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a, = (x+3y)^2
b, = (x-1/2)(x+1/2)
c, = (x-5)^2
d, = (2x+3y)(4x^2-6xy+9y^2)
e, = (x^3-y)^2
f,= (x+3y)^3
Bài 3:
a) ta có: \(A=x^2+4x+9\)
\(=x^2+4x+4+5=\left(x+2\right)^2+5\)
Ta có: \(\left(x+2\right)^2\ge0\forall x\)
\(\Rightarrow\left(x+2\right)^2+5\ge5\forall x\)
Dấu '=' xảy ra khi
\(\left(x+2\right)^2=0\Leftrightarrow x+2=0\Leftrightarrow x=-2\)
Vậy: GTNN của đa thức \(A=x^2+4x+9\) là 5 khi x=-2
b) Ta có: \(B=2x^2-20x+53\)
\(=2\left(x^2-10x+\frac{53}{2}\right)\)
\(=2\left(x^2-10x+25+\frac{3}{2}\right)\)
\(=2\left[\left(x-5\right)^2+\frac{3}{2}\right]\)
\(=2\left(x-5\right)^2+2\cdot\frac{3}{2}\)
\(=2\left(x-5\right)^2+3\)
Ta có: \(\left(x-5\right)^2\ge0\forall x\)
\(\Rightarrow2\left(x-5\right)^2\ge0\forall x\)
\(\Rightarrow2\left(x-5\right)^2+3\ge3\forall x\)
Dấu '=' xảy ra khi
\(2\left(x-5\right)^2=0\Leftrightarrow\left(x-5\right)^2=0\Leftrightarrow x-5=0\Leftrightarrow x=5\)
Vậy: GTNN của đa thức \(B=2x^2-20x+53\) là 3 khi x=5
c) Ta có : \(M=1+6x-x^2\)
\(=-x^2+6x+1\)
\(=-\left(x^2-6x-1\right)\)
\(=-\left(x^2-6x+9-10\right)\)
\(=-\left[\left(x-3\right)^2-10\right]\)
\(=-\left(x-3\right)^2+10\)
Ta có: \(\left(x-3\right)^2\ge0\forall x\)
\(\Rightarrow-\left(x-3\right)^2\le0\forall x\)
\(\Rightarrow-\left(x-3\right)^2+10\le10\forall x\)
Dấu '=' xảy ra khi
\(-\left(x-3\right)^2=0\Leftrightarrow\left(x-3\right)^2=0\Leftrightarrow x-3=0\Leftrightarrow x=3\)
Vậy: GTLN của đa thức \(M=1+6x-x^2\) là 10 khi x=3
Bài 2:
a) \(\left(x+y\right)^2+\left(x^2-y^2\right)\)
\(=\left(x+y\right)^2+\left(x-y\right).\left(x+y\right)\)
\(=\left(x+y\right).\left(x+y+x-y\right)\)
\(=\left(x+y\right).2x\)
c) \(x^2-2xy+y^2-z^2+2zt-t^2\)
\(=\left(x^2-2xy+y^2\right)-\left(z^2-2zt+t^2\right)\)
\(=\left(x-y\right)^2-\left(z-t\right)^2\)
\(=\left[x-y-\left(z-t\right)\right].\left(x-y+z-t\right)\)
\(=\left(x-y-z+t\right).\left(x-y+z-t\right)\)
Chúc bạn học tốt!
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 )\(2xy-x^2-y^2+9=-\left(x-y\right)^2+9=3^2-\left(x-y\right)^2=\left(3-x+y\right)\left(3+x-y\right)\)
b) 6x - 9 - x2
= - (x2 - 6x + 9 )
= - ( x2 - 2.x.3 + 32 )
= - ( x - 3 )2
c) x2 - 16
= x2 - 42
= ( x - 4 )( x + 4)
d) 9x2 - 25
= ( 3x )2 - 52
= ( 3x - 5 )( 3x + 5 )
e ) x4 - y4
= ( x2)2 - ( y2 )2
= ( x2 - y2 )( x2 + y2 )
f) x6 -y6
= ( x3 )2 - ( y3)2
=( x3 - y3 )( x3 + y3 )
g) 8x3 - \(\dfrac{1}{27}\)
= ( 2x )3 - ( \(\dfrac{1}{3}\))3
= ( 2x - \(\dfrac{1}{3}\) ) ( 2x + \(\dfrac{2}{3}\)x + \(\dfrac{1}{3}\))
a. \(\frac{5x-2}{3}=\frac{5x-3x}{2}\)
\(\Leftrightarrow2.\left(5x-2\right)=3.\left(5x-3x\right)
\)
\(\Leftrightarrow10x-4=15x-9x\)
\(\Leftrightarrow4x=4\)
\(\Leftrightarrow x=1\)
Vậy...
