giai phuong trinh \(\left(x^2-4x+3\right)\left(x^2-6x+8\right)=8\)
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\(\left(3x^2-x+1\right)\left(x-1\right)+x^2\left(4-3x\right)\)
= \(3x^2\left(x-1\right)-x\left(x-1\right)+1\left(x-1\right)+4x^2-3x^3\)
= \(3x^3-3x^2-x^2+x+x-1+4x^2-3x^3\)
= \(\left(3x^3-3x^3\right)+\left(-3x^2-x^2+4x^2\right)+\left(x+x\right)-1=2x-1\)
a) 5x2 + 10y2 - 6xy - 4x - 2y + 3
= ( x2 - 6xy + 9y2 ) + ( 4x2 - 4x + 1 ) + ( y2 - 2y + 1 ) + 1
= ( x - 3y )2 + ( 2x - 1 )2 + ( y - 1 )2 + 1 ≥ 1 > 0 ∀ x, y, z
=> đpcm
b) x2 + 4y2 + z2 - 2x - 6z + 8y + 15
= ( x2 - 2x + 1 ) + ( 4y2 + 8y + 4 ) + ( z2 - 6z + 9 ) + 1
= ( x - 1 )2 + ( 2y + 2 )2 + ( z - 3 )2 + 1 ≥ 1 > 0 ∀ x, y, z
=> đpcm
2x^2 + 3x - ( x^2 + 3x + 2 ) = 6
2x^2 + 3x - x^2 - 3x - 2 = 6
x^2 - 2 = 6
x^2 = 8
x = +- căn 8
Bài 1.
a) ( 3x + 4y )2 = ( 3x )2 + 2.3x.4y + ( 4y )2 = 9x2 + 24xy + 16y2
b) ( x2 + 1 )2 = ( x2 )2 + 2.x2.1 + 12 = x4 + 2x2 + 1
c) ( 3 - 2y )2 = 32 - 2.3.2y + ( 2y )2 = 9 - 12y + 4y2
d) ( xy2 - 2 )2 = ( xy2 )2 - 2.xy2.2 + 22 = x2y4 - 4xy2 + 4
Bài 2.
a) x2 - 9 = x2 - 32 = ( x - 3 )( x + 3 )
b) 25 - 4y2 = 52 - ( 2y )2 = ( 5 - 2y )( 5 + 2y )
c) 9x4 - 4y2 = ( 3x2 )2 - ( 2y )2 = ( 3x2 - 2y )( 3x2 + 2y )
d) ( x + 1 )2 - y2 = ( x - y + 1 )( x + y + 1 )
B1:
a) \(\left(3x+4y\right)^2=\left(3x\right)^2+2.3x.4y+\left(4y\right)^2=9x^2+24xy+16y^2\)
b) \(\left(x^2+1\right)^2=\left(x^2\right)^2+2.x^2.1+1^2=x^4+2x^2+1\)
c) \(\left(3-2y\right)^2=3^2-2.3.2y+\left(2y\right)^2=9-12y+4y^2\)
d) \(\left(xy^2-2\right)^2=\left(xy^2\right)^2-2.xy^2.2+2^2=xy^4-4xy^2+4\)
B2:
a) \(x^2-9=x^2-3^2=\left(x-3\right)\left(x+3\right)\)
b) \(25-4y^2=5^2-\left(2y\right)^2=\left(5-2y\right)\left(5+2y\right)\)
c) \(9x^4-4y^2=\left(3x^2\right)^2-\left(2y\right)^2=\left(3x^2-2y\right)\left(3x^2+2y\right)\)
d) \(\left(x+1\right)^2-y^2=\left(x+1-y\right)\left(x+1+y\right)\)
\(\left(x^2-4x+3\right)\left(x^2-6x+8\right)=8\)
\(\left(x^2-3x-x+3\right)\left(x^2-4x-2x+8\right)=8\)
\(\left[x\left(x-3\right)-1\left(x-3\right)\right]\left[x\left(x-4\right)-2\left(x-4\right)\right]=8\)
\(\left(x-1\right)\left(x-3\right)\left(x-2\right)\left(x-4\right)=8\)
\(\left(x-1\right)\left(x-4\right)\left(x-2\right)\left(x-3\right)=8\)
\(\left(x^2-5x+4\right)\left(x^2-5x+6\right)-8=0\)
Đặt \(t=x^2-5x+4\)
\(t\left(t+2\right)-8=0\)
\(t^2+2t-8=0\)
\(t^2+4t-2t-8=0\)
\(t\left(t+4\right)-2\left(t+4\right)=0\)
\(\left(t+4\right)\left(t-2\right)=0\)
\(\orbr{\begin{cases}t+4=0\\t-2=0\end{cases}}\)
\(\orbr{\begin{cases}t=-4\\t=2\end{cases}}\)
\(\orbr{\begin{cases}x^2-5x+4=-4\\x^2-5x+4=2\end{cases}}\)
\(\orbr{\begin{cases}x^2-5x+8=0\left(ptvn\right)\\x^2-5x+2=0\end{cases}}\)
\(x^2-5x+2=0\)
\(\orbr{\begin{cases}x=\frac{5+\sqrt{17}}{2}\\x=\frac{5-\sqrt{17}}{2}\end{cases}}\)