\(\sqrt{x-5}-3\sqrt{x^2-25}=0\)
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PN
7 tháng 10 2023
\(a,\dfrac{3}{\sqrt{x}-5}+\dfrac{20-2\sqrt{x}}{x-25}\\ =\dfrac{3}{\sqrt{x}-5}+\dfrac{20-2\sqrt{x}}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}\\ =\dfrac{3\left(\sqrt{x}+5\right)}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}+\dfrac{20-2\sqrt{x}}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}\\ =\dfrac{3\sqrt{x}+15+20-2\sqrt{x}}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}\\ =\dfrac{\sqrt{x}+35}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}\)
PN
7 tháng 10 2023
\(b,\dfrac{\sqrt{x}}{\sqrt{x}-3}+\dfrac{2\sqrt{x}-2}{x-9}\\ =\dfrac{\sqrt{x}}{\sqrt{x}-3}+\dfrac{2\sqrt{x}-2}{x-9}\\ =\dfrac{\sqrt{x}\left(\sqrt{x}+3\right)}{x-9}+\dfrac{2\sqrt{x}-2}{x-9}\\ =\dfrac{x+3\sqrt{x}+2\sqrt{x}-2}{x-9}\\ =\dfrac{x-5\sqrt{x}-2}{x-9}\)
1 tháng 7 2023
Sửa đề: căn x-5/căn x-3
a: \(A=\left(\dfrac{\sqrt{x}}{\sqrt{x}+5}-1\right):\dfrac{25-x-x+9+x-25}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{\sqrt{x}-\sqrt{x}-5}{\sqrt{x}+5}\cdot\dfrac{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}{-\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\dfrac{5}{\sqrt{x}+3}\)
b: x-5căn x+6=0
=>căn x=2 hoặc căn x=3
=>x=9(loại) hoặc x=4(nhận)
Khi x=4 thì A=5/(2+3)=5/5=1
NV
Nguyễn Việt Lâm
Giáo viên
23 tháng 8 2021
\(\sqrt{2-\sqrt{3}}\left(\sqrt{6}+\sqrt{2}\right)=\sqrt{4-2\sqrt{3}}\left(\sqrt{3}+1\right)=\sqrt{\left(\sqrt{3}-1\right)^2}\left(\sqrt{3}+1\right)\)
\(=\left|\sqrt{3}-1\right|\left(\sqrt{3}+1\right)=\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)=3-1=2\)
\(\dfrac{x-25}{\sqrt{x}-5}-\dfrac{x+4\sqrt{x}+4}{\sqrt{x}+2}=\dfrac{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}{\sqrt{x}-5}-\dfrac{\left(\sqrt{x}+2\right)^2}{\sqrt{x}+2}\)
\(=\sqrt{x}+5-\left(\sqrt{x}+2\right)=5-2=3\)
23 tháng 8 2021
a: Ta có: \(\sqrt{2-\sqrt{3}}\cdot\left(\sqrt{6}+\sqrt{2}\right)\)
\(=\sqrt{4-2\sqrt{3}}\cdot\left(\sqrt{3}+1\right)\)
\(=\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)\)
=3-1
=2
b: Ta có: \(\dfrac{x-25}{\sqrt{x}-5}-\dfrac{x+4\sqrt{x}+4}{\sqrt{x}+2}\)
\(=\sqrt{x}+5-\sqrt{x}-2\)
=3
7 tháng 10 2023
a) \(\dfrac{\sqrt{x}}{\sqrt{x}+3}+\dfrac{2\sqrt{x}}{\sqrt{x}-3}-\dfrac{3x+9}{x-9}\left(x\ge0;x\ne0\right)\)
\(=\dfrac{\sqrt{x}.\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right).\left(\sqrt{x-3}\right)}+\dfrac{2\sqrt{x}.\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right).\left(\sqrt{x}+3\right)}-\dfrac{3x+9}{\left(\sqrt{x}-3\right).\left(\sqrt{x+3}\right)}\)
\(=\dfrac{x-3\sqrt{x}+2x+6\sqrt{x}-3x-9}{\left(\sqrt{x}+3\right).\left(\sqrt{x-3}\right)}\)
\(=\dfrac{3\sqrt{x}-9}{\left(\sqrt{x}+3\right).\left(\sqrt{x}-3\right)}\)
