6-x2-x+22
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Ta có: x( x 2 -6 ) – x - 2 2 = x + 1 3
⇔ x 3 – 6x – x 2 +4x -4 = x 3 + 3 x 2 +3x +1
⇔ 4 x 2 +5x +5 =0
∆ = 5 2 -4.4.5 = 25 - 80 = -55 < 0
Vậy phương trình vô nghiệm
1) \(206^2-36=206^2-6^2=\left(206-6\right)\left(206+6\right)\)
\(=200.212=42400\)
2) \(\left(x-4\right)^2.2-\left(12x+x^2\right).2=6\)
\(\Rightarrow x^2-16x+32-24x-2x^2=6\)
\(\Rightarrow40x=26\Rightarrow x=\dfrac{13}{20}\)
Bài 1:
a) \(8xy^2+24x^2y-32x^3y^2=8xy\left(y+3x-4x^2y\right)\)
b) \(x^2-16x-y^2+64=\left(x-8\right)^2-y^2=\left(x-8-y\right)\left(x-8+y\right)\)
Bài 2:
\(\left(x-4\right)^2-\left(12x+x^2\right)=6\)
\(\Rightarrow x^2-8x+16-12x-x^2=6\)
\(\Rightarrow20x=10\Rightarrow x=\dfrac{1}{2}\)
\(1,\\ =8xy\left(y+3x-4x^2y\right)\\ =\left(x-8\right)^2-y^2=\left(x-y-8\right)\left(x+y-8\right)\)
\(2,\Leftrightarrow x^2-8x+16-12x-x^2=6\\ \Leftrightarrow-20x=-10\\ \Leftrightarrow x=2\)
Bài 3:
a) \(4x^2+4x+1=\left(2x+1\right)^2\)
b) \(9x^2-12x+4=\left(3x-2\right)^2\)
c) \(ab^2+\dfrac{1}{4}a^2b^4+1=\left(\dfrac{1}{2}ab^2+1\right)^2\)
\(1,=6xy\left(x^2-2xy+y^2\right)=6xy\left(x-y\right)^2\\ 2,=\left(x^2+4-4\right)\left(x^2+4+4\right)=x^2\left(x^2+8\right)\\ 3,=5x\left(x-y\right)-10\left(x-y\right)=5\left(x-2\right)\left(x-y\right)\\ 4,=\left(a-b\right)\left(a^2+ab+b^2\right)-3\left(a-b\right)=\left(a-b\right)\left(a^2+ab+b^2-3\right)\\ 5,=\left(x-1\right)^2-y^2=\left(x+y-1\right)\left(x-y-1\right)\\ 6,Sửa:x^2-x-2=x^2+x-2x-2=\left(x+1\right)\left(x-2\right)\\ 7,=x^4-4x^2-x^2+4=\left(x^2-4\right)\left(x^2-1\right)\\ =\left(x-2\right)\left(x+2\right)\left(x-1\right)\left(x+1\right)\\ 8,=-x^3-x^2-x=-x\left(x^2+x+1\right)\\ 9,=\left(a-3\right)\left(a^2+3a+9\right)+\left(a-3\right)\left(6a+9\right)\\ =\left(a-3\right)\left(a^2+9a+18\right)\\ =\left(a-3\right)\left(a^2+3a+6a+18\right)\\ =\left(a-3\right)\left(a+3\right)\left(a+6\right)\)
\(10,=x^2y-x^2z+y^2z-xy^2+z^2\left(x-y\right)\\ =xy\left(x-y\right)-z\left(x-y\right)\left(x+y\right)+z^2\left(x-y\right)\\ =\left(x-y\right)\left(xy-xz-yz+z^2\right)\\ =\left(x-y\right)\left(x-z\right)\left(y-z\right)\)
\(\dfrac{x+1}{x-3}-\dfrac{41}{x+3}+\dfrac{x^2+22}{9-x^2}=0\left(ĐKXĐ:x\ne3;x\ne-3\right)\\ \Leftrightarrow\dfrac{x+1}{x-3}-\dfrac{41}{x+3}-\dfrac{x^2+22}{x^2-9}=0\\ \Leftrightarrow\dfrac{\left(x+1\right)\left(x+3\right)-41\left(x-3\right)-x^2-22}{x^2-9}=0\\ \Leftrightarrow x^2+4x+3-41x+123-x^2-22=0\\ \Leftrightarrow-37x+104=0\\ \Leftrightarrow-37x=-104\\ \Leftrightarrow x=\dfrac{104}{37}\left(tmđk\right)\)
Vậy \(x=\dfrac{104}{37}\) là nghiệm của pt.
⇔ [( x 2 +x +1) + (4x -1 )] [( x 2 +x +1) - (4x -1 )]=0
∆ = - 3 2 -4.2.1 = 9 -8 =1 > 0
∆ = 1 =1
x 2 + 3 x + 2 2 = 6.( x 2 +3x +2)
⇔ x 2 + 3 x + 2 2 - 6.( x 2 +3x +2)=0
⇔ ( x 2 +3x + 2)[ ( x 2 +3x + 2) -6] =0
⇔ ( x 2 +3x + 2) .( x 2 +3x -4 )=0
x 2 +3x + 2 =0
Phương trình có dạng a –b +c =0 nên x 1 = -1 , x 2 =-2
x 2 +3x -4 =0
Phương trình có dạng a +b +c =0 nên x 1 = 1 , x 2 = -4
Vậy phương trình đã cho có 4 nghiệm :
x 1 = -1 , x 2 =-2 ; x 3 = 1 , x 4 =-4
\(=-x^2-x+2\)