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a ) \(\left(2a+b\right)^2-\left(2a+b\right)\left(2a-b\right)-2a\left(b-a\right)\)
\(=4a^2+4ab+b^2-\left(4a^2-b^2\right)-2ab+2a^2\)
\(=4a^2+4ab+b^2-4a^2+b^2-2ab+2a^2\)
\(=2a^2+2ab+2b^2\)
\(=\left(a^2+2ab+b^2\right)+a^2+b^2\)
\(=\left(a+b\right)^2+a^2+b^2\)
b ) \(\left(a+b-c\right)^2-\left(a+b\right)^2+2c\left(a+b\right)\)
\(=\left(a+b\right)^2-2\left(a+b\right)c+c^2-\left(a+b\right)^2+2c\left(a+b\right)\)
\(=\left[\left(a+b\right)^2-\left(a+b\right)^2\right]+\left[2c\left(a+b\right)-2\left(a+b\right)c\right]+c^2\)
\(=c^2\)
@Khôi Bùi
\(\left(2a+b\right)^2-\left(2a+b\right)\left(2a-b\right)-2a\left(b-a\right)\)
\(=\left(2a+b\right)\left[\left(2a+b\right)-\left(2a-b\right)\right]-2a\left(b-a\right)\)
\(=2b\left(2a+b\right)-2a\left(b-a\right)\)
\(=4ab+2b^2-2ab+2a^2=2\left(a^2+ab+b^2\right)\)
\(\left(a+b-c\right)^2-\left(a+b\right)^2+2c\left(a+b\right)\)
\(=\left(a+b-c+a+b\right)\left(a+b-c-a-b\right)+2c\left(a+b\right)\)
\(=-c\left(2a+2b-c\right)+2c\left(a+b\right)=\)
\(-2c\left(a+b\right)+c^2+2c\left(a+b\right)=c^2\)
a: \(=a^2-b^4\)
b: \(=\left(a^2+2a\right)^2-9\)
c: \(=a^2-\left(2a+3\right)^2\)
d: \(=a^4-\left(2a-3\right)^2\)
e: \(=\left(-a^2-2a+3\right)^2\)
g: \(=4a^2-a^4\)
a) A=(4-5x)2-(3+5x)2=(4-5x-3-5x)(4-5x+3+5x)=(-25x+1)1=-25x+1
B=(3x-1)(1+3x)-(3x+1)2=9x2-1-(3x+1)2=9x2-1-(9x2+6x+1)=9x2-1-9x2-6x-1=-6x-2=-2(3x+1)
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:
$M=(2a+b)^2-(b-2a)^2=[(2a+b)-(b-2a)][(2a+b)+(b-2a)]$
$=4a.2b=8ab$
$N=(3a+1)^2+2a(1-2b)+(2b-1)^2$
$=(9a^2+6a+1)+2a-4ab+(4b^2-4b+1)$
$=9a^2+8a+4b^2-4b-4ab+2$
$A=(m-n)^2+4mn=m^2-2mn+n^2+4mn$
$=m^2+2mn+n^2=(m+n)^2$
2.
\(P=\left(\dfrac{a+6}{3\left(a+3\right)}-\dfrac{1}{a+3}\right).\dfrac{27a}{a+2}=\left(\dfrac{a+3}{3\left(a+3\right)}\right).\dfrac{27a}{a+2}=\dfrac{27a}{3\left(a+2\right)}=\dfrac{9a}{a+2}\)
ĐKXĐ là :
\(a\ne0;-3;-2\)
Vs a = 1 ta có:
=> P=3
1.
\(M=\left(\dfrac{2a}{2a+b}-\dfrac{4a^2}{\left(2a+b\right)^2}\right):\left(\dfrac{2a}{\left(2a-b\right)\left(2a+b\right)}-\dfrac{1}{2a-b}\right)=\left(\dfrac{4a^2+2ab-4a^2}{\left(2a+b\right)^2}\right).\left(\dfrac{\left(2a+b\right)\left(2a-b\right)}{b}\right)=\dfrac{2a.\left(2a-b\right)}{\left(2a+b\right)}\)
\(B=\left(2a+b\right)^2-\left(2a-b\right)^2\)
\(=\left(2a+b-2a+b\right)\left(2a+b+2a-b\right)\)
\(=8ab\)