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a)\(3\left(\sqrt{2x^2+1}-1\right)=x\left(1+3x+8\sqrt{2x^2+1}\right)\)
\(\Leftrightarrow3\left(\dfrac{2x^2+1-1}{\sqrt{2x^2+1}+1}\right)-x\left(1+3x+8\sqrt{2x^2+1}\right)=0\)
\(\Leftrightarrow\dfrac{6x^2}{\sqrt{2x^2+1}+1}-x\left(1+3x+8\sqrt{2x^2+1}\right)=0\)
\(\Leftrightarrow x\left(\dfrac{6x}{\sqrt{2x^2+1}+1}-\left(1+3x+8\sqrt{2x^2+1}\right)\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\\dfrac{6x}{\sqrt{2x^2+1}+1}=1+3x+8\sqrt{2x^2+1}\end{matrix}\right.\)
Đặt \(\left\{{}\begin{matrix}a=\sqrt{2x^2+1}\\b=3x\end{matrix}\right.\left(a>0\right)\) thì
\(pt\left(2\right)\Leftrightarrow\)\(\dfrac{2b}{a+1}=1+b+8a\)
\(\Rightarrow\left\{{}\begin{matrix}a=-17\\b=120\end{matrix}\right.;\left\{{}\begin{matrix}a=-8\\b=49\end{matrix}\right.;\left\{{}\begin{matrix}a=-5\\b=26\end{matrix}\right.;\left\{{}\begin{matrix}a=-2\\b=5\end{matrix}\right.;\left\{{}\begin{matrix}a=-0\\b=1\end{matrix}\right.\) (loại vì \(a>0\))
Hay pt vô nghiệm
phần a liên hợp nhưng cx có yếu tố đặt ẩn là done r` nhé ;v còn phần b dg nghĩ có lẽ liên hợp nốt mà chủ thớt khó quá:v
Đặt \(\sqrt{x^2+9}=a\) ( \(a\ge9\) ) => \(x^2+9=a^2\)
Đặt \(3x+5=b\) => \(2x+3=\dfrac{2}{3}a-\dfrac{1}{3}\)
Ta có; \(2\left(3x+5\right)\sqrt{x^2+9}=3x^2+2x+30\)
<=> \(2ab=3a^2+\left(\dfrac{2}{3}b-\dfrac{1}{3}\right)\)
<=> \(6ab=9a^2+2b-1\)
<=> \(\left(9a^2-1\right)-\left(6ab-2b\right)=0\)
<=> \(\left(3a-1\right)\left(3a+1\right)-2b\left(3a-1\right)=0\)
<=> \(\left(3a-1\right)\left(3a+1-2b\right)=0\)
<=> \(\left[{}\begin{matrix}3a=1\left(1\right)\\3a-2b=-1\left(2\right)\end{matrix}\right.\)
(1) => \(3\sqrt{x^2+9}=1\) => Vô nghiệm ( vì \(\sqrt{x^2+9}\ge9\) )
(2) => \(3\sqrt{x^2+9}-2\left(3x+5\right)=-1\)
=> \(x=0\) (TM)
P/s: Mk nghĩ vì bn khá giỏi nên mk sẽ lm hơi tắt!
\(2\left(3x+5\right)\sqrt{x^2+9}=3x^2+2x+30\)
\(\Leftrightarrow2\left(3x+5\right)\sqrt{x^2+9}-30=3x^2+2x\)
\(\Leftrightarrow\dfrac{4\left(3x+5\right)^2\left(x^2+9\right)-900}{2\left(3x+5\right)\sqrt{x^2+9}+30}=x\left(3x+2\right)\)
\(\Leftrightarrow\dfrac{36x^4+120x^3+424x^2+1080x}{2\left(3x+5\right)\sqrt{x^2+9}+30}-x\left(3x+2\right)=0\)
\(\Leftrightarrow\dfrac{4x\left(9x^3+30x^2+106x+270\right)}{2\left(3x+5\right)\sqrt{x^2+9}+30}-x\left(3x+2\right)=0\)
\(\Leftrightarrow x\left(\dfrac{4\left(9x^3+30x^2+106x+270\right)}{2\left(3x+5\right)\sqrt{x^2+9}+30}-\left(3x+2\right)\right)=0\)
Dễ thấy: \(\dfrac{4\left(9x^3+30x^2+106x+270\right)}{2\left(3x+5\right)\sqrt{x^2+9}+30}-\left(3x+2\right)>0\)
\(\Rightarrow x=0\)
\(VT\ge0=>VP=4-2x\ge0=>x\le2.=>ĐK:2\ge x\ge1.\)
\(\sqrt{x-1}+\sqrt{x+3}-\left(4-2x\right)+2\sqrt{\left(x-1\right)\left(x^2-3x+5\right)}=0.\)
\(\sqrt{x-1}\left(1+\frac{13-4x}{\sqrt{x+3}+\left(4-2x\right)}+2\sqrt{x^2-3x+5}\right)=0.\)
\(Vi:2\ge x\ge1< =>-8\le-4x\le-4< =>5\le13-4x\le9=>13-4x>0\)=> Cái trong kia >0
=> x=1.
\(\sqrt{x-1}+\sqrt{x+3}+2\sqrt{\left(x-1\right)\left(x^2-3x+5\right)}=4-2x\)
Điều kiện: \(x\ge1\)
\(\hept{\begin{cases}VT=\sqrt{x-1}+\sqrt{x+3}+2\sqrt{\left(x-1\right)\left(x^2-3x+5\right)}\ge0+2+0=2\\VP=4-2x\le4-2=2\end{cases}}\)
Dấu = xảy ra khi \(x=1\)