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\(1\))\(x^2+5x+8=3\sqrt{x^3+5x^2+7x+6}\left(1\right)\\ĐK:x\ge-\dfrac{3}{2} \\ \left(1\right)\Leftrightarrow x^2+5x+8=3\sqrt{\left(2x+3\right)\left(x^2+x+2\right)}\left(2\right)\)
Đặt \(b=\sqrt{2x+3};a=\sqrt{x^2+x+2}\)
\(\left(2\right)\Leftrightarrow\left(a-b\right)\left(a-2b\right)=0\Leftrightarrow\left[{}\begin{matrix}a=b\\a=2b\end{matrix}\right.\)\(\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{1\pm\sqrt{5}}{2}\\x=\dfrac{7\pm\sqrt{89}}{2}\end{matrix}\right.\)
4)\(ĐK:x\ge-\dfrac{1}{3}\)
\(x^2-7x+2+2\sqrt{3x+1}=0\\ \Leftrightarrow x^2-7x+6+2\sqrt{3x+1}-4=0\\ \Leftrightarrow\left(x-1\right)\left(x-6\right)+\dfrac{12\left(x-1\right)}{2\sqrt{3x+1}+4}=0\\ \Leftrightarrow\left(x-1\right)\left(x-6+\dfrac{12}{2\sqrt{3x+1}+4}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x-6+\dfrac{12}{2\sqrt{3x+1}+4}=0\left(1\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow\left(x-5\right)+\dfrac{6}{\sqrt{3x+1}+2}-1=0\\ \Leftrightarrow\left(x-5\right)+\dfrac{4-\sqrt{3x+1}}{\sqrt{3x+1}+2}=0\\ \Leftrightarrow\left(x-5\right)-\dfrac{3\left(x-5\right)}{\left(\sqrt{3x+1}+2\right)\left(4+\sqrt{3x+1}\right)}=0\\ \Leftrightarrow\left(x-5\right)\left(1-\dfrac{3}{\left(\sqrt{3x+1}+2\right)\left(4+\sqrt{3x+1}\right)}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=5\\\left(1-\dfrac{3}{\left(\sqrt{3x+1}+2\right)\left(4+\sqrt{3x+1}\right)}\right)=0\left(2\right)\end{matrix}\right.\)
\(\left(2\right)\Leftrightarrow\left(\sqrt{3x+1}+2\right)\left(4+\sqrt{3x+1}\right)=3\\ \Leftrightarrow3x+1+6\sqrt{3x+1}+8=3\\ \Leftrightarrow x+2\sqrt{3x+1}+2=0\\ \Leftrightarrow2\sqrt{3x+1}=-x-2\ge0\Leftrightarrow x\le-2\)
Vậy pt có 2 nghiệm là x=1 và x=5
Mình giải mẫu pt đầu thôi nhé, những pt sau ttự.
1,\(x^4-\frac{1}{2}x^3-x^2-\frac{1}{2}x+1=0\)
Ta thấy x=0 ko là nghiệm.
Chia cả 2 vế cho x2 >0:
pt\(\Leftrightarrow x^2-\frac{1}{2}x-1-\frac{1}{2x}+\frac{1}{x^2}=0\)
Đặt \(t=x-\frac{1}{x}\left(t\in R\right)\)
\(\Rightarrow x^2+\frac{1}{x^2}=t^2+2\)
pt\(\Leftrightarrow t^2-\frac{1}{2}t+1=0\)(vô n0)
Vậy pt vô n0.
#Walker
\(\Leftrightarrow x^2+3-\left(6x+1\right)\sqrt{x^2+3}+9x^2+3x-2=0\)
Đặt \(\sqrt{x^2+3}=t\)
\(\Rightarrow t^2-\left(6x+1\right)t+9x^2+3x-2=0\)
\(\Delta=\left(6x+1\right)^2-4\left(9x^2+3x-2\right)=9\)
\(\Rightarrow\left[{}\begin{matrix}t=\frac{6x+1+3}{2}=3x+2\\t=\frac{6x+1-3}{2}=3x-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x^2+2}=3x+2\left(x\ge-\frac{2}{3}\right)\\\sqrt{x^2+2}=3x-1\left(x\ge\frac{1}{3}\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2+2=\left(3x+2\right)^2\\x^2+2=\left(3x-1\right)^2\end{matrix}\right.\) \(\Leftrightarrow...\)
a)\(26x^3-12x^2+13x=6\)
\(\Rightarrow26x^3-12x^2+13x-6=0\)
\(\Rightarrow2x^2\left(13x-6\right)+\left(13x-6\right)=0\)
\(\Rightarrow\left(13x-6\right)\left(2x^2+1\right)=0\)
\(\Rightarrow\left[\begin{matrix}13x-6=0\\2x^2+1=0\end{matrix}\right.\)\(\Rightarrow\left[\begin{matrix}x=\frac{6}{13}\\2x^2+1>0\left(loai\right)\end{matrix}\right.\)
Lời giải:
\(x^3+6x^2+12x+6=3\sqrt[3]{3x+8}\)
\(\Leftrightarrow x^3+6x^2+12x=3(\sqrt[3]{3x+8}-2)\)
\(\Leftrightarrow x(x^2+6x+12)=\frac{3.3x}{\sqrt[3]{(3x+8)^2}+2\sqrt[3]{3x+8}+4}\)
\(\Leftrightarrow x\left[(x^2+6x+12)-\frac{9}{\sqrt[3]{(3x+8)^2+2\sqrt[3]{3x+8}+4}}\right]=0\)
TH1: \(x=0\) (thỏa mãn)
TH2: Biểu thức trong ngoặc vuông bằng 0
Ta thấy \(x^2+6x+12=(x+3)^2+3\geq 3\forall x\in\mathbb{R}\) (1)
\(\sqrt[3]{(3x+8)^2}+2\sqrt[3]{3x+8}+4=(\sqrt[3]{3x+8}+1)^2+3\geq 3\)
\(\Rightarrow \frac{9}{\sqrt[3]{(3x+8)^2}+2\sqrt[3]{3x+8}+4}\leq 3\) (2)
Từ (1), (2) suy ra \(x^2+6x+12-\frac{9}{\sqrt[3]{(3x+8)^2}+2\sqrt[3]{3x+8}+4}\geq 0\)
Dấu bằng xảy ra khi \(x^2+6x+12=\frac{9}{\sqrt[3]{(3x+8)^2}+2\sqrt[3]{3x+8}+4}=3\Leftrightarrow \left\{\begin{matrix} (x+3)^2=0\\ (\sqrt[3]{3x+8}+1)^2=0\end{matrix}\right.\)
\(\Leftrightarrow \left\{\begin{matrix} x=-3\\ x=-3\end{matrix}\right.\) (thỏa mãn)
Vậy \(x\in\left\{-3;0\right\}\)
Minh Hiếu Tô : Đó là phép liên hợp
\((a-b)(a^2+ab+b^2)=a^3-b^3\Rightarrow a-b=\frac{a^3-b^3}{a^2+ab+b^2}\)
Ở đây \(a=\sqrt[3]{3x+8}; b=2\)
Còn bài trên kia bạn đăng hẳn bài riêng lên hộ mình nhé.