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Q=\(-2\left(X^2-2.X.\frac{5}{2}+\frac{25}{4}\right)-\frac{25}{4}\)
Q=\(-2\left(X-\frac{5}{2}\right)^2-2.\frac{-25}{4}\)
Q=\(-2\left(X-\frac{5}{2}\right)^2+\frac{25}{2}\)
=>\(GTLN\) LÀ 25/2 TẠI X=5/2
N=
4.
= x\(^2\)-2.\(\dfrac{5}{2}\)x+\(\dfrac{25}{4}\)-\(\dfrac{13}{4}\)
= (x-\(\dfrac{5}{2}\))\(^2\)-\(\dfrac{13}{4}\)lớn hơn hoặc bằng -\(\dfrac{13}{4}\) với mọi x
=> min= -\(\dfrac{13}{4}\) <=> x = 5/2
5.
= 2( x\(^2\)-\(\dfrac{5}{2}\)x-\(\dfrac{1}{2}\))
=2( x\(^2\)-2.\(\dfrac{5}{4}\)+\(\dfrac{25}{4}\)-\(\dfrac{27}{4}\))
=2( x-\(\dfrac{5}{4}\))\(^2\)-\(\dfrac{27}{2}\) lớn hơn hoặc bằng -27/2 với mọi x
vậy min = -\(\dfrac{27}{2}\) <=> x= 5/4
b: \(=x^4+x^2+36-2x^3+12x^2-12x+x^2-6x+9\)
\(=x^4-2x^3+14x^2-18x+45\)
\(=x^4+9x^2-2x^3-18x+5x^2+45\)
\(=\left(x^2+9\right)\left(x^2-2x+5\right)\)
d: \(=2x^4+2x^3+6x^2-x^3-x^2-3x+x^2+x+3\)
\(=\left(x^2+x+3\right)\left(2x^2-x+1\right)\)
e: \(=3x^4-3x^3-3x^2-2x^3+2x^2+2x+2x^2-2x-2\)
\(=\left(x^2-x-1\right)\left(3x^2-2x+1\right)\)
a) Ta có: \(x^2+9x+20\)
\(=x^2+4x+5x+20\)
\(=x\left(x+4\right)+5\left(x+4\right)\)
\(=\left(x+4\right)\left(x+5\right)\)
b) Ta có: \(x^2+x-12\)
\(=x^2+4x-3x-12\)
\(=x\left(x+4\right)-3\left(x+4\right)\)
\(=\left(x+4\right)\left(x-3\right)\)
c) Ta có: \(6x^2-11x-16\)
\(=6\left(x^2-\frac{11}{6}x-\frac{16}{6}\right)\)
\(=6\left(x^2-2\cdot x\cdot\frac{11}{12}+\frac{121}{144}-\frac{505}{144}\right)\)
\(=6\left[\left(x-\frac{11}{12}\right)^2-\frac{505}{144}\right]\)
\(=6\left(x-\frac{11+\sqrt{505}}{12}\right)\left(x-\frac{11-\sqrt{505}}{12}\right)\)
d) Ta có: \(4x^2-8x-5\)
\(=4x^2-10x+2x-5\)
\(=2x\left(2x-5\right)+\left(2x-5\right)\)
\(=\left(2x-5\right)\left(2x+1\right)\)
e) Ta có: \(x^3-6x^2-x+30\)
\(=x^3+2x^2-8x^2-16x+15x+30\)
\(=x^2\left(x+2\right)-8x\left(x+2\right)+15\left(x+2\right)\)
\(=\left(x+2\right)\left(x^2-8x+15\right)\)
\(=\left(x+2\right)\left(x^2-3x-5x+15\right)\)
\(=\left(x+2\right)\left[x\left(x-3\right)-5\left(x-3\right)\right]\)
\(=\left(x+2\right)\left(x-3\right)\left(x-5\right)\)
g) Ta có: \(x^3+9x^2+23x+15\)
\(=x^3+x^2+8x^2+8x+15x+15\)
\(=x^2\left(x+1\right)+8x\left(x+1\right)+15\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2+8x+15\right)\)
