Giair pt: \(\sqrt{2x+2\sqrt{2x-1}}-\sqrt{2x-2\sqrt{2x-1}}=\sqrt{2x-1}-10\)
Những câu hỏi liên quan
Giải pt:\(\sqrt{2x^2+4x+6}+\sqrt{x^4-2x^2+10}=4-2x-x^2\)
\(\sqrt{x^2+2x^2+1}=\sqrt{x^2+10x+25}-10x-22\)
Giair bài nào cx được nha vì mình đang cần gấp
Giair pt sau:
a, \(x^2+\sqrt{2x^2+4x+3}=6-2x\)
b, \(\sqrt{x^2-9x+24}-\sqrt{6x^2-59x+149}=5-x\)
c, \(2x^2+4x+3\sqrt{3-2x-x^2}=1\)
Giair pt : 2x2 + 2x + 1 = ( 2x + 3 )(\(\sqrt{x^2+x+2}\) - 1 )
giải pt: \(\sqrt{x-\sqrt{2x-1}}+\sqrt{x+\sqrt{2x-1}}=\sqrt{2}x\)
Từ pt suy ra \(x\ge0\).
PT \(\Leftrightarrow\sqrt{2x-2\sqrt{2x-1}}+\sqrt{2x+2\sqrt{2x-1}}=2x\)
\(\Leftrightarrow\left|\sqrt{2x-1}-1\right|+\left|\sqrt{2x-1}+1\right|=2x\). (*)
+) \(\sqrt{2x-1}-1\ge0\Leftrightarrow x\ge1\): Khi đó (*) tương đương \(2\sqrt{2x-1}=2x\Leftrightarrow x^2-2x+1=0\Leftrightarrow x=1\) (thoả mãn)
+) \(\sqrt{2x-1}-1< 0\Leftrightarrow x< 1\): Khi đó (*) tương đương \(2=2x\Leftrightarrow x=1\), vô lí.
Vậy x = 1
Đúng 1
Bình luận (0)
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
Giair cacs pt sau:
a. \(x-\sqrt{x^4-2x^2+1}=1\)
b. \(\sqrt{x-2}+\sqrt{x-3}=-5\)
c. \(\sqrt{x^2-2x+1}+\sqrt{x^2-6x+9}=1\)
d. \(\sqrt{x+5}+\sqrt{2-x}=x^2-25\)
e. \(\sqrt{x+3-4\sqrt{x-1}}+\sqrt{x+8-6\sqrt{x-1}}=1\)
f. \(\sqrt{8x+1}+\sqrt{3x-5}=\sqrt{7x+4}+\sqrt{2x-2}\)
Giải PT: \(\sqrt{2x^2-9x+4}+3\sqrt{2x-1}=\sqrt{2x^2+21x-11}\)
\(ĐK:\left\{{}\begin{matrix}x\le\dfrac{1}{2};4\le x\\\dfrac{1}{2}\le x\\x\le-11;\dfrac{1}{2}\le x\end{matrix}\right.\Leftrightarrow x\le-11;4\le x\)
\(PT\Leftrightarrow\sqrt{\left(x-4\right)\left(2x-1\right)}+3\sqrt{2x-1}-\sqrt{\left(2x-1\right)\left(x+11\right)}=0\\ \Leftrightarrow\sqrt{2x-1}\left(\sqrt{x-4}-\sqrt{x+11}+3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}2x-1=0\\\sqrt{x-4}-\sqrt{x+11}=-3\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\left(tm\right)\\x-4+x+11-2\sqrt{x^2+7x-44}=9\left(1\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow2\sqrt{x^2+7x-44}=2x-2\\ \Leftrightarrow\sqrt{x^2+7x-44}=x-1\\ \Leftrightarrow x^2+7x-44=x^2-2x+1\\ \Leftrightarrow9x=45\Leftrightarrow x=5\left(tm\right)\)
Vậy \(S=\left\{\dfrac{1}{2};5\right\}\)
Đúng 5
Bình luận (0)
Giải PT: \(\sqrt{2x^2-9x+4}+3\sqrt{2x-1}=\sqrt{2x^2+21x-11}\)
https://hoc24.vn/cau-hoi/giai-pt-sqrt2x2-9x43sqrt2x-1sqrt2x221x-11.2005877637936
làm r nha :vv
Đúng 2
Bình luận (0)
cho g(x)=\(\dfrac{2}{3}\left(\sqrt{x+3}\right)^3\) giair pt g'(x)+\(\sqrt{2x-1}\)=3
\(g'=2\left(\sqrt{x+3}\right)^2.\left(\sqrt{x+3}\right)'=2\left(x+3\right).\dfrac{1}{2\sqrt{x+3}}=\sqrt{x+3}\)
\(g'\left(x\right)+\sqrt{2x-1}=3\Leftrightarrow\sqrt{x+3}+\sqrt{2x-1}=3\)
\(DKXD:x\ge\dfrac{1}{2}\)
\(pt\Leftrightarrow x+3+2x-1+2\sqrt{\left(x+3\right)\left(2x-1\right)}=9\)
\(\Leftrightarrow2\sqrt{\left(x+3\right)\left(2x-1\right)}=7-3x\)
\(\Leftrightarrow4\left(2x^2+5x-3\right)=49-42x+9x^2\)
\(\Leftrightarrow x^2-62x+61=0\Leftrightarrow\left[{}\begin{matrix}x=61\left(loai\right)\\x=1\end{matrix}\right.\)
Đúng 0
Bình luận (0)
g'(x) = \(\sqrt{x+3}\)
ta có phương trình : \(\sqrt{x+3}\) + \(\sqrt{2x-1}\) =3 ( ĐK : x\(\ge\)\(\dfrac{1}{2}\))
\(\Leftrightarrow\) x+3 +2x-1 +\(2\sqrt{\left(x+3\right)\left(2x-1\right)}\) = 9
\(\Leftrightarrow\) \(2\sqrt{\left(x+3\right)\left(2x-1\right)}\) = 7-3x
\(\Leftrightarrow\) 4(2x2 +5x -3) = 49 - 42x +9x2
\(\Leftrightarrow\) x2 - 62x +61 = 0 \(\left\{{}\begin{matrix}x=61\\x=1\end{matrix}\right.\)
Đúng 0
Bình luận (1)