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a, \(\left(8a^3-27b^3\right)-2a\left(4a^2-9b^2\right)\)
\(=\left(2a-3b\right)\left[\left(2a\right)^2+2a.3b+\left(3b\right)^2\right]-2a\left(2a-3b\right)\left(2a+3b\right)\)
\(=\left(2a-3b\right)\left[4a^2+6ab+9b^2-2a\left(2a+3b\right)\right]\)
\(=\left(2a-3b\right)\left(4a^2+6ab+9b^2-4a^2-6ab\right)\)
\(=\left(2a-3b\right).9b^2\)
b, \(\left(x^3-y^3\right)+\left(x-y\right)^2\)
\(=\left(x-y\right)\left(x^2+xy+y^2\right)+\left(x-y\right)^2\)
\(=\left(x-y\right)\left[\left(x^2+xy+y^2\right)+\left(x-y\right)\right]\)
\(=\left(x-y\right)\left(x^2+xy+y^2+x-y\right)\)
c, \(\left(m^3+n^3\right)+\left(m+n\right)^2\)
\(=\left(m+n\right)\left(m^2-mn+n^2\right)+\left(m+n\right)^2\)
\(=\left(m+n\right)\left(m^2-mn+n^2+m+n\right)\)
Chúc bạn học tốt!!!
a) 4a2b3 - 6a3b2 = 2a2b2( 2b - 3a )
b) ( a - b )2 - ( b - a ) = ( a - b )2 + ( a - b ) = ( a - b )( a - b + 1 )
c) ( 8a3 - 27b3 ) - 2a( 4a2 - 9b2 ) = 8a3 - 27b3 - 8a3 + 18ab2 = 18ab2 - 27b3 = 9b2( 2a - 3b )
d) 10x2 + 10xy + 5x + 5y = 10x( x + y ) + 5( x + y ) = ( x + y )( 10x + 5 ) = 5( x + y )( 2x + 1 )
e) 5ay - 3bx + ax - 15by = 5y( a - 3b ) + x( a - 3b ) = ( a - 3b )( 5y + x )
a) \(4a^2.b^3-6a^3.b^2=2a^2.b^2\left(2b-3a\right)\)
b) \(\left(a-b\right)^2-\left(b-a\right)=\left(a-b\right)^2+\left(a-b\right)\)
\(=\left(a-b\right).\left(a-b+1\right)\)
c) \(8a^3-27b^3-2a.\left(4a^2-9b^2\right)=8a^3-27b^3-8a^3+18ab^2\)
\(=-27b^3+18ab^2=18ab^2-27b^3=9b^2.\left(2a-3b\right)\)
d) \(10x^2+10xy+5x+5y=5.\left(2x^2+2xy+x+y\right)\)
\(=5.\left[\left(2x^2+2xy\right)+\left(x+y\right)\right]=5.\left[2x\left(x+y\right)+\left(x+y\right)\right]\)
\(=5\left(x+y\right)\left(2y+1\right)\)
e) \(5ay-3bx+ax-15by=\left(5ay-15by\right)-\left(3bx-ax\right)\)
\(=5y\left(a-3b\right)-x\left(3b-a\right)=5y\left(a-3b\right)+x\left(a-3b\right)\)
\(=\left(a-3b\right)\left(x+5y\right)\)
a) 5ay - 3bx + ax - 15by
= (5ay + ax) - (3bx + 15by)
= a (5y + x) - 3b (x + 5y)
= (5y + x) (a - 3b)
b) x^3 + x^2 - x - 1
= (x^3 + x^2) - (x + 1)
= x^2 (x + 1) - (x + 1)
= (x + 1) (x^2 - 1)
c) (2a + b)^2 - (2b + a)^2
= 4a^2 + 4ab + b^2 - 4b^2 - 4ab - a^2
= 3a^2 - 3b^2
= 3 (a^2 - b^2)
d) (8a^3 - 27b^3) - 2a (4a^2 - 9b^2)
= 8a^3 - 27b^3 - 8a^3 + 18ab^2
= 27b^3 + 18ab^2
= 9b^2 (3b + 2a)
a,8a-8a2+3
=-8(a2-a)+3
=-8[a2-2a\(\dfrac{1}{2}\)+\(\left(\dfrac{1}{2}\right)^2\)-\(\dfrac{1}{4}\)]+3
=-8[(a-\(\dfrac{1}{2}\))2-\(\dfrac{1}{4}\)]+3
=-8(a-\(\dfrac{1}{2}\))2+2+3
=-8(a-\(\dfrac{1}{2}\))2+5
mà (a-\(\dfrac{1}{2}\))2\(\ge\)0
=>-8(a-\(\dfrac{1}{2}\))2\(\le\)0
=>-8(a-\(\dfrac{1}{2}\))2+5\(\le\)5
