Hỏi ngu tí!

Chứ bạn này là ai mà sống lâu vầy !!

\(\dfrac{x}{x^2-x+1}=\dfrac{1}{5}\)

\(\Rightarrow\dfrac{x^2-x+1}{x}=5\)

\(\Leftrightarrow x-1+\dfrac{1}{x}=5\)

\(\Rightarrow x+\dfrac{1}{x}=6\)

\(\dfrac{x^4+x^2+1}{x^2}=x^2+\dfrac{1}{x^2}+1=\left(x+\dfrac{1}{x}\right)^2-1=6-1=5\)Vậy \(\dfrac{x^2}{x^4+x^2+1}=\dfrac{1}{5}\)

\(D=5x^2-10x-2\)

\(=5\left(x^2-2x+1\right)-7\)

\(=5\left(x-1\right)^2-7\ge-7\)

Vậy \(min_D=-7\)

Để D = -7 thì \(x-1=0\Rightarrow x=1\)

\(E=x^2-2xy+2y^2+y-3\)

\(=\left(x^2-2xy+y^2\right)+\left(y^2+y+\dfrac{1}{4}\right)-\dfrac{13}{4}\)

\(=\left(x-y\right)^2+\left(y+\dfrac{1}{2}\right)^2-\dfrac{13}{4}\ge\dfrac{13}{4}\)

Vậy \(min_E=\dfrac{-13}{4}\)

Để \(E=-\dfrac{13}{4}\) thì \(\left\{{}\begin{matrix}x-y=0\\y+\dfrac{1}{2}=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=y=-\dfrac{1}{2}\\y=-\dfrac{1}{2}\end{matrix}\right.\)

\(x^2-2x^2+x-xy^2=-x^2+x-xy^2=-x\left(x+1-y^2\right)\)Mình k biết đề bài y cầu cái j hết:)

\(\left(x-3\right)^3+\left(x-2\right)^2+\left|x-1\right|+x=2013\)

Xét 2 trường hợp :

TH1: \(\left|x-1\right|\ge1\) với \(x\ge1\)

Ta có:

\(\left(x-3\right)^3+\left(x-2\right)^2+x+1+x=2013\) (1)

\(\Leftrightarrow x^3-9x^2+27x-27+x^2-4x+4+x-1+x=2013\)\(\Leftrightarrow x^3-8x^2+25x-2037=0\)

Đặt \(t=x-\dfrac{8}{3}\) ta được:

\(\left(1\right)\Leftrightarrow\left(t+\dfrac{8}{3}\right)^3-8\left(t+\dfrac{8}{3}\right)^2+25\left(t+\dfrac{8}{3}\right)-2037=0\)\(\Leftrightarrow t^3+\dfrac{11}{3}t-\dfrac{54223}{27}=0\) (2)

Đặt \(y=\dfrac{t}{2.\dfrac{\sqrt{11}}{3}}=\dfrac{t}{\dfrac{2\sqrt{11}}{3}}\Rightarrow t=\dfrac{2\sqrt{11}}{3}y\)

Khi đó: \(\left(2\right)\Leftrightarrow\left(\dfrac{2\sqrt{11}}{3}y\right)^3+\dfrac{11}{3}.\dfrac{2\sqrt{11}}{3}y-\dfrac{54223}{27}=0\)\(\Leftrightarrow\dfrac{22\sqrt{11}}{27}\left(4y^3+3y-743,129529\right)=0\)

\(\Leftrightarrow4y^3+3y-743,129529=0\) (3)

Giả sử \(y_0\) là nghiệm của phương trình , khi đó:

- Với \(y>y_0\) thì :

\(\left\{{}\begin{matrix}4y^3>4y_0^3\\3y>3y_0\end{matrix}\right.\Rightarrow4y^3+3y>4y_0^3+3y_0=743,129529\)\(\Rightarrow\) pt vô nghiệm

- Với \(y< y_0\) thì

\(\left\{{}\begin{matrix}4y^3< 4y^3_0\\3y< 3y_0\end{matrix}\right.\Rightarrow4y^3+3y< 4y_0^3+3y_0=743,129529\)\(\Rightarrow\) pt vô nghiệm

Vậy (3) nếu có nghiệm \(y_0\) thì nghiệm đó là duy nhất.