\(=\dfrac{3.\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right).\left(\sqrt{x}-3\right)}\)
\(=\dfrac{3}{\sqrt{x}+3}\)
7 tháng 10 2023
b) \(\dfrac{3}{\sqrt{x}-1}-\dfrac{\sqrt{x}+5}{x-1}\left(x\ge0;x\ne1\right)\)
\(=\dfrac{3.\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right).\left(\sqrt{x}+1\right)}-\dfrac{\sqrt{x}+5}{\left(\sqrt{x}-1\right).\left(\sqrt{x}+1\right)}\)
\(=\dfrac{3\sqrt{x}+3-\sqrt{x}-5}{\left(\sqrt{x}-1\right).\left(\sqrt{x}+1\right)}\)
\(=\dfrac{2\sqrt{x}-2}{\left(\sqrt{x}-1\right).\left(\sqrt{x}+1\right)}\)
\(=\dfrac{2.\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right).\left(\sqrt{x}+1\right)}\)
\(=\dfrac{2}{\sqrt{x}+1}\)
KV
19 tháng 6 2023
√(x² + x + 1) = 1
⇔ x² + x + 1 = 1
⇔ x² + x = 0
⇔ x(x + 1) = 0
⇔ x = 0 hoặc x + 1 = 0
*) x + 1 = 0
⇔ x = -1
Vậy x = 0; x = -1
--------------------
√(x² + 1) = -3
Do x² ≥ 0 với mọi x
⇒ x² + 1 > 0 với mọi x
⇒ x² + 1 = -3 là vô lý
Vậy không tìm được x thỏa mãn yêu cầu
--------------------
√(x² - 10x + 25) = 7 - 2x
⇔ √(x - 5)² = 7 - 2x
⇔ |x - 5| = 7 - 2x (1)
*) Với x ≥ 5, ta có
(1) ⇔ x - 5 = 7 - 2x
⇔ x + 2x = 7 + 5
⇔ 3x = 12
⇔ x = 4 (loại)
*) Với x < 5, ta có:
(1) ⇔ 5 - x = 7 - 2x
⇔ -x + 2x = 7 - 5
⇔ x = 2 (nhận)
Vậy x = 2
--------------------
√(2x + 5) = 5
⇔ 2x + 5 = 25
⇔ 2x = 20
⇔ x = 20 : 2
⇔ x = 10
Vậy x = 10
-------------------
√(x² - 4x + 4) - 2x +5 = 0
⇔ √(x - 2)² - 2x + 5 = 0
⇔ |x - 2| - 2x + 5 = 0 (2)
*) Với x ≥ 2, ta có:
(2) ⇔ x - 2 - 2x + 5 = 0
⇔ -x + 3 = 0
⇔ x = 3 (nhận)
*) Với x < 2, ta có:
(2) ⇔ 2 - x - 2x + 5 = 0
⇔ -3x + 7 = 0
⇔ 3x = 7
⇔ x = 7/3 (loại)
Vậy x = 3
GH
18 tháng 6 2023
1)
\(\Leftrightarrow x^2+x+1=1^2=1\\ \Leftrightarrow x^2+x=0\\ \Leftrightarrow x\left(x+1\right)=0\\ \Rightarrow\left\{{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)
2) Do \(x^2+1>0\forall x\) nên \(x\in\varnothing\)
3)
\(\Leftrightarrow\sqrt{\left(x-5\right)^2}=7-2x\\ \Leftrightarrow\left|x-5\right|=7-2x\)
Nếu \(x\ge5\) thì
\(\Leftrightarrow x-5-7+2x=0\\ \Leftrightarrow3x-12=0\\ \Leftrightarrow3x=12\\ \Rightarrow x=4\)
=> Loại trường hợp này
Nếu \(x< 5\) thì
\(\Leftrightarrow5-x-7+2x=0\\ \Leftrightarrow x-2=0\\ \Rightarrow x=2\)
=> Nhận trường hợp này
Vậy x = 2
4)
\(\Leftrightarrow2x+5=5^2=25\\ \Leftrightarrow2x=25-5=20\\ \Rightarrow x=\dfrac{20}{2}=10\)
5)
\(\Leftrightarrow\sqrt{\left(x-2\right)^2}-2x+5=0\\ \Leftrightarrow\left|x-2\right|-2x+5=0\)
Nếu \(x\ge2\) thì
\(\Leftrightarrow x-2-2x+5=0\\ \Leftrightarrow3-x=0\\ \Rightarrow x=3\)
=> Nhận trường hợp này
Nếu \(x< 2\) thì
\(\Leftrightarrow2-x-2x+5=0\\ \Leftrightarrow7-3x=0\\ \Leftrightarrow3x=7\\ \Rightarrow x=\dfrac{7}{3}\)
=> Loại trường hợp này
Vậy x = 3
1 tháng 11 2023
a: ĐKXĐ: x>=5