=> Gía trị lớn nhất biểu thức trên đạt được là 5( khi (a-\(\dfrac{1}{2}\))2=0\(\Leftrightarrow\)a=\(\dfrac{1}{2}\))
1. a. \(A=8a-8a^2+3=-8\left(a-\frac{1}{2}\right)^2+5\)
Vì \(\left(a-\frac{1}{2}\right)^2\ge0\forall a\)\(\Rightarrow-8\left(a-\frac{1}{2}\right)^2+5\le5\)
Dấu "=" xảy ra \(\Leftrightarrow-8\left(a-\frac{1}{2}\right)^2=0\Leftrightarrow a-\frac{1}{2}=0\Leftrightarrow a=\frac{1}{2}\)
Vậy Amax = 5 <=> a = 1/2
b. \(B=b-\frac{9b^2}{25}=-\frac{9}{25}\left(b-\frac{25}{18}\right)^2+\frac{25}{36}\)
Vì \(\left(b-\frac{25}{18}\right)^2\ge0\forall b\)\(\Rightarrow-\frac{9}{25}\left(b-\frac{25}{18}\right)^2+\frac{25}{36}\le\frac{25}{36}\)
Dấu "=" xảy ra \(\Leftrightarrow-\frac{9}{25}\left(b-\frac{25}{18}\right)^2=0\Leftrightarrow b-\frac{25}{18}=0\Leftrightarrow b=\frac{25}{18}\)
Vậy Bmax = 25/36 <=> b = 25/18
a,\(A=8a-8a^2+3\)
\(=-8\left(a^2-a\right)+3\)
\(=-8\left(a^2-2a\frac{1}{2}+\frac{1}{4}-\frac{1}{4}\right)+3\)
\(=-8\left[\left(a-\frac{1}{2}\right)^2-\frac{1}{4}\right]+3\)
\(=-8\left(a-\frac{1}{2}\right)^2+2+3\)
\(=-8\left(a-\frac{1}{2}\right)^2+5\le5\forall a\)
Dấu"=" xảy ra khi \(\left(a-\frac{1}{2}\right)^2=0\Rightarrow a=\frac{1}{2}\)
Vậy \(Max_A=5\)khi\(a=\frac{1}{2}\)
bài 2:
b,\(D=d^2+10e^2-6de-10e+26\)
\(=d^2-23de+\left(3e\right)^2+e^2-2.5e+5^2+1\)
\(=\left(d-3e\right)^2+\left(e-5\right)^2+1\ge1\forall d,e\)
Dấu"=" xảy ra khi\(\orbr{\begin{cases}\left(d-3e\right)^2=0\\\left(e-5\right)^2=0\end{cases}\Rightarrow\orbr{\begin{cases}d=15\\e=5\end{cases}}}\)
vậy \(D_{min}=1\)khi \(d=15;e=5\)
c,:\(E=4x^4+12x^2+11\)
\(=\left(2x^2\right)^2+2.2x^2.3+3^2+2\)
\(=\left(2x^2+3\right)^2+2\ge2\forall x\)
còn 1 đoạn nx bạn tự lm tiếp,lm giống như D
P/s : 8b-9a=31
Vì \(\frac{11}{7}>\frac{a}{b}>\frac{23}{29}\)
\(8b-9a=31\)(1)
\(\Rightarrow9a=8b-31\)
\(a=\frac{8b-31}{9}\)vì \(a\in N\)
\(8b-31\ge9\)
\(\Leftrightarrow8b\ge40\Leftrightarrow b\ge5\)
\(\Rightarrow\frac{11}{7}>\frac{8b-31}{9b}>\frac{23}{29}\)
\(\Leftrightarrow\frac{11}{7}>\frac{8}{9}>\frac{23}{29}\)
Mà \(7>\frac{8}{9}-\frac{31}{9b}< \frac{11}{7}\)
\(\frac{8}{9}-\frac{11}{7}< \frac{31}{9b}\)
...... \(\frac{-43}{63}< \frac{31}{9b}\)
\(\frac{-43}{7}< \frac{31}{b}\)
\(\Leftrightarrow-43b< 31.7\)
\(b>\frac{31.7}{-43}=\frac{-217}{43}\)
\(\Rightarrow b\in N\Leftrightarrow b>0\)
Mà \(\frac{8}{9}-\frac{31}{9b}>\frac{23}{29}\Leftrightarrow\frac{8}{9}-\frac{23}{29}>\frac{31}{9b}\)
\(\Leftrightarrow\frac{25}{261}>\frac{31}{9b}\Rightarrow25.9b>31.261\)
\(\Leftrightarrow b>\frac{31.261}{25.9}=\frac{899}{25}=35,9\)
Vậy \(5< b< \frac{899}{25}\)
\(\Rightarrow5< b< 35\)
Đến đây bạn lập bảng .