a/ \(2\left(x^2-3x+2\right)=3\sqrt{x^3+8}\)

\(\Rightarrow2x^2-6x+4=3\sqrt{\left(x+2\right)\left(x^2-2x+4\right)}\)

\(\Rightarrow\left(-2\right)\left(x+2\right)+2\left(x^2-2x+4\right)=3\sqrt{\left(x+2\right)\left(x^2-2x+4\right)}\)

Chia 2 vế cho x² - 2x + 4 ta được:

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

Đặt \(a=\sqrt{\frac{x+2}{x^2-2x+4}}\left(a\ge0\right)\) ta được:

\(-2a^2-3a+2=0\Rightarrow\left(1-2a\right)\left(a+2\right)=0\Rightarrow\orbr{\begin{cases}a=\frac{1}{2}\left(n\right)\\a=-2\left(l\right)\end{cases}}\)

\(a=\frac{1}{2}\Leftrightarrow\sqrt{\frac{x+2}{x^2-2x+4}}=\frac{1}{2}\Rightarrow\frac{x+2}{x^2-2x+4}=\frac{1}{4}\)

\(\Rightarrow x^2-6x-4=0\Rightarrow\orbr{\begin{cases}x=3+\sqrt{13}\\x=3-\sqrt{13}\end{cases}}\) (cái này tính denta là ra kết quả thôi)

                                                        Vậy có 2 nghiệm trên

câu b, c tương tự thôi

1/ Đặt \(\hept{\begin{cases}\sqrt{x-2013}=a\\\sqrt{x-2014}=b\end{cases}}\)

Thì ta có:

\(\frac{\sqrt{x-2013}}{x+2}+\frac{\sqrt{x-2014}}{x}=\frac{a}{a^2+2015}+\frac{b}{b^2+2014}\)

\(\le\frac{a}{2a\sqrt{2015}}+\frac{b}{2b\sqrt{2014}}=\frac{1}{2\sqrt{2015}}+\frac{1}{2\sqrt{2014}}\)

2/ \(\frac{x}{2x+y+z}+\frac{y}{x+2y+z}+\frac{z}{x+y+2z}\)

\(\le\frac{1}{4}\left(\frac{x}{x+y}+\frac{x}{x+z}+\frac{y}{y+x}+\frac{y}{y+z}+\frac{z}{z+x}+\frac{z}{z+y}\right)\)

\(=\frac{3}{4}\)

Bài 1:

1.

\((x^2-6x)^2-2(x-3)^2+2=0\)

\(\Leftrightarrow (x^2-6x)^2-2(x^2-6x+9)+2=0\)

\(\Leftrightarrow (x^2-6x)^2-2(x^2-6x)-16=0\)

Đặt $x^2-6x=a$ thì pt trở thành:

$a^2-2a-16=0$

$\Leftrightarrow a=1\pm \sqrt{17}$

Nếu $a=1+\sqrt{17}$

$\Leftrightarrow x^2-6x=1+\sqrt{17}$

$\Leftrightarrow (x-3)^2=10+\sqrt{17}$

$\Rightarrow x=3\pm \sqrt{10+\sqrt{17}}$

Nếu $a=1-\sqrt{17}$

$\Rightarrow x=3\pm \sqrt{10-\sqrt{17}}$

Vậy.........

2.

$x^4-2x^3+x=2$

$\Leftrightarrow x^3(x-2)+(x-2)=0$

$\Leftrightarrow (x-2)(x^3+1)=0$

$\Leftrightarrow (x-2)(x+1)(x^2-x+1)=0$

Thấy rằng $x^2-x+1=(x-\frac{1}{2})^2+\frac{3}{4}>0$ nên $(x-2)(x+1)=0$

$\Rightarrow x=2$ hoặc $x=-1$

Vậy.......

Bài 2:

1.