\(\sqrt{4x-20}+\sqrt{x-5}-\dfrac{1}{3}\cdot\sqrt{9x-45}=4\)
=>\(2\sqrt{x-5}+\sqrt{x-5}-\dfrac{1}{3}\cdot3\sqrt{x-5}=4\)
=>\(2\sqrt{x-5}=4\)
=>\(\sqrt{x-5}=2\)
=>x-5=4
=>x=9(nhận)
b: ĐKXĐ: x>=1/2
\(\sqrt{2x-1}-\sqrt{8x-4}+5=0\)
=>\(\sqrt{2x-1}-2\sqrt{2x-1}+5=0\)
=>\(5-\sqrt{2x-1}=0\)
=>\(\sqrt{2x-1}=5\)
=>2x-1=25
=>2x=26
=>x=13(nhận)
c: \(\sqrt{x^2-10x+25}=2\)
=>\(\sqrt{\left(x-5\right)^2}=2\)
=>\(\left|x-5\right|=2\)
=>\(\left[{}\begin{matrix}x-5=2\\x-5=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=7\\x=3\end{matrix}\right.\)
d: \(\sqrt{x^2-14x+49}-5=0\)
=>\(\sqrt{x^2-2\cdot x\cdot7+7^2}=5\)
=>\(\sqrt{\left(x-7\right)^2}=5\)
=>|x-7|=5
=>\(\left[{}\begin{matrix}x-7=5\\x-7=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=12\\x=2\end{matrix}\right.\)
H
1 tháng 11 2023
\(a,\sqrt{4x-20}+\sqrt{x-5}-\dfrac{1}{3}\sqrt{9x-45}=4\left(đkxđ:x\ge5\right)\\ \Leftrightarrow\sqrt{4\left(x-5\right)}+\sqrt{x-5}-\dfrac{1}{3}\sqrt{9\left(x-5\right)}=4\\ \Leftrightarrow2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}=4\\ \Leftrightarrow2\sqrt{x-5}=4\\ \Leftrightarrow\sqrt{x-5}=2\\ \Leftrightarrow x-5=4\\ \Leftrightarrow x=9\left(tm\right)\)
\(b,\sqrt{2x-1}-\sqrt{8x-4}+5=0\left(đkxđ:x\ge\dfrac{1}{2}\right)\\ \Leftrightarrow\sqrt{2x-1}-\sqrt{4\left(2x-1\right)}=-5\\ \Leftrightarrow\sqrt{2x-1}-2\sqrt{2x-1}=-5\\ \Leftrightarrow-\sqrt{2x-1}=-5\\ \Leftrightarrow\sqrt{2x-1}=5\\ \Leftrightarrow2x-1=25\\ \Leftrightarrow2x=26\\ \Leftrightarrow x=13\left(tm\right)\)
\(c,\sqrt{x^2-10x+25}=2\\ \Leftrightarrow\sqrt{\left(x-5\right)^2}=2\\ \Leftrightarrow\left|x-5\right|=2\\ \Leftrightarrow\left[{}\begin{matrix}x-5=2\\x-5=-2\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=7\\x=3\end{matrix}\right.\)
\(d,\sqrt{x^2-14x+49}-5=0\\ \Leftrightarrow\sqrt{\left(x-7\right)^2}=5\\ \Leftrightarrow\left|x-7\right|=5\\ \Leftrightarrow\left[{}\begin{matrix}x-7=5\\x-7=-5\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=12\\x=2\end{matrix}\right.\)
28 tháng 9 2021
1) \(\Leftrightarrow\sqrt{\left(x+5\right)^2}=4\)
\(\Leftrightarrow\left|x+5\right|=4\)
\(\Leftrightarrow\left[{}\begin{matrix}x+5=4\\x+5=-4\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-9\end{matrix}\right.\)
2) \(ĐK:x\ge2\)
\(\Leftrightarrow\sqrt{x-2}=2\)
\(\Leftrightarrow x-2=4\Leftrightarrow x=6\left(tm\right)\)
3) \(\Leftrightarrow\left(x^2-x+4\right)-\sqrt{x^2-x+4}+\dfrac{1}{4}=\dfrac{9}{4}\)
\(\Leftrightarrow\left(\sqrt{x^2-x+4}-\dfrac{1}{2}\right)^2=\dfrac{9}{4}\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x^2-x+4}-\dfrac{1}{2}=\dfrac{3}{2}\\\sqrt{x^2-x+4}-\dfrac{1}{2}=-\dfrac{3}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x^2-x+4}=2\\\sqrt{x^2-x+4}=-1\left(VLý\right)\end{matrix}\right.\)
