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a/ Giải rồi
b/ ĐKXĐ: \(x\ge-1\)
Đặt \(\sqrt{2x+3}+\sqrt{x+1}=t>0\)
\(\Rightarrow t^2=3x+4+2\sqrt{2x^2+5x+3}\) (1)
Pt trở thành:
\(t=t^2-6\Leftrightarrow t^2-t-6=0\Rightarrow\left[{}\begin{matrix}t=3\\t=-2\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\sqrt{2x+3}+\sqrt{x+1}=3\)
\(\Leftrightarrow3x+4+2\sqrt{2x^2+5x+3}=9\)
\(\Leftrightarrow2\sqrt{2x^2+5x+3}=5-3x\left(x\le\frac{5}{3}\right)\)
\(\Leftrightarrow4\left(2x^2+5x+3\right)=\left(5-3x\right)^2\)
\(\Leftrightarrow...\)
e/ ĐKXD: \(x>0\)
\(5\left(\sqrt{x}+\frac{1}{2\sqrt{x}}\right)=2\left(x+\frac{1}{4x}\right)+4\)
Đặt \(\sqrt{x}+\frac{1}{2\sqrt{x}}=t\ge\sqrt{2}\)
\(\Rightarrow t^2=x+\frac{1}{4x}+1\)
Pt trở thành:
\(5t=2\left(t^2-1\right)+4\)
\(\Leftrightarrow2t^2-5t+2=0\Rightarrow\left[{}\begin{matrix}t=2\\t=\frac{1}{2}\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\sqrt{x}+\frac{1}{2\sqrt{x}}=2\)
\(\Leftrightarrow2x-4\sqrt{x}+1=0\)
\(\Rightarrow\sqrt{x}=\frac{2\pm\sqrt{2}}{2}\)
\(\Rightarrow x=\frac{3\pm2\sqrt{2}}{2}\)
Hung nguyen, Trần Thanh Phương, Sky SơnTùng, @tth_new, @Nguyễn Việt Lâm, @Akai Haruma, @No choice teen
help me, pleaseee
Cần gấp lắm ạ!
minh ghi nhầm, dấu căn dưới mẫu là bao trùm luôn -7x+3 nhen
\(x=1+\sqrt[3]{2}+\sqrt[3]{4}=\frac{1}{\sqrt[3]{2}-1}.\)
\(\Rightarrow\sqrt[3]{2}x=x+1\)
\(\Rightarrow x^3-3x^2-3x-1=0\)
\(\Rightarrow\hept{\begin{cases}x^3+x^2+5x+3=4\left(x+1\right)^2\\x^3-2x^2-7x+3=\left(x-2\right)^2\end{cases}}\)
Khi đó:
\(P=\frac{2\left(x+1\right)-6}{x-2}=2\)(do x>0)
7.
ĐKXĐ: ...
\(\Leftrightarrow10\sqrt{\left(x+1\right)\left(x^2-x+1\right)}=3\left(x^2+2\right)\)
Đặt \(\left\{{}\begin{matrix}\sqrt{x^2-x+1}=a>0\\\sqrt{x+1}=b\ge0\end{matrix}\right.\)
\(\Rightarrow10ab=3\left(a^2+b^2\right)\)
\(\Leftrightarrow3a^2-10ab+3b^2=0\)
\(\Leftrightarrow\left(a-3b\right)\left(3b-a\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}a=3b\\3a=b\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\sqrt{x^2-x+1}=3\sqrt{x+1}\\3\sqrt{x^2-x+1}=\sqrt{x-1}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-x+1=9x+9\\9x^2-9x+9=x-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-10x-8=0\\9x^2-10x+10=0\end{matrix}\right.\) (casio)
6.
ĐKXĐ: ...
\(\Leftrightarrow2x^2+4=3\sqrt{\left(x+1\right)\left(x^2-x+1\right)}\)
Đặt \(\left\{{}\begin{matrix}\sqrt{x^2-x+1}=a>0\\\sqrt{x+1}=b\ge0\end{matrix}\right.\)
\(\Rightarrow2a^2+2b^2=3ab\)
\(\Leftrightarrow2a^2-3ab+2b^2=0\)
Phương trình vô nghiệm (vế phải là \(5\sqrt{x^3+1}\) sẽ hợp lý hơn)
mình ghi nhầm, dấu căn dưới mẫu là bao trùm luôn -7x+3 nhen
\(x=1+\sqrt[3]{2}+\sqrt[3]{4}\)
\(\Leftrightarrow x-1=\sqrt[3]{2}+\sqrt[3]{4}\)
\(\Leftrightarrow\left(x-1\right)^3=\left(\sqrt[3]{2}+\sqrt[3]{4}\right)^3\)
\(\Leftrightarrow x^3-3x^2+3x-1=2+4+3\sqrt[3]{2.4}\left(\sqrt[3]{2}+\sqrt[3]{4}\right)\)
\(\Leftrightarrow x^3-3x^2+3x-1=6+6\left(\sqrt[3]{2}+\sqrt[3]{4}\right)\)
\(\Leftrightarrow x^3-3x^2+3x-1=6+6\left(x-1\right)\)( vì \(x-1=\sqrt[3]{2}+\sqrt[3]{4}\))
\(\Leftrightarrow x^3-3x^2-3x-1=0\)
Ta có \(M=\frac{\sqrt{x^3+x^2+5x+3}-6}{\sqrt{x^3-2x^2-7x+3}}\)
\(=\frac{\sqrt{x^3-3x^2-3x-1+4x^2+8x+4}-6}{\sqrt{x^3-3x^2-3x-1+x^2-4x+4}}\)
\(=\frac{\sqrt{4x^2+8x+4}-6}{\sqrt{x^2-4x+4}}\)( vì \(x^3-3x^2-3x-1=0\))
\(=\frac{\sqrt{\left(2x+2\right)^2}-6}{\sqrt{\left(x-2\right)^2}}\)
\(=\frac{\left|2x+2\right|-6}{\left|x-2\right|}\)
Từ điều kiện đề bài \(\Rightarrow x>2\)
\(\Rightarrow M=\frac{2x+2-6}{x-2}=2\)
Vậy \(M=2\)\(\Leftrightarrow x=1+\sqrt[3]{2}+\sqrt[3]{4}\)