Cho a-3b=1, 2ab=-4. Tính:
A=(2a+6b)2-2
B=3a2+27b2-ab-1
C=a3-27b3+a2+9b2+2
D=a4+81b4-1
Bài 1:Phân tích đa thức thành nhân tử
a)x4+2x2y+y2
b)(2a+b)2-(2b+a)2
c) 8a2-27b2-2a(4a2-9b2)
`a)x^4+2x^2y+y^2`
`=(x^2+y)^2`
`b)(2a+b)^2-(2b+a)^2`
`=(2a+b-2b-a)(2a+b+2b+a)`
`=(a-b)(3a+3b)`
`=3(a-b)(a+b)`
`c)8a^3-27b^3-2a(4a^2-9b^2)`
`=(2a-3b)(4a^2+6ab+9b^2)-2a(2a-3b)(2a+3b)`
`=(2a-3b)(4a^2+6ab+9b^2-3a^2-6ab)`
`=9b^2(2a-3b)`
a) Ta có: \(x^4+2x^2y+y^2\)
\(=\left(x^2\right)^2+2\cdot x^2\cdot y+y^2\)
\(=\left(x^2+y\right)^2\)
b) Ta có: \(\left(2a+b\right)^2-\left(2b+a\right)^2\)
\(=\left(2a+b-2b-a\right)\left(2a+b+2b+a\right)\)
\(=\left(a-b\right)\left(3a+3b\right)\)
\(=3\left(a+b\right)\left(a-b\right)\)
Bài 1: Rút gọn
A=(7-2x)(7+2x)+(2x+7)2
B=(4x-5)2-(2x-1)(8x-5)
C=(5x-3)2-2(5x-3)(5-5x)+(5x-5)2
D=(2a+3b-c)(2a-3b+c)-(4a2-9b2-c2)
A=(7-2x)(7+2x)+(2x+7)2
=49-4x2+4x2+28x+49
= 98+28x
B=(4x-5)2-(2x-1)(8x-5)
= 16x2-25-((8x(2x-1))-(5(2x-1)))
= 16x2-25-((16x2+8x)-(10x+5))
= 16x2-25-(16x2+8x-10x-5)
= 16x2-25-16x2-8x+10x+5
= -20+2x
Bài 1: Rút gọn
A=(7-2x)(7+2x)+(2x+7)2
B=(4x-5)2-(2x-1)(8x-5)
C=(5x-3)2-2(5x-3)(5-5x)+(5x-5)2
D=(2a+3b-c)(2a-3b+c)-(4a2-9b2-c2)
a) Ta có: \(A=\left(7-2x\right)\left(7+2x\right)+\left(2x+7\right)^2\)
\(=7-4x^2+4x^2+28x+49\)
\(=28x+56\)
b) Ta có: \(B=\left(4x-5\right)^2-\left(2x-1\right)\left(8x-5\right)\)
\(=16x^2-40x+25-\left(16x^2-10x-8x+5\right)\)
\(=16x^2-40x+25-16x^2+18x-5\)
\(=-22x+20\)
c) Ta có: \(C=\left(5x-3\right)^2-2\left(5x-3\right)\left(5-5x\right)+\left(5x-5\right)^2\)
\(=\left(5x-3\right)^2+2\cdot\left(5x-3\right)\left(5x-5\right)+\left(5x-5\right)^2\)
\(=\left(5x-3+5x-5\right)^2\)
\(=\left(10x-8\right)^2\)
\(=100x^2-160x+64\)
d) Ta có: \(D=\left(2a+3b-c\right)\left(2a-3b+c\right)-\left(4a^2-9b^2-c^2\right)\)
\(=\left[\left(2a+\left(3b-c\right)\right)\left(2a-\left(3b-c\right)\right)\right]-\left(4a^2-9b^2-c^2\right)\)
\(=4a^2-\left(3b-c\right)^2-4a^2+9b^2+c^2\)
\(=-9b^2+6bc-c^2+9b^2+c^2\)
=6bc
Chứng minh đẳng thức:
a) a 2 − 3 a a 2 + 9 − 6 a 2 27 − 9 a + 3 a 2 − a 3 . 1 − 2 a − 3 a 2 = a + 1 a với a ≠ 0 ; 3 ;
b) 2 5 b − 2 b + 1 . b + 1 5 b − 3 5 b − 3 5 : b − 1 b = 6 b 5 ( b − 1 ) với b ≠ 0 ; ± 1 .
Thực hiện phép tính đối với vế trái của mỗi đẳng thức.