Đặt : \(a=\sqrt[3]{743,129529+\sqrt{743,129529^2+1}}\)

\(\alpha=\dfrac{1}{2}\left(a-\dfrac{1}{a}\right)\) ta được:

\(4\alpha^3+3\alpha=743,129529\) \(\Rightarrow\alpha=y\)là nghiệm của pt

\(\Rightarrow y=\dfrac{1}{2}\cdot\left(\sqrt[3]{743,129529+\sqrt{743,129529^2+1}}+\dfrac{1}{\sqrt[3]{743,129529+\sqrt{743,129529^2+1}}}\right)\)\(=5,662228051\)

\(\Rightarrow\) \(t=\dfrac{2\sqrt{11}}{3}.5662228051=12,51965728\)

\(\Rightarrow x=12,51965728+\dfrac{8}{3}=15,18632395\)

TH2: \(\left|x-1\right|< 1\) với x < 1

Ta có:

\(\left(x-3\right)^3+\left(x-2\right)^2-x+1+x=2013\) (1)

\(\Leftrightarrow x^3-9x^2+27x-27+x^2-4x+4-x+1+x-2013=0\)\(\Leftrightarrow x^3-8x^2+23x-2035=0\)

Đặt \(t=x-\dfrac{8}{3}\Rightarrow x=t+\dfrac{8}{3}\) , ta đươc:

\(\left(1\right)\Leftrightarrow\left(t+\dfrac{8}{3}\right)^3-8\left(t+\dfrac{8}{3}\right)^2+23\left(t+\dfrac{8}{3}\right)-2035=0\)\(\Leftrightarrow t^3+\dfrac{5}{3}t-\dfrac{54313}{27}=0\) (2)

Đặt \(y=\dfrac{t}{2.\sqrt{\dfrac{5}{9}}}=\dfrac{t}{\dfrac{2\sqrt{5}}{3}}\Rightarrow t=\dfrac{2\sqrt{5}}{3}y\)

Khi đó :

\(\left(2\right)\Leftrightarrow\left(\dfrac{2\sqrt{5}}{3}y\right)^3+\dfrac{5}{3}.\dfrac{2\sqrt{5}}{3}-\dfrac{54313}{27}=0\)

\(\Leftrightarrow4y^3+3y=2428,951201\) (3)

Chứng minh (3) có 1 nghiệm duy nhất ( cmtren)

Đặt \(a=\sqrt[3]{2428,951201+\sqrt{2428,951201^2+1}}\)

\(\alpha=\dfrac{1}{2}\left(a-\dfrac{1}{a}\right)\) , ta được:

\(4\alpha^3-3\alpha=2428,951201\Leftrightarrow y=\alpha\) là nghiệm của pt

Khi đó:

\(y=\dfrac{1}{2}\left(\sqrt[3]{2428,951201+\sqrt{2428,951201^2+1}}+\dfrac{1}{\sqrt[3]{2428,951201+\sqrt{2428,951201^2+1}}}\right)\)=\(8,438583197\)

\(\Rightarrow t=\dfrac{2\sqrt{5}}{3}.8,438583197=12,57949826\)

\(\Rightarrow x=12,57949826+\dfrac{8}{3}=15,24616493\)

Bài 2:

\(a+\dfrac{1}{a}\ge2\)

\(\Leftrightarrow\dfrac{a^2+1}{a}\ge2\)

\(\Leftrightarrow\dfrac{\left(a^2-2a+1\right)+2a}{a}\ge2\)

\(\Leftrightarrow\dfrac{\left(a-1\right)^2+2a}{a}\ge2\)

\(\Leftrightarrow\dfrac{\left(a-1\right)^2}{a}+2\ge2\)

\(\Leftrightarrow\dfrac{\left(a-1\right)^2}{a}\ge0\)

Vì \(a>0\Rightarrow\dfrac{\left(a+1\right)^2}{a}\ge0\Rightarrowđpcm\)

b, \(\sqrt{\dfrac{a+b}{2}}\ge\dfrac{\sqrt{a}+\sqrt{b}}{2}\)

\(\Leftrightarrow\dfrac{a+b}{2}\ge\dfrac{\left(\sqrt{a}+\sqrt{b}\right)^2}{4}\)

\(\Leftrightarrow\dfrac{a+b}{2}-\dfrac{a+2\sqrt{ab}+b}{4}\ge0\)

\(\Leftrightarrow\dfrac{2a+2b-a-2\sqrt{ab}-b}{4}\ge0\)

\(\Leftrightarrow\dfrac{a-2\sqrt{ab}+b}{a}\ge0\)

