\(\sqrt{-x}\)
\(\sqrt{16x^2-25}\)
\(\sqrt{4x^2-49}\)
\(\sqrt{8-x^2}\)
\(\sqrt{x^2-12}\)
làm hộ với , mik mới học lên chưa hiểu lắm
Tìm điều kiện có nghĩa:
1) \(\sqrt{16x^2-25}\)
2) \(\sqrt{4x^2-49}\)
3) \(\sqrt{8-x^2}\)
4)\(\sqrt{x^2-12}\)
5) \(\sqrt{x^2+4}\)
LÀM CHI TIẾT GIÚP MK NHÉ!
1) ĐKXĐ: \(16x^2-25\ge0\)
\(\Leftrightarrow x^2\ge\dfrac{25}{16}\)
\(\Leftrightarrow\left[{}\begin{matrix}x\ge\dfrac{5}{4}\\x\le-\dfrac{5}{4}\end{matrix}\right.\)
2) ĐKXĐ: \(4x^2-49\ge0\Leftrightarrow x^2\ge\dfrac{49}{4}\)
\(\Leftrightarrow\left[{}\begin{matrix}x\ge\dfrac{7}{2}\\x\le-\dfrac{7}{2}\end{matrix}\right.\)
3) ĐKXĐ: \(8-x^2\ge0\Leftrightarrow x^2\le8\)
\(\Leftrightarrow-2\sqrt{2}\le x\le2\sqrt{2}\)
4) ĐKXĐ: \(x^2-12\ge0\Leftrightarrow x^2\ge12\)
\(\Leftrightarrow\left[{}\begin{matrix}x\ge2\sqrt{3}\\x\le-2\sqrt{3}\end{matrix}\right.\)
5) ĐKXĐ: \(x^2+4\ge0\left(đúng\forall x\right)\)
Tìm điều kiện có nghĩa:
1) \(\sqrt{2x^2}\)
2) \(\sqrt{-x}\)
3) \(\sqrt{-x^2-3}\)
4) \(\sqrt{x^2+2x+3}\)
5) \(\sqrt{-a^2+8a-16}\)
6) \(\sqrt[]{16x^2-25}\)
7) \(\sqrt{4x^2-49}\)
8) \(\sqrt{8-x^2}\)
9) \(\sqrt{x^2-12}\)
10) \(\sqrt{x^2+2x-3}\)
11) \(\sqrt{2x^2+5x+3}\)
12) \(\sqrt{\dfrac{4}{x-1}}\)
13) \(\sqrt{\dfrac{-1}{x-3}}\)
14) \(\sqrt{\dfrac{-3}{x+2}}\)
15) \(\sqrt{\dfrac{1}{2a-1}}\)
16) \(\sqrt{\dfrac{2}{3-2a}}\)
17) \(\sqrt{\dfrac{-1}{2a-5}}\)
18) \(\sqrt{\dfrac{-2}{3-5a}}\)
19) \(\sqrt{\dfrac{-a}{5}}\)
20) \(\dfrac{1}{\sqrt{-3a}}\)
1) \(ĐK:x\in R\)
2) \(ĐK:x< 0\)
3) \(ĐK:x\in\varnothing\)
4) \(=\sqrt{\left(x+1\right)^2+2}\)
\(ĐK:x\in R\)
5) \(=\sqrt{-\left(a-4\right)^2}\)
\(ĐK:x\in\varnothing\)
Giải PT:
a) -5x+7\(\sqrt{x}\) +12=0
b) \(\dfrac{1}{3}\)\(\sqrt{4x^2-20}\) +2\(\sqrt{\dfrac{x^2-5}{9}}\) -3\(\sqrt{x^2-5}=0\)
c) \(\sqrt{9x+27}+5\sqrt{x+3}-\dfrac{3}{4}\sqrt{16x+48}=5\)
d) \(\sqrt{49x-98}-14\sqrt{\dfrac{x-2}{49}}=3\sqrt{x-2}+8\)
a. ĐKXĐ: $x\geq 0$
PT $\Leftrightarrow -5x-5\sqrt{x}+12\sqrt{x}+12=0$
$\Leftrightarrow -5\sqrt{x}(\sqrt{x}+1)+12(\sqrt{x}+1)=0$
$\Leftrightarrow (\sqrt{x}+1)(12-5\sqrt{x})=0$
Dễ thấy $\sqrt{x}+1>1$ với mọi $x\geq 0$ nên $12-5\sqrt{x}=0$
$\Leftrightarrow \sqrt{x}=\frac{12}{5}$
$\Leftrightarrow x=5,76$ (thỏa mãn)
d. ĐKXĐ: $x\geq 2$
PT $\Leftrightarrow \sqrt{49}.\sqrt{x-2}-14\sqrt{\frac{1}{49}}\sqrt{x-2}=3\sqrt{x-2}+8$
$\Leftrightarrow 7\sqrt{x-2}-2\sqrt{x-2}=3\sqrt{x-2}+8$
