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tôi bt làm 1 câu à mấy câu kia khó quá *-*
1. 5x2+4x-2=0
\(\Leftrightarrow x\left(5x+4\right)=2\)
\(\Leftrightarrow\orbr{\begin{cases}x=2\\5x+4=2\end{cases}\Leftrightarrow\orbr{\begin{cases}x=2\\x=\frac{-2}{5}\end{cases}}}\)
\(\Rightarrow\) Nghiệm pt là :\(S=\left\{\frac{-2}{5};2\right\}\)
chúc bn sớm làm dc bài này ha
Bài 3a)
\(a+b+c=0\Leftrightarrow a+b=-c\Leftrightarrow a^3+b^3+3ab\left(a+b\right)=-c^3\)
\(\Leftrightarrow a^3+b^3+c^3=-3ab\left(a+b\right)\)
mà \(a+b=-c\Rightarrow a^3+b^3+c^3=3abc\)
3, \(=x^4-x^2+3x^2-3\)
\(=x^2\left(x^2-1\right)+3\left(x^2-1\right)\)
\(=\left(x^2+3\right)\left(x-1\right)\left(x+1\right)\)
5, nhận xét : \(\left(a-b\right)^3=a^3-3a^2b+3ab^2-b^3\Rightarrow a^3-b^3=\left(a-b\right)^3+3a^2b-3ab^2\)
thay vào đầu bài ta có: \(\left(a-b\right)^3+c^3+3a^2b-3ab^2+3abc\)
\(=\left(a-b+c\right)\left[\left(a-b\right)^2-\left(a-b\right)c+c^2\right]+3ab\left(a-b+c\right)\)
\(=\left(a-b+c\right)\left(a^2-2ab+b^2-ac+bc+c^2+3ab\right)\)
\(=\left(a-b+c\right)\left(a^2+b^2+c^2+ab-ac+bc\right)\)
3x^2+2x-1
=3x^2+3x-x-1
=3x(x+1)-(x+1)
=(x+1)(3x-1)
x^3+6x^2+11x+6
=x^3+5x^2+6x+x^2+5x+6
=x(x^2+5x+6)+(x^2+5x+6)
=(x+1)(x^2+5x+6)
=(x+1)(x^2+3x+2x+6)
=(x+1)(x+2)(x+3)
x^4+2x^2-3
=x^4-x^2+3x^2-3
=x^2(x^2-1)+3(x^2-1)
=(x^2-1)(x^2+3)
=(x+1)(x-1)(x^2+3)
ab+ac+b^2+2bc+c^2
=a(b+c)+(b+c)^2
=(b+c)(a+b+c)
a^3-b^3+c^3+3abc
=(a-b)^3+3ab(a-b)+c^3+3abc
=(a-b+c)^3-3(a-b)c(a-b+c)+3ab(a-b+c)
=(a-b+c)(a^2+b^2+c^2-2ab+2ac-2bc-3ac+3...
=(a-b+c)(a^2+b^2+c^2+ab+bc-ca)
=1/2.(a-b+c)(a^2+2ab+b^2+b^2+2bc+c^2+c...
=1/2.(a-b+c)[(a+b)^2+(b+c)^2+(c-a)^2]
1)\(3x^2+2x-1=3x^2+3x-x-1=3x\left(x+1\right)-\left(x+1\right)=\left(3x-1\right)\left(x+1\right)\)
2)\(x^3+6x^2+11x+6=x^3+3x^2+3x^2+9x+2x+6\)
\(=x^2\left(x+3\right)+3x\left(x+3\right)+2\left(x+3\right)\)\(=\left(x^2+3x+2\right)\left(x+3\right)\)
\(=\left(x^2+2x+x+2\right)\left(x+3\right)\)\(=\left[x\left(x+2\right)+\left(x+2\right)\right]\left(x+3\right)\)
\(=\left(x+1\right)\left(x+2\right)\left(x+3\right)\)
3)\(x^4+2x^2-3=x^4+3x^2-x^2-3=x^2\left(x^2+3\right)-\left(x^2+3\right)=\left(x^2-1\right)\left(x^3+3\right)\)
\(=\left(x-1\right)\left(x+1\right)\left(x^2+3\right)\)
4)\(ab+ac+b^2+2bc+c^2=a\left(b+c\right)+\left(b+c\right)^2=\left(b+c\right)\left(a+b+c\right)\)
5) câu này sau khi phân tích được (a-b+c)(a2+b2+c2+ab+bc-ac)