ĐKXĐ: $x\neq 1$. Ta có:

\(x^2+(\frac{x}{x-1})^2=8\)

\(\Leftrightarrow x^2+(\frac{x}{x-1})^2+\frac{2x^2}{x-1}=8+\frac{2x^2}{x-1}\)

\(\Leftrightarrow (x+\frac{x}{x-1})^2=8+\frac{2x^2}{x-1}\)

\(\Leftrightarrow (\frac{x^2}{x-1})^2=8+\frac{2x^2}{x-1}\)

Đặt $\frac{x^2}{x-1}=a$ thì pt trở thành:

$a^2=8+2a$

$\Leftrightarrow (a-4)(a+2)=0$

Nếu $a=4\Leftrightarrow \frac{x^2}{x-1}=4$

$\Rightarrow x^2-4x+4=0\Leftrightarrow (x-2)^2=0\Rightarrow x=2$ (tm)

Nếu $a=-2\Leftrightarrow \frac{x^2}{x-1}=-2$

$x^2+2x-2=0\Rightarrow x=-1\pm \sqrt{3}$ (tm)

Vậy........

2. ĐKXĐ: $x\neq 0; 2$

$(\frac{x-1}{x})^2+(\frac{x-1}{x-2})^2=\frac{40}{49}$

$\Leftrightarrow (\frac{x-1}{x}+\frac{x-1}{x-2})^2-\frac{2(x-1)^2}{x(x-2)}=\frac{40}{49}$

$\Leftrightarrow 4\left[\frac{(x-1)^2}{x(x-2)}\right]^2-\frac{2(x-1)^2}{x(x-2)}=\frac{40}{49}$

Đặt $\frac{(x-1)^2}{x(x-2)}=a$ thì pt trở thành:

$4a^2-2a=\frac{40}{49}$

$\Rightarrow 2a^2-a-\frac{20}{49}=0$

$\Rightarrow a=\frac{7\pm \sqrt{209}}{28}$

$\Leftrightarrow 1+\frac{1}{x(x-2)}=\frac{7\pm \sqrt{209}}{28}$

$\Leftrightarrow \frac{1}{x(x-2)}=\frac{-21\pm \sqrt{209}}{28}$

$\Rightarrow x(x-2)=\frac{28}{-21\pm \sqrt{209}}$

$\Rightarrow (x-1)^2=\frac{7\pm \sqrt{209}}{-21\pm \sqrt{209}}$.

Dễ thấy $\frac{7+\sqrt{209}}{-21+\sqrt{209}}< 0$ nên vô lý

Do đó $(x-1)^2=\frac{7-\sqrt{209}}{-21-\sqrt{209}}$

$\Leftrightarrow x=1\pm \sqrt{\frac{7-\sqrt{209}}{-21-\sqrt{209}}}$

Vậy........

Bài 3:

ĐKXĐ: $x\neq 0$

PT $\Leftrightarrow (x-\frac{1}{x})^2+2=4-(x-\frac{1}{x})$

Đặt $x-\frac{1}{x}=a$ thì pt trở thành:

$a^2+2=4-a$

$\Leftrightarrow a^2+a-2=0$

$\Leftrightarrow (a-1)(a+2)=0\Rightarrow a=1$ hoặc $a=-2$

Nếu $a=1\Leftrightarrow x^2-1=x$

$\Leftrightarrow x^2-x-1=0\Rightarrow x=\frac{1\pm \sqrt{5}}{2}$

Nếu $a=-2\Leftrightarrow x^2-1=-2x$

$\Leftrightarrow x^2+2x-1=0\Rightarrow x=-1\pm \sqrt{2}$

Vậy............

2. ĐKXĐ: $x\neq 0$

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

$\Rightarrow 4x^2+\frac{1}{x^2}=8x-\frac{4}{x}+4$

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

Đặt $2x-\frac{1}{x}=a$ thì pt trở thành:

