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18 tháng 9 2020
B1:
\(C=\left(3-\sqrt{5}\right)\sqrt{3+\sqrt{5}}+\left(3+\sqrt{5}\right)\sqrt{3-\sqrt{5}}\)
\(=\sqrt{3-\sqrt{5}}.\sqrt{3+\sqrt{5}}\left(\sqrt{3-\sqrt{5}}+\sqrt{3+\sqrt{5}}\right)\)
\(=\sqrt{3^2-\left(\sqrt{5}\right)^2}\left(\sqrt{3-\sqrt{5}}+\sqrt{3+\sqrt{5}}\right)\)
\(=\sqrt{2}\left(\sqrt{3-\sqrt{5}}.\sqrt{2}+\sqrt{3+\sqrt{5}}.\sqrt{2}\right)\)
\(=\sqrt{2}\left(\sqrt{6-2\sqrt{5}}+\sqrt{6+2\sqrt{5}}\right)\)
\(=\sqrt{2}\left(\sqrt{\left(\sqrt{5}-1\right)^2}+\sqrt{\left(\sqrt{5}+1\right)^2}\right)\)
\(=\sqrt{2}\left(\sqrt{5}-1+\sqrt{5}+1\right)=2\sqrt{10}\)
2 tháng 10 2016
a)\(\sqrt{\frac{2x-3}{x-1}}=2\RightarrowĐk:\frac{2x-3}{x-1}\ge0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x\ge\frac{3}{2}\\x< 1\end{array}\right.\)
\(\sqrt{\frac{2x-3}{x-1}}=2\Rightarrow\frac{2x-3}{x-1}=4\)
\(\Leftrightarrow2x-3=4\left(x-1\right)\Leftrightarrow2x-3=4x-4\)
\(\Leftrightarrow2x=1\Leftrightarrow x=\frac{1}{2}\)(nhận)
b)\(\frac{\sqrt{2x-3}}{\sqrt{x-1}}=2\RightarrowĐk:\begin{cases}2x-3\ge0\\x-1>0\end{cases}\)
\(\Leftrightarrow x\ge\frac{3}{2}\)
\(\frac{\sqrt{2x-3}}{\sqrt{x-1}}=2\Leftrightarrow\sqrt{2x-3}=2\sqrt{x-1}\)
\(\Leftrightarrow2x-3=4x-4\)\(\Leftrightarrow x=\frac{1}{2}\)(loại)
c)\(\sqrt{\frac{4x+3}{x+1}}=3\RightarrowĐk:\frac{4x+3}{x+1}\ge0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x\ge\frac{-3}{4}\\x< -1\end{array}\right.\)
\(\sqrt{\frac{4x+3}{x+1}}=3\Rightarrow\frac{4x+3}{x+1}=9\)
\(\Leftrightarrow4x+3=9\left(x+1\right)\Leftrightarrow4x+3=9x+9\)
\(\Leftrightarrow5x=-6\Leftrightarrow x=\frac{-6}{5}\)(nhận)
c)\(\frac{\sqrt{4x+3}}{\sqrt{x+1}}=3\RightarrowĐk:\begin{cases}4x+3\ge0\\x+1>0\end{cases}\)
\(\Rightarrow x\ge\frac{-3}{4}\)
\(\frac{\sqrt{4x+3}}{\sqrt{x+1}}=3\Rightarrow\sqrt{4x+3}=3\sqrt{x+1}\)
\(\Leftrightarrow4x+3=9\left(x+1\right)\Leftrightarrow4x+3=9x+9\)
\(\Leftrightarrow x=\frac{-6}{5}\)(loại)
S
2 tháng 4 2019
Bài 3 \(\hept{\begin{cases}x+y+xy=2+3\sqrt{2}\\x^2+y^2=6\end{cases}}\)
\(\hept{\begin{cases}\left(x+y\right)+xy=2+3\sqrt{2}\\\left(x+y\right)^2-2xy=6\end{cases}}\)
\(\hept{\begin{cases}S+P=2+3\sqrt{2}\left(1\right)\\S^2-2P=6\left(2\right)\end{cases}}\)
Từ (1)\(\Rightarrow P=2+3\sqrt{2}-S\)Thế P vào (2) rồi giải tiếp nhé. Mình lười lắm ^.^
25 tháng 8 2017
Điều kiện:\(x,y\ge0\)
\(2x+y-2\sqrt{xy}-4\sqrt{x}+2\sqrt{y}+2=0\)
\(\Leftrightarrow\left(x-2\sqrt{xy}+y\right)-2\left(\sqrt{x}-\sqrt{y}\right)+1+\left(x-2\sqrt{x}+1\right)=0\)
\(\Leftrightarrow\left(\sqrt{x}-\sqrt{y}-1\right)^2+\left(\sqrt{x}-1\right)^2=0\)
\(Do\hept{\begin{cases}\left(\sqrt{x}-\sqrt{y}-1\right)^2\\\left(\sqrt{x}-1\right)^2\ge0\end{cases}\ge0}\)
\(Nên\hept{\begin{cases}\sqrt{x}-\sqrt{y}-1=0\\\sqrt{x}-1=0\end{cases}}\)
Tự tìm x,y nha!!
