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\(a,Đk:1\le x\le4\)
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 :((
\(a,\left(x^2-4x+11\right)\left(x^4-8x^2+21\right)=35\)
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 .............................
ĐKXĐ: \(x\ge-3\)
\(\Leftrightarrow\left(2x-1\right)x-\left(2x-1\right)\sqrt{x+3}-x^2+x+3=0\)
\(\Leftrightarrow\left(2x-1\right)\left(x-\sqrt{x+3}\right)-\left(x^2-x-3\right)=0\)
\(\Rightarrow\frac{\left(2x-1\right)\left(x^2-x-3\right)}{x+\sqrt{x+3}}-\left(x^2-x-3\right)=0\)
\(\Leftrightarrow\left(x^2-x-3\right)\left(\frac{2x-1}{x+\sqrt{x+3}}-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x^2-x-3=0\\\frac{2x-1}{x+\sqrt{x+3}}=1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x^2-x-3=0\\x-1=\sqrt{x+3}\left(1\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow\left\{{}\begin{matrix}x\ge1\\\left(x-1\right)^2=x+3\end{matrix}\right.\)
Bạn tự giải nốt
\(2x-5a\sqrt{x-a}+2a\left(a-1\right)=0\)
Đặt \(\sqrt{x-a}=b\ge0\)
\(\Rightarrow2b^2-5ab+2a^2=0\)
\(\Leftrightarrow\left(b-2a\right)\left(2b-a\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=2b\\b=2a\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}a=2\sqrt{x-a}\\\sqrt{x-a}=2a\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{a^2}{4}+a\\x=4a^2+a\end{matrix}\right.\)
nếu bn đặt là b thì tại sao x bn lại cho là b2