\(\sqrt{x^2+2x-1}=x-1\)
Những câu hỏi liên quan
Pleft(frac{sqrt{x}+1}{sqrt{2x}+1}+frac{sqrt{2x}+sqrt{x}}{sqrt{2x}-1}-1right):left(1+frac{sqrt{x}+1}{sqrt{2x}+1}-frac{sqrt{2x}-sqrt{x}}{sqrt{2x}-1}right)frac{left(sqrt{x}+1right)left(sqrt{2x}-1right)}{left(sqrt{2x}+1right)left(sqrt{2x}-1right)}+frac{left(sqrt{2x}+sqrt{x}right)left(sqrt{2x}+1right)}{MTC}-frac{2x-1}{MTC}frac{xsqrt{2}-sqrt{x}+sqrt{2x}-1+2x+sqrt{2x}+xsqrt{2}+sqrt{x}-2x+1}{MTC}frac{2xsqrt{2}+2sqrt{2x}}{MTC}
Đọc tiếp
P=\(\left(\frac{\sqrt{x}+1}{\sqrt{2x}+1}+\frac{\sqrt{2x}+\sqrt{x}}{\sqrt{2x}-1}-1\right):\left(1+\frac{\sqrt{x}+1}{\sqrt{2x}+1}-\frac{\sqrt{2x}-\sqrt{x}}{\sqrt{2x}-1}\right)\)
=\(\frac{\left(\sqrt{x}+1\right)\left(\sqrt{2x}-1\right)}{\left(\sqrt{2x}+1\right)\left(\sqrt{2x}-1\right)}+\frac{\left(\sqrt{2x}+\sqrt{x}\right)\left(\sqrt{2x}+1\right)}{MTC}-\frac{2x-1}{MTC}\)
=\(\frac{x\sqrt{2}-\sqrt{x}+\sqrt{2x}-1+2x+\sqrt{2x}+x\sqrt{2}+\sqrt{x}-2x+1}{MTC}\)
=\(\frac{2x\sqrt{2}+2\sqrt{2x}}{MTC}\)
\(\frac{2x-1}{MTC}+\frac{\left(\sqrt{x}+1\right)\left(2\sqrt{x}-1\right)}{MTC}-\frac{\left(\sqrt{2x}+\sqrt{x}\left(\sqrt{2x}+1\right)\right)}{MTC}\)
=\(\frac{2x-1+x\sqrt{2}-\sqrt{x}+\sqrt{2x}-1-2x-\sqrt{2x}-x\sqrt{2}-\sqrt{x}}{MTC}\)
=\(\frac{-2\sqrt{x}-2}{\left(\sqrt{2x}-1\right)\left(\sqrt{2x+1}\right)}\)
14 tháng 8 2021 lúc 8:33
Rút gọn : A=\(\dfrac{\sqrt{x+2\sqrt{x-1}}+\sqrt{x-2\sqrt{x-1}}}{\sqrt{x+\sqrt{2x-1}+\sqrt{x-\sqrt{2x-1}}}}.\left(\sqrt{2x-1}\right)\)
Giải phương trình
a) \(\sqrt{x^2-2x+4}=2x-2\)
b) \(\sqrt{x+2\sqrt{x-1}}=2\)
c) \(\sqrt{2x^2-2x+1}=2x-1\)
d) \(\sqrt{x+4\sqrt{x-4}}=2\)
a.
PT \(\Leftrightarrow \left\{\begin{matrix} 2x-2\geq 0\\ x^2-2x+4=(2x-2)^2\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\geq 1\\ 3x^2-6x=0\end{matrix}\right.\)
\(\Leftrightarrow \left\{\begin{matrix} x\geq 1\\ 3x(x-2)=0\end{matrix}\right.\Leftrightarrow x=2\)
b. ĐK: $x\geq 1$
PT $\Leftrightarrow \sqrt{(x-1)+2\sqrt{x-1}+1}=2$
$\Leftrightarrow \sqrt{(\sqrt{x-1}+1)^2}=2$
$\Leftrightarrow |\sqrt{x-1}+1|=2$
$\Leftrightarrow \sqrt{x-1}+1=2$
$\Leftrightarrow \sqrt{x-1}=1$
$\Leftrightarrow x=2$ (tm)
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c.
