Phân thức đại số - Hỏi đáp

\(A=\dfrac{3x^2-6x+17}{x^2-2x+5}=3+\dfrac{2}{x^2-2x+5}\)

Để A đạt giá trị lớn nhất thì \(\dfrac{2}{x^2-2x+5}\) phải đạt giá trị lớn nhất

Để \(\dfrac{2}{x^2-2x+5}\) đạt GTLN thì \(x^2-2x+5\) đạt GTNN

Mà \(x^2-2x+5=\left(x-1\right)^2+4\ge4\)

\(\Rightarrow\dfrac{2}{\left(x-1\right)^2+4}\ge\dfrac{2}{4}=\dfrac{1}{2}\)

\(\Rightarrow A=3+\dfrac{2}{\left(x-1\right)^2+4}\ge3+\dfrac{1}{2}=3.5\)

Vậy Max A =3.5 khi\(\left(x-1\right)^2=0\Rightarrow x=1\)

\(\left(8-5x\right)\left(x+2\right)+4\left(x-2\right)\left(x+1\right)=\left(x-2\right)\left(x+2\right)\)

\(\Leftrightarrow\left(8-5x\right)\left(x+2\right)+4\left(x-2\right)\left(x+1\right)-\left(x-2\right)\left(x+2\right)=0\)

\(\Leftrightarrow\left(8x+16-5x^2-10x\right)+\left(4x-8\right)\left(x+1\right)-\left(x^2-4\right)=0\)

\(\Leftrightarrow\left(8x+16-5x^2-10x\right)+\left(4x^2+4x-8x-8\right)-\left(x^2-4\right)=0\)

\(\Leftrightarrow8x+16-5x^2-10x+4x^2+4x-8x-8-x^2+4=0\)

\(\Leftrightarrow-2x^2-6x+12=0\)

\(\Leftrightarrow x=1,372281323\)

a) \(2x^2+3\left(x-1\right)\left(x+1\right)=5x\left(5-1\right)\)

\(\Leftrightarrow2x^2+3\left(x-1\right)\left(x+1\right)-5x\left(5-1\right)=0\)

\(\Leftrightarrow2x^2+3\left(x^2-1\right)-5x.4=0\)

\(\Leftrightarrow2x^2+3x^2-3-20x=0\)

\(\Leftrightarrow5x^2-20x-3=0\)

\(\Leftrightarrow x=4,144761059\)

a) \(\dfrac{x}{x-3}-\dfrac{x^2+3x}{2x+3}\left(\dfrac{x+3}{x^2-3x}-\dfrac{x}{x^2-9}\right)\)

ĐKXĐ:\(\left\{{}\begin{matrix}x-3\ne0\\2x +3\ne0\\x^2-3x\ne0\\x^2-9\ne0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x\ne3\\x\ne-\dfrac{3}{2}\\x\ne0\\x\ne\pm3\end{matrix}\right.\)

\(=\dfrac{x}{x-3}-\dfrac{x\left(x+3\right)}{2x+3}\left(\dfrac{x+3}{x\left(x-3\right)}-\dfrac{x}{\left(x-3\right)\left(x+3\right)}\right)\)

\(=\dfrac{x}{x-3}-\dfrac{x\left(x+3\right)}{2x+3}.\dfrac{\left(x+3\right)^2-x^2}{x\left(x-3\right)\left(x+3\right)}\)

\(=\dfrac{x}{x-3}-\dfrac{x\left(x+3\right)}{2x+3}.\dfrac{\left(x+3-x\right)\left(x+3+x\right)}{x\left(x-3\right)\left(x+3\right)}\)

\(=\dfrac{x}{x-3}-\dfrac{x\left(x+3\right).3\left(2x+3\right)}{\left(2x+3\right)x\left(x-3\right)\left(x+3\right)}\)

\(=\dfrac{x}{x-3}-\dfrac{3}{x-3}\)

\(=\dfrac{x-3}{x-3}\)

=1

\(\Rightarrow\) ĐPCM

TBR: x² - 4x + 1 = 0
⇔ x² - 4x + 4 = 3
⇔ ( x - 2 )² = 3
⇔ x - 2 = +-\(\sqrt{3}\)
Xét với x = \(\sqrt{3}\):
\(\dfrac{x^4+x^2+1}{x^2}\) = \(\dfrac{9+3+1}{3}\) = \(\dfrac{13}{3}\)
Xét với x = -\(\sqrt{3}\) :
\(\dfrac{x^4+x^2+1}{x^2}\) = \(\dfrac{9+3+1}{3}\) = \(\dfrac{13}{3}\)

Vậy A = \(\dfrac{13}{3}\)

a,\(\Leftrightarrow9x^2+4x-3-9x^2-12x-4>0\)

\(\Leftrightarrow-8x-7>0\)

\(\Leftrightarrow-8x>7\)\(\Leftrightarrow x< -\dfrac{7}{8}\)

c,\(\Leftrightarrow x-1>0\Leftrightarrow x>1\)

Vậy....

Để biểu thức đạt giá trị nguyên thì x +3 \(⋮\) 2x-1

=> 2(x+3) \(⋮\)2x-1

=> 2x -1 + 7 \(⋮\) 2x+1

=> 7 \(⋮\) 2x +1 hay 2x+1 là ước của 7

Vậy x \(\in\left\{3;0;-1;-4\right\}\)

3x. (x²-1)= 3x³-3x

(3x-1).(2x² - x + 1)= 3x(2x² - x + 1) - 1(2x² - x + 1)

= 6x³- 3x²+ 3x - 2x² +x - 1= bước này dễ, tự làm nha

5x.(x²-2x+7) = 5x³- 10x²+ 35x

(3x+4)(x²-2)= 3x(x²-2) +4(x²-2)= 3x³- 6x +4x²-8

Ở câu đầu là \(3x.\left(x^2-1\right)\) nha mọi người

\(\Leftrightarrow x^2-2+\dfrac{1}{x^2}+y^2-2+\dfrac{1}{y^2}=4-2-2\)

\(\Leftrightarrow\left(x-\dfrac{1}{x}\right)^2+\left(y-\dfrac{1}{y}\right)^2=0\)

Với mọi x, y ta luôn có \(\left\{{}\begin{matrix}\left(x-\dfrac{1}{x}\right)^2\ge0\\\left(y-\dfrac{1}{y}\right)^2\ge0\end{matrix}\right.\)

=> \(\left(x-\dfrac{1}{x}\right)^2+\left(y-\dfrac{1}{y}\right)^2\ge0\)

mà \(\left(x-\dfrac{1}{x}\right)^2+\left(y-\dfrac{1}{y}\right)^2=0\)

\(\Rightarrow\left\{{}\begin{matrix}x-\dfrac{1}{x}=0\\y-\dfrac{1}{y}=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x^2-1}{x}=0\\\dfrac{y^2-1}{y}=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1;x=-1\\y=1;y=-1\end{matrix}\right.\)

Vậy....

mk giải luôn đó nha

Giải:

Áp dụng BĐT AM-GM cho hai số dương, ta có:

\(x^2+\dfrac{1}{x^2}\ge2\sqrt{x^2.\dfrac{1}{x^2}}=2\)

\(y^2+\dfrac{1}{y^2}\ge2\sqrt{y^2.\dfrac{1}{y^2}}=2\)

\(\Leftrightarrow x^2+\dfrac{1}{x^2}+y^2+\dfrac{1}{y^2}\ge4\)

Dấu "=" xảy ra khi:

\(x=y=\pm1\)

Vậy ...