bài 1:giải phương trình
a)\(\sqrt{9x^2+12x+4}-4\) = 0
b)\(3\sqrt{x+3}-\sqrt{x-5}\) = 0
c)\(x-7+\sqrt{x-1}\) = 0
giải cụ thể chi tiết giúp mk vớiiiiii ạ
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14 tháng 12 2021
\(M=6\sqrt{2x}-\sqrt{2x}+2\sqrt{2x}-4\sqrt{2x}=3\sqrt{2x}\)
TN
2
28 tháng 8 2021
â) \(\sqrt{x+9}=7\\ \Rightarrow x+9=49\\ \Rightarrow x=40\)
b) \(\sqrt{x-4}=4-x\\ \Rightarrow x-4=16-8x+x^2\\ \Rightarrow x^2-9x+20=0\\ \Rightarrow\left(x-4\right)\left(x-5\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=4\\x=5\end{matrix}\right.\)
c) \(\sqrt{x^2-12x+36}=81\\ \Rightarrow x-6=81\\ \Rightarrow x=87\)
28 tháng 8 2021
a: Ta có: \(\sqrt{x+9}=7\)
\(\Leftrightarrow x+9=49\)
hay x=40
b: Ta có: \(\sqrt{x-4}=4-x\)
\(\Leftrightarrow\left(x-4\right)^2-\left(x-4\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(x-5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=4\left(nhận\right)\\x=5\left(loại\right)\end{matrix}\right.\)
c: Ta có: \(\sqrt{x^2-12x+36}=81\)
\(\Leftrightarrow\left|x-6\right|=81\)
\(\Leftrightarrow\left[{}\begin{matrix}x-6=81\\x-6=-81\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=87\\x=-75\end{matrix}\right.\)
6 tháng 2 2022
a: \(\Leftrightarrow\dfrac{1}{3}\sqrt{x-2}-\dfrac{2}{3}\cdot3\sqrt{x-2}+6\cdot\dfrac{\sqrt{x-2}}{9}=-4\)
\(\Leftrightarrow\sqrt{x-2}=4\)
=>x-2=16
hay x=18
b: \(\Leftrightarrow\left|3x+2\right|=4x\)
\(\Leftrightarrow\left[{}\begin{matrix}3x+2=4x\left(x>=-\dfrac{2}{3}\right)\\3x+2=-4x\left(x< -\dfrac{2}{3}\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\left(nhận\right)\\x=-\dfrac{2}{7}\left(nhận\right)\end{matrix}\right.\)
c: \(\Leftrightarrow3\sqrt{x-2}-2\sqrt{x-2}+3\sqrt{x-2}=40\)
\(\Leftrightarrow4\sqrt{x-2}=40\)
=>x-2=100
hay x=102
d: =>5x-6=9
hay x=3
MH
6 tháng 2 2022
\(\dfrac{1}{3}\sqrt{x-2}-\dfrac{2}{3}\sqrt{9x-18}+6\sqrt{\dfrac{x-2}{81}}=-4\) (đk: x≥2)
\(\dfrac{1}{3}\sqrt{x-2}-\dfrac{2}{3}\sqrt{9\left(x-2\right)}+6\sqrt{\dfrac{1}{81}\left(x-2\right)}=-4\)
\(\dfrac{1}{3}\sqrt{x-2}-2\sqrt{x-2}+\dfrac{2}{3}\sqrt{x-2}=-4\)
\(\dfrac{1}{3}\sqrt{x-2}-\dfrac{4}{3}\sqrt{x-2}=-4\)
\(-\sqrt{x-2}=-4\)
\(\sqrt{x-2}=4\)
\(\left|x-2\right|=16\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=16\\x-2=-16\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=18\left(TM\right)\\x=-14\left(L\right)\end{matrix}\right.\)
2
20 tháng 5 2023
`a)\sqrt{3x}-5\sqrt{12x}+7\sqrt{27x}=12` `ĐK: x >= 0`
`<=>\sqrt{3x}-10\sqrt{3x}+21\sqrt{3x}=12`
`<=>12\sqrt{3x}=12`
`<=>\sqrt{3x}=1`
`<=>3x=1<=>x=1/3` (t/m)
`b)5\sqrt{9x+9}-2\sqrt{4x+4}+\sqrt{x+1}=36` `ĐK: x >= -1`
`<=>15\sqrt{x+1}-4\sqrt{x+1}+\sqrt{x+1}=36`
`<=>12\sqrt{x+1}=36`
`<=>\sqrt{x+1}=3`
`<=>x+1=9`
`<=>x=8` (t/m)
28 tháng 8 2021
3: Ta có: \(\sqrt{4x+1}=x+1\)
\(\Leftrightarrow x^2+2x+1=4x+1\)
\(\Leftrightarrow x\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(nhận\right)\\x=2\left(nhận\right)\end{matrix}\right.\)
4: Ta có: \(2\sqrt{x-1}+\dfrac{1}{3}\sqrt{9x-9}=15\)
\(\Leftrightarrow3\sqrt{x-1}=15\)
\(\Leftrightarrow x-1=25\)
hay x=26
5: Ta có: \(\sqrt{4x^2-12x+9}=7\)
\(\Leftrightarrow\left|2x-3\right|=7\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-3=7\\2x-3=-7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=10\\2x=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-2\end{matrix}\right.\)
6 tháng 7 2021
