Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(a,\left(x^2-4x+11\right)\left(x^4-8x^2+21\right)=35\)
Phương trình trên tương đương với:
\(\left[\left(x-2\right)^2+7\right]\left[\left(x^2-4\right)^2+5\right]=35\left(1\right)\)
Do: \(\hept{\begin{cases}\left(x-2\right)^2+7\ge7\forall x\\\left(x^2-4\right)^2+5\ge5\forall x\end{cases}}\Rightarrow\left[\left(x+2\right)^2+7\right]\left[\left(x^2+4\right)^2+5\right]\ge35\forall x\)
Nên: \(\left(1\right)\Leftrightarrow\hept{\begin{cases}\left(x-2\right)^2+7=7\\\left(x^2-4\right)^2+5=5\end{cases}\Leftrightarrow}x=2\)
Vậy ..................................
\(b,\sqrt{x}+\sqrt{1-x}+\sqrt{x\left(1-x\right)}=1\)
\(Đkxđ:0\le x\le1\) Đặt: \(0< a=\sqrt{x}+\sqrt{1-x}\Rightarrow\frac{a^2-1}{2}=\sqrt{x\left(1-x\right)}\)
\(+)\) Phương trình mới là: \(a+\frac{a^2-1}{2}=1\Leftrightarrow a^2+2a-3=0\Leftrightarrow\left(a-1\right)\left(a+3\right)=0\)
\(\Leftrightarrow a=\left\{-3;1\right\}\Rightarrow a=1>0\)
\(\sqrt{x}+\sqrt{1-x}=1\)
\(+)\) Nếu \(a=1\Leftrightarrow x+1-x+2\sqrt{x\left(1-x\right)}=1\Leftrightarrow\sqrt{x\left(1-x\right)}=0\)
\(\Rightarrow x=\left\{0;1\right\}\left(tm\right)\)
Vậy .............................
\(a,Đk:1\le x\le4\)
Đặt \(y=\sqrt{4-x}+\sqrt{2x-2}\)Ta có: \(y^2=4-x+2x-2+2\sqrt{\left(4-x\right)\left(2x-2\right)}\)
\(\Leftrightarrow x+2+2\sqrt{\left(4-x\right)\left(2x-2\right)}=y^2\Leftrightarrow x+2\sqrt{\left(4-x\right)\left(2x-2\right)}=y^2-2\)
Phương trình trở thành: \(5+y^2-2=4y\)
\(\Leftrightarrow y^2-4y+3=0\)
\(\Leftrightarrow\orbr{\begin{cases}y=1\\y=3\end{cases}}\) ( Vì \(a+b+c=0\))
- \(y=1.\) Ta có: \(\sqrt{4-x}+\sqrt{2x-2}=1\Leftrightarrow\sqrt{2x-2}=1-\sqrt{4-x}\)
\(\Leftrightarrow\hept{\begin{cases}1-\sqrt{4-x}\ge0\\2x-2=\left(1-\sqrt{4-x}\right)^2\end{cases}}\) \(\Leftrightarrow\hept{\begin{cases}\sqrt{4-x}\le1\\2x-2=1-2\sqrt{4-x}+4-x\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}0\le4-x\le1\\2\sqrt{4-x}=7-3x\end{cases}}\) \(\Leftrightarrow\hept{\begin{cases}3\le x\le4;7-3x\ge0\\4\left(4-x\right)=\left(7-3x\right)^2\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x\in\varnothing\\4\left(4-x\right)=\left(7-3x\right)^2\end{cases}}\) \(\Leftrightarrow x\in\varnothing\)
- \(y=3\)Ta có: \(\sqrt{4-x}+\sqrt{2x-2}=3\Leftrightarrow\sqrt{2x-2}=3-\sqrt{4-x}\)
\(\Leftrightarrow\hept{\begin{cases}3-\sqrt{4-x}\ge0\\2x-2=\left(3-\sqrt{4-x}\right)^2\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}\sqrt{4-x}\le3\\2x-2=9-6\sqrt{4-x}+4-x\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}\sqrt{4-x}\le3\\2\sqrt{4-x}=5-x\end{cases}}\) \(\Leftrightarrow\hept{\begin{cases}0\le4-x\le9;5-x\ge0\\4\left(4-x\right)=\left(5-x\right)^2\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}-5\le x\le4\\x^2-6x+9=0\end{cases}}\) \(\Leftrightarrow\hept{\begin{cases}-5\le x\le4\\\left(x-3\right)^2=0\end{cases}}\Leftrightarrow x=3\)
Vậy pt có nghiệm duy nhất là \(x=3\)
(Làm xong hoa mắt :((
Lời giải:
a. ĐKXĐ: $x\geq 0$
$2\sqrt{2x}-5\sqrt{8x}+7\sqrt{18x}=28$
$\Leftrightarrow 2\sqrt{2x}-10\sqrt{2x}+21\sqrt{2x}=28$
$\Leftrightarrow 13\sqrt{2x}=28$
$\Leftrightarrow \sqrt{2x}=\frac{28}{13}$
$\Leftrightarrow 2x=\frac{784}{169}$
$\Leftrightarrow x=\frac{392}{169}$
b. ĐKXĐ: $x\geq 5$
PT $\Leftrightarrow \sqrt{4}.\sqrt{x-5}+\sqrt{x-5}-\frac{1}{3}.\sqrt{9}.\sqrt{x-5}=4$
$\Leftrightarrow 2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}=4$
$\Leftrightarrow 2\sqrt{x-5}=4$
$\Leftrightarrow \sqrt{x-5}=2$
$\Leftrightarrow x-5=4$
$\Leftrightarrow x=9$ (tm)
c. ĐKXĐ: $x\geq \frac{2}{3}$ hoặc $x< -1$
PT $\Leftrightarrow \frac{3x-2}{x+1}=9$
$\Rightarrow 3x-2=9(x+1)$
$\Leftrightarrow x=\frac{-11}{6}$ (tm)
\(a,\sqrt{5x^2+10x+1}=7-\left(x^2+2x\right)\)
Đặt: \(\sqrt{5x^2+10x+1}=t\ge0\) ta được:
\(t=7-\frac{t^2-1}{5}\)
\(\Rightarrow t^2+5t-36=0\)
\(\Rightarrow t=4\)
\(\Rightarrow\hept{\begin{cases}x_1=-3\\x_2=1\end{cases}}\)
Vậy .................