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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:
\(\Leftrightarrow2x-1=4-4x^2+x^2=0\)
\(\Leftrightarrow2x-1-4+4x-x^2=0\)
\(\Leftrightarrow6x-5-x^2=0\)
\(\Leftrightarrow\left(x-5\right)\left(x-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-5=0\\x-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=5\left(ktm\right)\\x=1\left(tm\right)\end{cases}}\)
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
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:
\(A=\sqrt{\left(2-\sqrt{3}\right)^2}-\sqrt{\left(2+\sqrt{3}\right)^2}\)
\(=\left|2-\sqrt{3}\right|-\left|2+\sqrt{3}\right|\)
\(=2-\sqrt{3}-2-\sqrt{3}=-2\sqrt{3}\)
\(B=\sqrt{\left(3-\sqrt{6}\right)^2}+\sqrt{\left(2\sqrt{6}-3\right)^2}\)
\(=\left(3-\sqrt{6}\right)+\left(2\sqrt{6}-3\right)\)
\(=\sqrt{6}\)
\(C=\left(\frac{3+\sqrt{5}-3+\sqrt{5}}{\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)}\right).\frac{\left(\sqrt{5}-1\right)}{\sqrt{5}\left(\sqrt{5}-1\right)}\)
\(=\frac{2\sqrt{5}}{4}.\frac{1}{\sqrt{5}}=\frac{1}{2}\)
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
\(b,\sqrt{4x-20}+3\sqrt{\frac{x-5}{9}}-\frac{1}{3}\sqrt{9x-45}=4\)
\(\Leftrightarrow\sqrt{4\left(x-5\right)}+3.\frac{\sqrt{x-5}}{\sqrt{9}}-\frac{1}{3}\sqrt{9\left(x-5\right)}=4\)
\(\Leftrightarrow2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}=4\)
\(\Leftrightarrow\sqrt{x-5}\left(2+1-1\right)=4\)
\(\Leftrightarrow2\sqrt{x-5}=4\)
\(\Leftrightarrow\sqrt{x-5}=2\)
\(\Leftrightarrow x-5=4\)
\(\Leftrightarrow x=9\)
2T
12 tháng 8 2019
\(\sqrt{\frac{-6}{1+x}}=5\)
\(\Leftrightarrow\sqrt{\frac{-6}{1+x}}^2=5^2\)
\(\Leftrightarrow\frac{-6}{1+x}=25\)
\(\Leftrightarrow x+1=\frac{-6}{25}\)
\(\Leftrightarrow x=\frac{-6}{25}-1=\frac{-31}{25}\)
2T
12 tháng 8 2019
\(\sqrt{\left(\sqrt{x}-7\right)\left(\sqrt{x}+7\right)}=2\)
\(\Leftrightarrow\sqrt{x-49}=2\)
\(\Leftrightarrow x-49=4\Leftrightarrow x=53\)
ND
8 tháng 8 2019
b,
+ Với \(x=0\) \(\Rightarrow PTVN\)
+ Với \(x\ne0\), chia cả 2 vế cho \(x^2\) :
\(PT\Leftrightarrow x^2-16x+46+\frac{144}{x}+\frac{81}{x^2}=0\)
\(\Leftrightarrow\left(x^2+\frac{81}{x^2}\right)-16\left(x-\frac{9}{x}\right)+46=0\)
Đặt \(x-\frac{9}{x}=t\Rightarrow t^2=x^2+\frac{81}{x^2}-18\)
\(\Leftrightarrow t^2+18-16t+46=0\)
\(\Leftrightarrow t^2-16t+64=0\Rightarrow t=8\)
\(\Leftrightarrow x-\frac{9}{x}=8\Leftrightarrow x^2-8x-9=0\) \(\Rightarrow\left[{}\begin{matrix}x=-1\\x=9\end{matrix}\right.\) (t/m)
9 tháng 8 2019
cậu xem làm được mấy bài kia không làm giùm với (đang gấp) :))
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 : \(\hept{\begin{cases}x>0\\x\ne1\\x\ne4\end{cases}}\)
\(=\left(\frac{\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}-\frac{\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x-1}\right)}\right)\div\left(\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}-\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\right)\)
\(=\left(\frac{\sqrt{x}-\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}\right)\div\left(\frac{x-1}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}-\frac{x-4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\right)\)
\(=\frac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}\div\left(\frac{x-1-x+4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\right)\)
\(=\frac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}\div\frac{3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\)
\(=\frac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}\times\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}{3}\)
\(=\frac{\sqrt{x}-2}{3\sqrt{x}}\)
\(\sqrt{\frac{1}{4}x^2+x+1}=\sqrt{\left(\frac{1}{2}x\right)^2+2.\frac{1}{2}x.1+1^2}=\sqrt{\left(\frac{1}{2}x+1\right)^2}=\left|\frac{1}{2}x+1\right|\)
\(\sqrt{6-2\sqrt{5}}=\sqrt{5-2\sqrt{5}+1}=\sqrt{\left(\sqrt{5}-1\right)^2}=\left|\sqrt{5}-1\right|=\sqrt{5}-1\)
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 \(x=2\sqrt{5}+4\); \(x=-2\sqrt{5}\)