\(\Leftrightarrow\dfrac{\left(\sqrt{a}-\sqrt{b}\right)^2}{4}\ge0\)

\(\Leftrightarrow\left(\dfrac{\sqrt{a}-\sqrt{b}}{2}\right)^2\ge0\)

Đúng với \(a\ge0;b\ge0\)

d, \(\dfrac{a^2+2}{\sqrt{a^2+1}}\)

Ta có:

\(a^2+2=\left(\sqrt{a^2+1}\right)^2+1\) nên

\(\Rightarrow\dfrac{a^2+2}{\sqrt{a^2+1}}=\dfrac{\left(\sqrt{a^2+1}\right)^2+1}{\sqrt{a^2+1}}=\sqrt{a^2+1}+\dfrac{1}{\sqrt{a^2+1}}\ge2\)( chứng minh phần a)

\(P=\left(\dfrac{\sqrt{x}+2}{\sqrt{x}-1}-\dfrac{x+4}{x-1}\right):\dfrac{1}{\sqrt{x}-1}\)

ĐKXĐ:

\(x\ne1\)

\(P=\left(\dfrac{\sqrt{x}+2}{\sqrt{x}-1}-\dfrac{x+4}{\left(\sqrt{x}\right)^2-1}\right):\dfrac{1}{\sqrt{x}-1}\)

\(P=\left(\dfrac{\sqrt{x}+2}{\sqrt{x}-1}-\dfrac{x+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right):\dfrac{1}{\sqrt{x}-1}\)\(P=\left(\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\dfrac{x+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right):\dfrac{1}{\sqrt{x}-1}\)\(P=\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}+1\right)-x-4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}.\left(\sqrt{x}-1\right)\)

\(P=\dfrac{x+3\sqrt{x}+2-x-4}{\sqrt{x}+1}\)

\(P=\dfrac{3\sqrt{x}-4}{\sqrt{x}+1}\)

\(=\dfrac{3\sqrt{x}+3-5}{\sqrt{x}+1}\)

\(=\dfrac{3\left(\sqrt{x}+1\right)-5}{\sqrt{x}+1}=3-\dfrac{5}{\sqrt{x}+1}\)

Với mọi giá trị của x ta có:

\(\sqrt{x}\ge0\Rightarrow\sqrt{x}+1\ge1\Rightarrow\dfrac{5}{\sqrt{x}+1}\le5\)

\(\Rightarrow P\ge3-5=-2\)

Vậy \(Min_P=-2\)

Để P = -2 thì \(\sqrt{x}+1=1\Rightarrow\sqrt{x}=0\Rightarrow x=0\)

\(a,4x^2-\left(3x+1\right)\left(2x-1\right)=2\left(x-3\right)^2\)

\(\Leftrightarrow4x^2-\left(6x^2-3x+2x-1\right)=2\left(x^2-6x+9\right)\)

\(\Leftrightarrow4x^2-6x^2+x+1-2x^2+12x-18=0\)

\(\Leftrightarrow-4x^2+13x-17=0\)

\(\Leftrightarrow-4\left(x^2-\dfrac{13}{4}x+\dfrac{169}{64}\right)-\dfrac{103}{16}=0\)

\(\Leftrightarrow-4\left(x-\dfrac{13}{8}\right)^2=\dfrac{103}{16}\)

\(\Leftrightarrow\left(x-\dfrac{13}{8}\right)^2=\dfrac{-103}{64}\Rightarrow\) pt vô nghiệm

\(b,\left(5x-1\right)\left(x+1\right)-\left(2x-1\right)\left(2x+1\right)=x.\left(x+1\right)\)\(\Leftrightarrow5x^2+5x-x-1-\left(4x^2-1\right)=x^2+x\)

\(\Leftrightarrow5x^2+5x-x-1-4x^2+1-x^2-x=0\) \(\Leftrightarrow3x=0\Rightarrow x=0\)

\(c,7x^2-\left(2x-3\right)^2=1+3\left(x+2\right)^2\)

\(\Leftrightarrow7x^2-\left(4x^2-12x+9\right)=1+3\left(x^2+4x+4\right)\)

\(\Leftrightarrow7x^2-4x^2+12x-9=1+3x^2+12x+12\)\(\Leftrightarrow7x^2-4x^2+12x-9-1-3x^2-12x-12=0\)\(\Leftrightarrow-22=0\) ( vô lí)

Vậy phương trình vô nghiệm