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a, x2 + y2
= x2 + 2y + y2 - 2xy
= (x + y)2 - 2xy
b, x3 + y3
= x3 + 3x2y + 3xy2 + y3 - 3x2y - 3xy2
= (x + y)3 - 3xy(x + y)
a) Biến đổi VP :
\(\left(x+y\right)^2-2xy\)
\(=x^2+2xy+y^2-2xy\)
\(=x^2+y^2\left(=VT\right)\left(đpcm\right)\)
b) Biến đổi vế phải :
\(\left(x+y\right)^3-3xy\left(x+y\right)\)
\(=x^3+3x^2y+3xy^2+y^3-3xy\left(x+y\right)\)
\(=x^3+3xy\left(x+y\right)+y^3-3xy\left(x+y\right)\)
\(=x^3+y^3\left(=VT\right)\left(đpcm\right)\)
a, 2x + 4 = 2( x + 2)
b, 5x - 20 = 5x - 5.4 = 5(x - 4)
c, x^2 + x = x.x + x = x( x + 1)
d, 3x^2y + 6xy^2 = 3xy( x + 2y)
a ) \(4a^2b^2-\left(a^2+b^2-c^2\right)^2\)
\(=\left(2ab+a^2+b^2-c^2\right)\left(2ab-a^2-b^2+c^2\right)\)
\(=\left[\left(a+b\right)^2-c^2\right]\left[c^2-\left(a^2-2ab+b^2\right)\right]\)
\(=\left(a+b-c\right)\left(a+b+c\right)\left[c^2-\left(a-b\right)^2\right]\)
\(=\left(a+b-c\right)\left(a+b+c\right)\left(c-a+b\right)\left(c+a-b\right)\)
b ) \(x^2\left(x+4\right)^2-\left(x+4\right)^2-\left(x^2-1\right)\)
\(=\left(x^2-1\right)\left(x+4\right)^2-\left(x^2-1\right)\)
\(=\left(x^2-1\right)\left[\left(x+4\right)^2-1\right]\)
\(=\left(x-1\right)\left(x+1\right)\left(x+4-1\right)\left(x+4+1\right)\)
\(=\left(x-1\right)\left(x+1\right)\left(x+3\right)\left(x+5\right)\)
A) x2 - 4x + 4 = (x - 2)2 (hằng đẳng thức số 2)
Cm : x2 - 4x + 4 = x2 - 2x - 2x + 4 = x(x - 2) - 2(x - 2) = (x - 2)(x - 2) = (x - 2)2
b tương tự
1/x-1/y-1/z = 1
<=>(1/x-1/y-1/z)^2 = 1
<=> 1/x^2+1/y^2+1/z^2+2.(-1/xy+1/yz-1/zx) = 1
<=> 1/x^2+1/y^2+1/z^2 = 1-2.(-z+x-y/xyz) = 1-2.(x-y-z/xyz) = 1-2.0 = 1
=> ĐPCM
k mk nha
a, 3x3 . 5x2 = 15x5
b, 2x . ( 3x2 + 2x ) = 6x3 + 4x2
c, -3xy . ( 2x + 5y ) = -6x2y +-15xy2
d, 3x2. ( 6 - x2 + 2x ) = 18x2 - 3x3 + 6x3
e, ( x + 2 ) . ( x + 3 ) = x2 + 5x + 6
i, ( x - 4 ) . ( x + 4 ) = x2 - 16
h, ( 1 - 2x ) . ( 3x + 2 ) = 2 - 6x2 - x
k, ( x - y ) . ( x + y ) = x2 - y2
t, ( 2x + 1 ) . ( 4x2 - 2x + 1 ) = 8x3 - 1
a, 3x3 . 5x2 = 15x5
b, 2x . ( 3x2 + 2x ) = 6x3 + 4x2
c, -3xy . ( 2x + 5y ) = -6x2y +-15xy2
d, 3x2. ( 6 - x2 + 2x ) = 18x2 - 3x3 + 6x3
e, ( x + 2 ) . ( x + 3 ) = x2 + 5x + 6
i, ( x - 4 ) . ( x + 4 ) = x2 - 16
h, ( 1 - 2x ) . ( 3x + 2 ) = 2 - 6x2 - x
k, ( x - y ) . ( x + y ) = x2 - y2
t, ( 2x + 1 ) . ( 4x2 - 2x + 1 ) = 8x3 - 1
\(x^2=x-20\)
\(\Leftrightarrow x^2-x+20=0\)
\(\Leftrightarrow x^2-2.x.\frac{1}{2}+\frac{1}{4}+\frac{79}{4}\)
\(\Leftrightarrow\left(x-\frac{1}{2}\right)^2=-\frac{79}{4}\)
Do \(\left(x-\frac{1}{2}\right)^2\ge0\text{ }\forall\text{ }x\in R\)
Nên pt vô nghiệm
thank s đủ r