a) Ta có: \(\sqrt{\left(x+1\right)^2}=3\)
\(\Leftrightarrow\left|x+1\right|=3\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=3\\x+1=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-4\end{matrix}\right.\)
b) Ta có: \(3\sqrt{4x+4}-\sqrt{9x-9}-8\sqrt{\dfrac{x+1}{16}}=5\)
\(\Leftrightarrow6\sqrt{x+1}-3\sqrt{x-3}-2\sqrt{x+1}=5\)
\(\Leftrightarrow4\sqrt{x+1}=5+3\sqrt{x-3}\)
\(\Leftrightarrow16\left(x+1\right)=25+30\sqrt{x-3}+9\left(x-3\right)\)
\(\Leftrightarrow16x+16=25+9x-27+30\sqrt{x-3}\)
\(\Leftrightarrow30\sqrt{x-3}=16x+16+2-9x\)
\(\Leftrightarrow30\sqrt{x-3}=7x+18\)
\(\Leftrightarrow x-3=\left(\dfrac{7x+18}{30}\right)^2\)
\(\Leftrightarrow x-3=\dfrac{49x^2}{900}+\dfrac{7}{25}x+\dfrac{9}{25}\)
\(\Leftrightarrow\dfrac{49}{900}x^2-\dfrac{18}{25}x+\dfrac{84}{25}=0\)
\(\Delta=\left(-\dfrac{18}{25}\right)^2-4\cdot\dfrac{49}{900}\cdot\dfrac{84}{25}=-\dfrac{16}{75}< 0\)
Vậy: Phương trình vô nghiệm
6 tháng 7 2021
a)Pt\(\Leftrightarrow\left|x+1\right|=3\Leftrightarrow\left[{}\begin{matrix}x+1=3\\x+1=-3\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-4\end{matrix}\right.\)
b)Đk:\(x\ge-1\)
Sửa đề: \(3\sqrt{4x+4}-\sqrt{9x+9}-8\sqrt{\dfrac{x+1}{16}}=5\)
Pt \(\Leftrightarrow6\sqrt{x+1}-3\sqrt{x+1}-2\sqrt{x+1}=5\)
\(\Leftrightarrow\sqrt{x+1}=5\)
\(\Leftrightarrow x=24\left(tm\right)\)
HP
17 tháng 9 2021
d. \(\sqrt{9x^2+12x+4}=4\)
<=> \(\sqrt{\left(3x+2\right)^2}=4\)
<=> \(|3x+2|=4\)
<=> \(\left[{}\begin{matrix}3x+2=4\\3x+2=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=2\\3x=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-2\end{matrix}\right.\)
17 tháng 9 2021
c: Ta có: \(\dfrac{5\sqrt{x}-2}{8\sqrt{x}+2.5}=\dfrac{2}{7}\)
\(\Leftrightarrow35\sqrt{x}-14=16\sqrt{x}+5\)
\(\Leftrightarrow x=1\)
3 tháng 8 2023
a) \(\sqrt{1-6x+9x^2}=9\)
\(\Leftrightarrow\sqrt{\left(1-3x\right)^2}=9\)
\(\Leftrightarrow\left|1-3x\right|=9\)
\(\Leftrightarrow\left[{}\begin{matrix}1-3x=9\\1-3x=-9\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}3x=1-9\\3x=1+9\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}3x=-8\\3x=10\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{8}{3}\\x=\dfrac{10}{3}\end{matrix}\right.\)
b) \(\sqrt{2x-3}-\sqrt{x+1}=0\) (\(x\ge\dfrac{3}{2}\))
\(\Leftrightarrow\sqrt{2x-3}=\sqrt{x+1}\)
\(\Leftrightarrow2x-3=x+1\)
\(\Leftrightarrow2x-x=1+3\)
\(\Leftrightarrow x=4\left(tm\right)\)
c) \(\sqrt{9x^2+12+4}-2=3x\)
\(\Leftrightarrow\sqrt{\left(3x+2\right)^2}=3x+2\)
\(\Leftrightarrow\left|3x+2\right|=3x+2\)
\(\Leftrightarrow3x+2\ge0\)
\(\Leftrightarrow3x\ge-2\)
\(\Leftrightarrow x\ge-\dfrac{2}{3}\)
3 tháng 8 2023
a: =>|3x-1|=9
=>3x-1=9 hoặc 3x-1=-9
=>x=-8/3 hoặc x=10/3
b: =>căn 2x-3=căn x+1
=>2x-3=x+1
=>x=4
c: =>|3x+2|=3x+2
=>3x+2>=0
=>x>=-2/3
29 tháng 7 2021
Bài 4:
a, \(\sqrt{3x+4}-\sqrt{2x+1}=\sqrt{x+3}\) (ĐK: \(x\ge\dfrac{-1}{2}\))
\(\Rightarrow\) \(\left(\sqrt{3x+4}-\sqrt{2x+1}\right)^2\) = x + 3
\(\Leftrightarrow\) \(3x+4+2x+1-2\sqrt{\left(3x+4\right)\left(2x+1\right)}=x+3\)
\(\Leftrightarrow\) \(4x+2=2\sqrt{6x^2+11x+4}\)
\(\Leftrightarrow\) \(2x+1=\sqrt{6x^2+11x+4}\)
\(\Rightarrow\) \(4x^2+4x+1=6x^2+11x+4\)
\(\Leftrightarrow\) \(2x^2+7x+3=0\)
\(\Delta=7^2-4.2.3=25\); \(\sqrt{\Delta}=5\)
Vì \(\Delta\) > 0; theo hệ thức Vi-ét ta có:
\(x_1=\dfrac{-7+5}{4}=\dfrac{-1}{2}\)(TM); \(x_2=\dfrac{-7-5}{4}=-3\) (KTM)
Vậy ...
