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Bài 1:
a. A = x^2 - 5x - 1
\(=x^2-5x+\frac{25}{4}-\frac{29}{4}\)
\(=x^2-5x+\left(\frac{5}{2}\right)^2-\frac{29}{4}\)
\(=\left(x-\frac{5}{2}\right)^2-\frac{29}{4}\ge0-\frac{29}{4}=-\frac{29}{4}\)
Dấu = khi x=5/2
Vậy MinC=-29/4 khi x=5/2
2. Tìm x:
a. ( 2x - 3 )^2 - ( 4x + 1 )( 4x - 1 ) = ( 2x - 1 ).( 3 - 7x )
=>4x2-12x+9+1-16x2=-14x2+13x-3
=>-12x2-12x+10=-14x2+13x-3
=>2x2-25x+13=0
\(\Rightarrow2\left(x-\frac{25}{4}\right)^2-\frac{521}{8}=0\)
\(\Rightarrow\left(x-\frac{25}{4}\right)^2=\frac{521}{16}\)
\(\Rightarrow x-\frac{25}{4}=\pm\sqrt{\frac{521}{16}}\)
\(\Rightarrow x=\frac{25}{4}\pm\frac{\sqrt{521}}{4}\)
c. 4.( x - 3 ) - ( x + 2 ) = 0
=>4x-12-x-2=0
=>3x-14=0
=>3x=14
=>x=14/3
\(e,-5x+x^2-14\)
\(=x^2+2x-7x-14\)
\(=x\left(x+2\right)-7\left(x+2\right)\)
\(=\left(x+2\right)\left(x-7\right)\)
\(f,x^3+8+6x\left(x+2\right)\)
\(=\left(x+2\right)\left(x^2+2x+4\right)+6x\left(x+2\right)\)
\(=\left(x+2\right)\left(x^2+8x+4\right)\)
\(g,15x^2-7xy-2y^2\)
\(=15x^2+3xy-10xy-2y^2\)
\(=3\left(5x+y\right)-2y\left(5x+y\right)\)
\(=\left(5x+y\right)\left(3-2y\right)\)
\(h,3x^2-16x+5\)
\(=3x^2-x-15x+5\)
\(=x\left(3x-1\right)+5\left(3x-1\right)\)
\(=\left(3x-1\right)\left(x+5\right)\)
\(a,x^3+2x^2y+xy^2=x\left(x^2+2xy+y^2\right)\)
\(=x\left(x+y\right)^2\)
\(b,4x^2-9y^2+4x-6y\)
\(=4x^2+4x+1-\left(9y^2+6y+1\right)\)
\(=\left(2x+1\right)^2-\left(3y+1\right)^2\)
\(=\left(2x-3y\right)\left(2x+3y+2\right)\)
\(c,-x^2+5x+2xy-5y-y^2\)
\(=-\left(x^2-2xy+y^2\right)+5\left(x-y\right)\)
\(=-\left(x-y\right)^2+5\left(x-y\right)\)
\(=\left(x-y\right)\left(y-x+5\right)\)
\(d,x^2+4x-12\)
\(=x^2-2x+6x-12\)
\(=x\left(x-2\right)+6\left(x-2\right)\)
\(=\left(x-2\right)\left(x+6\right)\)
a) \(x^2+4x+3=\left(x^2+4x+4\right)-1=\left(x+2\right)^2-1^2=\left(x+1\right)\left(x+3\right)\) (mình sửa lại)
b) \(x^2+8x-9=\left(x^2+8x+16\right)-25=\left(x+4\right)^2-5^2=\left(x-1\right)\left(x+9\right)\)
c) \(3x^2+6x-9=3\left[\left(x^2+2x+1\right)-4\right]=3\left[\left(x+1\right)^2-2^2\right]=3\left(x-1\right)\left(x+3\right)\)
d) \(2x^2+x-3=2x^2-4x+2+5x-5=2\left(x^2-2x+1\right)+5\left(x-1\right)=2\left(x-1\right)^2+5\left(x-1\right)=\left(x-1\right)\left(2x+3\right)\)
A=\(x^3-2x^2+x\)
=x.(x2-2x+1)
=x(x-1)2
B=\(2x^2+4x+2-2y^2\)
=\(2\left(x^2+2x+1-y^2\right)\)
=\(2.\left[\left(x+1\right)^1-y^2\right]\)
