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a: \(\Leftrightarrow2\sqrt{3x}+12-4x+5\sqrt{3}=0\)
\(\Leftrightarrow-4x+2\sqrt{3}\cdot\sqrt{x}+12+5\sqrt{3}=0\)
Đặt \(\sqrt{x}=a\left(a>=0\right)\)
Phương trình trở thành \(-4a^2+2\sqrt{3}a+12+5\sqrt{3}=0\)
\(\Delta=\left(2\sqrt{3}\right)^2-4\cdot\left(-4\right)\cdot\left(12+5\sqrt{3}\right)\)
\(=12+16\left(12+5\sqrt{3}\right)\)
\(=12+192+80\sqrt{3}=204+80\sqrt{3}\)
Vì Δ>0 nên phương trình có hai nghiệm phân biệt là:
\(\left\{{}\begin{matrix}a_1=\dfrac{-2\sqrt{3}-\sqrt{204+80\sqrt{3}}}{-8}=\dfrac{2\sqrt{3}+\sqrt{204+80\sqrt{3}}}{8}\left(nhận\right)\\a_2=\dfrac{-2\sqrt{3}+\sqrt{204+80\sqrt{3}}}{-8}\left(loại\right)\end{matrix}\right.\)
\(\Leftrightarrow a=\dfrac{2\sqrt{3}+2\sqrt{26+20\sqrt{3}}}{8}=\dfrac{\sqrt{3}+\sqrt{26+20\sqrt{3}}}{4}\)
\(\Leftrightarrow x=a^2\simeq5,66\)
c: \(\Leftrightarrow x\sqrt{2}+5\sqrt{2}-4x-5-4\sqrt{2}=0\)
\(\Leftrightarrow x\left(\sqrt{2}-4\right)+\sqrt{2}-5=0\)
\(\Leftrightarrow x=\dfrac{5-\sqrt{2}}{\sqrt{2}-4}=\dfrac{-18-\sqrt{2}}{14}\)
d: \(\Leftrightarrow\dfrac{7x+1-4x-4002}{2001}=\dfrac{3x+2}{2003}-1\)
\(\Leftrightarrow3x-4001=0\)
hay x=4001/3
c/\(P=\frac{\frac{2\left(\sqrt{x}-1\right)}{x\sqrt{x}-1}}{1-\frac{x+2}{x+\sqrt{x}+1}}\)\(\Leftrightarrow P=\frac{2\left(\sqrt{x}-1\right)}{x\sqrt{x}-1}:\frac{\sqrt{x}-1}{x+\sqrt{x}+1}\)
\(\Leftrightarrow\frac{2\left(x+\sqrt{x}+1\right)}{x\sqrt{x}-1}\)
Xét P-1 ta có \(\frac{2x+2\sqrt[]{x}+2-x\sqrt{x}+1}{x\sqrt{x}-1}=\frac{2x+2\sqrt{x}-x\sqrt{x}+3}{x\sqrt{x}-1}\)
với x<1 thì tử dương, mẫu âm, với x>1 thì tử âm và mẫu dương
Từ đó ta luuon có P-1\(\le0\RightarrowĐPCM\)
a/\(\Leftrightarrow x=\frac{5-\sqrt{5}}{1-\sqrt{5}}+\frac{5+\sqrt{5}}{1+\sqrt{5}}-\frac{25-5}{1-5}-1\)
\(\Leftrightarrow x=0+5-1\Leftrightarrow x=4\)
Thay vào B đc \(B=\frac{4+2}{4+2+1}=\frac{6}{7}\)
b/
a/ \(2x^2-3x+1>0\Rightarrow\left[{}\begin{matrix}x>1\\x< \frac{1}{2}\end{matrix}\right.\)
b/ \(-3x^2+2x+1< 0\Rightarrow-\frac{1}{3}< x< 1\)
c/ \(\frac{x+3}{x-2}\ge0\Rightarrow\left[{}\begin{matrix}x>2\\x\le-3\end{matrix}\right.\)
