đề dài v~

1.

a) \(f\left(x\right)=5x^2-2x+1\)

\(5f\left(x\right)=25x^2-10x+5\)

\(5f\left(x\right)=\left(25x^2-10x+1\right)+4\)

\(5f\left(x\right)=\left(5x-1\right)^2+4\)

Mà  \(\left(5x-1\right)^2\ge0\)

\(\Rightarrow5f\left(x\right)\ge4\)

\(\Leftrightarrow f\left(x\right)\ge\frac{4}{5}\)

Dấu " = " xảy ra khi :

\(5x-1=0\Leftrightarrow x=\frac{1}{5}\)

Vậy ....

b)  \(P\left(x\right)=3x^2+x+7\)

\(3P\left(x\right)=9x^2+3x+21\)

\(3P\left(x\right)=\left(9x^2+3x+\frac{1}{4}\right)+\frac{83}{4}\)

\(3P\left(x\right)=\left(3x+\frac{1}{2}\right)^2+\frac{83}{4}\)

Mà  \(\left(3x+\frac{1}{2}\right)^2\ge0\)

\(\Rightarrow3P\left(x\right)\ge\frac{83}{4}\)

\(\Leftrightarrow P\left(x\right)\ge\frac{83}{12}\)

Dấu "=" xảy ra khi :

\(3x+\frac{1}{2}=0\Leftrightarrow x=-\frac{1}{6}\)

Vậy ...

c)  \(Q\left(x\right)=5x^2-3x-3\)

\(5Q\left(x\right)=25x^2-15x-15\)

\(\Leftrightarrow5Q\left(x\right)=\left(25x^2-15x+\frac{9}{4}\right)-\frac{69}{4}\)

\(\Leftrightarrow5Q\left(x\right)=\left(5x-\frac{3}{2}\right)^2-\frac{69}{4}\)

Mà  \(\left(5x-\frac{3}{2}\right)^2\ge0\)

\(\Rightarrow5Q\left(x\right)\ge\frac{-69}{4}\)

\(\Leftrightarrow Q\left(x\right)\ge-\frac{69}{20}\)

Dấu "=" xảy ra khi :

\(5x-\frac{3}{2}=0\Leftrightarrow x=0,3\)

Vậy ...

2.

a)  \(f\left(x\right)=-3x^2+x-2\)

\(-3f\left(x\right)=9x^2-3x+6\)

\(-3f\left(x\right)=\left(9x^2-3x+\frac{1}{4}\right)+\frac{23}{4}\)

\(-3f\left(x\right)=\left(3x-\frac{1}{2}\right)^2+\frac{23}{4}\)

Mà  \(\left(3x-\frac{1}{2}\right)^2\ge0\)

\(\Rightarrow-3f\left(x\right)\ge\frac{23}{4}\)

\(\Leftrightarrow f\left(x\right)\le\frac{23}{12}\)

Dấu "=" xảy ra khi :

\(3x-\frac{1}{2}=0\Leftrightarrow x=\frac{1}{6}\)

Vậy ...

b)  \(P\left(x\right)=-x^2-7x+1\)

\(-P\left(x\right)=x^2+7x-1\)

\(-P\left(x\right)=\left(x^2+7x+\frac{49}{4}\right)-\frac{53}{4}\)

\(-P\left(x\right)=\left(x+\frac{7}{2}\right)^2-\frac{53}{4}\)

Mà  \(\left(x+\frac{7}{2}\right)^2\ge0\)

\(\Rightarrow-P\left(x\right)\ge-\frac{53}{4}\)

\(\Leftrightarrow P\left(x\right)\le\frac{53}{4}\)

Dấu "=" xảy ra khi :

\(x+\frac{7}{2}=0\Leftrightarrow x=-\frac{7}{2}\)

Vậy ...

c)  \(Q\left(x\right)=-2x^2+x-8\)

\(-2Q\left(x\right)=4x^2-2x+16\)

\(-2Q\left(x\right)=\left(4x^2-2x+\frac{1}{4}\right)+\frac{63}{4}\)

\(-2Q\left(x\right)=\left(2x-\frac{1}{2}\right)^2+\frac{63}{4}\)

Mà :  \(\left(2x-\frac{1}{2}\right)^2\ge0\)

\(\Rightarrow-2Q\left(x\right)\ge\frac{63}{4}\)

\(\Leftrightarrow Q\left(x\right)\le-\frac{63}{8}\)

Dấu "=" xảy ra khi :

\(2x-\frac{1}{2}=0\Leftrightarrow x=\frac{1}{4}\)

Vậy ...

\(A=\frac{x^2-4x+1}{x^2}=\frac{x^2-4x+4-3}{x^2}=\frac{\left(x-2\right)^2-3}{x^2}\)

Ta có: \(x^2>0\Rightarrow GTNN\) của (x-2)²-3 có giá trị âm

=> (x-2)² > hoặc = 0 => GTNN của tử số  là - 3

Khi đó: (x-2)² = 0 <=> x - 2 = 0 <=> x = 2

=> Mẫu số: 2² = 4

Vậy GTNN_A = -3/4 khi x = 2

\(A=\frac{x^2-4x+1}{x^2}\)

\(A=\frac{x^2}{x^2}-\frac{4x}{x^2}+\frac{1}{x^2}\)

\(A=1-\frac{4}{x}+\frac{1}{x^2}\)

\(A=\left(\frac{1}{x^2}-\frac{4}{x}+4\right)-3\)

\(A=\left(\frac{1}{x}-2\right)^2-3\)

Mà  \(\left(\frac{1}{x}-2\right)^2\ge0\forall x\)

\(\Rightarrow A\ge-3\)

Dấu "=" xảy ra khi :

\(\frac{1}{x}-2=0\Leftrightarrow\frac{1}{x}=2\Leftrightarrow x=\frac{1}{2}\)

Vậy  \(A_{Min}=-3\Leftrightarrow x=\frac{1}{2}\)

Câu hỏi của Hoàng Lê Minh - Toán lớp 8 - Học toán với OnlineMath

\(P=\left(\frac{x}{x^2-25}-\frac{x-5}{x^2+5x}\right):\frac{10x-25}{x^2+5x}+\frac{x}{5-x}\)

\(=\left[\frac{x}{\left(x-5\right)\left(x+5\right)}-\frac{x-5}{x\left(x+5\right)}\right]:\frac{10x-25}{x^2+5x}+\frac{x}{5-x}\)

\(=\left[\frac{x^2}{x\left(x-5\right)\left(x+5\right)}-\frac{\left(x-5\right)^2}{x\left(x-5\right)\left(x+5\right)}\right]:\frac{10x-25}{x^2+5x}+\frac{x}{5-x}\)

\(=\frac{x^2-\left(x^2-10x+25\right)}{x\left(x-5\right)\left(x+5\right)}:\frac{10x-25}{x\left(x+5\right)}+\frac{x}{5-x}\)

\(=\frac{10x-25}{x\left(x-5\right)\left(x+5\right)}.\frac{x\left(x+5\right)}{10x-25}+\frac{x}{5-x}\)

\(=\frac{1}{x-5}-\frac{x}{x-5}\)

\(=\frac{1-x}{x-5}=-\frac{x-1}{x-5}=-\frac{x-5+4}{x-5}=-1-\frac{4}{x-5}\)

Để P nguyên <=> x - 5 thuộc Ư(4) = {1;-1;2;-2;4;-4}

Ta có bảng:

Vậy....

\(ĐKXĐ:x\ne0;x\ne\pm5;x\ne\frac{5}{2}\)