=\(2\left(x+1-y\right)\left(x+1+y\right)\)
C=\(2xy-x^2-y^2+16\)
=\(-\left(-2xy+x^2+y^2-16\right)\)
=\(-\left[\left(x-y\right)^2-4^2\right]\)
=-(x-y-4)(x-y+4)
D=\(x^3+2x^2y+xy^2-9x\)
=\(x\left(x^2+2xy-y^2-9\right)\)
=\(x.\left[\left(x-y\right)^2-3^2\right]\)
=x.(x-y-3)(x-y+3)
E=\(2x-2y-x^2+2xy-y^2\)
\(=\left(2x-2y\right)-\left(x^2-2xy+y^2\right)\)
=\(2\left(x-y\right)-\left(x-y\right)\left(x-y\right)\)
=(x-y)(2x-2y-x+y)
=(x-y)(x+y)
e)
$x^3+6x^2+12x+8=x^3+3.2.x^2+3.2^2.x+2^3=(x+2)^3$
f)
$a^3-2a^2-ab^2+2b^2=(a^3-ab^2)-(2a^2-2b^2)$
$=a(a^2-b^2)-2(a^2-b^2)=(a^2-b^2)(a-2)=(a-b)(a+b)(a-2)$
g)
$2a^2x-2a^2-2abx+4ab-2b^2=(2a^2x-2abx)-(2a^2-4ab+2b^2)$
$=2ax(a-b)-2(a-b)^2=2(a-b)(ax-a+b)$
h)
\(x^2-2xy+y^2-25=(x-y)^2-25=(x-y)^2-5^2=(x-y+5)(x-y-5)\)
a)
$4x^2-40x^4+100x^3=4x^2(1-10x^2+25x)$
b)
\(3xy(x-5)-7x+35=3xy(x-5)-7(x-5)\)
\(=(x-5)(3xy-7)\)
c)
\(a^2-am-b^2-bm=(a^2-b^2)-(am+bm)=(a-b)(a+b)-m(a+b)\)
\(=(a+b)(a-b-m)\)
d)
\(x^3-4x-x^2y+4y=(x^3-x^2y)-(4x-4y)\)
\(=x^2(x-y)-4(x-y)=(x^2-4)(x-y)=(x-2)(x+2)(x-y)\)
\(a,A=6x^2-6x+1\)
\(=6\left(x^2-x+\frac{1}{4}\right)-\frac{1}{2}\)
\(=6\left(x-\frac{1}{2}\right)^2-\frac{1}{2}\ge-\frac{1}{2}\)
Dấu = xảy ra \(\Leftrightarrow x=\frac{1}{2}\)
Vậy \(Min_A=-\frac{1}{2}\Leftrightarrow x=\frac{1}{2}\)
\(b,B=3+2x+3x^2\)
\(=3\left(x^2+\frac{2}{3}x+\frac{1}{9}\right)+\frac{8}{3}\)
\(=3\left(x+\frac{1}{3}\right)^2+\frac{8}{3}\ge\frac{8}{3}\)
Dấu = xảy ra \(\Leftrightarrow x=-\frac{1}{3}\)
Vậy \(Min_B=\frac{8}{3}\Leftrightarrow x=-\frac{1}{3}\)
\(c,C=4x+2x^2-3\)
\(=2\left(x^2+2x+1\right)-5\)
\(=2\left(x+1\right)^2-5\ge-5\)
Dấu = xảy ra \(\Leftrightarrow x=-1\)
Vậy \(Min_C=-5\Leftrightarrow x=-1\)
\(d,D=10x+6+x^2\)
\(=\left(x^2+10x+25\right)-19\)
\(=\left(x+5\right)^2-19\ge-19\)
Dấu = xảy ra \(\Leftrightarrow x=-5\)
Vậy \(Min_D=-19\Leftrightarrow x=-5\)
\(e,E=8x^2-6x+3\)
\(=8\left(x^2-\frac{3}{4}x+\frac{9}{64}\right)+\frac{15}{8}\)
\(=8\left(x-\frac{3}{8}\right)^2+\frac{15}{8}\ge\frac{15}{8}\)
Dấu = xảy ra \(\Leftrightarrow x=\frac{3}{8}\)
Vậy \(Min_E=\frac{15}{8}\Leftrightarrow x=\frac{3}{8}\)
Bài 1: Phân tích đa thức thành nhân tử
a) Ta có: \(8a^3-6a^2-1+3a\)
\(=\left[\left(2a\right)^3-1^3\right]-3a\left(2a-1\right)\)
\(=\left(2a-1\right)\left(4a^2+2a+1\right)-3a\left(2a-1\right)\)
\(=\left(2a-1\right)\left(4a^2+2a+1-3a\right)\)
\(=\left(2a-1\right)\left(4a^2-a+1\right)\)
b) Ta có: \(x^3-2x^2y+xy^2-9x\)
