Tìm x: \(x^3\)+\(2x^2\)-13x+10=0
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\(a,x^2+\frac{1}{x^2}=\left(x+\frac{1}{x}\right)^2-2=a^2-2\)
\(x^3+\frac{1}{x^3}=\left(x+\frac{1}{x}\right)^3-3\left(x+\frac{1}{x}\right)=a^3-3a\)
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Làm mẫu 1 phần nếu ko bít thì hỏi
Ta có: \(x-y=m\)
\(\Rightarrow\left(x-y\right)^2=m^2\)
\(\Leftrightarrow x^2-2xy+y^2=m^2\)
\(\Leftrightarrow x^2+y^2-2n=m^2\)
\(\Leftrightarrow x^2+y^2=m^2+2n\)
a) 4x4 - 37x2 + 9 = (4x4 - 36x2) - (x2 - 9)
= 4x2(x2 - 9) - (x2 - 9)
= (4x2 - 1)(x2 - 9)
= (2x - 1)(2x + 1)(x - 3)(x + 3)
b) x4 - 13x2 + 36
= x4 - 4x2 - 9x2 + 36
= x2(x2 - 4) - 9(x2 - 4)
= (x2 - 9)(X2 - 4)
= (x - 3)(x + 3)(x - 2)(x + 2)
c) x4 - 8x2 + 7
= x4 - 7x2 - x2 + 7
= x2(x2 - 7) - (x2 - 7)
= (x2 - 1)(x2 - 7)
= (x - 1)(x + 1)(x2 - 7)
d) x4 - 7x2y2 + 12y4
= x4 - 3x2y2 - 4x2y2 + 12y4
= x2(x2 - 3y2) - 4y2(x2 - 3y2)
= (x2 - 4y2)(x2 - 3y2)
= (x - 2y)(x + 2y)(x2 - 3y2)
Bài làm :
a) 4x4 - 37x2 + 9 = (4x4 - 36x2) - (x2 - 9)
= 4x2(x2 - 9) - (x2 - 9)
= (4x2 - 1)(x2 - 9)
= (2x - 1)(2x + 1)(x - 3)(x + 3)
b) x4 - 13x2 + 36
= x4 - 4x2 - 9x2 + 36
= x2(x2 - 4) - 9(x2 - 4)
= (x2 - 9)(X2 - 4)
= (x - 3)(x + 3)(x - 2)(x + 2)
c) x4 - 8x2 + 7
= x4 - 7x2 - x2 + 7
= x2(x2 - 7) - (x2 - 7)
= (x2 - 1)(x2 - 7)
= (x - 1)(x + 1)(x2 - 7)
d) x4 - 7x2y2 + 12y4
= x4 - 3x2y2 - 4x2y2 + 12y4
= x2(x2 - 3y2) - 4y2(x2 - 3y2)
= (x2 - 4y2)(x2 - 3y2)
= (x - 2y)(x + 2y)(x2 - 3y2)
a) x2 - 6x + 8 = x2 - 2x - 4x + 8 = x(x - 2) - 4(x - 2) = (x - 4)(x - 2)
b) x2 - 5x - 14 = x2 - 7x + 2x - 14 = x(x - 7) + 2(x - 7) = (x + 2)(x - 7)
c) 4x2 - 36x + 56 = 4(x2 - 9x + 14) = 4(x2 - 2x - 7x + 14) = 4[x(x - 2) - 7(x - 2)] = 4(x - 7)(x - 2)
d) 3x2 - 16x + 5 = 3x2 - 15x - x + 5 = 3x(x - 5) - (x - 5) = (3x - 1)(x - 5)
f) 8x2 + 30x + 7 = 8x2 + 16x + 14x + 7 = 8x(x + 2) + 7(x + 2) = (8x + 8)(x + 2)
g) 2x2 - 5xy - 12y2 = 2x2 - 8xy + 3xy - 12y2 = 2x(x - 4y) + 3y(x - 4y) = (2x + 3y)(x - 4y)
h) x2 - 7xy + 10y2 = x2 - 2xy - 5xy + 10y2 = x(x - 2y) - 5y(x - 2y) = (x - 5y)(x - 2y)
\(x^3+2x^2-13x+10=0\)
\(\Rightarrow x^3+2x^2-13x=-10\)
\(\Rightarrow x\times\left(x^2+2x-13\right)=-10\)
\(\Rightarrow x;x^2+2x-13\inƯ\left(10\right)=\left\{\pm1;\pm2;\pm5;\pm10\right\}\)
Mà \(x^2+2x-13\)lẻ
\(\Rightarrow x^2+2x-13\in\left\{\pm5\right\}\)
Lập bảng làm tiếp nhé, em ms lớp 7 nên có gì sai sót mong chị bỏ qua.
~Std well~
#Dư Khả
\(x^3+2x^2-13x+10=0\)
\(\Rightarrow x^3-x^2+3x^2-3x-10x+10=0\)
\(\Rightarrow x^2\left(x-1\right)+3x\left(x-1\right)-10\left(x-1\right)=0\)
\(\Rightarrow\left(x-1\right)\left(x^2+3x-10\right)=0\)