Q = (x + 2z)(3x2 + 5x2y) – (7x2 – 3x2y)(2z + x)
Bài toán 1.2: Phân tích đa thức
P = 3a(b2 – 2c) – (a – 4)(2c – b2)
Bài toán 1.3: Phân tích đa thức
H = 3xmy – 9xny2 + 15xn+1 với m, n N, m > n.
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Bài toán 1.2 :
\(3a\left(b^2-2c\right)-\left(a-4\right)\left(2c-b^2\right)\)
\(=3a\left(b^2-2c\right)+\left(a-4\right)\left(b^2-2c\right)\)
\(=\left(b^2-2c\right)\left(3a+a-4\right)\)
\(=\left(b^2-2c\right)\left(4a-4\right)\)
a: \(x^2-9-x^2\left(x^2-9\right)\)
\(=\left(x^2-9\right)-x^2\left(x^2-9\right)\)
\(=\left(x^2-9\right)\left(1-x^2\right)\)
\(=\left(1-x\right)\left(1+x\right)\left(x-3\right)\left(x+3\right)\)
b: \(x^2\left(x-y\right)+y^2\left(y-x\right)\)
\(=x^2\left(x-y\right)-y^2\left(x-y\right)\)
\(=\left(x-y\right)\left(x^2-y^2\right)\)
\(=\left(x-y\right)\left(x-y\right)\left(x+y\right)=\left(x-y\right)^2\cdot\left(x+y\right)\)
c: \(x^3+27+\left(x+3\right)\left(x-9\right)\)
\(=\left(x+3\right)\left(x^2-3x+9\right)+\left(x+3\right)\left(x-9\right)\)
\(=\left(x+3\right)\left(x^2-3x+9+x-9\right)\)
\(=\left(x+3\right)\left(x^2-2x\right)=x\left(x-2\right)\left(x+3\right)\)
d: \(x^2+5x+6\)
\(=x^2+2x+3x+6\)
\(=x\left(x+2\right)+3\left(x+2\right)=\left(x+2\right)\left(x+3\right)\)
e: \(3x^2-4x-4\)
\(=3x^2-6x+2x-4\)
\(=3x\left(x-2\right)+2\left(x-2\right)\)
\(=\left(x-2\right)\left(3x+2\right)\)
g: \(x^4+64y^4\)
\(=x^4+16x^2y^2+64y^4-16x^2y^2\)
\(=\left(x^2+8y^2\right)^2-\left(4xy\right)^2\)
\(=\left(x^2+8y^2-4xy\right)\left(x^2+8y^2+4xy\right)\)
h: \(a^2+b^2+2a-2b-2ab\)
\(=a^2-2ab+b^2+2a-2b\)
\(=\left(a-b\right)^2+2\left(a-b\right)=\left(a-b\right)\left(a-b+2\right)\)
i: \(\left(x+1\right)^2-2\left(x+1\right)\left(y-3\right)+\left(y-3\right)^2\)
\(=\left(x+1-y+3\right)^2\)
\(=\left(x-y+4\right)^2\)
k: \(x^2\left(x+1\right)-2x\left(x+1\right)+\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2-2x+1\right)\)
\(=\left(x+1\right)\left(x-1\right)^2\)
Bài 4:
Ta có: \(\left(x^3-x^2\right)-4x^2+8x-4=0\)
\(\Leftrightarrow x^2\left(x-1\right)-4\left(x-1\right)^2=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-2\right)^2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
A = \(4x^2-3x+7x^2+2x-5\)
\(11x^2-3x+2x-5\)
\(11x^2-x-5\)
B = \(3x+7y-6x-8+y-2\)
\(3x+7y-6x-10+y\)
\(- 3x+7y-10+y\)
\(3x+8y-10\)
C = chịu
D= \(6x^4-3x^2+x^2-4x+3.4-x+2\)
\(6x^4-3x^2+x^2-4x;12-x+2\\ \)
\(6x^4-3x^2+x^2-4x+14-x\)
\(6x^4-2x^2-4x+14-x\)
\(6x^4-2x^2-5x+14\)
b: \(xy\left(x+y\right)-yz\left(y+z\right)+xz\left(x-z\right)\)
\(=x^2y+xy^2-y^2z-yz^2+x^2z-xz^2\)
\(=x^2y-yz^2+xy^2-y^2z+x^2z-xz^2\)
\(=y\left(x-z\right)\left(x+z\right)+y^2\left(x-z\right)+xz\left(x-z\right)\)
\(=\left(x-z\right)\left(xy+yz+y^2+xz\right)\)
\(=\left(x-z\right)\left(x+y\right)\left(x+z\right)\)
\(=\dfrac{2\left(x-2y+z\right)^3+4\left(x-2y+z\right)^2}{2\left(x-2y+z\right)}=\left(x-2y+z\right)^2+2\left(x-2y+z\right)\)
Anh chỉ em pp học Toán giỏi như anh được ko ạ rất thích môn Toán nên cũng rất hâm mộ anh
a ) \(x^3z+x^2yz-x^2z^2-xyz^2=\left(x^3z-x^2z^2\right)+\left(x^2yz-xyz^2\right)\)
\(=\left(x-z\right)\left(x^2z+xyz\right)\)
\(=xz\left(x-z\right)\left(x+y\right)\)
b ) \(p^{m+2}.q-p^{m+1}q^3-p^2q^{n+1}+pq^{n+3}\)
\(=p^{m+1}q\left(p-q^2\right)-pq^{n+1}\left(p-q^2\right)\)
\(=\left(p-q^2\right)\left(p^{m+1}q-pq^{n+1}\right)\)
\(=pq\left(p-q^2\right)\left(p^m-q^n\right)\)