\(\sqrt{x^2+2}=\sqrt{1-2x}\)
giúp tui vs
\(\dfrac{2+\sqrt{2}}{\sqrt{2}+1}+\dfrac{1}{\sqrt{3}+\sqrt{2}}\)
giúp tui vs pls tui đag cần gấp lắm
\(\dfrac{2+\sqrt{2}}{\sqrt{2}+1}+\dfrac{1}{\sqrt{3}+\sqrt{2}}\)
\(=\dfrac{\sqrt{2}\left(\sqrt{2}+1\right)}{\sqrt{2}+1}+\dfrac{\sqrt{3}-\sqrt{2}}{3-2}\)
\(=\sqrt{2}+\sqrt{3}-\sqrt{2}=\sqrt{3}\)
☘☘☘ CỨU TUI VỚI ☘☘☘
GIẢI PT:\(\left\{{}\begin{matrix}x-y=\sqrt{x-y+2}\\\sqrt{3x-y-3}+4\sqrt{x+3}=x^2+y^2+2x+6\end{matrix}\right.\)
ai biết làm ko ạ ,Giúp mình vs
Rút gọn biểu thức sau: Q=(\(\dfrac{1}{2\sqrt{x}+1}+\dfrac{1}{2\sqrt{x}-1}\)):\(\dfrac{1}{1-4x}\) với x≥0,x≠\(\dfrac{1}{4}\)
giúp tui giải bài này vs tui c.ơn
\(Q=\left(\dfrac{1}{2\sqrt{x}+1}+\dfrac{1}{2\sqrt{x}-1}\right):\dfrac{1}{1-4x}\)
\(=\left(\dfrac{2\sqrt{x}-1}{\left(2\sqrt{x}+1\right)\left(2\sqrt{x}-1\right)}+\dfrac{2\sqrt{x}+1}{\left(2\sqrt{x}+1\right)\left(2\sqrt{x}-1\right)}\right).\left(1-4x\right)\)
\(=\left(\dfrac{2\sqrt{x}-1+2\sqrt{x}+1}{4x-1}\right)\left(1-4x\right)\)
\(=\dfrac{-4\sqrt{x}.\left(4x-1\right)}{4x-1}=-4\sqrt{x}\)
\(Q=\left(\dfrac{1}{2\sqrt{x}+1}+\dfrac{1}{2\sqrt{x}-1}\right):\dfrac{1}{1-4x}\left(dkxd:x\ge0;x\ne\dfrac{1}{4}\right)\)
\(=\left[\dfrac{2\sqrt{x}-1}{\left(2\sqrt{x}-1\right)\left(2\sqrt{x}+1\right)}+\dfrac{2\sqrt{x}+1}{\left(2\sqrt{x}-1\right)\left(2\sqrt{x}+1\right)}\right]\cdot\left(1-4x\right)\)
\(=\dfrac{2\sqrt{x}-1+2\sqrt{x}+1}{4x-1}\cdot\left[-\left(4x-1\right)\right]\)
\(=4\sqrt{x}\cdot\left(-1\right)\)
\(=-4\sqrt{x}\)
1) giải pt:
a) \(\sqrt{3x+10}=4\)
b) \(\sqrt{9x^2-6x+1}=\sqrt{x^2+8x+16}\)
c) \(\sqrt{2x+1}=3\)
d) \(\sqrt{2x+1}+1=x\)
giúp mk vs ah
a) \(\sqrt{3x+10}=4\left(đk:x\ge-\dfrac{10}{3}\right)\Leftrightarrow3x+10=16\Leftrightarrow x=2\)
b) \(\sqrt{9x^2-6x+1}=\sqrt{x^2+8x+16}\Leftrightarrow\sqrt{\left(3x-1\right)^2}=\sqrt{\left(x+4\right)^2}\Leftrightarrow3x-1=x+4\Leftrightarrow2x=5\Leftrightarrow x=\dfrac{5}{2}\)
c) \(\sqrt{2x+1}=3\left(đk:x\ge-\dfrac{1}{2}\right)\Leftrightarrow2x+1=9\Leftrightarrow x=4\)
d) \(\sqrt{2x+1}+1=x\left(đk:x\ge1\right)\Leftrightarrow\sqrt{2x+1}=x-1\Leftrightarrow2x+1=x^2-2x+1\Leftrightarrow x^2-4x=0\Leftrightarrow x\left(x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\)\(\Leftrightarrow x=4\)(do \(x\ge1\))
a: Ta có: \(\sqrt{3x+10}=4\)
\(\Leftrightarrow3x+10=16\)
\(\Leftrightarrow3x=6\)
hay x=2
b: Ta có: \(\sqrt{9x^2-6x+1}=\sqrt{x^2+8x+16}\)
\(\Leftrightarrow\left|3x-1\right|=\left|x+4\right|\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-1=x+4\\3x-1=-x-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=5\\4x=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-\dfrac{3}{4}\end{matrix}\right.\)
c: Ta có: \(\sqrt{2x+1}=3\)
\(\Leftrightarrow2x+1=9\)
\(\Leftrightarrow x=4\)
Giúp tui.
