Cho a-3b=1, 2ab=-4. Tính:
A=2a+(7ab)/2-6b+2
B=(2a+6b)2-2
C= 3a2+27b2-ab-1
D= a3-27b3+a2+9b2+2
E=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: \(\left(2a+b\right)^2-\left(2b+a\right)^2\)
\(=\left(2a+b-2b-a\right)\left(2a+b+2b+a\right)\)
\(=\left(a-b\right)\left(3a+3b\right)\)
\(=3\left(a+b\right)\left(a-b\right)\)
Cho a-3b=1, 2ab=-4. Tính:
A=2a+(7ab)/2-6b+2
B= (2a+6b)2-2
C+ 3a2+27b2-ab-1
D=a3-27b3+a2+9b2+2
E=a4+81b4-1
Cho a-3b=1, 2ab=-4. Tính:
A= 2a+(7ab)/2-6b+2
B=(2a+6b)2-2
C=3a2+27b2-ab-1
D= a3-27b3+a2+9b2+2
E= a4+81b4-1
Giúp mik vs, tik cho!
Thu gọn biểu thức:
a) 2.( a-3b+5b) + (-3a-7c+5c) -4b b)7.(-a-2b+c)+3.(-2c-6b+a)
c)2.(-3c-6b+7b)-4.(2a-3b+8c) d) -3.(2a+3b-4c)+7.(-2c+8a-2c)+20a+2a+24b
e)-(5a-6b+c)+3.(-2c-6b+a)
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 S = a + b + c + d + e
a) c = 2a ; d = 2b ; 6b + 5c = 6e ; 2a + 3b = 2e ; d - a = 40
b) 3b = 4a ; 3d = 4c ; 3b + c = 2e ; 6d - a = 5e ; c - b = 1
c) 3b = 4a ; 3d = 4c ; 3b + c = 2e ; 4e + b = 5d ; d - a = 5
1. Rút gọn các biểu thức sau:
M = (2a+b)2-(b-2a)2
N = (3a+2)2+2a(1-2b)+(2b-1)2
A = (m-n)2+4mn
2. Tính:
a) (x+5)2 b) (5/2-t)2
c) (2u+3v)2 d) (-1/8 a+2/3 bc)2
e) (x/y-1/z)2 f) (mn/4-x/6)(mn/4+x/6)
Bài 2:
a) \(\left(x+5\right)^2=x^2+10x+25\)
b) \(\left(\dfrac{5}{2}-t\right)^2=\dfrac{25}{4}-5t+t^2\)
c) \(\left(2u+3v\right)^2=4u^2+12uv+9v^2\)
d) \(\left(-\dfrac{1}{8}a+\dfrac{2}{3}bc\right)^2=\dfrac{1}{64}a^2-\dfrac{1}{6}abc+\dfrac{4}{9}b^2c^2\)
e) \(\left(\dfrac{x}{y}-\dfrac{1}{z}\right)^2=\dfrac{x^2}{y^2}-\dfrac{2x}{yz}+\dfrac{1}{z^2}\)
f) \(\left(\dfrac{mn}{4}-\dfrac{x}{6}\right)\left(\dfrac{mn}{4}+\dfrac{x}{6}\right)=\dfrac{m^2n^2}{16}-\dfrac{x^2}{36}\)
Bài 1:
$M=(2a+b)^2-(b-2a)^2=[(2a+b)-(b-2a)][(2a+b)+(b-2a)]$
$=4a.2b=8ab$
$N=(3a+1)^2+2a(1-2b)+(2b-1)^2$
$=(9a^2+6a+1)+2a-4ab+(4b^2-4b+1)$
$=9a^2+8a+4b^2-4b-4ab+2$
$A=(m-n)^2+4mn=m^2-2mn+n^2+4mn$
$=m^2+2mn+n^2=(m+n)^2$
Bài 1:
a: Ta có: \(M=\left(2a+b\right)^2-\left(b-2a\right)^2\)
\(=4a^2+4ab+b^2-b^2+4ab-4a^2\)
\(=8ab\)
b: Ta có: \(N=\left(3a+2\right)^2+2a\left(1-2b\right)+\left(2b-1\right)^2\)
\(=\left(3a+2+1-2b\right)^2\)
\(=\left(3a-2b+3\right)^2\)
\(=9a^2+4b^2+9-12ab+18a-12b\)
c: Ta có: \(A=\left(m-n\right)^2+4nm\)
\(=m^2-2mn+n^2+4mn\)
\(=m^2+2mn+n^2\)
\(=\left(m+n\right)^2\)
2:
a: \(\left(x+5\right)^2=x^2+10x+25\)
b: \(\left(\dfrac{5}{2}-t\right)^2=\dfrac{25}{4}-5t+t^2\)
Cho a,b,c là các số dương thỏa mãn a^2+b^2+c^2 +6=2(a+2b+c).Tính K=√2a+3b+c
A,K=6
B,K=2
C,K=3
D,K=8
Lời giải:
$a^2+b^2+c^2+6=2(a+2b+c)$
$\Leftrightarrow (a^2-2a+1)+(b^2-4b+4)+(c^2-2c+1)=0$
$\Leftrightarrow (a-1)^2+(b-2)^2+(c-1)^2=0$
Vì $(a-1)^2\geq 0; (b-2)^2\geq 0; (c-1)^2\geq 0$ với mọi $a,b,c\in\mathbb{R}$ nên để tổng của chúng bằng $0$ thì:
$(a-1)^2=(b-2)^2=(c-1)^2=0$
$\Rightarrow a=c=1; b=2$
$\Rightarrow K=3$
Đáp án C.
cho a>b>0 và a2 - 6b2= -ab
tính M= (2ab)/(2a2 - 3b2)
Từ \(a^2-6b^2=-ab\Rightarrow a^2-6b^2+ab=0\)
\(\Rightarrow a^2+3ab-2ab-6b^2=0\)
\(\Rightarrow a\left(a+3b\right)-2b\left(a+3b\right)=0\)
\(\Rightarrow\left(a+3b\right)\left(a-2b\right)=0\)
\(\Rightarrow\orbr{\begin{cases}a+3b=0\\a-2b=0\end{cases}}\Rightarrow\orbr{\begin{cases}a=-3b\\a=2b\end{cases}}\)
Xét \(a=-3b\) thay vào M ta có:\(M=\frac{2\cdot3\left(-b\right)\cdot b}{2\left(-3b\right)^2-3b^2}=\frac{-6b^2}{15b^2}=-\frac{2}{5}\)
Xét \(a=2b\) thay vào M ta có:\(M=\frac{2\cdot2b\cdot b}{2\cdot\left(2b\right)^2-3b^2}=\frac{4b^2}{8b^2-3b^2}=\frac{4b^2}{5b^2}=\frac{4}{5}\)