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19 tháng 9 2021
1) \(\sqrt{5-2x}=6\left(đk:x\le\dfrac{5}{2}\right)\)
\(\Leftrightarrow5-2x=36\)
\(\Leftrightarrow2x=-31\Leftrightarrow x=-\dfrac{31}{2}\left(tm\right)\)
2) \(\sqrt{2-x}=\sqrt{x+1}\left(đk:2\ge x\ge-1\right)\)
\(\Leftrightarrow2-x=x+1\)
\(\Leftrightarrow2x=1\Leftrightarrow x=\dfrac{1}{2}\left(tm\right)\)
3) \(\Leftrightarrow\sqrt{\left(2x+1\right)^2}=6\)
\(\Leftrightarrow\left|2x+1\right|=6\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+1=6\\2x+1=-6\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-\dfrac{7}{2}\end{matrix}\right.\)
4) \(\sqrt{x^2-10x+25}=x-2\left(đk:x\ge2\right)\)
\(\Leftrightarrow\sqrt{\left(x-5\right)^2}=x-2\)
\(\Leftrightarrow\left|x-5\right|=x-2\)
\(\Leftrightarrow\left[{}\begin{matrix}x-5=x-2\left(x\ge5\right)\\x-5=2-x\left(2\le x< 5\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}5=2\left(VLý\right)\\x=\dfrac{7}{2}\left(tm\right)\end{matrix}\right.\)
5 tháng 6 2018
a/ \(x+\sqrt{x+\frac{1}{2}+\sqrt{x+\frac{1}{4}}}=4\)
\(\Leftrightarrow x+\sqrt{\left(\sqrt{x+\frac{1}{4}}+\frac{1}{2}\right)^2}=4\)
\(\Leftrightarrow x+\sqrt{x+\frac{1}{4}}+\frac{1}{2}=4\)
Làm nốt
5 tháng 6 2018
b/ \(\sqrt{2x+4-6\sqrt{2x-5}}+\sqrt{2x-4+2\sqrt{2x-5}}=4\)
\(\sqrt{\left(\sqrt{2x-5}-3\right)^2}+\sqrt{\left(\sqrt{2x-5}-1\right)^2}=4\)
Làm nốt
7 tháng 3 2016
Bài 2 giải như sau (sau khi tác giả đã sửa): Điều kiện \(x,y>0.\)
Từ hệ ta suy ra \(1+\frac{3}{x+3y}=\frac{2}{\sqrt{x}},1-\frac{3}{x+3y}=\frac{4\sqrt{2}}{\sqrt{7y}}.\) Cộng và trừ hai phương trình, chia cả hai vế cho 2, ta sẽ được 2 phương trình \(1=\frac{1}{\sqrt{x}}+\frac{2\sqrt{2}}{\sqrt{7y}},\frac{3}{x+3y}=\frac{1}{\sqrt{x}}-\frac{2\sqrt{2}}{\sqrt{7y}}.\) Nhân hai phương trình với nhau, vế theo vế, ta được
\(\frac{3}{x+3y}=\frac{1}{x}-\frac{8}{7y}\to21xy=\left(x+3y\right)\left(7y-8x\right)\to21y^2-38xy-8x^2=0\to x=\frac{y}{2},x=-\frac{21}{4}y.\)
Đến đây ta được y=2x (trường hợp kia loại). Từ đó thế vào ta được \(1+\frac{3}{7x}=\frac{2}{\sqrt{x}}\to7x-14\sqrt{x}+3=0\to\sqrt{x}=\frac{7\pm2\sqrt{7}}{2}\to...\)
16 tháng 9 2019