$a^2+4=4a+4$

$\Leftrightarrow a(a-4)=0\Rightarrow a=0$ hoặc $a=4$

Nếu $a=0\Leftrightarrow 2x-\frac{1}{x}=0$

$\Rightarrow 2x=\frac{1}{x}\Rightarrow x=\pm \sqrt{\frac{1}{2}}$

Nếu $a=4\Rightarrow 2x^2-4x-1=0\Rightarrow x=\frac{2\pm \sqrt{6}}{2}$

a/ \(\left\{{}\begin{matrix}x^2+y+xy\left(x^2+y\right)+xy+1=-\frac{1}{4}\\x^4+y^2+2x^2y+xy+1=-\frac{1}{4}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(x^2+y+1\right)\left(xy+1\right)=-\frac{1}{4}\\\left(x^2+y\right)^2+xy+1=-\frac{1}{4}\end{matrix}\right.\)

Đặt \(\left\{{}\begin{matrix}x^2+y=a\\xy+1=b\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\left(a+1\right)b=-\frac{1}{4}\\a^2+b=-\frac{1}{4}\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\left(a+1\right)b=-\frac{1}{4}\\b=-\frac{1}{4}-a^2\end{matrix}\right.\)

\(\Rightarrow\left(a+1\right)\left(-\frac{1}{4}-a^2\right)=-\frac{1}{4}\)

\(\Leftrightarrow4a^3+4a^2+a=0\Leftrightarrow a\left(2a+1\right)^2=0\)

\(\Rightarrow\left[{}\begin{matrix}a=0\Rightarrow b=-\frac{1}{4}\\a=-\frac{1}{2}\Rightarrow b=-\frac{1}{2}\end{matrix}\right.\)

TH1: \(\left\{{}\begin{matrix}x^2+y=0\\xy+1=-\frac{1}{4}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}y=-x^2\\-x^3=-\frac{5}{4}\end{matrix}\right.\) \(\Rightarrow...\)

TH2: \(\left\{{}\begin{matrix}x^2+y=-\frac{1}{2}\\xy+1=-\frac{1}{2}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}y=-\frac{1}{2}-x^2\\x\left(-\frac{1}{2}-x^2\right)=-\frac{5}{4}\end{matrix}\right.\) \(\Rightarrow...\)

b/ ĐKXĐ; ...

\(\Leftrightarrow\left\{{}\begin{matrix}x^3+3x^2+3x+1-16x-16=\frac{8}{y^3}-\frac{8}{y}\\5\left(x^2+2x+2\right)=1+\frac{4}{y^2}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(x+1\right)^3-16\left(x+1\right)=\frac{8}{y^3}-\frac{8}{y}\\5\left(x+1\right)^2=\frac{4}{y^2}-4\end{matrix}\right.\)

Đặt \(\left\{{}\begin{matrix}x+1=a\\\frac{1}{y}=b\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a^3-16a=8b^3-8b\\5a^2=4b^2-4\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a^3-8b^3=16a-8b\\4=-5a^2+4b^2\end{matrix}\right.\)

Nhân vế với vế:

\(4\left(a^3-8b^3\right)=4\left(4a-2b\right)\left(-5a^2+4b^2\right)\)

\(\Leftrightarrow21a^3-10a^2b-16ab^2=0\)

\(\Leftrightarrow a\left(21a^2-10ab-16b^2\right)=0\)

\(\Leftrightarrow a\left(7a-8b\right)\left(3a+2b\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}a=0\\7a=8b\\3a=-2b\end{matrix}\right.\) \(\Rightarrow...\)

\(5\sqrt{2x^3+16}=2\left(x^2+8\right)\left(x>-2\right)\)

\(\Leftrightarrow20\sqrt{\left(x+2\right)\left(x^2-2x+4\right)}=2\left(x^2+8\right)\)

\(\Leftrightarrow2\left(x^2+8\right)-20\sqrt{\left(x+2\right)\left(x^2-2x+4\right)}=0\)

\(\Leftrightarrow x^2+8-10\sqrt{x+2}\sqrt{x^2-2x+4}=0\)

\(\Leftrightarrow x^2-2x+4+2x+4-10\sqrt{x+2}\sqrt{x^2-2x+4}=0\)

Đặt a = \(\sqrt{x^2-2x+4}\left(a>0\right)\)

      b = \(\sqrt{x+2}\left(b\ge0\right)\)

=> pt có dạng:

\(a^2-10ab+b^2=0\)

bạn phân tích rồi làm tiếp nhá

b, \(\sqrt[3]{24+x}+\sqrt{12-x}=6\) (đk \(-24\le x\le12\)) (*)

Đặt \(\sqrt[3]{24+x}=a\) , \(\sqrt{12-x}=b\left(b\ge0\right)\)

Có \(a^3+b^2=24+x+12-x=36\)(1)

a+b=6 => b=6-a

Thay b=6-a vào (1) có:

\(a^3+\left(6-a\right)^2=36\)

<=> \(a^3+a^2-12a+36=36\)

<=> \(a^3+a^2-12a=0\)

<=> \(a\left(a^2+a-12\right)=0\)

<=> \(a\left(a^2-3a+4a-12\right)=0\)

<=> \(a\left(a+4\right)\left(a-3\right)=0\)

=>\(\left[{}\begin{matrix}a=0\\a=-4\\a=3\end{matrix}\right.\)<=> \(\left[{}\begin{matrix}24+x=0\\24+x=-4^3=-64\\24+x=3^3=27\end{matrix}\right.\)<=>\(\left[{}\begin{matrix}x=-24\\x=-88\\x=3\end{matrix}\right.\)(tm pt(*))

Vậy pt (*) có tập nghiệm \(S=\left\{-24,-88,3\right\}\)

a) \(4\sqrt{x}+\frac{2}{\sqrt{x}}< 2x+\frac{1}{2x}+2\)

hay \(2\sqrt{x}+\frac{1}{\sqrt{x}}< x+\frac{1}{4x}+1\)

\(\Leftrightarrow0< x+\frac{1}{4x}+1-2\sqrt{x}-\frac{1}{\sqrt{x}}\)

\(\Leftrightarrow0< \left(\sqrt{x}\right)^2-2\sqrt{x}-2\sqrt{x}\cdot1+1+\frac{1}{\left(2\sqrt{x}\right)^2}-2\cdot\frac{1}{2\sqrt{x}}\)

\(\Leftrightarrow1< \left(\sqrt{x}-1\right)^2+\left(\frac{1}{2\sqrt{x}}-1\right)^2\)

\(\Rightarrow\hept{\begin{cases}x>0\\\sqrt{x}>1\\2\sqrt{x}>1\end{cases}\Rightarrow\hept{\begin{cases}x>1\\x>\frac{1}{4}\end{cases}\Rightarrow}x>1}\)

b) \(\frac{1}{1-x^2}>\frac{3}{\sqrt{1-x^2}}-1\left(1\right)\left(ĐK:-1< x< 1\right)\)

Ta có (1) <=> \(\frac{1}{1-x^2}-1-\frac{3x}{\sqrt{1-x^2}}+2>0\)\(\Leftrightarrow\frac{x^2}{1-x^2}-\frac{3x}{\sqrt{1-x^2}}+2>0\)

Đặt \(t=\frac{x}{\sqrt{1-x^2}}\)ta được

\(t^2-3t+2>0\Leftrightarrow\orbr{\begin{cases}\frac{x}{\sqrt{1-x^2}}< 1\\\frac{x}{\sqrt{1-x^2}}>2\end{cases}\Leftrightarrow\orbr{\begin{cases}\sqrt{1-x^2}>x\left(a\right)\\2\sqrt{1-x^2}< x\left(b\right)\end{cases}}}\)

(a) <=> \(\hept{\begin{cases}x< 0\\1-x^2>0\end{cases}\Leftrightarrow\hept{\begin{cases}x\ge0\\1-x^2>x^2\end{cases}}}\)

\(\Leftrightarrow-1< x< 0\)hoặc \(\hept{\begin{cases}x\ge0\\x^2< \frac{1}{2}\end{cases}}\)

\(\Leftrightarrow-1< x< 0\)hoặc \(0\le x\le\frac{\sqrt{2}}{2}\Leftrightarrow-1< x< \frac{\sqrt{2}}{2}\)

