\(Let\)\(\left|x\right|< 1\)\(.\)\(Find\)\(Min\)\(:\)
\(P=\frac{5x+6\sqrt{1-x^2}+7}{\sqrt{1-x^2}}\)
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\(P=\left(1-\frac{\sqrt{x}}{\sqrt{x}+1}\right):\left(\frac{\sqrt{x}+3}{\sqrt{x}-2}+\frac{\sqrt{x}+2}{3-\sqrt{x}}+\frac{\sqrt{x}+2}{x-5\sqrt{x}+6}\right)\\ \)\(=\left(\frac{\sqrt{x}+1}{\sqrt{x}+1}-\frac{\sqrt{x}}{\sqrt{x}+1}\right):\left(\frac{\left(\sqrt{x}+3\right).\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-2\right).\left(\sqrt{x}-3\right)}-\frac{\left(\sqrt{x}+2\right).\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-3\right).\left(\sqrt{x-2}\right)}+\frac{\sqrt{x}+2}{\left(\sqrt{x}-2\right).\left(\sqrt{x}-3\right)}\right)\)
\(=\frac{1}{\sqrt{x}+1}:\left(\frac{x-9}{\left(\sqrt{x}-2\right).\left(\sqrt{x}-3\right)}-\frac{x-4}{\left(\sqrt{x}-2\right).\left(\sqrt{x}-3\right)}+\frac{\sqrt{x}+2}{\left(\sqrt{x}-2\right).\left(\sqrt{x}-3\right)}\right)\)
\(=\frac{1}{\sqrt{x}+1}:\frac{\sqrt{x}-3}{\left(\sqrt{x}-2\right).\left(\sqrt{x}-3\right)}=\frac{1}{\sqrt{x}+1}:\frac{1}{\sqrt{x}-2}=\frac{\sqrt{x}-2}{\sqrt{x}+1}\)
b.
\(P=\frac{\sqrt{x}-2}{\sqrt{x}+1}=\frac{\left(\sqrt{x}+1\right)-3}{\sqrt{x}+1}=1-\frac{3}{\sqrt{x}+1}\)
Vì \(\sqrt{x}\ge0\Rightarrow\sqrt{x}+1\ge1\Rightarrow\frac{3}{\sqrt{x}+1}\le3\Rightarrow1-\frac{3}{\sqrt{x}+1}\ge1-3=-2\Rightarrow P\ge-2\)
Dấu "=" xảy ra <=> x=0
vậy Min (P) = -2 <=> x=0
Rút gọn: \(P=\left(1-\frac{\sqrt{x}}{\sqrt{x}+1}\right):\left(\frac{\sqrt{x}+3}{\sqrt{x}-2}+\frac{\sqrt{x}+2}{3-\sqrt{x}}+\frac{\sqrt{x}+2}{x-5\sqrt{x}+6}\right)\)
\(=\left(\frac{\sqrt{x}+1-\sqrt{x}}{\sqrt{x}+1}\right):\left(\frac{\sqrt{x}+3}{\sqrt{x}-2}-\frac{\sqrt{x}+2}{\sqrt{x}-3}+\frac{\sqrt{x}+2}{x-5\sqrt{x}+6}\right)\)
\(=\frac{1}{\sqrt{x}+1}:\left(\frac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)-\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)+\sqrt{x}+2}{x-5\sqrt{x}+6}\right)\)
\(=\frac{1}{\sqrt{x}+1}:\left(\frac{x-9-x+4+\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\right)=\frac{1}{\sqrt{x}+1}:\frac{\sqrt{x}-3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(=\frac{1}{\sqrt{x}+1}.\left(\sqrt{x}-2\right)=\frac{\sqrt{x}-2}{\sqrt{x}+1}\)
3.
ĐKXĐ: \(x\ge-1;x\ne13\)
\(\left(x+2\right)\left(\sqrt{x+1}-2\right)=\sqrt[3]{2x+1}-3\)
\(\Leftrightarrow\left(x+2\right)\sqrt{x+1}-2x-4=\sqrt[3]{2x+1}-3\)
\(\Leftrightarrow\left(x+1\right)\sqrt{x+1}+x+1-\left(2x+1\right)-\sqrt[3]{2x+1}=0\)
Đặt \(\left\{{}\begin{matrix}\sqrt{x+1}=a\\\sqrt[3]{2x+1}=b\end{matrix}\right.\)
\(\Rightarrow a^3+a-b^3-b=0\)
\(\Leftrightarrow\left(a-b\right)\left(a^2+ab+b^2+1\right)=0\)
\(\Leftrightarrow a=b\)
\(\Leftrightarrow\sqrt{x+1}=\sqrt[3]{2x+1}\) (\(x\ge-\frac{1}{2}\))
\(\Leftrightarrow\left(x+1\right)^3=\left(2x+1\right)^2\)
\(\Leftrightarrow x=?\)
2.
ĐKXĐ: \(x\ge-\frac{1}{2}\)
\(\Leftrightarrow8x^3+2x-\left(2x+2\right)\sqrt{2x+1}=0\)
Đặt \(\left\{{}\begin{matrix}2x=a\\\sqrt{2x+1}=b\end{matrix}\right.\)
\(\Rightarrow a^3+a-\left(b^2+1\right)b=0\)
\(\Leftrightarrow a^3-b^3+a-b=0\)
\(\Leftrightarrow\left(a-b\right)\left(a^2+ab+b^2+1\right)=0\)
\(\Leftrightarrow a=b\)
\(\Leftrightarrow2x=\sqrt{2x+1}\) (\(x\ge0\))
\(\Leftrightarrow4x^2=2x+1\)
\(\Leftrightarrow x=?\)
Hung nguyen, Trần Thanh Phương, Sky SơnTùng, @tth_new, @Nguyễn Việt Lâm, @Akai Haruma, @No choice teen
help me, pleaseee
Cần gấp lắm ạ!
từ dòng cuối là sai rồi bạn à
Bạn bỏ dòng cuối đi còn lại đúng rồi
Ở tử đặt nhân tử chung căn x chung rồi lại đặt căn x +1 chung
Ở mẫu tách 3 căn x ra 2 căn x +căn x rồi đặt nhân tử 2 căn x ra
rút gọn được \(\frac{3\sqrt{x}-5}{2\sqrt{x}+1}\)
tích mình đi
ai tích mình
mình tích lại
thanks
Ok, mình k rồi đấy. k mình đi