\(=\left(x+1\right)\left(x^2+3x+5x+15\right)\)
\(=\left(x+1\right)\left[x\left(x+3\right)+5\left(x+3\right)\right]\)
\(=\left(x+1\right)\left(x+3\right)\left(x+5\right)\)
h) Ta có: \(2x^4-x^3-9x^2+13x\)
\(=x\left(2x^3-x^2-9x+13\right)\)
i) Ta có: \(x^4+2x^3-16x^2-2x+15\)
\(=x^4-3x^3+5x^3-15x^2-x^2+3x-5x+15\)
\(=x^3\left(x-3\right)+5x^2\left(x-3\right)-x\left(x-3\right)-5\left(x-3\right)\)
\(=\left(x-3\right)\left(x^3+5x^2-x-5\right)\)
\(=\left(x-3\right)\left[x^2\left(x+5\right)-\left(x+5\right)\right]\)
\(=\left(x-3\right)\left(x+5\right)\left(x^2-1\right)\)
\(=\left(x-3\right)\left(x+5\right)\left(x-1\right)\left(x+1\right)\)
Bài 1:
a) \(\dfrac{15xy}{10x^2y}\)
= \(\dfrac{3.5xy}{2.5xyx}\)
= \(\dfrac{3}{2x}\)
d) \(\dfrac{6x\left(x+5\right)^3}{2x^2\left(x+5\right)}\)
= \(\dfrac{3.2x\left(x+5\right)\left(x+5\right)^2}{x.2x\left(x+5\right)}\)
= \(\dfrac{3\left(x+5\right)^2}{x}\)
Dài dữ trời :V Về sau gửi từng bài một thôi, nhìn hoa mắt quá @@
B1: Phân tích thành nhân tử:
a) \(6x^2+9x=3x\left(2x+3\right)\)
b) \(4x^2+8x=4x\left(x+2\right)\)
c) \(5x^2+10x=5x\left(x+2\right)\)
d) \(2x^2-8x=2x\left(x-4\right)\)
e) \(5x-15y=5\left(x-3y\right)\)
f) \(x\left(x^2-1\right)+3\left(x^2-1\right)=\left(x^2-1\right)\left(x+3\right)\)
\(=\left(x-1\right)\left(x+1\right)\left(x+3\right)\)
g) \(x^2-2x+1-4y^2=\left(x-1\right)^2-4y^2\)
\(=\left(x-1-2y\right)\left(x-1+2y\right)\)
h) \(x^2-100=\left(x-10\right)\left(x+10\right)\)
i) \(9x^2-18x+9=\left(3x-3\right)^2\)
k) \(x^3-8=\left(x-2\right)\left(x^2+2x+4\right)\)
l) \(x^2+6xy^2+9y^4=\left(x+3y\right)^2\)
m) \(4xy-4x^2-y^2=-\left(4x^2-4xy+y^2\right)\)
\(=-\left(2x-y\right)^2\)
n) \(\left(x-15\right)^2-16=\left(x-15-16\right)\left(x-15+16\right)\)
\(=\left(x-31\right)\left(x+1\right)\)
o) \(25-\left(3-x\right)^2=\left(5-3+x\right)\left(5+3+x\right)\)
\(=\left(2+x\right)\left(8+x\right)\)
p) \(\left(7x-4\right)^2-\left(2x+1\right)^2\)
\(=\left(7x-4-2x-1\right)\left(7x-4+2x+1\right)\)
\(=\left(5x-5\right)\left(9x-3\right)\)
Bài 1 :
a ) \(6x^2+9x=3x\left(x+3\right)\)
b ) \(4x^2+8x=4x\left(x+2\right)\)
c ) \(5x^2+10x=5x\left(x+2\right)\)
d ) \(2x^2-8x=2x\left(x-4\right)\)