(b) \(\Leftrightarrow\hept{\begin{cases}1-x^2>0\\x>0\\4\left(1-x^2\right)< x^2\end{cases}\Leftrightarrow\hept{\begin{cases}0< x< 1\\x^2>\frac{4}{5}\end{cases}\Leftrightarrow}\frac{2}{\sqrt{5}}< x< 1}\)

ok đợi nấu ăn xong r làm cho

a) điều kiện x>0

khi đó

\(\left(a\right)\Leftrightarrow4\left(\sqrt{4}+\frac{1}{2\sqrt{x}}\right)< 2\left(\sqrt{x}+\frac{1}{2\sqrt{x}}\right)^2\)

\(\Leftrightarrow\sqrt{x}+\frac{1}{2\sqrt{x}}>2\Leftrightarrow2x-4\sqrt{x}+1>0\)

\(\Leftrightarrow\orbr{\begin{cases}\sqrt{x}< \frac{2-\sqrt{2}}{2}\\\sqrt{x}>\frac{2+\sqrt{2}}{2}\end{cases}}\)

1) ĐK: \(x\ge-2012\)

Đặt \(\sqrt{x+2012}=t\left(t\ge0\right)\Rightarrow x=t^2-2012\)

Ta có hệ \(\hept{\begin{cases}x^2+t=2012\\-x+t^2=2012\end{cases}}\)

\(\Rightarrow x^2+t-t^2+x=0\Rightarrow\left(x+t\right)\left(x-t+1\right)=0\)

Với \(x+t=0\Leftrightarrow\sqrt{x+2012}=x\Rightarrow x^2-x-2012=0\Rightarrow x=\frac{\sqrt{8049}+1}{2}\)

Với \(x-t+1=0\Leftrightarrow\sqrt{x+2012}=x+1\Rightarrow x^2+x-2011=0\Rightarrow x=\frac{\sqrt{8045}-1}{2}\)

2) ĐK \(\orbr{\begin{cases}x< -\frac{1}{3}\\x>1\end{cases}}\)

Đặt \(\sqrt{\frac{3x+1}{x-1}}=t\), phương trình trở thành \(4t+\frac{1}{t}=4\Rightarrow\frac{4t^2-4t+1}{t}=0\Rightarrow t=\frac{1}{2}\)

Khi đó ta có \(\sqrt{\frac{3x+1}{x-1}}=\frac{1}{2}\Rightarrow\frac{3x+1}{x-1}=\frac{1}{4}\Rightarrow11x+5=0\)

\(\Rightarrow x=-\frac{5}{11}\left(tm\right)\)

c) TH1: \(x\le-1\), phương trình trở thành \(\left(x-3\right)\left(x+1\right)-4\sqrt{\left(x-3\right)\left(x+1\right)}+3=0\)

Đặt \(\sqrt{\left(x-3\right)\left(x+1\right)}=t\left(t\ge0\right)\) thì \(t^2-4t+3=0\Rightarrow\orbr{\begin{cases}t=1\\t=3\end{cases}}\)

Với \(t=1\Rightarrow\left(x-3\right)\left(x+1\right)=1\Rightarrow x^2-2x-4=0\Rightarrow\orbr{\begin{cases}x=1+\sqrt{5}\left(l\right)\\x=1-\sqrt{5}\left(tm\right)\end{cases}}\)

Với \(t=3\Rightarrow\left(x-3\right)\left(x+1\right)=9\Rightarrow x^2-2x-12=0\Rightarrow\orbr{\begin{cases}x=1+\sqrt{13}\left(l\right)\\x=1-\sqrt{13}\left(tm\right)\end{cases}}\)

Với \(x>3\), phương trình trở thành \(\left(x-3\right)\left(x+1\right)+4\sqrt{\left(x-3\right)\left(x+1\right)}+3=0\)

Đặt \(\sqrt{\left(x-3\right)\left(x+1\right)}=t\left(t\ge0\right)\) thì \(t^2+4t+3=0\Rightarrow\orbr{\begin{cases}t=-1\\t=-3\end{cases}\left(l\right)}\)

Vậy pt có 2 nghiệm \(x=1-\sqrt{5}\) hoặc \(x=1-\sqrt{13}\)