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Bậc thầy Hóa học
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Giải Danh dự "Thử thách cùng Box Hóa 2017"
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Tìm x, biết:
a/
x 2 x^{2} x 2 +x = 6
b/ 6
x 3 x^{3} x 3 +
x 2 x^{2} x 2 = 2x
a)
x 2 + x = 6 ⇔ x 2 + x − 6 = 0 ⇔ x 2 + 3 x − 2 x − 6 = 0 ⇔ x ( x + 3 ) − 2 ( x + 3 ) = 0 ⇔ ( x − 2 ) ( x + 3 ) = 0 ⇔ x = 2 h o a c x = − 3 x^2+x = 6\\\Leftrightarrow x^2+x-6=0\\\Leftrightarrow x^2+3x-2x-6=0\\\Leftrightarrow x(x+3)-2(x+3)=0\\\Leftrightarrow (x-2)(x+3)=0 \\\Leftrightarrow x=2 hoac x=-3 x 2 + x = 6 ⇔ x 2 + x − 6 = 0 ⇔ x 2 + 3 x − 2 x − 6 = 0 ⇔ x ( x + 3 ) − 2 ( x + 3 ) = 0 ⇔ ( x − 2 ) ( x + 3 ) = 0 ⇔ x = 2 h o a c x = − 3
Vậy........
b)
6 x 3 + x 2 = 2 x ⇔ 6 x 3 + x 2 − 2 x = 0 ⇔ 6 x 3 − 3 x 2 + 4 x 2 − 2 x = 0 ⇔ 3 x 2 ( 2 x − 1 ) + 2 x ( 2 x − 1 ) = 0 ⇔ ( 3 x 2 + 2 x ) ( 2 x − 1 ) = 0 ⇔ x ( 3 x + 2 ) ( 2 x − 1 ) = 0 ⇔ x = 0 h o a c x = − 2 3 h o a c x = 1 2 6x^3 +x^2 = 2x\\\Leftrightarrow 6x^3+x^2-2x=0\\\Leftrightarrow 6x^3-3x^2+4x^2-2x=0\\\Leftrightarrow 3x^2(2x-1)+2x(2x-1)=0\\\Leftrightarrow (3x^2+2x)(2x-1)=0\\\Leftrightarrow x(3x+2)(2x-1)=0\\\Leftrightarrow x=0 hoac x=\frac{-2}{3} hoac x=\frac{1}{2} 6 x 3 + x 2 = 2 x ⇔ 6 x 3 + x 2 − 2 x = 0 ⇔ 6 x 3 − 3 x 2 + 4 x 2 − 2 x = 0 ⇔ 3 x 2 ( 2 x − 1 ) + 2 x ( 2 x − 1 ) = 0 ⇔ ( 3 x 2 + 2 x ) ( 2 x − 1 ) = 0 ⇔ x ( 3 x + 2 ) ( 2 x − 1 ) = 0 ⇔ x = 0 h o a c x = 3 − 2 h o a c x = 2 1
Vậy.......
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