Viết biểu thức sau thành hằng đẳng thức
a. (x-3)^2 + 2(x-3)(x+2)+(x+2)^2
b. (x+5)^2 - (2x+10) (x - 6) + (x - 6)^2
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Bài 2:
a: \(A=\left(x+1\right)^3+5=20^3+5=8005\)
b: \(B=\left(x-1\right)^3+1=10^3+1=1001\)
a) (x + 3)2 - 2(x + 3)(x - 2) + (x - 2)2
= (x + 3 - x + 2)2 = 52 = 25
b) (2x + 5)2 + 2(2x + 5)(3x - 1) + (3x - 1)2
= (2x + 5 + 3x - 1)2 = (5x + 4)2
a) Ta có: \(a^3y^3+125\)
\(=\left(ay+5\right)\left(a^2y^2-5ay+25\right)\)
b) Ta có: \(8x^3-y^3-6xy\cdot\left(2x-y\right)\)
\(=\left(2x-y\right)\left(4x^2+2xy+y^2\right)-6xy\left(2x-y\right)\)
\(=\left(2x-y\right)\left(4x^2+2xy-6xy+y^2\right)\)
\(=\left(2x-y\right)^3\)
1:
a: \(\left(2x-5\right)^2-4x\left(x+3\right)\)
\(=4x^2-20x+25-4x^2-12x\)
=-32x+25
b: \(\left(x-2\right)^3-6\left(x+4\right)\left(x-4\right)-\left(x-2\right)\left(x^2+2x+4\right)\)
\(=x^3-6x^2+12x-8-\left(x^3-8\right)-6\left(x^2-16\right)\)
\(=-6x^2+12x-6x^2+96=-12x^2+12x+96\)
c: \(\left(x-1\right)^2-2\left(x-1\right)\left(x+2\right)+\left(x+2\right)^2+5\left(2x-3\right)\)
\(=\left(x-1-x-2\right)^2+5\left(2x-3\right)\)
\(=\left(-3\right)^2+5\left(2x-3\right)\)
\(=9+10x-15=10x-6\)
2:
a: \(\left(2-3x\right)^2-5x\left(x-4\right)+4\left(x-1\right)\)
\(=9x^2-12x+4-5x^2+20x+4x-4\)
\(=4x^2+12x\)
b: \(\left(3-x\right)\left(x^2+3x+9\right)+\left(x-3\right)^3\)
\(=27-x^3+x^3-9x^2+27x-27\)
\(=-9x^2+27x\)
c: \(\left(x-4\right)^2\left(x+4\right)-\left(x-4\right)\left(x+4\right)^2+3\left(x^2-16\right)\)
\(=\left(x-4\right)\left(x+4\right)\left(x-4-x-4\right)+3\left(x^2-16\right)\)
\(=\left(x^2-16\right)\left(-8\right)+3\left(x^2-16\right)\)
\(=-5\left(x^2-16\right)=-5x^2+80\)
Bài 1:
\(1,\left(x+y\right)^2-\left(x-y\right)^2=\left[\left(x+y\right)-\left(x-y\right)\right]\left[\left(x+y\right)+\left(x-y\right)\right]=\left(x+y-x+y\right)\left(x+y+x-y\right)=2y.2x=4xy\)
\(2,\left(x+y\right)^3-\left(x-y\right)^3-2y^3\)
\(=x^3+3x^2y+3xy^2+y^3-x^3+3x^2y-3xy^2+y^3-2y^3\)
\(=6x^2y\)
\(3,\left(x+y\right)^2-2\left(x+y\right)\left(x-y\right)+\left(x-y\right)^2\\ =\left[\left(x+y\right)-\left(x-y\right)\right]^2\\ =\left(x+y-x+y\right)^2\\ =4y^2\)
\(4,\left(2x+3\right)^2-2\left(2x+3\right)\left(2x+5\right)+\left(2x+5\right)^2\\ =\left[\left(2x+3\right)-\left(2x+5\right)\right]^2\\ =\left(2x+3-2x-5\right)^2\\ =\left(-2\right)^2\\ =4\)
\(5,9^8.2^8-\left(18^4+1\right)\left(18^4-1\right)\\ =18^8-\left[\left(18^4\right)^2-1\right]\\ =18^8-18^8+1\\ =1\)
1: =x^2+2xy+y^2-x^2+2xy-y^2=4xy
2: =x^3+3x^2y+3xy^2+y^3-x^3+3x^2y-3xy^2+y^3-2y^3
=6x^2y
3: =(x+y-x+y)^2=(2y)^2=4y^2
4: =(2x+3-2x-5)^2=(-2)^2=4
5: =18^8-18^8+1=1
Lời giải:
a.