$\Leftrightarrow 2\sqrt{x-2}=8$
$\Leftrightarrow \sqrt{x-2}=4$
$\Leftrightarrow x=4^2+2=18$ (tm)
b. ĐKXĐ: $x^2\geq 5$
PT $\Leftrightarrow \frac{1}{3}\sqrt{4}.\sqrt{x^2-5}+2\sqrt{\frac{1}{9}}\sqrt{x^2-5}-3\sqrt{x^2-5}=0$
$\Leftrightarrow \frac{2}{3}\sqrt{x^2-5}+\frac{2}{3}\sqrt{x^2-5}-3\sqrt{x^2-5}=0$
$\Leftrightarrow -\frac{5}{3}\sqrt{x^2-5}=0$
$\Leftrightarrow \sqrt{x^2-5}=0$
$\Leftrightarrow x=\pm \sqrt{5}$
\(\sqrt{4x}\) - \(\sqrt{9x}\) + \(\sqrt{16x}\) = 2
b, \(\sqrt{4x}\) + \(2\sqrt{16x}\) - \(\sqrt{25x}\) = 1,2
c, \(\sqrt{16\left(x-1\right)}\) - \(\sqrt{x-1}\) + \(\sqrt{49\left(x-1\right)}\) =5
\(\sqrt{X+2}\)+\(\sqrt{16x+32}\)-\(\sqrt{4x+8}\)=12 tìm x
\(\sqrt{x+2}\) + \(\sqrt{16x+32}\) - \(\sqrt{4x+8}\) = 16 (đk \(x\ge\) -2)
\(\sqrt{x+2}\) + \(\sqrt{16\left(x+2\right)}\) - \(\sqrt{4\left(x+2\right)}\) = 16
\(\sqrt{x+2}\) + 4\(\sqrt{x+2}\) - 2\(\sqrt{x+2}\) = 16
( 1 + 4 - 2)\(\sqrt{x+2}\) = 16
3\(\sqrt{x+2}\) = 16
\(\sqrt{x+2}\) = \(\dfrac{16}{3}\)
\(x+2\) = \(\dfrac{256}{9}\)
\(x\) = \(\dfrac{256}{9}\) - 2
\(x\) = \(\dfrac{238}{9}\) (thỏa mãn)
Vậy \(x=\dfrac{238}{9}\)
a) \(\sqrt{4x^2-9}=2\sqrt{x+3}\)
b) \(\sqrt{4x+20}+3\sqrt{\dfrac{x-5}{9}}-\dfrac{1}{3}\sqrt{9x-45}=4\)
c) \(\dfrac{2}{3}\sqrt{9x-9}-\dfrac{1}{4}\sqrt{16x-16}+27\sqrt{\dfrac{x-1}{81}}=4\)
d)\(5\sqrt{\dfrac{9x-27}{25}}-7\sqrt{\dfrac{4x-12}{9}}-7\sqrt{x^2-9}+18\sqrt{\dfrac{9x^2-81}{81}}=0\)
\(a) \sqrt{4x^2− 9} = 2\sqrt{x + 3}\)
\(ĐK:x\ge\dfrac{3}{2}\)
\(pt\Leftrightarrow4x^2-9=4\left(x+3\right)\)
\(\Leftrightarrow4x^2-9=4x+12\)
\(\Leftrightarrow4x^2-4x-21=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1-\sqrt{22}}{2}\left(l\right)\\x=\dfrac{1+\sqrt{22}}{2}\left(tm\right)\end{matrix}\right.\)
\(b)\sqrt{4x-20}+3.\sqrt{\dfrac{x-5}{9}}-\dfrac{1}{3}\sqrt{9x-45}=4\)
\(ĐK:x\ge5\)
\(pt\Leftrightarrow2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}=4\)
\(\Leftrightarrow2\sqrt{x-5}=4\Leftrightarrow\sqrt{x-5}=2\)
\(\Leftrightarrow x-5=4\Leftrightarrow x=9\left(tm\right)\)
\(c)\dfrac{2}{3}\sqrt{9x-9}-\dfrac{1}{4}\sqrt{16x-16}+27.\sqrt{\dfrac{x-1}{81}}=4\)
ĐK:x>=1
\(pt\Leftrightarrow2\sqrt{x-1}-\sqrt{x-1}+3\sqrt{x-1}=4\)
\(\Leftrightarrow4\sqrt{x-1}=4\Leftrightarrow\sqrt{x-1}=1\)
\(\Leftrightarrow x-1=1\Leftrightarrow x=2\left(tm\right)\)