15 tháng 8 2020
ĐKXĐ: x \(\ge\)0; x \(\ne\)1
a) P = \(\left(\frac{2}{\sqrt{x}-1}-\frac{5}{x+\sqrt{x}-2}\right):\left(1+\frac{3-x}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\right)\)
P = \(\left(\frac{2\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}-\frac{5}{x+2\sqrt{x}-\sqrt{x}-2}\right):\frac{x+\sqrt{x}-2+3-x}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)
P = \(\frac{2\sqrt{x}+4-5}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\cdot\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}{\sqrt{x}+1}\)
P = \(\frac{2\sqrt{x}+1}{\sqrt{x}+1}\)
b) P = \(\frac{1}{\sqrt{x}}\) <=> \(\frac{2\sqrt{x}+1}{\sqrt{x}+1}=\frac{1}{\sqrt{x}}\)
=> \(\sqrt{x}\left(2\sqrt{x}+1\right)-\sqrt{x}-1=0\)
<=> \(2x+\sqrt{x}-\sqrt{x}-1=0\)
<=> \(x=\frac{1}{2}\)(tm)
c)Với đk: x \(\ge\)0 và x \(\ne\)1
\(x-2\sqrt{x-1}=0\) (đk: \(x\ge1\))
<=> \(x-1-2\sqrt{x-1}+1=0\)
<=> \(\left(\sqrt{x-1}-1\right)^2=0\)
<=> \(\sqrt{x-1}-1=0\)
<=> \(\sqrt{x-1}=1\)
<=> \(\left(\sqrt{x-1}\right)^2=1\)
<=> \(\left|x-1\right|=1\)
<=> \(\orbr{\begin{cases}x=0\left(ktm\right)\\x=2\left(tm\right)\end{cases}}\)
Với x = 2 => P = \(\frac{2\sqrt{2}+1}{\sqrt{2}+1}=\frac{\left(2\sqrt{2}+1\right)\left(\sqrt{2}-1\right)}{\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}=\frac{4-2\sqrt{2}+\sqrt{2}-1}{2-1}=3-\sqrt{2}\)
15 tháng 8 2020
a) P = \(\frac{2\sqrt{x}-1}{\sqrt{x}+1}\)(sửa lại)
b) \(\frac{2\sqrt{x}-1}{\sqrt{x}+1}=\frac{1}{\sqrt{x}}\) => \(2x-\sqrt{x}-\sqrt{x}-1=0\)
<=> \(2x-2\sqrt{x}-1=0\)<=> \(2\left(x-\sqrt{x}+\frac{1}{4}\right)-\frac{3}{4}=0\)
<=> \(2\left(\sqrt{x}-\frac{1}{2}\right)^2=\frac{3}{4}\) <=> \(\left(\sqrt{x}-\frac{1}{2}\right)^2=\frac{3}{8}\)....(tiếp tự lm)
Đk: \(x\ge\frac{-3}{2}\)
Bất pt <=> \(2x+3+x+2+2\sqrt{2x^2+7x+6}\le1\)
<=> \(2\sqrt{2x^2+7x+6}\le-4-3x\)
<=> \(\hept{\begin{cases}-3-4x\ge0\\4\left(2x^2+7x+6\right)\le16+24x+9x^2\end{cases}}\)
<=> \(\hept{\begin{cases}x\le-\frac{3}{4}\\x^2-4x-8\ge0\end{cases}}\)
<=> \(\hept{\begin{cases}x\le-\frac{3}{4}\\\left(x-2\right)^2\ge12\end{cases}}\)
<=> \(x\le2-\sqrt{12}\)
Đối chiếu đk: \(-\frac{3}{2}\le x\le2-\sqrt{12}\)