$A=(x+6)^2-(x+2)^2+2[(x-5)^2-(x-3)^2]$
$=(x+6-x-2)(x+6+x+2)+2[(x-5-x+3)(x-5+x-3)]$
$=4(2x+8)+2(-2)(2x-8)$
$=4(2x+8)-4(2x-8)=4[(2x+8)-(2x-8)]=4.16=64$ không phụ thuộc vào $x$
b.
$B=(x^3-2^3)-(x^3+2^3)=-16$ không phụ thuộc vào $x$
c.
$C=x^4+2x^2-[(x^2+3)^2-(2x)^2]$
$=x^4+2x^2-(x^4+6x^2-4x^2)$
$=x^4+2x^2-(x^4+2x^2)=0$ không phụ thuộc vào $x$
a) Ta có: \(A=\left(x+6\right)^2+2\left(x-5\right)^2-\left(x+2\right)^2-2\left(x-3\right)^2\)
\(=x^2+12x+36+2\left(x^2-10x+25\right)-\left(x^2+4x+4\right)-2\left(x^2-6x+9\right)\)
\(=x^2+12x+36+2x^2-20x+50-x^2-4x-4-2x^2+12x-18\)
\(=34\)
b) Ta có: \(B=\left(x-2\right)\left(x^2+2x+4\right)-\left(x+2\right)\left(x^2-2x+4\right)\)
\(=x^3-8-x^3-8\)
=-16
c) Ta có: \(C=x^4+2x^2-\left(x^2-2x+3\right)\left(x^2+2x+3\right)\)
\(=x^4+2x^2-\left[\left(x^2+3\right)^2-4x^2\right]\)
\(=x^4+2x^2-\left(x^4+6x^2+9\right)+4x^2\)
\(=-9\)
a) \(\left(x-3\right)^2+2\left(x-3\right)\left(x+2\right)+\left(x+2\right)^2\)
\(=\left(x-3+x+2\right)^2\)
\(=\left(2x-1\right)^2\)
Hằng đẳng thức: \(\left(a+b\right)^2=a^2+2ab+b^2\).
b) \(\left(x+5\right)^2-\left(2x+10\right)\left(x-6\right)+\left(x-6\right)^2\)
\(=\left(x+5\right)^2-2\left(x+5\right)\left(x-6\right)+\left(x-6\right)^2\)
\(=\left[\left(x+5\right)-\left(x-6\right)\right]^2\)
\(=11^2=121\)
Hằng đẳng thức: \(\left(a-b\right)^2=a^2-2ab+b^2\).
a.\(\left(x-3\right)^2+2\left(x-3\right)\left(x+2\right)+\left(x+2\right)^2\)
\(=\left[\left(x-3\right)+\left(x+2\right)\right]^2\)
\(=\left(x-3+x+2\right)^2\)
\(=\left(2x-1\right)^2\)
b.\(\left(x+5\right)^2-\left(2x+10\right)\left(x-6\right)+\left(x-6\right)^2\)
\(=\left(x+5\right)^2-2\left(x+5\right)\left(x-6\right)+\left(x-6\right)^2\)
\(=\left[\left(x+5\right)-\left(x-6\right)\right]^2\)
\(=\left(x+5-x+6\right)^2\)