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Câu 1:
PT \(\Leftrightarrow x^2+3x+8=(x+5)\sqrt{x^2+x+2}\)
\(\Leftrightarrow (x^2+x+2)+2(x+5)-4=(x+5)\sqrt{x^2+x+2}\)
Đặt \(\sqrt{x^2+x+2}=a; x+5=b(a\geq 0)\)
\(PT\Leftrightarrow a^2+2b-4=ba\)
\(\Leftrightarrow (a^2-4)-b(a-2)=0\)
\(\Leftrightarrow (a-2)(a+2-b)=0\Rightarrow \left[\begin{matrix} a=2\\ a+2=b\end{matrix}\right.\)
Nếu \(a=2\Rightarrow x^2+x+2=a^2=4\)
\(\Leftrightarrow x^2+x-2=0\Leftrightarrow (x-1)(x+2)=0\Rightarrow x=1; x=-2\) (đều thỏa mãn)
Nếu \(a+2=b\Leftrightarrow \sqrt{x^2+x+2}+2=x+5\)
\(\Leftrightarrow \sqrt{x^2+x+2}=x+3\)
\(\Rightarrow \left\{\begin{matrix} x+3\geq 0\\ x^2+x+2=(x+3)^2\end{matrix}\right.\Rightarrow \left\{\begin{matrix} x+3\geq 0\\ 5x+7=0\end{matrix}\right.\Rightarrow x=\frac{-7}{5}\) (thỏa mãn)
Vậy..........
Câu 2:
ĐKXĐ: \(x\geq 1\) hoặc \(x\leq \frac{1}{2}\)
\(10x^2-9x-8x\sqrt{2x^2-3x+1}+3=0\)
\(\Leftrightarrow 3(2x^2-3x+1)-8x\sqrt{2x^2-3x+1}+4x^2=0\)
Đặt \(\sqrt{2x^2-3x+1}=a(a\geq 0)\)
Khi đó PT \(\Leftrightarrow 3a^2-8xa+4x^2=0\)
\(\Leftrightarrow (a-2x)(3a-2x)=0\) \(\Rightarrow \left[\begin{matrix} a=2x\\ 3a=2x\end{matrix}\right.\)
Nếu \(a=\sqrt{2x^2-3x+1}=2x\Rightarrow \left\{\begin{matrix} x\geq 0\\ 2x^2-3x+1=4x^2\end{matrix}\right.\)
\(\Rightarrow \left\{\begin{matrix} x\geq 0\\ 2x^2+3x-1=0\end{matrix}\right.\Rightarrow x=\frac{-3+\sqrt{17}}{4}\) (t/m)
Nếu \(3a=3\sqrt{2x^2-3x+1}=2x\Rightarrow \left\{\begin{matrix} x\geq 0\\ 9(2x^2-3x+1)=4x^2\end{matrix}\right.\)
\(\Rightarrow \left\{\begin{matrix} x\geq 0\\ 14x^2-27x+9=0\end{matrix}\right.\Rightarrow x=\frac{3}{2}; x=\frac{3}{7}\) (t/m)
Vậy...........
a: Sửa đề; \(P=\left(\dfrac{3x+3\sqrt{x}-3}{x+\sqrt{x}-2}-\dfrac{\sqrt{x}+1}{\sqrt{x}+2}+\dfrac{\sqrt{x}-2}{\sqrt{x}-1}\right)\cdot\left(\dfrac{1}{1-\sqrt{x}}-1\right)\)
\(=\dfrac{3x+3\sqrt{x}-3-x+1+x-4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\cdot\dfrac{1-1+\sqrt{x}}{1-\sqrt{x}}\)
\(=\dfrac{3x+3\sqrt{x}-6}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\cdot\dfrac{\sqrt{x}}{1-\sqrt{x}}=\dfrac{3\sqrt{x}}{1-\sqrt{x}}\)
b: Để \(P=\sqrt{x}\) thì \(3\sqrt{x}=\sqrt{x}-x\)
\(\Leftrightarrow x+2\sqrt{x}=0\)
hay x=0
Câu 1:
\(\sqrt[3]{\left(3x+1\right)^2}+\sqrt[3]{\left(3x-1\right)^2}+\sqrt[3]{9x^2-1}=1\)
\(\Leftrightarrow\left(\sqrt[3]{3x+1}\right)^2+\left(\sqrt[3]{3x-1}\right)^2+\sqrt[3]{\left(3x-1\right)\left(3x+1\right)}=1\)
Đặt \(\left\{{}\begin{matrix}\sqrt[3]{3x+1}=a\\\sqrt[3]{3x-1}=m\end{matrix}\right.\), ta có hpt:
\(\left\{{}\begin{matrix}a^2+m^2+am=1\\a^3-m^3=2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a^2+am+m^2=1\\\left(a-m\right)\left(a^2+am+m^2\right)=2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a^2+am+m^2=1\left(1\right)\\a-m=2\left(2\right)\end{matrix}\right.\)
\(\left(2\right)\Rightarrow a=m+2\). Thay vào (1)
\(\Rightarrow\left(m+2\right)^2+\left(m+2\right)m+m^2=1\)
\(\Leftrightarrow3m^2+6m+3=0\)
\(\Leftrightarrow3\left(m+1\right)^2=0\)
\(\Leftrightarrow m=-1\)
\(\Rightarrow\sqrt[3]{3x-1}=-1\)
\(\Leftrightarrow3x-1=-1\)
\(\Leftrightarrow x=0\)
Câu 2: Đặt ẩn phụ và giải hpt như câu 1 >v<"