PT \(\Leftrightarrow \left\{\begin{matrix} 2x-1\geq 0\\ 2x^2-2x+1=(2x-1)^2\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\geq \frac{1}{2}\\ 2x^2-2x+1=4x^2-4x+1\end{matrix}\right.\)
\(\Leftrightarrow \left\{\begin{matrix} x\geq \frac{1}{2}\\ 2x^2-2x=2x(x-1)=0\end{matrix}\right.\Leftrightarrow x=1\) (tm)
d.
ĐKXĐ: $x\geq 4$
PT $\Leftrightarrow \sqrt{(x-4)+4\sqrt{x-4}+4}=2$
$\Leftrightarrow \sqrt{(\sqrt{x-4}+2)^2}=2$
$\Leftrightarrow |\sqrt{x-4}+2|=2$
$\Leftrightarrow \sqrt{x-4}+2=2$
$\Leftrightarrow \sqrt{x-4}=0$
$\Leftrightarrow x=4$ (tm)
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a: Ta có: \(\sqrt{x^2-2x+4}=2x-2\)
\(\Leftrightarrow x^2-2x+4=4x^2-8x+4\)
\(\Leftrightarrow-3x^2+6x=0\)
\(\Leftrightarrow-3x\left(x-2\right)=0\)
\(\Leftrightarrow x=2\)
b: Ta có: \(\sqrt{x+2\sqrt{x-1}}=2\)
\(\Leftrightarrow\left|\sqrt{x-1}+1\right|=2\)
\(\Leftrightarrow\sqrt{x-1}+1=2\)
\(\Leftrightarrow x-1=1\)
hay x=2
c: Ta có: \(\sqrt{2x^2-2x+1}=2x-1\)
\(\Leftrightarrow2x^2-2x+1=4x^2-4x+1\)
\(\Leftrightarrow-2x^2+2x=0\)
\(\Leftrightarrow-2x\left(x-1\right)=0\)
hay x=1
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Xem thêm câu trả lời
1) \(x+\sqrt{1-x^2}< x\sqrt{1-x^2}\)
2)\(\dfrac{1}{\sqrt{2x^2+3x-3}}>\dfrac{1}{2x-1}\)
3)\(5\sqrt{x}+\dfrac{5}{2\sqrt{x}}< 2x+\dfrac{1}{2x}+4\)
giúp mình ạ
giải các PT sau :a) left|2x+3right|-left|xright|+left|x-1right|2x+4b) sqrt{x}-dfrac{4}{sqrt{x+2}}+sqrt{x+2}0c) sqrt{x+sqrt{2x-1}}+sqrt{x-sqrt{2x-1}}sqrt{2}d) x+sqrt{x+dfrac{1}{2}+sqrt{x+dfrac{1}{4}}}4e) sqrt{4x+3}+sqrt{2x+1}6x+sqrt{8x^2+10x+3}-16f)sqrt[3]{x-2}+sqrt{x+1}3GIÚP MÌNH VỚI MÌNH ĐANG CẦN GẤP
Đọc tiếp
giải các PT sau :
a) \(\left|2x+3\right|-\left|x\right|+\left|x-1\right|=2x+4\)
b) \(\sqrt{x}-\dfrac{4}{\sqrt{x+2}}+\sqrt{x+2}=0\)
c) \(\sqrt{x+\sqrt{2x-1}}+\sqrt{x-\sqrt{2x-1}}=\sqrt{2}\)
d) \(x+\sqrt{x+\dfrac{1}{2}+\sqrt{x+\dfrac{1}{4}}}=4\)
e) \(\sqrt{4x+3}+\sqrt{2x+1}=6x+\sqrt{8x^2+10x+3}-16\)
f)\(\sqrt[3]{x-2}+\sqrt{x+1}=3\)
GIÚP MÌNH VỚI MÌNH ĐANG CẦN GẤP
giai cac phuong trinh
a)\(2x^4+5x^3+x^2+5x+2=0\)