\(d)5\sqrt{\dfrac{9x-27}{25}}-7\sqrt{\dfrac{4x-12}{9}}-7\sqrt{x^2-9}+18\sqrt{\dfrac{9x^2-81}{81}}=0\)
\(ĐK:x\ge3\)
\(pt\Leftrightarrow3\sqrt{x-3}-\dfrac{14}{3}\sqrt{x-3}-7\sqrt{x^2-9}+6\sqrt{x^2-9}=0\)
\(\Leftrightarrow-\dfrac{5}{3}\sqrt{x-3}-\sqrt{x^2-9}=0\Leftrightarrow\dfrac{5}{3}\sqrt{x-3}+\sqrt{x^2-9}=0\)
\(\Leftrightarrow(\dfrac{5}{3}+\sqrt{x+3})\sqrt{x-3}=0\)
\(\Leftrightarrow\sqrt{x-3}=0\) (vì \(\dfrac{5}{3}+\sqrt{x+3}>0\))
\(\Leftrightarrow x-3=0\Leftrightarrow x=3\left(nhận\right)\)
Giaỉ phương trình:
1, x + y + 12= 4\(\sqrt{x}+6\sqrt{y-1}\)
2, \(x+y+z=2\sqrt{x-1}+2\sqrt{y-5}+2\sqrt{z+3}\)
3, \(\sqrt{3x^2+12x+13}+\sqrt{4x^2+16x+25}=-x^2-4x\\\)
4, \(\sqrt{x+3-4\sqrt{x-1}}+\sqrt{x+8+6\sqrt{x-1}}=5\)
a) 2sqrt(25(x - 3)) - 1/2 * sqrt(4x - 12) + 1/7 * sqrt(49(x - 3)) = 20 b) sqrt(x ^ 2 - 6x + 9) = 2
Giải phương trình:
a. \(3\sqrt{8x}-\sqrt{32x}+\sqrt{50x}=21\)
b. \(\sqrt{25x+50}+3\sqrt{4x+8}-2\sqrt{16x+32}=15\)
c. \(\sqrt{\left(x-2\right)^2}=12\)
d. \(\sqrt{x^2-6x+9}-3=5\)
e.\(\sqrt{\left(2x-1\right)^2}-x=3\)
f. \(\sqrt{3x-6}-x=-2\)
h. \(\sqrt{3-2x}-2=x\)
a.
ĐKXĐ: $x\geq 0$
PT $\Leftrightarrow 6\sqrt{2x}-4\sqrt{2x}+5\sqrt{2x}=21$
$\Leftrightarrow 7\sqrt{2x}=21$
$\Leftrightarrow \sqrt{2x}=3$
$\Leftrightarrow 2x=9$
$\Leftrightarrow x=\frac{9}{2}$ (tm)
b.
ĐKXĐ: $x\geq -2$
PT $\Leftrightarrow \sqrt{25(x+2)}+3\sqrt{4(x+2)}-2\sqrt{16(x+2)}=15$
$\Leftrightarrow 5\sqrt{x+2}+6\sqrt{x+2}-8\sqrt{x+2}=15$
$\Leftrightarrow 3\sqrt{x+2}=15$
$\Leftrightarrow \sqrt{x+2}=5$
$\Leftrightarrow x+2=25$
$\Leftrightarrow x=23$ (tm)
c.
$\sqrt{(x-2)^2}=12$
$\Leftrightarrow |x-2|=12$
$\Leftrightarrow x-2=12$ hoặc $x-2=-12$
$\Leftrightarrow x=14$ hoặc $x=-10$
e.
PT $\Leftrightarrow |2x-1|-x=3$
Nếu $x\geq \frac{1}{2}$ thì $2x-1-x=3$
$\Leftrightarrow x=4$ (tm)
Nếu $x< \frac{1}{2}$ thì $1-2x-x=3$
$\Leftrightarrow x=\frac{-2}{3}$ (tm)
f.
ĐKXĐ: $x\geq 2$
PT $\Leftrightarrow \sqrt{3(x-2)}-(x-2)=0$
$\Leftrightarrow \sqrt{x-2}(\sqrt{3}-\sqrt{x-2})=0$
$\Leftrightarrow \sqrt{x-2}=0$ hoặc $\sqrt{3}-\sqrt{x-2}=0$
$\Leftrightarrow x=2$ hoặc $x=5$ (tm)
h. ĐKXĐ: $x\leq \frac{3}{2}$
PT $\Leftrightarrow \sqrt{3-2x}=x+2$
\(\Rightarrow \left\{\begin{matrix} x+2\geq 0\\ 3-2x=(x+2)^2=x^2+4x+4\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\geq -2\\ x^2+6x+1=0\end{matrix}\right.\)
\(\Leftrightarrow x=-3+2\sqrt{2}\) (tm)
Vậy.......