Cho biểu thức: \(M=x-\dfrac{2x-2\sqrt{x}}{\sqrt{x}-1}+\dfrac{x\sqrt{x+1}}{x-\sqrt{x}+1}+1\)
a. Rút gọn biểu thức M.
b. Tính giá trị nhỏ nhất của biểu thức M
a:
ĐKXĐ: x>=0; x<>1
Sửa đề: \(M=x-\dfrac{2x-2\sqrt{x}}{\sqrt{x}-1}+\dfrac{x\sqrt{x}+1}{x-\sqrt{x}+1}+1\)
\(=x-\dfrac{2\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}-1}+\dfrac{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{x-\sqrt{x}+1}+1\)
\(=x-2\sqrt{x}+1+\sqrt{x}+1=x-\sqrt{x}+2\)
b: \(M=x-\sqrt{x}+2\)
\(=x-\sqrt{x}+\dfrac{1}{4}+\dfrac{7}{4}\)
\(=\left(\sqrt{x}-\dfrac{1}{2}\right)^2+\dfrac{7}{4}>=\dfrac{7}{4}\forall x\) thỏa mãn ĐKXĐ
Dấu '=' xảy ra khi \(\sqrt{x}-\dfrac{1}{2}=0\)
=>\(\sqrt{x}=\dfrac{1}{2}\)
=>x=1/4
Mng giúp mình vs ạ rút gọn bth này nha:
P=\(\dfrac{2x+2}{\sqrt{x}}+\dfrac{x\sqrt{x}-1}{x-\sqrt{x}}-\dfrac{x\sqrt{x+1}}{x+\sqrt{x}}\)
\(P=\dfrac{2x+2+x+\sqrt{x}+1-x+\sqrt{x}-1}{\sqrt{x}}\)
\(=\dfrac{2x+2\sqrt{x}+2}{\sqrt{x}}\)
ai có thể giúp mình giải bài này vs đc không mình đang cần rất gấp (làm chi tiết hộ mình nhé, xin cảm ơn)
Bài 4:
a, \(\sqrt{3x+4}-\sqrt{2x+1}=\sqrt{x+3}\)
b, \(\sqrt{2x-5}+\sqrt{x+2}=\sqrt{2x+1}\)
c, \(\sqrt{x+4}-\sqrt{1-x}=\sqrt{1-2x}\)
d,\(\sqrt{x+9}=5-\sqrt{2x+4}\)
Bài 5:
a, \(\sqrt{x+4\sqrt{x}+4}=5x+2\)
b, \(\sqrt{x^2-2x+1}+\sqrt{x^2+4x+4}=4\)
c, \(\sqrt{x+2\sqrt{x-1}}+\sqrt{x-2\sqrt{x-1}}=2\)
d,\(\sqrt{x-2+\sqrt{2x-5}}+\sqrt{x+2+3\sqrt{2x-5}}=7\sqrt{2}\)
Ví Dụ 1:
a, \(\sqrt{2x-1}=\sqrt{2}-1\)
b, \(\sqrt{x+5}=3-\sqrt{2}\)
c, \(\sqrt{3x^2}-\sqrt{12}=0\)
d, \(\sqrt{2}\left(x-1\right)-\sqrt{50}=0\)
Vd1:
d) Ta có: \(\sqrt{2}\left(x-1\right)-\sqrt{50}=0\)
\(\Leftrightarrow\sqrt{2}\left(x-1-5\right)=0\)
\(\Leftrightarrow x=6\)
Giải phương trình:
a) \(\sqrt{x+3-4\sqrt{x+1}}+\sqrt{x+8-6\sqrt{x-1}}=1\)
b) \(\sqrt{x+\sqrt{x-11}}+\sqrt{x-\sqrt{x-11}}=4\)
c) \(\sqrt{x+2+3\sqrt{2x-5}}+\sqrt{x-2\sqrt{2x-5}}=2\sqrt{2}\)
d) \(\sqrt{x-4}+\sqrt{6-x}=x^2-10x+27\)
e) \(\sqrt{2x+1}+\sqrt{17-2x}=x^4-8x^3+17x^2-8x+22\)
f) \(\sqrt{3x^2+12x+16}+\sqrt{y^2-4y+13}=5\)
g) \(\sqrt{x+x^2}+\sqrt{x-x^2}=x+1\)
ai lmmm giúp tui ikkk
\(\sqrt{x+3-4\sqrt{x-1}}+\sqrt{x+8-6\sqrt{x-1}}=1\) ( SỬA ĐỀ)
\(\sqrt{x-1-2.2.\sqrt{x-1}+4}+\sqrt{x-1-2.3.\sqrt{x-1}+9}=1\)
\(|x-1-2|+|x-1-3|=1\)
\(|x-3|+|x-4|=1\)
Với \(x\le3\)thì PT thành \(3-x+4-x=1\) \(\Rightarrow-2x=-6\Rightarrow x=3\)(thõa mãn)
Với \(3\le x< 4\)thì PT thành \(x-3+4-x=1\Leftrightarrow0x=0\Rightarrow\)Đúng với mọi x từ \(3\le x< 4\)
Với \(x\ge4\)thì PT thành \(x-3+x-4=1\Leftrightarrow2x=8\Leftrightarrow x=4\)(thõa mãn)
Vậy \(3\le x\le4\)
\(\sqrt{x^2-2x+2018}+2019.\sqrt{x^4+2x^2+2020}=2018\)
Giúp mik vs ạ
cái . ở giữa 2019 . \(\sqrt{x^4}\) là x hay bài khác vậy ?