a) \(\sqrt{x-1}+\sqrt{2x-1}=5\)
\(\Leftrightarrow3x-2+2\sqrt{\left(x-1\right)\left(2x-1\right)}=25\)
\(\Leftrightarrow2\sqrt{\left(x-1\right)\left(2x-1\right)}=25-3x+2\)
\(\Leftrightarrow2\sqrt{\left(x-1\right)\left(2x-1\right)}=-3x+27\)
Bình phương 2 vế, ta được:
\(\Leftrightarrow4\left(x-1\right)\left(2x-1\right)=9\left(x-9\right)^2\)
\(\Leftrightarrow8x^2-4x-8x+4=9x^2-162x+729\)
\(\Leftrightarrow8x^2-12x+4-9x^2+162x-729=0\)
\(\Leftrightarrow-x^2+150x-725=0\)
\(\Leftrightarrow x^2-150x+725=0\)
\(\Leftrightarrow\left(x-145\right)\left(x-5\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-145=0\\x-5=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=145\left(ktm\right)\\x=5\left(tm\right)\end{cases}}\)
\(\Rightarrow x=5\)
b) \(x+\sqrt{2x-1}-2=0\)
\(\Leftrightarrow\sqrt{2x-1}=2-x\)
Bình phương 2 vế, ta được:
\(\Leftrightarrow2x-1=4-4x^2+x^2=0\)
\(\Leftrightarrow2x-1-4+4x-x^2=0\)
\(\Leftrightarrow6x-5-x^2=0\)
\(\Leftrightarrow\left(x-5\right)\left(x-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-5=0\\x-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=5\left(ktm\right)\\x=1\left(tm\right)\end{cases}}\)
14 tháng 7 2019
\(a,\frac{9x-7}{\sqrt{7x+5}}=\sqrt{7x+5}\)\(ĐKXĐ:x\ge-\frac{5}{7}\)
\(\Leftrightarrow9x-7=7x+5\)
\(\Leftrightarrow9x-7x=5+7\)
\(\Leftrightarrow2x=12\)
\(\Leftrightarrow x=6\)
14 tháng 7 2019
\(b,\sqrt{4x-20}+3\sqrt{\frac{x-5}{9}}-\frac{1}{3}\sqrt{9x-45}=4\)
\(\Leftrightarrow\sqrt{4\left(x-5\right)}+3.\frac{\sqrt{x-5}}{\sqrt{9}}-\frac{1}{3}\sqrt{9\left(x-5\right)}=4\)
\(\Leftrightarrow2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}=4\)
\(\Leftrightarrow\sqrt{x-5}\left(2+1-1\right)=4\)
\(\Leftrightarrow2\sqrt{x-5}=4\)
\(\Leftrightarrow\sqrt{x-5}=2\)
\(\Leftrightarrow x-5=4\)
\(\Leftrightarrow x=9\)
NV
Nguyễn Việt Lâm
Giáo viên
16 tháng 9 2019
Bài 1:
a/ ĐKXĐ: \(x\ge1\)
\(\Leftrightarrow\sqrt{x-1}-2+\sqrt{2x-1}-3=0\)
\(\Leftrightarrow\frac{x-5}{\sqrt{x-1}+2}+\frac{2\left(x-5\right)}{\sqrt{2x-1}+3}=0\)
\(\Leftrightarrow\left(x-5\right)\left(\frac{1}{\sqrt{x-1}+2}+\frac{2}{\sqrt{2x-1}+3}\right)=0\)
\(\Rightarrow x=5\)
b/ĐKXĐ:...
