Giải nhanh giùm mình với cả nhà ơi
1. x2+5x+8=3\(\sqrt{2x^3+5x^2+7x+6}\)
2. x2+\(\dfrac{25x^2}{\left(x+5\right)^2}\)=11
3. x2-7x+2+2\(\sqrt{3x+1}\)=0
4. x2+2(2-x)\(\sqrt{x-1}\)+2-3x=0
5. \(\sqrt{5x-5}\)+\(\sqrt{7-3x}\)=9x2-36x+38
6.\(\left\{{}\begin{matrix}x^2+y^2=5\\x^3+2y^3=10x-10y\end{matrix}\right.\)
7. \(\left\{{}\begin{matrix}x^3-2y^3=x+4y\\6x^2-19xy+15y^2=1\end{matrix}\right.\)
\(1\))\(x^2+5x+8=3\sqrt{x^3+5x^2+7x+6}\left(1\right)\\ĐK:x\ge-\dfrac{3}{2} \\ \left(1\right)\Leftrightarrow x^2+5x+8=3\sqrt{\left(2x+3\right)\left(x^2+x+2\right)}\left(2\right)\)
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
3)\(x^2-7x+2+2\sqrt{3x+1}=0\\ \Leftrightarrow\left\{{}\begin{matrix}x^2-7x+2\le0\\\left(x^2-7x+2\right)^2=4\left(3x+1\right)\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{7-\sqrt{41}}{2}\le x\le\dfrac{7+\sqrt{41}}{2}\\x^4-14x^3+53x^2-40x=0\left(1\right)\end{matrix}\right.\\ \left(1\right)\Leftrightarrow x\left(x-1\right)\left(x^2-13x+40\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\left(loai\right)\\x=1\left(nhan\right)\\x=5\left(nhan\right)\\x=8\left(loai\right)\end{matrix}\right.\)