b. \(\frac{10x+3}{12}=1+\frac{6+8x}{9}\left(1\right)\)
MC = 36.
pt (1) <=>
\(\frac{3\left(10x+3\right)}{36}=\frac{36}{36}+\frac{4\left(6+8x\right)}{36}\)
=> 3.(10x+3) = 36 + 4(6+8x)
<=> 30x+9 = 36+24+32x
<=> -2x = 51
<=> x = \(\frac{-51}{2}\)
Vậy...
c. \(\frac{7x-1}{6}+2=\frac{16-x}{5}\left(2\right)\)
MC = 30.
pt (2) <=>
\(\frac{5\left(7x-1\right)}{30}+\frac{60x}{30}=\frac{6\left(16-x\right)}{30}\)
=> 5(7x-1) + 60x = 6(16-x)
<=> 35x-5 + 60x = 96-6x
<=> 101x = 101
<=> x = 1
Vậy...
d. \(\frac{3x+2}{2}-\frac{3x+1}{6}=5\) (3)
MC = 12.
pt (3)<=>
\(\frac{6\left(3x+2\right)}{12}-\frac{2\left(3x+1\right)}{12}=\frac{60}{12}\)
=> 6(3x+2) - 2(3x+1) = 60
<=> 18x+12 - 6x-2 = 60
<=> 12x = 50
<=> x = \(\frac{25}{6}\)
Vậy...
e. \(\frac{x+4}{5}-x+4=\frac{x}{3}-\frac{x-2}{2}\) (4)
MC = 30.
pt (4) <=>
\(\frac{6\left(x+4\right)}{30}-\frac{30x}{30}+\frac{120}{30}=\frac{10x}{30}-\frac{15\left(x-2\right)}{30}\)
=> 6(x+4) - 30x + 120 = 10x - 15(x-2)
<=> 6x+24 - 30x + 120 = 10x - 15x+30
<=> -19x = -114
<=> x = \(\frac{114}{19}=6\)
Vậy...
Bài 1:
a) \(\dfrac{15xy}{10x^2y}\)
= \(\dfrac{3.5xy}{2.5xyx}\)
= \(\dfrac{3}{2x}\)
d) \(\dfrac{6x\left(x+5\right)^3}{2x^2\left(x+5\right)}\)
= \(\dfrac{3.2x\left(x+5\right)\left(x+5\right)^2}{x.2x\left(x+5\right)}\)
= \(\dfrac{3\left(x+5\right)^2}{x}\)
B1:a)(3x-5)2-(3x+1)2=8
[(3x-5)+(3x+1)].[(3x-5)-(3x+1)]=8
(3x-5+3x+1)(3x-5-3x-1)=8
9x2-15x-9x2-3x-15x+25+15x+5+9x2-15x-9x2-3x+3x-5-3x-1=8
-36x+24=8
-36x=8-24=16
x=16:(-36)=\(\dfrac{-4}{9}\)
Bài 5:
a: \(=\left(xy-u^2v^3\right)\left(xy+u^2v^3\right)\)
b: \(=\left(2xy^2-3xy^2+1\right)\left(2xy^2+3xy^2-1\right)\)
\(=\left(1-xy^2\right)\left(5xy^2-1\right)\)
Bài 6:
a: \(\left(a+b+c-d\right)\left(a+b-c+d\right)\)
\(=\left(a+b\right)^2+\left(c-d\right)^2\)
\(=a^2+2ab+b^2+c^2-2cd+d^2\)
b: \(\left(a+b-c-d\right)\left(a-b+c-d\right)\)
\(=\left(a-d\right)^2-\left(b-c\right)^2\)
\(=a^2-2ad+d^2-b^2+2bc-c^2\)
1.
