(3c+ab\right)x+ac\)
Ta có:
\(\begin{cases}3b+a=-14\\3c+ab=4\\ac=3\end{cases}\)\(\Rightarrow\begin{cases}a=1\\b=-5\\c=3\end{cases}\)
Vậy \(A=\left(3x+1\right)\left(x^2-5x+3\right)\)
câu này mih biết làm nhưng pp nhẩm nghiệm là sao bạn
bạn có thể cho mih vd đi\ược ko
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)\(x\left(2x-1\right)\left(1-2x\right)=0\)
\(\Leftrightarrow-x\left(2x-1\right)\left(2x-1\right)=0\)
\(\Leftrightarrow-x\left(2x-1\right)^2=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}-x=0\\\left(2x-1\right)^2=0\end{array}\right.\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\2x=1\end{array}\right.\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\x=\frac{1}{2}\end{array}\right.\)
\(2x^3+3x^2+2x+3=0\)
\(2x\left(x^2+1\right)+3\left(x^2+1\right)=0\)
\(\left(2x+3\right)\left(x^2+1\right)=0\)
\(2x+3=0\left(x^2+1\ge1>0\right)\)
\(2x=-3\)
\(x=-\frac{3}{2}\)
\(x\left(2x-1\right)\left(1-2x\right)=0\)
\(\left[\begin{array}{nghiempt}x=0\\2x-1=0\\1-2x=0\end{array}\right.\)
\(\left[\begin{array}{nghiempt}x=0\\2x=1\\2x=1\end{array}\right.\)
\(\left[\begin{array}{nghiempt}x=0\\x=\frac{1}{2}\end{array}\right.\)
a) 2x + 2y - x2 - xy
= 2(x + y) + x(x + y)
= (x + y) (x + 2)
mk ko bít phân tích đúng ko đúng thì t i c k nhé!! 245433463463564564574675687687856856846865855476457
a)\(2x+2y-x^2-xy=2\left(x+y\right)-x\left(x+y\right)=\left(2-x\right)\left(x+y\right)\)
b)\(\left(x+3\right)^2-\left(2x-5\right)\left(x+3\right)\)
\(=\left(x+3\right)\left[\left(x+3\right)-\left(2x-5\right)\right]\)
\(=\left(x+3\right)\left(8-x\right)\)
c)\(\left(3x+2\right)^2+\left(3x-2\right)^2-2\left(9x^2-4\right)\)
\(=\left(3x+2\right)^2+\left(3x-2\right)^2-2\left(3x-2\right)^2\)
\(=\left(3x+2\right)\left[\left(3x+2\right)-\left(3x-2\right)\right]+\left(3x-2\right)\left[\left(3x-2\right)-\left(3x+2\right)\right]\)
\(=4\left(3x+2\right)-4\left(3x-2\right)\)
\(=4\left(3x+2-3x+2\right)\)
=4.4=16
\(2x^3-3x^2+3x-1=x^3+x^3-3x^2+3x-1\)
=\(x^3+\left(x^3-3x^2+3x-1\right)\)=\(x^3+\left(x-1\right)^3\)
=\(\left(x+x-1\right)\left(x^2-x\left(x-1\right)+\left(x-1\right)^2\right)\)
=\(\left(2x-1\right)\left(x^2-x^2+x+x^2-2x+1\right)\)
=\(\left(2x-1\right)\left(x^2-x+1\right)\)