\(x-1+\sqrt{2x-1}-1=0\)
\(\Leftrightarrow x-1+\frac{2\left(x-1\right)}{\sqrt{2x-1}+1}=0\)
\(\Leftrightarrow\left(x-1\right)\left(1+\frac{2}{\sqrt{2x-1}+1}\right)=0\)
\(\Rightarrow x=1\)
NV
Nguyễn Việt Lâm
Giáo viên
16 tháng 9 2019
Bài 2:
\(A=\sqrt{\left(2-\sqrt{3}\right)^2}-\sqrt{\left(2+\sqrt{3}\right)^2}\)
\(=\left|2-\sqrt{3}\right|-\left|2+\sqrt{3}\right|\)
\(=2-\sqrt{3}-2-\sqrt{3}=-2\sqrt{3}\)
\(B=\sqrt{\left(3-\sqrt{6}\right)^2}+\sqrt{\left(2\sqrt{6}-3\right)^2}\)
\(=\left(3-\sqrt{6}\right)+\left(2\sqrt{6}-3\right)\)
\(=\sqrt{6}\)
\(C=\left(\frac{3+\sqrt{5}-3+\sqrt{5}}{\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)}\right).\frac{\left(\sqrt{5}-1\right)}{\sqrt{5}\left(\sqrt{5}-1\right)}\)
\(=\frac{2\sqrt{5}}{4}.\frac{1}{\sqrt{5}}=\frac{1}{2}\)
21 tháng 7 2018
ĐKXĐ:\(\hept{\begin{cases}x-2>0\\y-1>0\\z-5>0\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}x>2\\y>1\\z>5\end{cases}}\)
pt\(\Leftrightarrow\frac{4}{\sqrt{x-2}}+\frac{1}{\sqrt{y-1}}+\frac{25}{\sqrt{z-5}}+\sqrt{x-2}+\sqrt{y-1}+\sqrt{z-5}=16\)
Áp dụng BĐT Cauchy:
\(\frac{4}{\sqrt{x-2}}+\sqrt{x-2}+\frac{1}{\sqrt{y-1}}+\sqrt{y-1}+\frac{25}{\sqrt{z-5}}+\sqrt{z-5}\)
\(\ge2\sqrt{\frac{4}{\sqrt{x-2}}.\sqrt{x-2}}+2\sqrt{\frac{1}{\sqrt{y-1}}.\sqrt{y-1}}+2\sqrt{\frac{25}{\sqrt{z-5}}.\sqrt{z-5}}\)
\(=2\sqrt{4}+2\sqrt{1}+2\sqrt{25}=2.2+2.1+2.5\)
\(=4+2+10=16\)
Dấu "=" xảy ra khi \(\hept{\begin{cases}x-2=4\\y-1=1\\z-5=25\end{cases}}\Leftrightarrow\hept{\begin{cases}x=6\\y=2\\z=30\end{cases}}\)
9 tháng 10 2020
1.\(\frac{5+\sqrt{5}}{5-\sqrt{5}}+\frac{5-\sqrt{5}}{5+\sqrt{5}}=\frac{\left(5+\sqrt{5}\right)\left(5+\sqrt{5}\right)}{\left(5-\sqrt{5}\right)\left(5+\sqrt{5}\right)}+\frac{\left(5-\sqrt{5}\right)\left(5-\sqrt{5}\right)}{\left(5-\sqrt{5}\right)\left(5+\sqrt{5}\right)}\)
\(=\frac{25+10\sqrt{5}+5}{25-5}+\frac{25-10\sqrt{5}+5}{25-5}\)
\(=\frac{25+10\sqrt{5}+5+25-10\sqrt{5}+5}{20}\)
\(=\frac{60}{20}=3\)
2.
a) \(\sqrt{45x}-2\sqrt{20x}+2\sqrt{80x}=21\)
ĐK : x ≥ 0
<=> \(\sqrt{5x\cdot9}-2\sqrt{5x\cdot4}+2\sqrt{5x\cdot16}=21\)
<=> \(\sqrt{5x\cdot3^2}-2\sqrt{2^2\cdot5x}+2\sqrt{5x\cdot4^2}=21\)
<=> \(\left|3\right|\sqrt{5x}-2\cdot\left|2\right|\sqrt{5x}+2\cdot\left|4\right|\sqrt{5x}=21\)
<=> \(\sqrt{5x}\cdot\left(3-4+8\right)=21\)
<=> \(\sqrt{5x}\cdot7=21\)
<=> \(\sqrt{5x}=3\)
<=> \(5x=9\)
<=> \(x=\frac{9}{5}\left(tm\right)\)
9 tháng 10 2020
ơ đang làm lại bấm " Gửi trả lời " ._.