d/ \(\frac{2x+1}{x+2}\ge1\Leftrightarrow\frac{2x+1}{x+2}-1\ge0\Leftrightarrow\frac{x-1}{x+2}\ge0\Rightarrow\left[{}\begin{matrix}x\ge1\\x< -2\end{matrix}\right.\)
e/ \(\frac{\sqrt{x}+3}{2-\sqrt{x}}\le0\Rightarrow\left\{{}\begin{matrix}x\ge0\\2-\sqrt{x}< 0\end{matrix}\right.\) \(\Rightarrow x>4\)
g/\(\frac{\sqrt{x}-3}{\sqrt{x}-2}\ge0\Rightarrow\left\{{}\begin{matrix}x\ge0\\\left[{}\begin{matrix}x\ge9\\x< 4\end{matrix}\right.\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x\ge0\\0\le x< 4\end{matrix}\right.\)
h/ \(\frac{\sqrt{x}-3}{\sqrt{x}-1}-\frac{1}{3}< 0\Rightarrow\frac{2\left(\sqrt{x}-4\right)}{3\left(\sqrt{x}-1\right)}< 0\Rightarrow1< x< 16\)
\(A=\left(\frac{1}{\sqrt{x-1}}+\frac{1}{\sqrt{x+1}}\right)^2\cdot\frac{x^2-1}{2}-\sqrt{x^2-1}.\)
\(=\left(\frac{1}{x-1}+\frac{1}{x+1}+\frac{2}{\sqrt{\left(x-1\right)\left(x+1\right)}}\right)\cdot\frac{\left(x-1\right)\left(x+1\right)}{2}-\sqrt{\left(x-1\right)\left(x+1\right)}\)
\(=\frac{x+1}{2}+\frac{x-1}{2}+\sqrt{\left(x-1\right)\left(x+1\right)}-\sqrt{\left(x-1\right)\left(x+1\right)}\)
\(=x\)
KO hiểu thì ib nha!! bùn ngủ quá lm tắt
1) Nhìn cái pt hết ham, nhưng bấm nghiệm đẹp v~`~
\(\left(\sqrt{2}+2\right)\left(x\sqrt{2}-1\right)=2x\sqrt{2}-\sqrt{2}\)
\(\Leftrightarrow\left(\sqrt{2}+2\right)\left(x\sqrt{2}-1\right)-2x\sqrt{2}+\sqrt{2}=0\)
\(\Leftrightarrow2x-\sqrt{2}+2x\sqrt{2}-2-2x\sqrt{2}+\sqrt{2}=0\)
\(\Leftrightarrow2x-2=0\Leftrightarrow2x=2\Rightarrow x=1\)
\(P=\frac{\frac{1}{a^2}}{\frac{1}{b}+\frac{1}{c}}+\frac{\frac{1}{b^2}}{\frac{1}{a}+\frac{1}{c}}+\frac{\frac{1}{c^2}}{\frac{1}{a}+\frac{1}{b}}\)
Đặt \(\hept{\begin{cases}x=\frac{1}{a}\\y=\frac{1}{b}\\z=\frac{1}{c}\end{cases}}\Rightarrow xyz=1\Rightarrow P=\frac{x^2}{y+z}+\frac{y^2}{x+z}+\frac{z^2}{x+y}\)
Áp dụng BĐT Cauchy-Schwarz dạng Engel ta có:
\(P\ge\frac{\left(x+y+z\right)^2}{y+z+x+z+x+y}=\frac{x+y+z}{2}\ge\frac{3\sqrt[3]{xyz}}{2}=\frac{3}{2}\)
Dấu "=" xảy ra khi \(x=y=z\Leftrightarrow a=b=c=1\)
Cần cách khác thì nhắn cái
\(A=\left(\frac{1}{\sqrt{x}-2}+\frac{1}{\sqrt{x}+2}\right):\frac{5x}{x-5}\)
\(=\left(\frac{\sqrt{x}+2}{x-4}+\frac{\sqrt{x}-2}{x-4}\right):\frac{5x}{x-5}\)
\(=\frac{2\sqrt{x}}{x-4}.\frac{x-5}{5x}=\frac{2\sqrt{x}\left(x-5\right)}{5x^2-20x}\)