b)\(\sqrt{x-1}-\sqrt[3]{2-x}=1\)
c)\(x-\sqrt{x}+1=\sqrt{2x^2-30x+2}\)
d)\(2x^2+3x+7=\left(x-5\right)\sqrt{2x^2+1}\)
e)\(\sqrt{x-2}+\sqrt{4-x}=2x^2-5x-1\)
giải pt :
a, (x+5)(2-x)=3\(\sqrt{x^2+3x}\)
b, \(\sqrt[3]{\dfrac{2x}{x+1}}+\sqrt[3]{\dfrac{1}{2}+\dfrac{1}{2x}}=2\)
c,\(\sqrt[5]{\dfrac{16x}{x-1}}+\sqrt[5]{\dfrac{x-1}{16x}}=\dfrac{5}{2}\)
d, \(\sqrt{5x^2+10x+1}=7-2x-x^2\)
e, \(\sqrt{2x^2+4x+1}=1-2x-x^2\)
Giải các pt sau:
1, \(\sqrt{x^2+x+1}=2x+\sqrt{x^2-x+1}\)
2, \(2x^2+2x+6=2x\sqrt{x^2-x+1}+4\sqrt{3x+1}\)
3, \(\left(\sqrt{x+3}-\sqrt{x}\right)\left(1+\sqrt{x^2+3x}\right)=3\)
4, \(\sqrt{2x^2-1}+\sqrt{x^2-3x-2}=\sqrt{2x^2-2x+3}+\sqrt{x^2-x+2}\)
5, \(13\sqrt{x-1}+9\sqrt{x+1}=16x\)
8 tháng 8 2021 lúc 8:57
\(\sqrt{x-\sqrt{x^2-1}}+\sqrt{x+\sqrt{x^2-1}}=2\)
\(2\left(1-x\right)\sqrt{x^2+2x-1}=x^2-2x-1\)
a, ĐK: \(x\ge1\)
\(\sqrt{x-\sqrt{x^2-1}}+\sqrt{x+\sqrt{x^2-1}}=2\)
\(\Leftrightarrow\sqrt{2x-2\sqrt{x^2-1}}+\sqrt{2x+2\sqrt{x^2-1}}=2\sqrt{2}\)
\(\Leftrightarrow\sqrt{x-1+x+1-2\sqrt{\left(x-1\right)\left(x+1\right)}}+\sqrt{x-1+x+1+2\sqrt{\left(x-1\right)\left(x+1\right)}}=2\sqrt{2}\)
\(\Leftrightarrow\sqrt{\left(\sqrt{x-1}-\sqrt{x+1}\right)^2}+\sqrt{\left(\sqrt{x-1}+\sqrt{x+1}\right)^2}=2\sqrt{2}\)
\(\Leftrightarrow\sqrt{x+1}-\sqrt{x-1}+\sqrt{x-1}+\sqrt{x+1}=2\sqrt{2}\)
\(\Leftrightarrow2\sqrt{x+1}=2\sqrt{2}\)
\(\Leftrightarrow x+1=2\)
\(\Leftrightarrow x=1\left(tm\right)\)
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b, ĐK: \(x\ge-1+\sqrt{2},x\le-1-\sqrt{2}\)
Đặt \(\sqrt{x^2+2x-1}=t\left(t\ge0\right)\)
\(pt\Leftrightarrow2\left(1-x\right)t=t^2-4x\)
\(\Leftrightarrow t^2-4x+2xt-2t=0\)
\(\Leftrightarrow\left(t-2\right)\left(2x+t\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}t=2\\t=-2x\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x^2+2x-1}=2\\\sqrt{x^2+2x-1}=-2x\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2+2x-5=0\\\sqrt{x^2+2x-1}=-2x\left(vn\right)\end{matrix}\right.\)
\(\Leftrightarrow x=-1\pm\sqrt{6}\left(tm\right)\)
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