Viết đề mà ko ai đọc được vậy :v
a) \(3x^2+2x+3=\left(3x+1\right)\sqrt{x^2+3}\)
\(\Leftrightarrow3x^2+2x+3-3x\sqrt{x^2+3}-\sqrt{x^2+3}=0\)
\(\Leftrightarrow x^2+3-x\sqrt{x^2+3}-\sqrt{x^2+3}-2x\sqrt{x^2+3}+2x^2+2x=0\)
\(\Leftrightarrow\sqrt{x^2+3}\cdot\left(\sqrt{x^2+3}-x-1\right)-2x\cdot\left(\sqrt{x^2+3}-x-1\right)=0\)
\(\Leftrightarrow\left(\sqrt{x^2+3}-x-1\right)\left(\sqrt{x^2+3}-2x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x^2+3}=x+1\left(x\ge-1\right)\\\sqrt{x^2+3}=2x\left(x\ge0\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=1\end{matrix}\right.\)\(\Leftrightarrow x=1\) ( thỏa mãn )
Vậy...
\(\left(4x-1\right)\sqrt{x^2+1}=2x^2+2x+1\) (1)
<=>\(\left(4x-1\right)\left[\sqrt{x^2+1}-\left(3-x\right)\right]=6x^2-11x+4\)
Xét \(\sqrt{x^2+1}+3-x=0\)
<=> \(x^2+1=x^2-6x+9\) <=>\(x=\frac{4}{3}\)(tm phương trình (1))
Xét \(\sqrt{x^2+1}+3-x\ne0\)
pt <=>\(\frac{\left(4x-1\right)\left(x^2+1-x^2+6x-9\right)}{\sqrt{x^2+1}+3-x}=\left(3x-4\right)\left(2x-1\right)\)
<=> \(\frac{\left(4x-1\right)\left(6x-8\right)}{\sqrt{x^2+1}+3-x}-\left(3x-4\right)\left(2x-1\right)=0\)
<=>\(\left(3x-4\right)\left(\frac{2\left(4x-1\right)}{\sqrt{x^2+1}+3-x}-2x+1\right)=0\)
<=>\(\left[{}\begin{matrix}x=\frac{4}{3}\left(tm\right)\\\frac{8x-2}{\sqrt{x^2+1}+3-x}-2x+1=0\left(2\right)\end{matrix}\right.\)
pt (2) <=>\(8x-2=\left(2x-1\right)\sqrt{x^2+1}-2x^2+7x-3\)
<=>\(2x^2+x+1=\left(2x-1\right)\sqrt{x^2+1}\)( đk: \(x\ge\frac{1}{2}\))
=>\(4x^4+x^2+1+4x^3+2x+4x^2=\left(2x-1\right)^2\left(x^2+1\right)\)
<=>\(4x^4+4x^3+5x^2+2x+1=4x^4-4x^3+5x^2-4x+1\)
<=>\(8x^3+6x=0\) <=> \(x\left(8x^2+6\right)=0\) <=>x=0 (do 8x2+6>0) (không t/m (2))
=>(2) vô nghiệm
Vậy pt có tập nghiệm \(S=\left\{\frac{4}{3}\right\}\)
P/s: Hơi dài :)
Mấy anh chị khác god phân tích lắm nên em đành làm cách khác:(
\(2x^2+2x+1=\left(4x-1\right)\sqrt{x^2+1}\)
Đặt \(\sqrt{x^2+1}=a\ge1\)
\(PT\Leftrightarrow-2a^2+\left(4x-1\right)a-2x+1=0\)
\(\Leftrightarrow\left(2a-1\right)\left(2x-a-1\right)=0\) \(\Leftrightarrow\left[{}\begin{matrix}a=\frac{1}{2}\left(L\right)\\2x=a+1\left(1\right)\end{matrix}\right.\)
Xét (1): Do \(a\ge1\rightarrow a+1\ge2\Rightarrow x\ge1\)
(1) \(\Leftrightarrow2x=\sqrt{x^2+1}+1\)
\(\Leftrightarrow\frac{5}{4}x-\sqrt{x^2+1}+\frac{3}{4}\left(x-\frac{4}{3}\right)=0\)
\(\Leftrightarrow\left(x-\frac{4}{3}\right)\left[\frac{\frac{3}{16}\left(3x+4\right)}{\frac{5}{4}x+\sqrt{x^2+1}}+\frac{3}{4}\right]=0\)
\(\Leftrightarrow x=\frac{4}{3}\) (vì cái ngoặc to luôn > 0 với mọi \(x\ge1\))
Vậy...
b) \(x^2+3x+4=\left(x+3\right)\sqrt{x^2+x+2}\)
\(\Leftrightarrow\left(\sqrt{x^2+x+2}-x-1\right)\left(\sqrt{x^2+x+2}-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x^2+x+2}=x+1\\\sqrt{x^2+x+2}=2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\) ( thỏa )
Vậy...
Bài cuối:
\(15x^2+2\left(x+1\right)\sqrt{x+2}=2-5x\)
ĐKXĐ: \(x\ge-2\). Đặt \(\sqrt{x+2}=a\ge0\)
PT \(\Leftrightarrow-a^2+2\left(x+1\right)a+15x^2+6x=0\)
\(\Leftrightarrow\left(a+3x\right)\left(5x+2-a\right)=0\Leftrightarrow\left[{}\begin{matrix}a=-3x\left(1\right)\\a=5x+2\left(2\right)\end{matrix}\right.\)
(1) \(\Leftrightarrow\left\{{}\begin{matrix}x\le0\\x+2=9x^2\end{matrix}\right.\Leftrightarrow x=\frac{1-\sqrt{73}}{18}\) (nghiệm còn lại loại)
(2) \(\Leftrightarrow\left\{{}\begin{matrix}x\ge-\frac{2}{5}\\x+2=\left(5x+2\right)^2\end{matrix}\right.\Leftrightarrow x=\frac{\sqrt{161}-19}{50}\) (nghiệm kia loại)
Vậy....
