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Bài 1 :
a, \(A=x\left(x-6\right)+10\)
=x^2 - 6x + 10
=x^2 - 2.3x+9+1
=(x-3)^2 +1 >0 Với mọi x dương
a) \(x^2\left(x-3\right)+12-4x=x^2\left(x-3\right)-4x+12\)
\(=x^2\left(x-3\right)-4\left(x-3\right)\)
\(=\left(x-3\right)\left(x^2-4\right)\)
\(=\left(x-3\right)\left(x^2-2^2\right)\)
\(=\left(x+3\right)\left(x-2\right)\left(x+2\right)\)
b)\(x^2-4+\left(x-2\right)^2=x^2-2^2+\left(x-2\right)^2\)
\(=\left(x-2\right)\left(x+2\right)+\left(x-2\right)^2\)
\(=\left(x-2\right)\left(x+2+x-2\right)\)
\(=\left(x-2\right)2x\)
c)\(x^3-4x^2-12x+27=x^3+3x^2-7x^2-21x+9x+27\)
\(=x^2\left(x+3\right)-7x\left(x+3\right)+9\left(x+3\right)\)
\(=\left(x+3\right)\left(x^2-7x+9\right)\)
\(=\left(x+3\right)\left(x^2-7x+9\right)\)
a) => x2.(x-3)-4(x-3)=(x-3)(x2-4)=(x-3)(x-2)(x+2)
b) => (x+2)(x-2)+(x-2)2=(x-2)(x+2+x-2)=2x(x-2)
c) => x3+27-(4x2+12x)=(x+3)(x2-3x+3)-4x(x+3)=(x+3)(x2-3x+3-4x)=(x-3)(x2-7x+3)
Ta có :
\(4x\left(x-1\right)-3\left(x^2-5\right)-x^2=\left(x-3\right)-\left(x+4\right)\)
\(\Leftrightarrow\)\(4x^2-4x-3x^2+15=x-3-x-4\)
\(\Leftrightarrow\)\(x^2-4x+15=-7\)
\(\Leftrightarrow\)\(\left(x^2-2.x.2+2^2\right)+11=-7\)
\(\Leftrightarrow\)\(\left(x-2\right)^2=-18\)
Mà \(\left(x-2\right)^2\ge0\) \(\left(\forall x\inℝ\right)\)
\(\Rightarrow\)\(x\in\left\{\varnothing\right\}\)
Vậy không có giá trị nào của x thoã mãn đề bài
Chúc bạn học tốt ~
a) \(x^3-\frac{1}{4}x=x\left(x^2-\frac{1}{4}\right)=x\left(x-\frac{1}{2}\right)\left(x+\frac{1}{2}\right)\)
b) \(\left(2x-1\right)^2-\left(x+3\right)^2=\left(2x-1-x-3\right)\left(2x-1+x+3\right)=\left(x-4\right)\left(3x+2\right)\)
c) \(x^2-y^2-2y-1=x^2-\left(y^2+2y+1\right)=x^2-\left(y+1\right)^2=\left(x-y-1\right)\left(x+y+1\right)\)
d) \(x^2\left(x-3\right)+12-4x=x^2\left(x-3\right)-4\left(x-3\right)=\left(x-3\right)\left(x^2-2^2\right)=\left(x-3\right)\left(x-2\right)\left(x+2\right)\)
Phép tính b):
Đặt 2x - 1 = a ; x + 3 = b. Từ đầu bài suy ra:
\(\left(2x-1\right)^2-\left(x+3\right)^2\Rightarrow a^2-b^2\)
\(\Rightarrow a^2-b^2-\left(ab-ab\right)\Rightarrow\left(a^2-ab\right)-\left(b^2-ab\right)\)
\(\Rightarrow a\left(a-b\right)-b\left(b-a\right)\Rightarrow a\left(a-b\right)+b\left(a-b\right)\)
\(\Rightarrow\left(a+b\right)\left(a-b\right)\)
Thế lại vào ta có:
\(\orbr{\begin{cases}a+b=\left(2x-1\right)+\left(x+3\right)=\left(2x+x\right)-\left(1-3\right)=3x+2\\a-b=\left(2x-1\right)-\left(x-3\right)=\left(2x-x\right)-\left(1-3\right)=x+2\end{cases}}\)
\(\Rightarrow\left(a+b\right)\left(a-b\right)=\left(3x+2\right)\left(x+2\right)\)
\(\Leftrightarrow4x^2+x-8x-2-4x^2-27x=1\)
=>-34x=3
hay x=-3/34
\(\left(x-2\right)\left(4x+1\right)-4x\left(x+7\right)=1\\ \Rightarrow4x^2-8x+x-2-4x^2-28x=1\\ \Rightarrow-35x=3\\ \Rightarrow x=\dfrac{-3}{35}\)