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a ) \(\left(5x+2y\right)^2=25x^2+20xy+4y^2\)
b ) \(\left(-3x+2\right)^2=9x^2-12x+4\)
c ) \(\left(\dfrac{2}{3}x+\dfrac{1}{3}y\right)^2=\dfrac{4}{9}x^2+\dfrac{4}{9}xy+\dfrac{1}{9}y^2\)
d ) \(\left(2x-\dfrac{5}{2}y\right)^2=4x^2-10xy+\dfrac{25}{4}y^2\)
e ) \(\left(x+\dfrac{4}{3}y^2\right)^2=x^2+\dfrac{8}{3}xy^2+\dfrac{16}{9}y^4\)
f ) \(\left(2x^2+\dfrac{5}{3}y\right)^2=4x^4+\dfrac{20}{3}x^2y+\dfrac{25}{9}y^2\)
Bài 1:
a) -16 +(x-3)2
<=> (x-3)2-16
<=> (x-3)2 -42
<=> (x-3-4)(x-3+4)
<=> (x-7)(x+1)
b) 64+16y+y2
<=> y2 + 2.8.y + 82
<=> (y+8)2
c) \(\dfrac{1}{8}-8x^3\)
\(\Leftrightarrow\left(\dfrac{1}{2}\right)^3-\left(2x\right)^3\)
\(\Leftrightarrow\left(\dfrac{1}{2}-2x\right)\left(\dfrac{1}{4}+x+4x^2\right)\)
d)\(x^2-x+\dfrac{1}{4}\)
\(\Leftrightarrow x^2-2.\dfrac{1}{2}.x+\left(\dfrac{1}{2}\right)^2\)
\(\Leftrightarrow\left(x-\dfrac{1}{2}\right)^2\)
e) x4 + 4x2 + 4
<=> (x2)2 + 2.2.x2 +22
<=> (x2 + 2)2
g)\(8x^3+60x^2y+150xy^2+125y^3\)
\(\Leftrightarrow\left(2x+5y\right)^3\)
a: \(=\dfrac{x^2+2x+1-x^2+2x-1}{\left(x-1\right)\left(x+1\right)}:\left(\dfrac{1}{x+1}+\dfrac{x}{x-1}+\dfrac{2}{\left(x-1\right)\left(x+1\right)}\right)\)
\(=\dfrac{4x}{\left(x-1\right)\left(x+1\right)}:\dfrac{x-1+x^2+x+2}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{4x}{\left(x-1\right)\left(x+1\right)}\cdot\dfrac{\left(x-1\right)\left(x+1\right)}{x^2+2x+1}=\dfrac{4x}{x^2+2x+1}\)
b: \(=\dfrac{x+2}{-\left(x-2\right)}\cdot\dfrac{\left(x-2\right)^2}{4x^2}\cdot\left(\dfrac{2}{2-x}-\dfrac{4}{\left(x+2\right)\left(x^2-2x+4\right)}\cdot\dfrac{x^2-2x+4}{2-x}\right)\)
\(=\dfrac{-\left(x+2\right)\left(x-2\right)}{4x^2}\cdot\left(\dfrac{2}{2-x}-\dfrac{4}{\left(x+2\right)\left(2-x\right)}\right)\)
\(=\dfrac{-\left(x+2\right)\left(x-2\right)}{4x^2}\cdot\dfrac{2x+4-4}{\left(2-x\right)\left(x+2\right)}\)
\(=\dfrac{2x}{4x^2}=\dfrac{1}{2x}\)
\(\left(m-n\right)^6-6\left(m-n\right)^4+12\left(m-n\right)^2-8=\left[\left(m-n\right)^2-2\right]^3\)
\(\dfrac{8}{27}a^3-\dfrac{8}{3}a^2b+8b^2a-8b^3=\left(\dfrac{2}{3}a-2b\right)^3\)
Chúc bạn học tốt !!