\(=x\left(x^2-2xy+y^2-9\right)\)
\(=x\left[\left(x^2-2xy+y^2\right)-9\right]\)
\(=x\left[\left(x-y\right)^2-3^2\right]\)
\(=x\left(x-y-3\right)\left(x-y+3\right)\)
c) Ta có: \(5x^2-45\)
\(=5\left(x^2-9\right)\)
\(=5\left(x-3\right)\left(x+3\right)\)
d) Ta có: \(2x^3-4x^2+2x\)
\(=x\left(2x^2-4x+2\right)\)
\(=x\left(2x^2-2x-2x+2\right)\)
\(=x\left[2x\left(x-1\right)-2\left(x-1\right)\right]\)
\(=x\left(x-1\right)\left(2x-2\right)\)
\(=2x\left(x-1\right)^2\)
e) Ta có: \(6x\left(3x-2\right)-12\left(2-3x\right)\)
\(=6x\left(3x-2\right)+12\left(3x-2\right)\)
\(=\left(3x-2\right)\left(6x+12\right)\)
\(=6\left(3x-2\right)\left(x+2\right)\)
f) Ta có: \(4x^2-8xy+4y^2-10\)
\(=\left(2x\right)^2-2\cdot2x\cdot2y+\left(2y\right)^2-10\)
\(=\left(2x-2y\right)^2-10\)
\(=\left(2x-2y-\sqrt{10}\right)\left(2x-2y+\sqrt{10}\right)\)
g) Ta có: \(2x^2-8x+8\)
\(=2\left(x^2-4x+4\right)\)
\(=2\left(x-2\right)^2\)
h) Ta có: \(\left(2x+1\right)^2-\left(x-1\right)^2\)
\(=\left[\left(2x+1\right)-\left(x-1\right)\right]\left[\left(2x+1\right)+\left(x-1\right)\right]\)
\(=\left(2x+1-x+1\right)\left(2x+1+x-1\right)\)
\(=3x\left(x+2\right)\)
Bài 1: Phân tích đa thức thành nhân tử
a) Ta có: \(16x^2-y^2+6y-9\)
\(=16x^2-\left(y^2-6y+9\right)\)
\(=\left(4x\right)^2-\left(y-3\right)^2\)
\(=\left[4x-\left(y-3\right)\right]\left[4x+\left(y-3\right)\right]\)
\(=\left(4x-y+3\right)\left(4x+y-3\right)\)
b) Ta có: \(a^2-16a^2b^2+b^2+2ab\)
\(=\left(a^2+2ab+b^2\right)-\left(4ab\right)^2\)
\(=\left(a+b\right)^2-\left(4ab\right)^2\)
\(=\left(a+b-4ab\right)\left(a+b+4ab\right)\)
c) Ta có: \(x^3-6x^2-9x\)
\(=x\left(x^2-6x-9\right)\)
d) Ta có: \(mx^2+my^2-nx^2-ny^2\)
\(=m\left(x^2+y^2\right)-n\left(x^2+y^2\right)\)
\(=\left(x^2+y^2\right)\left(m-n\right)\)
e) Ta có: \(a^3+b^3+a^2c+b^2c-abc\)
\(=\left(a+b\right)\left(a^2-ab+b^2\right)+c\left(a^2-ab+b^2\right)\)
\(=\left(a^2-ab+b^2\right)\left(a+b+c\right)\)
f) Ta có: \(4x^2-y^2-4x+1\)
\(=\left(4x^2-4x+1\right)-y^2\)
\(=\left(2x-1\right)^2-y^2\)
\(=\left(2x-1-y\right)\left(2x-1+y\right)\)
g) Ta có: \(\left(2x+3\right)^2+5\cdot\left(2x+3\right)\)
\(=\left(2x+3\right)\left(2x+3+5\right)\)
\(=\left(2x+3\right)\left(2x+8\right)\)
\(=2\left(2x+3\right)\left(x+4\right)\)
h) Ta có: \(3x^2-10x-8\)
\(=3x^2-12x+2x-8\)
\(=3x\left(x-4\right)+2\left(x-4\right)\)
\(=\left(x-4\right)\left(3x+2\right)\)
cảm ơn nhiều ạ :)))