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Câu 4:
a: ĐKXĐ: \(x\notin\left\{0;-5\right\}\)
b: \(A=\dfrac{x^2+2x}{2\left(x+5\right)}+\dfrac{x-5}{x}+\dfrac{50-5x}{2x\left(x+5\right)}\)
\(=\dfrac{x^3+2x^2}{2x\left(x+5\right)}+\dfrac{2\left(x^2-25\right)}{2x\left(x+5\right)}+\dfrac{50-5x}{2x\left(x+5\right)}\)
\(=\dfrac{x^3+2x^2+2x^2-50+50-5x}{2x\left(x+5\right)}\)
\(=\dfrac{x^3+4x^2-5x}{2x\left(x+5\right)}=\dfrac{x\left(x^2+4x-5\right)}{2x\left(x+5\right)}\)
\(=\dfrac{x\left(x+5\right)\left(x-1\right)}{2x\left(x+5\right)}=\dfrac{x-1}{2}\)
c: Để A=-3 thì x-1=-6
hay x=-5(loại)
\(a,\left(x+1\right)^2-\left(x-1\right)^2-3\left(x+1\right)\left(x-1\right)\)
\(=x^2+2x+1-\left(x^2-2x+1\right)-3\left(x^2-1\right)\)
\(=x^2+2x+1-x^2+2x-1-3x^2+2=-3x^2+4x+2\)\(b,5\left(x+2\right)\left(x-2\right)-\left(2x-3\right)^2-x^2+17\)
\(=5\left(x^2-4\right)-\left(4x^2-12x+9\right)-x^2+17\)
\(=5x^2-20-4x^2+12x-9-x^2+17=12x-12\)
\(a,2x^2+8x+5\)
\(=\left(\sqrt{2}x\right)^2+2.\sqrt{2}x.\dfrac{8}{2\sqrt{2}}+\left(\dfrac{8}{2\sqrt{2}}\right)^2-\left(\dfrac{8}{2\sqrt{2}}\right)^2+5\)
\(=\left[\left(\sqrt{2}x\right)^2+2.\sqrt{2}x.\dfrac{8}{2\sqrt{2}}+\left(\dfrac{8}{2\sqrt{2}}\right)^2\right]-\left(\dfrac{8}{2\sqrt{2}}\right)^2+5\)
\(=\left(\sqrt{2}x+\dfrac{8}{2\sqrt{2}}\right)^2-3\)
Ta có :
\(\left(\sqrt{2}x+\dfrac{8}{2\sqrt{2}}\right)^2\ge0\forall x\)
\(\Rightarrow\left(\sqrt{2}x+\dfrac{8}{2\sqrt{2}}\right)^2-3\ge-3>0\)
Dấu = xảy ra khi \(\sqrt{2}x+\dfrac{8}{2\sqrt{2}}=0\Rightarrow x=-2\)
Các câu còn lại dễ rồi mk ko lm nx nha bn ,bn ko bt lm cỗ nào thì hỏi mk
\(z^4-4z^3+z^2+4z^2-4z+1\)
\(=z^4-4z^3+z^2+4z^2-4z+1\)
\(=\left(z^4-4z^3+z^2\right)+\left(4z^2-4z+1\right)\)
\(=z^2\left(z^2-4z+1\right)+\left(4z^2-4z+1\right)\)
\(=z^2\left(z^2-4z+1\right)+\left[\left(2z\right)^2-2.2z.1+1^2\right]\)
\(=z^2\left(z-1\right)^2+\left(2z-1\right)^2\)
Ta có :
\(z^2\left(z-1\right)^2\ge0;\left(2z-1\right)^2\ge0\)
\(\Rightarrow z^2\left(z-1\right)^2+\left(2z-1\right)^2\ge0\) Dấu = xảy ra khi \(\left\{{}\begin{matrix}z-1=0\\2z-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}z=1\\z=\dfrac{1}{2}\end{matrix}\right.\)
Đặt tính \(2n^2-n+2\) : \(2n+1\) sẽ bằng n - 1 dư 3
Để chia hết thì 3 phải chia hết cho 2n + 1 hay 2n + 1 là ước của 3
Ư(3) = {\(\pm\) 3; \(\pm\) 1}
\(2n+1=1\Leftrightarrow2n=0\Leftrightarrow n=0\)
\(2n+1=-1\Leftrightarrow2n=-2\Leftrightarrow n=-1\)
\(2n+1=3\Leftrightarrow2n=2\Leftrightarrow n=1\)
\(2n+1=-3\Leftrightarrow2n=-4\Leftrightarrow n=-2\)
Vậy \(n=\left\{0;-2;\pm1\right\}\)
\(A=3x^2-12x+10\\ A=3x^2-12x+12-2\\ A=\left(3x^2-12x+12\right)-2\\ A=3\left(x^2-4x+4\right)-2\\ A=3\left(x^2-2\cdot x\cdot2+2^2\right)-2\\ A=3\left(x-2\right)^2-2\\ Do\left(x-2\right)^2\ge0\forall x\\ \Rightarrow3\left(x-2\right)^2\ge0\forall x\\ \Rightarrow A=3\left(x-2\right)^2-2\ge-2\forall x\\ \text{Dấu “=” xảy ra khi : }\\ \left(x-2\right)^2=0\\ \Leftrightarrow x-2=0\\ \Leftrightarrow x=2\\ \text{ Vậy }A_{\left(Min\right)}=-2\text{ khi }x=2\)
A=3x2 - 12x + 10
A= (3x2- 2.3x.2+22)-22+10
A= (3x-2)2+6 \(\ge\) +6
Vậy min A = 6 . Dấu = xảy ra khi 3x -2 = 0
3x= 2
x= \(\dfrac{2}{3}\)