\(\Leftrightarrow x^2-x+4=4\Leftrightarrow x\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
4) \(ĐK:x\ge0\)
\(\Leftrightarrow3\sqrt{x}-3=\sqrt{x}+2\)
\(\Leftrightarrow\sqrt{x}=\dfrac{5}{2}\Leftrightarrow x=\dfrac{25}{4}\left(tm\right)\)
18 tháng 10 2021
\(d,ĐK:x\ge0\\ PT\Leftrightarrow\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=2\\\sqrt{x}=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\left(tm\right)\\x=9\left(tm\right)\end{matrix}\right.\\ e,ĐK:x\ge1\\ PT\Leftrightarrow\sqrt{x-1}+\dfrac{3}{2}\cdot2\sqrt{x-1}-\dfrac{2}{5}\cdot5\sqrt{x-1}=4\\ \Leftrightarrow2\sqrt{x-1}=4\Leftrightarrow\sqrt{x-1}=2\\ \Leftrightarrow x-1=4\Leftrightarrow x=5\left(tm\right)\\ f,ĐK:x\ge5\\ PT\Leftrightarrow\sqrt{x-5}+2\sqrt{x-5}-\dfrac{1}{3}\cdot3\sqrt{x-5}=6\\ \Leftrightarrow2\sqrt{x-5}=6\Leftrightarrow\sqrt{x-5}=3\\ \Leftrightarrow x-5=9\Leftrightarrow x=14\left(tm\right)\)
19 tháng 9 2021
1) \(\sqrt{5-2x}=6\left(đk:x\le\dfrac{5}{2}\right)\)
\(\Leftrightarrow5-2x=36\)
\(\Leftrightarrow2x=-31\Leftrightarrow x=-\dfrac{31}{2}\left(tm\right)\)
2) \(\sqrt{2-x}=\sqrt{x+1}\left(đk:2\ge x\ge-1\right)\)
\(\Leftrightarrow2-x=x+1\)
\(\Leftrightarrow2x=1\Leftrightarrow x=\dfrac{1}{2}\left(tm\right)\)
3) \(\Leftrightarrow\sqrt{\left(2x+1\right)^2}=6\)
\(\Leftrightarrow\left|2x+1\right|=6\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+1=6\\2x+1=-6\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-\dfrac{7}{2}\end{matrix}\right.\)
4) \(\sqrt{x^2-10x+25}=x-2\left(đk:x\ge2\right)\)
\(\Leftrightarrow\sqrt{\left(x-5\right)^2}=x-2\)
\(\Leftrightarrow\left|x-5\right|=x-2\)
\(\Leftrightarrow\left[{}\begin{matrix}x-5=x-2\left(x\ge5\right)\\x-5=2-x\left(2\le x< 5\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}5=2\left(VLý\right)\\x=\dfrac{7}{2}\left(tm\right)\end{matrix}\right.\)
ĐKXĐ : \(x\ge\pm5\)
\(\sqrt{x-5}-3\sqrt{x^2-25}=0\)
\(\Leftrightarrow\sqrt{x-5}\left(1-3\sqrt{x+5}\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}\sqrt{x-5}=0\\1-3\sqrt{x+5}=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x-5=0\\3\sqrt{x+5}=1\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=5\\\sqrt{x+5}=\frac{1}{3}\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=5\\x+5=\frac{1}{9}\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=5\\x=-\frac{44}{9}\end{cases}\left(tm\right)}\)
Vậy ....
đk: \(x\ge5\)
Ta có: \(\sqrt{x-5}-3\sqrt{x^2-25}=0\)
\(\Leftrightarrow\sqrt{x-5}=3\sqrt{x^2-25}\)
\(\Leftrightarrow x-5=9\left(x^2-25\right)\)
\(\Leftrightarrow9x^2-x-220=0\)
\(\Leftrightarrow\left(x-5\right)\left(9x+44\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=5\left(tm\right)\\x=-\frac{44}{9}\left(ktm\right)\end{cases}}\)
Vậy x = 5