a) x2 +4x+4 = ( x + 2 )2
b) 16x2 - 8xy + y2 = ( 4x - y )2
c)9a2 +16b2 - 24ab = ( 3a - 4y ) 2
d) x2 - x + \(\dfrac{1}{4}\)= ( x - \(\dfrac{1}{2}\))2
e) y2 + \(\dfrac{1}{2}y\) + \(\dfrac{1}{16}\) = ( y + \(\dfrac{1}{4}\))2
\(1.\)
\(a.\)
\(\dfrac{8}{\left(x^2+3\right)\left(x^2-1\right)}+\dfrac{2}{x^2+3}+\dfrac{1}{x+1}\)
\(=\dfrac{8}{\left(x^2+3\right)\left(x^2-1\right)}+\dfrac{2\left(x^2-1\right)}{\left(x^2+3\right)\left(x^2-1\right)}+\dfrac{1\left(x-1\right)\left(x^2+3\right)}{\left(x^2-1\right)\left(x^2+3\right)}\)
\(=\dfrac{8}{\left(x^2+3\right)\left(x^2-1\right)}+\dfrac{2x^2-2}{\left(x^2+3\right)\left(x^2-1\right)}+\dfrac{x^3-x^2+3x-3}{\left(x^2-1\right)\left(x^2+3\right)}\)
\(=\dfrac{8+2x^2-2+x^3-x^2+3x-3}{\left(x^2+3\right)\left(x^2-1\right)}\)
\(=\dfrac{x^3+x^2+3x+3}{\left(x^2+3\right)\left(x^2-1\right)}\)
\(=\dfrac{x^2\left(x+1\right)+3\left(x+1\right)}{\left(x^2+3\right)\left(x^2-1\right)}\)
\(=\dfrac{\left(x^2+3\right)\left(x+1\right)}{\left(x^2+3\right)\left(x^2-1\right)}\)
\(=x-1\)
\(b.\)
\(\dfrac{x+y}{2\left(x-y\right)}-\dfrac{x-y}{2\left(x+y\right)}+\dfrac{2y^2}{x^2-y^2}\)
\(=\dfrac{x+y}{2\left(x-y\right)}-\dfrac{x-y}{2\left(x+y\right)}+\dfrac{2y^2}{\left(x-y\right)\left(x+y\right)}\)
\(=\dfrac{\left(x+y\right)^2}{2\left(x^2-y^2\right)}-\dfrac{\left(x-y\right)^2}{2\left(x^2-y^2\right)}+\dfrac{4y^2}{2\left(x^2-y^2\right)}\)
\(=\dfrac{x^2+2xy+y^2}{2\left(x^2-y^2\right)}-\dfrac{x^2-2xy+y^2}{2\left(x^2-y^2\right)}+\dfrac{4y^2}{2\left(x^2-y^2\right)}\)
\(=\dfrac{x^2+2xy+y^2-x^2+2xy-y^2+4y^2}{2\left(x^2-y^2\right)}\)
\(=\dfrac{4xy+4y^2}{2\left(x^2-y^2\right)}\)
\(=\dfrac{4y\left(x+y\right)}{2\left(x^2-y^2\right)}\)
\(=\dfrac{2y}{\left(x-y\right)}\)
Tương tự các câu còn lại
Bài 2 .