e ) \(5x-15y=5\left(x-3y\right)\)
f ) \(x\left(x^2-1\right)+3\left(x^2-1\right)=\left(x^2-1\right)\left(x+3\right)\)
g ) \(x^2-2x+1-4y^2=\left(x-1\right)^2-\left(2y\right)^2=\left(x-1-2y\right)\left(x-1+2y\right)\)
h ) \(x^2-100=x^2-10^2=\left(x-10\right)\left(x+10\right)\)
i ) \(9x^2-18x+9=\left(3x-3\right)^2\)
k ) \(x^3-8=\left(x-2\right)\left(x^2+2x+2^2\right)\)
l ) \(x^2+6xy^2+9y^4=\left(x+3y^2\right)^2\)
m ) \(4xy-4x^2-y^2=-\left(2x-y\right)^2\)
n ) \(\left(x-15\right)^2=x^2-30x+15^2\)
o ) \(25-\left(3-x\right)^2=\left(5-3+x\right)\left(5+3-x\right)=\left(2+x\right)\left(8-x\right)\)
p ) \(\left(7x-4\right)^2-\left(2x+1\right)^2=\left(7x-4-2x-1\right)\left(7x-4+2x+1\right)=\left(5x-5\right)\left(9x-3\right)\)
Bài 2 :
a ) \(3x^3-6x^2+3x^2y-6xy=3x\left(x^2-2x+xy-2y\right)\)
b ) \(x^2-2x+xy-2y=x\left(x-2\right)+y\left(x-2\right)=\left(x-2\right)\left(x+y\right)\)
c ) \(2x+x^2-2y-2xy=......................\)
d ) \(x^2-2xy+y^2-9=\left(x-y\right)^2-3^2=\left(x-y-3\right)\left(x-y+3\right)\)
e ) \(x^2+y^2-2xy-4=\left(x-y\right)^2-2^2=\left(x-y-2\right)\left(x-y+2\right)\)
f )\(2xy-x^2-y^2+9=-\left(x-y\right)^2+9=3^2-\left(x-y\right)^2=\left(3-x+y\right)\left(3+x-y\right)\)
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: \(\left(2x+1\right)^2-\left(x-1\right)^2\)
\(=\left[\left(2x+1\right)-\left(x-1\right)\right]\left[\left(2x+1\right)+\left(x-1\right)\right]\)
\(=\left(2x+1-x+1\right)\left(2x+1+x-1\right)\)
\(=3x\left(x+2\right)\)
a, (x-2)(3x-2)
b, (x+5)(x+6)
c, (x+1)(x+4)
d (x-5y)(x-2y)
e, (x-6)(x-3)
f, (x-2)(x-1)
g (x-2)(3x+1)
h (2x-3)(x+2)
i
\(Q=-2x^2+10x=\left(-2x^2+10x-\frac{25}{2}\right)+\frac{25}{2}=-2\left(x-\frac{5}{2}\right)^2+\frac{25}{2}\le\frac{25}{2}\)
Dấu = xảy ra khi \(x=\frac{5}{2}\)
\(N=-5x^2+6x+3=\left(-5x^2+6x-\frac{9}{5}\right)+\frac{9}{5}+3=-\left(\sqrt{5}x-\frac{3}{\sqrt{5}}\right)^2+\frac{24}{5}\le\frac{24}{5}\)
Dấu = xảy ra khi \(x=\frac{3}{5}\)
\(P=4-x^2+2x=\left(-x^2+2x-1\right)+5=-\left(x-1\right)^2+5\le5\)
Dấu = xảy ra khi \(x=1\)
\(H=-9x^2+6x-2=\left(-9x^2+6x-1\right)-1=-\left(3x-1\right)^2-1\le-1\)
Dấu = xảy ra khi \(x=\frac{1}{3}\)
Cac ban giai thich that chat che nhe. Cam on nhieu