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22 tháng 12 2023
Câu 61:
a: \(B=\dfrac{3}{\sqrt{x}-2}+\dfrac{4}{\sqrt{x}+2}-\dfrac{12}{x-4}\)
\(=\dfrac{3}{\sqrt{x}-2}+\dfrac{4}{\sqrt{x}+2}-\dfrac{12}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{3\left(\sqrt{x}+2\right)+4\left(\sqrt{x}-2\right)-12}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{3\sqrt{x}+6+4\sqrt{x}-8-12}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{7\sqrt{x}-14}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{7\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{7}{\sqrt{x}+2}\)
b: \(A=\dfrac{\sqrt{x}+1}{\sqrt{x}-1}+\dfrac{\sqrt{x}-1}{\sqrt{x}+1}+\dfrac{3\sqrt{x}+1}{1-x}\)
\(=\dfrac{\left(\sqrt{x}+1\right)}{\sqrt{x}-1}+\dfrac{\sqrt{x}-1}{\sqrt{x}+1}-\dfrac{3\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{\left(\sqrt{x}+1\right)^2+\left(\sqrt{x}-1\right)^2-3\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{x+2\sqrt{x}+1+x-2\sqrt{x}+1-3\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{2x-3\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{\left(\sqrt{x}-1\right)\left(2\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{2\sqrt{x}-1}{\sqrt{x}+1}\)
Câu 60
Khi a=2 thì hệ phương trình sẽ trở thành:
\(\left\{{}\begin{matrix}\left(2^2-1\right)x+y=3\\2x-y=7\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}3x+y=3\\2x-y=7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}5x=10\\2x-y=7\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=2\\y=2x-7=2\cdot2-7=-3\end{matrix}\right.\)
17 tháng 9 2021
\(\sqrt{9x+9}-2\sqrt{\dfrac{x+1}{4}}=4\left(đk:x\ge-1\right)\)
\(\Leftrightarrow3\sqrt{x+1}-\sqrt{x+1}=4\)
\(\Leftrightarrow2\sqrt{x+1}=4\)
\(\Leftrightarrow\sqrt{x+1}=2\Leftrightarrow x+1=4\Leftrightarrow x=3\left(tm\right)\)
17 tháng 9 2021
Bài 2:
e) \(\sqrt{4x-8}-12\sqrt{\dfrac{x-2}{9}}=\sqrt{x-2}-12\left(đk:x\ge2\right)\)
\(\Leftrightarrow\sqrt{4}.\sqrt{x-2}-12.\sqrt{\dfrac{1}{9}}.\sqrt{x-2}=\sqrt{x-2}-12\)
\(\Leftrightarrow2\sqrt{x-2}-4\sqrt{x-2}=\sqrt{x-2}-12\)
\(\Leftrightarrow3\sqrt{x-2}=12\)
\(\Leftrightarrow\sqrt{x-2}=4\)
\(\Leftrightarrow x-2=16\Leftrightarrow x=18\left(tm\right)\)
NV
Nguyễn Việt Lâm
Giáo viên
17 tháng 9 2021
\(\Leftrightarrow\sqrt{4x^2-4x+1}=3x-1\)
\(\Leftrightarrow\left\{{}\begin{matrix}3x-1\ge0\\4x^2-4x+1=\left(3x-1\right)^2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{1}{3}\\5x^2-2x=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{1}{3}\\\left[{}\begin{matrix}x=0\left(loại\right)\\x=\dfrac{2}{5}\end{matrix}\right.\end{matrix}\right.\)
Vậy \(x=\dfrac{2}{5}\)
9 tháng 10 2021
\(2,\\ a,ĐK:x\in R\\ PT\Leftrightarrow\sqrt{\left(3x+1\right)^2}=1-2x\\ \Leftrightarrow\left|3x+1\right|=1-2x\Leftrightarrow\left[{}\begin{matrix}3x+1=1-2x\left(x\ge-\dfrac{1}{3}\right)\\3x+1=2x-1\left(x< -\dfrac{1}{3}\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(tm\right)\\x=-2\left(tm\right)\end{matrix}\right.\\ b,ĐK:x\ge4\\ PT\Leftrightarrow6\cdot\dfrac{1}{3}\sqrt{x-4}+\dfrac{2}{5}\cdot5\sqrt{x-4}=2\sqrt{x-4}+10\\ \Leftrightarrow2\sqrt{x-4}=10\Leftrightarrow\sqrt{x-4}=5\\ \Leftrightarrow x-4=25\Leftrightarrow x=29\left(tm\right)\)
18 tháng 10 2021
\(7,=\dfrac{\sqrt{3}\left(\sqrt{5}-3\right)}{\sqrt{5}-3}+\dfrac{4\left(\sqrt{3}-2\right)}{-1}-4+2\sqrt{3}\\ =\sqrt{3}-4\sqrt{3}+8-4+2\sqrt{3}=4-\sqrt{3}\)
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NV
Nguyễn Việt Lâm
Giáo viên
25 tháng 12 2020
\(a^3+b^3=\sqrt{\left(\sqrt{6}-\sqrt{2}\right)^2}-\dfrac{4\left(\sqrt{6}-\sqrt{2}\right)}{\left(\sqrt{6}+\sqrt{2}\right)\left(\sqrt{6}-\sqrt{2}\right)}\)
\(=\sqrt{6}-\sqrt{2}-\dfrac{4\left(\sqrt{6}-\sqrt{2}\right)}{4}=0\)
\(\Rightarrow a=-b\Rightarrow a^5+b^5=0\)
\(\sqrt{x^2+6x+9}-1=2x\)
\(\sqrt{\left(x^2+2.x.3+3^2\right)}-1=2x\)
\(\sqrt{\left(x+3\right)^2}-1=2x\)
\(x+3-1=2x\)
\(x+2=2x\)
\(x=2\)
\(\Leftrightarrow\sqrt{x^2+6x+9}=2x+1\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x+1\ge0\\x^2+6x+9=\left(2x+1\right)^2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge-\dfrac{1}{2}\\x^2+6x+9=4x^2+4x+1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge-\dfrac{1}{2}\\3x^2-2x-8=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge-\dfrac{1}{2}\\\left[{}\begin{matrix}x=2\\x=-\dfrac{4}{3}\left(loại\right)\end{matrix}\right.\end{matrix}\right.\)
Vậy \(x=2\)