a) \(\dfrac{2x}{x^2+2xy}+\dfrac{y}{xy-2y^2}+\dfrac{4}{x^2-4y^2}\)
\(=\dfrac{2x}{x\left(x+2y\right)}+\dfrac{y}{y\left(x-2y\right)}+\dfrac{4}{\left(x-2y\right)\left(x+2y\right)}\)
\(=\dfrac{2xy\left(x-2y\right)+xy\left(x+2y\right)+4xy}{xy\left(x+2y\right)\left(x-2y\right)}\)
\(=\dfrac{2x^2y-2xy^2+x^2y+2xy^2+4xy}{xy\left(x+2y\right)\left(x-2y\right)}\)
\(=\dfrac{3x^2y+4xy}{xy\left(x+2y\right)\left(x-2y\right)}\)
b) Sai đề hay sao ý
c) \(\dfrac{2x+y}{2x^2-xy}+\dfrac{16x}{y^2-4x^2}+\dfrac{2x-y}{2x^2+xy}\)
\(=\dfrac{2x+y}{x\left(2x-y\right)}+\dfrac{-16x}{\left(2x-y\right)\left(2x+y\right)}+\dfrac{2x-y}{x\left(2x+y\right)}\)
\(=\dfrac{\left(2x+y\right)^2-16x^2+\left(2x-y\right)^2}{x\left(2x-y\right)\left(2x+y\right)}\)
\(=\dfrac{4x^2+4xy+y^2-16x^2+4x^2-4xy+y^2}{x\left(2x-y\right)\left(2x+y\right)}\)
\(=\dfrac{-8x^2}{x\left(2x-y\right)\left(2x+y\right)}\)
d) \(\dfrac{1}{1-x}+\dfrac{1}{1+x}+\dfrac{2}{1+x^2}+\dfrac{4}{1+x^4}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)
\(=\dfrac{2}{1-x^2}+\dfrac{2}{1+x^2}+\dfrac{4}{1+x^4}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)
\(=\dfrac{4}{1-x^4}+\dfrac{4}{1+x^4}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)
.....
\(=\dfrac{16}{1-x^{16}}+\dfrac{16}{1+x^{16}}\)
\(=\dfrac{32}{1-x^{32}}\)
a) x2 + 2x + 1 = x2 + 2.x.1+ 12 = ( x + 1)2
b) 9x2 + y2 + 6xy = (3x)2 + 2.3.x.y + y2 = (3x + y)2
c) 25a2 + 4b2 – 20ab = (5a)2 – 2.5.a.2b. + (2b)2 = (5a – 2b)2
Hoặc 25a2 + 4b2 – 20ab = (2b)2 – 2.2b.5a. + (5a)2 = (2b – 5a)2
d) x2 – x + \(\dfrac{1}{4}\) = x2 – 2.x. \(\dfrac{1}{2}\) + ( \(\dfrac{1}{2}\))22 = ( x - \(\dfrac{1}{2}\) )2
Hoặc x2 – x + \(\dfrac{1}{4}\) = \(\dfrac{1}{4}\) - x + x2 = (\(\dfrac{1}{2}\))2 – 2. \(\dfrac{1}{2}\).x + x2 = (\(\dfrac{1}{2}\) - x)2
a) x2 + 2x + 1 = x2+ 2 . x . 1 + 12
= (x + 1)2
b) 9x2 + y2+ 6xy = (3x)2 + 2 . 3 . x . y + y2 = (3x + y)2
c) 25a2 + 4b2– 20ab = (5a)2 – 2 . 5a . 2b + (2b)2 = (5a – 2b)2
Hoặc 25a2 + 4b2 – 20ab = (2b)2 – 2 . 2b . 5a + (5a)2 = (2b – 5a)2
d) x2 – x + 1414 = x2 – 2 . x . 1212 + (12)2(12)2= (x−12)2(x−12)2
Hoặc x2 – x + 1414 = 1414 - x + x2 = (12)2(12)2 - 2 . 1212 . x + x2 = (12−x)2
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: \(\left(a+b-c-d\right)\left(a-b+c-d\right)\)
\(=\left(a-d\right)^2-\left(b-c\right)^2\)
\(=a^2-2ad+d^2-b^2+2bc-c^2\)
a/ \(\left(x+\dfrac{4}{3}y^2\right)^2\)
\(=x^2+2.x.\dfrac{4}{3}y^2+\left(\dfrac{4}{3}y^2\right)^2\)
\(=x^2+\dfrac{8}{3}xy^2+\dfrac{16}{9}y^4\)
b/ \(\left(2x^2+\dfrac{5}{3}y\right)^2\)
\(=\left(2x^2\right)^2+2.2x^2.\dfrac{5}{3}y+\left(\dfrac{5}{3}y\right)^2\)
\(=4x^4+\dfrac{20}{3}x^2y+\dfrac{25}{9}y^2\)