2b) \(\sqrt{x^2-10x+25}=4\)
<=> \(\sqrt{\left(x-5\right)^2}=4\)
<=> \(\left|x-5\right|=4\)
<=> \(\orbr{\begin{cases}x-5=4\\x-5=-4\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=9\\x=1\end{cases}}\)
3. \(A=\left(\frac{1}{\sqrt{x}-1}-\frac{1}{\sqrt{x}}\right)\div\left(\frac{\sqrt{x}+1}{\sqrt{x}-2}-\frac{\sqrt{x}+2}{\sqrt{x}-1}\right)\)
ĐK : \(\hept{\begin{cases}x>0\\x\ne1\\x\ne4\end{cases}}\)
\(=\left(\frac{\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}-\frac{\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x-1}\right)}\right)\div\left(\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}-\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\right)\)
\(=\left(\frac{\sqrt{x}-\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}\right)\div\left(\frac{x-1}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}-\frac{x-4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\right)\)
\(=\frac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}\div\left(\frac{x-1-x+4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\right)\)
\(=\frac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}\div\frac{3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\)
\(=\frac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}\times\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}{3}\)
\(=\frac{\sqrt{x}-2}{3\sqrt{x}}\)
28 tháng 9 2021
1) \(\Leftrightarrow\sqrt{\left(x+5\right)^2}=4\)
\(\Leftrightarrow\left|x+5\right|=4\)
\(\Leftrightarrow\left[{}\begin{matrix}x+5=4\\x+5=-4\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-9\end{matrix}\right.\)
2) \(ĐK:x\ge2\)
\(\Leftrightarrow\sqrt{x-2}=2\)
\(\Leftrightarrow x-2=4\Leftrightarrow x=6\left(tm\right)\)
3) \(\Leftrightarrow\left(x^2-x+4\right)-\sqrt{x^2-x+4}+\dfrac{1}{4}=\dfrac{9}{4}\)
\(\Leftrightarrow\left(\sqrt{x^2-x+4}-\dfrac{1}{2}\right)^2=\dfrac{9}{4}\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x^2-x+4}-\dfrac{1}{2}=\dfrac{3}{2}\\\sqrt{x^2-x+4}-\dfrac{1}{2}=-\dfrac{3}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x^2-x+4}=2\\\sqrt{x^2-x+4}=-1\left(VLý\right)\end{matrix}\right.\)
\(\Leftrightarrow x^2-x+4=4\Leftrightarrow x\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
4) \(ĐK:x\ge0\)
\(\Leftrightarrow3\sqrt{x}-3=\sqrt{x}+2\)
\(\Leftrightarrow\sqrt{x}=\dfrac{5}{2}\Leftrightarrow x=\dfrac{25}{4}\left(tm\right)\)
\(\sqrt{\frac{1}{4}x^2+x+1}=\sqrt{\left(\frac{1}{2}x\right)^2+2.\frac{1}{2}x.1+1^2}=\sqrt{\left(\frac{1}{2}x+1\right)^2}=\left|\frac{1}{2}x+1\right|\)
\(\sqrt{6-2\sqrt{5}}=\sqrt{5-2\sqrt{5}+1}=\sqrt{\left(\sqrt{5}-1\right)^2}=\left|\sqrt{5}-1\right|=\sqrt{5}-1\)
phương trình <=> \(\left|\frac{1}{2}x+1\right|=\sqrt{5}-1\)
<=> \(\frac{1}{2}x+1=\sqrt{5}-1\) hoặc \(\frac{1}{2}x+1=-\sqrt{5}+1\)
+) \(\frac{1}{2}x+1=\sqrt{5}-1\)<=> \(x=2\sqrt{5}+4\)
+) \(\frac{1}{2}x+1=-\sqrt{5}+1\) <=> \(x=-2\sqrt{5}\)
Vậy pt có 2 nghiệm \(x=2\sqrt{5}+4\); \(x=-2\sqrt{5}\)