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a) \(m^2-n^2=\left(m-n\right)\left(m+n\right)\)
b) \(\left(x^2+x-1\right)^2-\left(x^2+2x+3\right)^2\)
\(=\left(x^2+x-1+x^2+2x+3\right)\left(x^2+x-1-x^2-2x-3\right)\)
\(=\left(2x^2+3x+2\right)\left(-4-x\right)\)
c) \(-16+\left(x-3\right)^2=\left(x-3\right)^2-16=\left(x-3-4\right)+\left(x-3+4\right)=\left(x-7\right)\left(x+1\right)\)
d) \(64+16y+y^2\)
\(=8^2+2.8.y+y^2\)
\(=\left(8+y\right)^2\)
a: \(=\left(m-n\right)\left(m+n\right)\)
b: \(=\left(x^2+x-1-x^2-2x-3\right)\left(x^2+x-1+x^2+2x+3\right)\)
\(=\left(-x-4\right)\left(2x^2+3x+2\right)\)
d: \(=\left(y+8\right)^2\)
Giải:
a) \(\left(x^2+x-1\right)^2-\left(x^2+2x+3\right)^2\)
\(=\left[\left(x^2+x-1\right)-\left(x^2+2x+3\right)\right]\left[\left(x^2+x-1\right)+\left(x^2+2x+3\right)\right]\)
\(=\left(x^2+x-1-x^2-2x-3\right)\left(x^2+x-1+x^2+2x+3\right)\)
\(=\left(-x-4\right)\left(2x^2+3x+2\right)\)
Vậy ...
b) \(-16+\left(x-3\right)^2\)
\(=\left(x-3\right)^2-16\)
\(=\left(x-3\right)^2-4^2\)
\(=\left(x-3-4\right)\left(x-3+4\right)\)
\(=\left(x-7\right)\left(x+1\right)\)
Vậy ...
c) \(64+16y+y^2\)
\(=8^2+2.8.y+y^2\)
\(=\left(8+y\right)^2\)
Vậy ...
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)\(8x^3+60x^2y+150xy^2+125y^3\)
\(\Leftrightarrow\left(2x+5y\right)^3\)
\(a.9a^2-25b^4=\left(3a\right)^2-\left(5b^2\right)^2=\left(3a-5b^2\right)\left(3a+5b^2\right)\)
\(b.\left(2x+y\right)^2-1=\left(2x+y-1\right)\left(2x+y+1\right)\)
\(c.\left(x+y+z\right)^2-\left(x-y-z\right)^2=\left[\left(x+y+z\right)+\left(x-y-z\right)\right]\left[\left(x+y+z\right)\right]-\left(x-y-z\right)\\ =2x.\left(2y+2z\right)\)
a) \(9a^2-25b^4=\left(3a\right)^2-\left(5b^2\right)^2=\left(3a-5b^2\right)\left(3a+5b^2\right)\)
b) \(\left(2x+y\right)^2-1=\left(2x+y\right)^2-1^2=\left(2x+y+1\right)\left(2x+y-1\right)\)
c) \(\left(x+y+z\right)^2-\left(x-y-z\right)^2=\left(x+y+z+x-y-z\right)\left(x+y+z-x+y+z\right)\)
\(=2x\left(2y+2z\right)\)
\(x^2+6x+9=\left(x+3\right)^2\)
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\(x^2-x+\dfrac{1}{4}=\left(x-\dfrac{1}{2}\right)^2\)
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\(x^3+12x^2+48x+64=\left(x+4\right)^3\)
1) \(\dfrac{\left(x+5\right)^2+\left(x-5\right)^2}{x^2+25}\)
\(=\dfrac{x^2+10x+25+x^2-10x+25}{x^2+25}\)
\(=\dfrac{2x^2+50}{x^2+25}\)
\(=\dfrac{2\left(x^2+25\right)}{x^2+25}=2\)
2) \(\left(x+3\right)\left(x^2-3x+9\right)-\left(54+x^3\right)\)
\(=x^3+3^3-54-x^3\)
\(=27-54=-27\)
3) \(\left(2x+y\right)^2-\left(y+3x\right)^2\)
\(=4x^2+4xy+y^2-y^2-6xy-9x^2\)
\(=-5x^2-2xy\)
4) \(\left(2x+1\right)^3-\left(2x-1\right)^3-24x^2\)
\(=8x^3+12x^2+6x+1-8x^3+12x^2-6x+1-24x^2\)
\(=2\)
!!
b) \(-4x^2-4x-1\)
\(=-\left(4x^2+4x+1\right)\)
\(=-\left(2x+1\right)^2\)
c) \(\frac{4}{9}x^2-25y^2\)
\(=\left(\frac{2}{3}x+5y\right)\left(\frac{2}{3}x-5y\right)\)
d) \(\frac{1}{27}x^3-8\)
\(=\left(\frac{1}{3}x-2\right)\left(\frac{1}{9}x+\frac{2}{3}x+4\right)\)
a) \(m^2-n^2=\left(m-n\right)\left(m+n\right)\)
b) \(\left(x^2+x-1\right)^2-\left(x^2+2x+3\right)^2=\left(x^2+x+1+x^2+2x+3\right)\left(x^2+x+1-x^2-2x-3\right)\)
\(=-\left(2x^2+3x+4\right)\left(x+2\right)\)
d) \(64+16y+y^2=\left(8+y\right)^2\)
c) mk chỉnh đề:
\(16-\left(x-3\right)^2=\left(4+x-3\right)\left(4-x+3\right)=\left(x+1\right)\left(7-x\right)\)
ồ cuk dễ nhỉ
Nếu các bn thích thì ...........
cứ cho NTN này nhé !