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bài 1
a) \(7x\left(5x-1\right)+5x-1=\left(5x-1\right)\left(7x+1\right)\)
b) \(4xy-4x^2-y^2+25=25-\left(4x^2-4xy+y^2\right)\)
\(=5^2-\left(2x-y\right)=\left(5-2x+y\right)\left(5+2x-y\right)\)
c) \(2x^2-2y+xy-4x=\left(2x^2+xy\right)-\left(2y+4x\right)\)
\(=x^2\left(2x+y\right)-2\left(2x+y\right)=\left(2x+y\right)\left(x^2-2\right)\)
d) \(3x^2-7x+2=3x^2-6x-x+2\)
\(=3x\left(x-2\right)-\left(x-2\right)\)
\(=\left(x-2\right)\left(3x-1\right)\)
bài 2
a) * Rút gọn:
\(Q=3\left(2x-1\right)^2+2\left(2x+3\right)\left(x-1\right)-\left(x-3\right)\left(x+3\right)\)
\(Q=\left[3\left(4x^2-4x+1\right)\right]+\left[2\left(2x^2-2x+3x-3\right)\right]-\left(x^2-9\right)\)
\(Q=\left(12x^2-12x+3\right)+\left(4x^2-4x+6x-6\right)-\left(x^2-9\right)\)
\(Q=12x^2-12x+3+4x^2-4x+6x-6-x^2+9\)
\(Q=15x^2-10x+6=5x\left(3x-2\right)+6\)
Thế x = 2 vào biểu thức Q ta được:
\(Q=5\cdot2\left(3\cdot2-2\right)+6=46\)
b) \(Q=5x\left(3x-2\right)+6=6\)
\(\Leftrightarrow5x\left(3x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}5x=0\\3x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{2}{3}\end{matrix}\right.\)
Dễ mà bạn
\(P=\left(x-y\right)\left(x^2+xy+y^2\right)-2y^3=x^3-y^3-2y^3=x^3-3y^3=\left(\frac{1}{2}\right)^3-3.\left(\frac{2}{3}\right)^3=\frac{-55}{72}\)
1. a) = 16
b) = 29x^2 + 29 - 29x^2 = 29
2. =x^2-2x+1 + y^2 - 2y + 1 = (x-1)^2 + (y-1)^2
b) = a^2+4a+4 + b^2 + 4b + 4 = (a+2)^2 + (b+2)^2
bạn giải chi tiết cho mình đc k ? pls xin đáy và cảm ơn bạn vô cùng
Bài 1:
a) \(\left(6x+1\right)^2+\left(6x-1\right)^2-2\left(1+6x\right)\left(6x-1\right)\)
\(=36x^2+72x+1+36x^2-72x+1-2\left(36x^2-1\right)\)
\(=36x^2+72x+1+36x^2-72x+1-72x^2+2\)
\(=4\)
b) \(3\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^{16}-1\right)\left(2^{16}+1\right)\)
\(=2^{32}-1\)
c) \(x\left(2x^3-3\right)-x^2\left(5x+1\right)+x^2\)
\(=2x^4-3x-5x^3-x^2+x^2\)
\(=2x^4-5x^3-3x\)
d) \(3x\left(x-2\right)-5x\left(1-x\right)-8\left(x^2-3\right)\)
\(=3x^2-6x-5x+5x^2-8x^2+24\)
\(=-11x+24\)
A=[(x-y).(x2+xy+y2)] +2y3
= x3-y3+2y3=x3+y3
=(2/3)3+(1/3)3
=4/9 + 1/9 =5/9
5x(x2 - xy +1)-2y(x2-xy+1)
=5x3-5x2y+1 - 2x2y - 2xy2 +1
=5x^3-5x^2y+5x-2x^2y-2xy^2-2y