Bài 2: Tìm đa thức P biết
a)x2+5x+6/x2+4x+4=P/x+2
b)a+1/a-1=(a+1)2/P
c)P/2a-6=a2+3a+9/2
d)a3+b3=(a-b).P
e)x2+y2=(x+y).P
a) Ta có: \(\dfrac{P}{x+2}=\dfrac{x^2+5x+6}{x^2+4x+4}\)
\(\Leftrightarrow\dfrac{P}{x+2}=\dfrac{\left(x+2\right)\left(x+3\right)}{\left(x+2\right)^2}=\dfrac{x+3}{x+2}\)
hay P=x+3
b) Ta có: \(\dfrac{\left(a+1\right)^2}{P}=\dfrac{a+1}{a-1}\)
\(\Leftrightarrow P=\left(a+1\right)\left(a-1\right)\)
\(\Leftrightarrow P=a^2-1\)
Bài 1:Cho a+b=5 và a.b=-6 Tính:
a) a.(4a+b)+4b
b) a2+b2
c) a4+b4
Bài 2: 2a-b=5 và a.b=3
a) a.(b-2)+b
b) 4.a2+b2
Tìm GTLN của: M=-x2+12x+8
Tìm GTNN của: N=a2+9b2+5a-6b-3
Tìm GTNN của: Q=3a2-30a-7
Tìm GTLN của: M=-x2+12x+8
Tìm GTNN của: N=a2+9b2+5a-6b-3
Tìm GTNN của: Q=3a2-30a-7
\(M=-x^2+12x+8=-\left(x-6\right)^2+44\le44\)
\(M_{max}=44\) khi \(x=6\)
\(N=a^2+9b^2+5a-6b=\left(a+\dfrac{5}{2}\right)^2+\left(3b-1\right)^2-\dfrac{41}{4}\ge-\dfrac{41}{4}\)
\(N_{min}=-\dfrac{41}{4}\) khi \(\left(a;b\right)=\left(-\dfrac{5}{2};\dfrac{1}{3}\right)\)
\(Q=3\left(a-5\right)^2-82\ge-82\)
\(Q_{min}=-82\) khi \(a=5\)
bài 1: cho a,b,c thỏa mãn a+b+c=0
tính: (a+2b)2+(b+2c)2+(c+2a)2 / (a-2b)2+(b-2c)2+(c-2a)2
bài 2: cho số a,b,c có tổng khác 0 thỏa mãn: a3+b3+c3=3abc
tính: ab+2bc+3ca / 3a2+4b2+5c2
1.
\(a+b+c=0\)
\(\Rightarrow\left(a+b+c\right)^2=0\)
\(\Rightarrow a^2+b^2+c^2+2ab+2bc+2ca=0\)
\(\Rightarrow a^2+b^2+c^2=-2\left(ab+bc+ca\right)\)
Ta có:
\(\dfrac{\left(a+2b\right)^2+\left(b+2c\right)^2+\left(c+2a\right)^2}{\left(a-2b\right)^2+\left(b-2c\right)^2+\left(c-2a\right)^2}\)
\(=\dfrac{a^2+4b^2+4ab+b^2+4c^2+4bc+c^2+4a^2+4ca}{a^2+4b^2-4ab+b^2+4c^2-4bc+c^2+4a^2-4ca}\)
\(=\dfrac{5\left(a^2+b^2+c^2\right)+4\left(ab+bc+ca\right)}{5\left(a^2+b^2+c^2\right)-4\left(ab+bc+ca\right)}\)
\(=\dfrac{-10\left(ab+bc+ca\right)+4\left(ab+bc+ca\right)}{-10\left(ab+bc+ca\right)-4\left(ab+bc+ca\right)}\)
\(=\dfrac{-6}{-14}=\dfrac{3}{7}\)
b.
\(a^3+b^3+c^3=3abc\)
\(\Leftrightarrow a^3+b^3+3ab\left(a+b\right)-3ab\left(a+b\right)+c^3-3abc=0\)
\(\Leftrightarrow\left(a+b\right)^3+c^3-3ab\left(a+b+c\right)=0\)
\(\Leftrightarrow\left(a+b+c\right)\left(\left(a+b\right)^2-c\left(a+b\right)+c^2\right)-3abc\left(a+b+c\right)=0\)
\(\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)=0\)
\(\Leftrightarrow a^2+b^2+c^2-ab-bc-ca=0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}a-b=0\\b-c=0\\c-a=0\end{matrix}\right.\) \(\Leftrightarrow a=b=c\)
\(\Rightarrow\dfrac{ab+2bc+3ca}{3a^2+4b^2+5c^2}=\dfrac{a^2+2a^2+3a^2}{3a^2+4a^2+5a^2}=\dfrac{6}{12}=\dfrac{1}{2}\)
Tính:
a) 8 - 2 - 3
b) 7 - 4 - 1
c) 10 - 5 - 2
d) 3 + 6 - 4