Các phần còn lại bạn làm tương tự nha, phần d bạn chuyển \(-\sqrt{2x+4}\) sang vế trái rồi bình phương 2 vế như bình thường là được
Bài 5:
a, \(\sqrt{x+4\sqrt{x}+4}=5x+2\)
\(\Leftrightarrow\) \(\sqrt{\left(\sqrt{x}+2\right)^2}=5x+2\)
\(\Rightarrow\) \(\sqrt{x}+2=5x+2\)
\(\Leftrightarrow\) \(5x-\sqrt{x}=0\)
\(\Leftrightarrow\) \(\sqrt{x}\left(5\sqrt{x}-1\right)=0\)
\(\Leftrightarrow\) \(\left[{}\begin{matrix}\sqrt{x}=0\\5\sqrt{x}-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{25}\end{matrix}\right.\)
Vậy ...
Phần b cũng là hằng đẳng thức thôi nha \(\sqrt{x^2-2x+1}=\sqrt{\left(x-1\right)^2}=x-1\); \(\sqrt{x^2+4x+4}=\sqrt{\left(x+2\right)^2}=x+2\) rồi giải như bình thường là xong nha!
VD1:
a, \(\sqrt{2x-1}=\sqrt{2}-1\) (x \(\ge\) \(\dfrac{1}{2}\))
\(\Leftrightarrow\) \(2x-1=\left(\sqrt{2}-1\right)^2\) (Bình phương 2 vế)
\(\Leftrightarrow\) \(2x-1=2-2\sqrt{2}+1\)
\(\Leftrightarrow\) \(2x=4-2\sqrt{2}\)
\(\Leftrightarrow\) \(x=2-\sqrt{2}\) (TM)
Vậy ...
Phần b tương tự nha
c, \(\sqrt{3}x^2-\sqrt{12}=0\)
\(\Leftrightarrow\) \(\sqrt{3}x^2=\sqrt{12}\)
\(\Leftrightarrow\) \(x^2=2\)
\(\Leftrightarrow\) \(x=\pm\sqrt{2}\)
Vậy ...
d, \(\sqrt{2}\left(x-1\right)-\sqrt{50}=0\)
\(\Leftrightarrow\) \(\sqrt{2}\left(x-1\right)=\sqrt{50}\)
\(\Leftrightarrow\) \(x-1=5\)
\(\Leftrightarrow\) \(x=6\)
Vậy ...
VD2:
Phần a dễ r nha (Bình phương 2 vế rồi tìm x như bình thường)
b, \(\sqrt{x^2-x}=\sqrt{3-x}\) (\(x\le3\); \(x^2\ge x\))
\(\Leftrightarrow\) \(x^2-x=3-x\) (Bình phương 2 vế)
\(\Leftrightarrow\) \(x^2=3\)
\(\Leftrightarrow\) \(x=\pm\sqrt{3}\) (TM)
Vậy ...
c, \(\sqrt{2x^2-3}=\sqrt{4x-3}\) (x \(\ge\) \(\dfrac{\sqrt{3}}{2}\))
\(\Leftrightarrow\) \(2x^2-3=4x-3\) (Bình phương 2 vế)
\(\Leftrightarrow\) \(2x^2-4x=0\)
\(\Leftrightarrow\) \(2x\left(x-2\right)=0\)
\(\Leftrightarrow\) \(\left[{}\begin{matrix}2x=0\\x-2=0\end{matrix}\right.\)
\(\Leftrightarrow\) \(\left[{}\begin{matrix}x=0\left(KTM\right)\\x=2\left(TM\right)\end{matrix}\right.\)
Vậy ...
Chúc bn học tốt! (Có gì không biết cứ hỏi mình nha!)
a: \(\Leftrightarrow\left[{}\begin{matrix}3x+2=4\\3x+2=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-2\end{matrix}\right.\)