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21 tháng 7 2015

a/ = x4 + 36x2 +81 +12x3 + 108x + 18x2 + x4 +1 + 4x2 + 2x2 - 4x  - 4x3 -16 = 2x4 + 8x3 + 60x2 + 104x + 66 = 2(x4 + 4x3 + 30x2 + 52x + 33)

19 tháng 6 2016

a)x7+x5+1=x7+x6-x6+2x5-x5+x4-x4+x3-x3+x2-x2+1

=x7-x6+x5-x3+x2+x6-x5+x4-x2+x+x5-x4+x3-x+1

=x2(x5-x4+x3-x+1)+x(x5-x4+x3-x+1)+1(x5-x4+x3-x+1)

=(x2+x+1)(x5-x4+x3-x+1)

b)4x4-32x2+1=4x4+12x3+2x2-12x3-36x2-6x+2x2+6x+1

=2x2(2x2+6x+1)-6x(2x2+6x+1)+1(2x2+6x+1)

=(2x2-6x+1)(2x2+6x+1)

c)x6+27=(x2+3)(x2-3x+3)(x2+3x+3)

d)3(x4+x2+1)-(x2+x+1)

=3x4-3x3+2x2+3x3-3x2+2x+3x2-3x+2

=x2(3x2-3x+2)+x(3x2-3x+2)+1(3x2-3x+2)

=(x2+x+1)(3x2-3x+2)

e)bạn tự làm nhé

24 tháng 8 2016

b/ (x+ x + 1)(x- x+ x3 - x + 1)

24 tháng 8 2016

c/ (x - 1)(2x+ 1)

6 tháng 10 2021

\(a,=\left(3x-5\right)\left(3x+3\right)=3\left(x+1\right)\left(3x-5\right)\\ b,=\left(5x-4-7x\right)\left(5x-4+7x\right)=\left(-2x-4\right)\left(12x-4\right)\\ =-8\left(x+2\right)\left(x-3\right)\\ c,=\left(2x+5-x+9\right)\left(2x+5+x-9\right)\\ =\left(x+14\right)\left(3x-4\right)\\ d,=\left(3x+1-2x+4\right)\left(3x+1+2x-4\right)\\ =\left(x+5\right)\left(5x-3\right)\\ e,=\left(6x+9-2x-2\right)\left(6x+9+2x+2\right)\\ =\left(4x+7\right)\left(8x+11\right)\\ f,=\left(2bc-b^2-c^2+a^2\right)\left(2bc+b^2+c^2-a^2\right)\\ =\left[a^2-\left(b-c\right)^2\right]\left[\left(b+c\right)^2-a^2\right]\\ =\left(a-b+c\right)\left(a+b-c\right)\left(b+c-a\right)\left(b+c+a\right)\\ g,=\left(ax+by-ay-bx\right)\left(ax+by+ay+bx\right)\\ =\left(a-b\right)\left(x-y\right)\left(a+b\right)\left(x+y\right)\)

\(h,=\left(a^2+b^2-5-2ab-4\right)\left(a^2+b^2-5+2ab+4\right)\\ =\left[\left(a-b\right)^2-9\right]\left[\left(a+b\right)^2-1\right]\\ =\left(a-b-3\right)\left(a-b+3\right)\left(a+b-1\right)\left(a+b+1\right)\)

a: \(\left(3x-1\right)^2-16\)

\(=\left(3x-1-4\right)\left(3x-1+4\right)\)

\(=\left(3x+3\right)\left(3x-5\right)\)

\(=3\left(x+1\right)\left(3x-5\right)\)

b: \(\left(5x-4\right)^2-49x^2\)

\(=\left(5x-4-7x\right)\left(5x-4+7x\right)\)

\(=\left(-2x-4\right)\left(12x-4\right)\)

\(=-8\left(x+2\right)\left(3x-1\right)\)

a) Ta có: \(x^4+64\)

\(=x^4+16x^2+64-16x^2\)

\(=\left(x^2+8\right)^2-\left(4x\right)^2\)

\(=\left(x^2-4x+8\right)\left(x^2+4x+8\right)\)

b) Ta có: \(81x^4+4y^4\)

\(=81x^4+36x^2y^2+4y^4-36x^2y^2\)

\(=\left(9x^2+2y^2\right)^2-\left(6xy\right)^2\)

\(=\left(9x^2-6xy+2y^2\right)\left(9x^2+6xy+2y^2\right)\)

c) Ta có: \(x^5+x+1\)

\(=x^5+x^2-x^2+x-1\)

\(=x^2\left(x^3+1\right)-\left(x^2-x+1\right)\)

\(=x^2\left(x+1\right)\left(x^2-x+1\right)-\left(x^2-x+1\right)\)

\(=\left(x^2-x+1\right)\left(x^3+x^2-1\right)\)

 

28 tháng 9 2019

1) \(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24\)

\(=\left(x+2\right)\left(x+5\right)\left(x+3\right)\left(x+4\right)-24\)

\(=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\)

Đặt \(x^2+7x=t\)

\(\Rightarrow BT=\left(t+10\right)\left(t+12\right)-24\)

\(=t^2+22x+96=\left(t+11\right)^2-25\ge-25\)

Vậy GTNN của bt là - 25\(\Leftrightarrow x^2+7x+11=0\)

\(\Delta=7^2-4.11=5\)

\(\orbr{\begin{cases}x_1=\frac{-22+\sqrt{5}}{2}\\x_2=\frac{-22-\sqrt{5}}{2}\end{cases}}\)

28 tháng 9 2019

2) \(\left(x-1\right)\left(x-3\right)\left(x-5\right)\left(x-7\right)-20\)

\(=\left(x-1\right)\left(x-7\right)\left(x-3\right)\left(x-5\right)-20\)

\(=\left(x^2-8x+7\right)\left(x^2-8x+15\right)-20\)

Đặt \(x^2-8x=t\)

\(\RightarrowĐT=\left(t+7\right)\left(t+15\right)-20\)

\(=t^2+22t+85=\left(t+11\right)^2-36\ge-36\)

Vậy GTNN của bt là - 36\(\Leftrightarrow x^2-8x+11=0\)

\(\Delta=\left(-8\right)^2-4.11=20\)

\(\orbr{\begin{cases}x_1=\frac{-22-\sqrt{20}}{2}\\x_2=\frac{-22+\sqrt{20}}{2}\end{cases}}\)

a) Ta có: \(x^4-16x^2=0\)

\(\Leftrightarrow x^2\left(x^2-16\right)=0\)

\(\Leftrightarrow x^2\left(x-4\right)\left(x+4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\\x=-4\end{matrix}\right.\)

b) Ta có: \(x^8+36x^4=0\)

\(\Leftrightarrow x^4\left(x^4+36\right)=0\)

\(\Leftrightarrow x^4=0\)

hay x=0

c) Ta có: \(\left(x-5\right)^3-x+5=0\)

\(\Leftrightarrow\left(x-5\right)\cdot\left[\left(x-5\right)^2-1\right]=0\)

\(\Leftrightarrow\left(x-5\right)\left(x-4\right)\left(x-6\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=4\\x=6\end{matrix}\right.\)

d) Ta có: \(5\left(x-2\right)-x^2+4=0\)

\(\Leftrightarrow5\left(x-2\right)-\left(x-2\right)\left(x+2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(5-x-2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(3-x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=3\end{matrix}\right.\)