a) \(x\left(x+4\right)+x+4=0\)
\(\Leftrightarrow\left(x+1\right)\left(x+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+4=0\\x+1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-4\\x=-1\end{matrix}\right.\)
b) \(x\left(x-3\right)+2x-6=0\)
\(\Leftrightarrow\left(x+2\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=0\\x-3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-2\\x=3\end{matrix}\right.\)
Bài 1:
a, \(x\left(x+4\right)+x+4=0\)
\(\Leftrightarrow x\left(x+4\right)+\left(x+4\right)=0\)
\(\Leftrightarrow\left(x+4\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+4=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-4\\x=-1\end{matrix}\right.\)
Vậy \(x=-4\) hoặc \(x=-1\)
b, \(x\left(x-3\right)+2x-6=0\)
\(\Leftrightarrow x\left(x-3\right)+2\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)
Vậy \(x=3\) hoặc \(x=-2\)
Bài 1:
a, \(\left(2x+2\right)\left(2x-2\right)=4x^2-4\)
b, đề thiếu
Bài 2:
a, \(\left(x-1\right)\left(x^2+2x+4\right)=\left(x-1\right)^3\)
Thay x = -1 vào đa thức trên, ta được:
\(\left(x-1\right)^3=\left(-1-1\right)^3=-2^3=-8\)
b, \(x^2-2xy-9z^2+y^2=\left(x^2-2xy+y^2\right)-\left(3z\right)^2=\left(x-y\right)^2-\left(3z\right)^2=\left(x-y+3z\right)\left(x-y-3z\right)\)
Thay x = 6; y = -4; z = 20 vào đa thức, ta được:
\(\left(x-y+3z\right)\left(x-y-3z\right)=\left(6+4+3.20\right)\left(6+4-3.20\right)=70.\left(-50\right)=-3500\)
Bài 3:
a, \(x^3-2x^2+x=x^3-x^2-x^2+x=x^2\left(x-1\right)-x\left(x-1\right)=\left(x-1\right)\left(x^2-x\right)=x\left(x-1\right)^2\)
b, \(x+y-y^2-xy=\left(x+y\right)-x\left(y+x\right)=\left(x+y\right)\left(1-x\right)\)
1.
a)\(\left(2x+2\right)\left(2x-2\right)=4x^2-4\)
b) đề thiếu
c) đặt tính ra
2.
a)\(\left(x-1\right)\left(x^2+2x+4\right)=x^3-1\)
Giá trị của biểu thức trên tại x=-1 là:
\(\left(-1\right)^3-1=-2\)
b)\(x^2-2xy-9z^2+y^2=\left(x-y\right)^2-\left(3z\right)^2=\left(x-y+3z\right)\left(x-y-3z\right)\)
Giá trị của biểu thức trên tại x=6; y=-4 và z=20 là:
\(\left[6-\left(-4\right)+3.20\right]\left[6-\left(-4\right)-3.20\right]=\left(10+60\right)\left(10-60\right)=70.\left(-50\right)=-3500\)
3.
a)\(x^3-2x^2+x=x\left(x^2-2x+1\right)=x\left(x-1\right)^2\)
b)\(x+y-y^2-xy=x\left(1-y\right)+y\left(1-y\right)=\left(x+y\right)\left(1-y\right)\)
â. (A+B)2 = A2+2AB+B2
b. A2 – B2= (A-B)(A+B)
c. (A – B)2= A2 – 2AB+ B2
d. A3 + B3= (A+B)(A2- AB +B2)
e. cái này bạn phải chú ý cách sắp xếp mà sx nó lại \(x^6-2x^3y+y^2\) (A – B)2= A2 – 2AB+ B2
f. (A+B)3= A3+3A2B +3AB2+B3
a) x2+6xy+9y2 = x2+2.x.3y+(3y)2 = (x+3y)2
b) x2-\(\dfrac{1}{4}\)= x2- (\(\dfrac{1}{2}\))2 = (x-\(\dfrac{1}{2}\))(x+\(\dfrac{1}{2}\))
c) x2 -10x+25 = x2 -2.x.5+52 = (x-5)2
d) 8x3+27y3 = (2x)3+(3y)3 = (2x+3y)[(2x)2 -2x.3y+(3y)2]
e) x6 +y2 -2x3y = x6-2x3y +y2 = (x3)2 -2x3y +y2 = (x3 -y)2
f) x3 +9x2y +27xy2 +27y3 = x3 +3.x2.3y +3.x.(3y)2 +(3y)3 = (x+3y)3