Tìm x:
\(\sqrt{4x-4}\)+\(\sqrt{25x-25}\)+\(\sqrt{81x-81}\)=\(-\)1
Giải chi tiết nha!
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Những câu hỏi liên quan
15 tháng 10 2023
a: ĐKXĐ: x-5>=0
=>x>=5
\(\sqrt{4x-20}+\sqrt{x-5}-\dfrac{1}{3}\cdot\sqrt{9x-45}=4\)
=>\(2\sqrt{x-5}+\sqrt{x-5}-\dfrac{1}{3}\cdot3\sqrt{x-5}=4\)
=>\(2\sqrt{x-5}=4\)
=>x-5=4
=>x=9(nhận)
b: ĐKXĐ: x-1>=0
=>x>=1
\(\sqrt{x-1}+\sqrt{4x-4}-\sqrt{25x-25}=4\)
=>\(\sqrt{x-1}+2\sqrt{x-1}-5\sqrt{x-1}=4\)
=>\(-2\sqrt{x-1}=4\)
=>\(\sqrt{x-1}=-2\)(vô lý)
Vậy: Phương trình vô nghiệm
c: ĐKXĐ: x-2>=0
=>x>=2
\(\dfrac{1}{3}\sqrt{x-2}-\dfrac{2}{3}\cdot\sqrt{9x-18}+6\cdot\sqrt{\dfrac{x-2}{81}}=-4\)
=>\(\dfrac{1}{3}\sqrt{x-2}-\dfrac{2}{3}\cdot3\sqrt{x-2}+6\cdot\dfrac{\sqrt{x-2}}{9}=-4\)
=>\(\sqrt{x-2}\left(\dfrac{1}{3}-2+\dfrac{2}{3}\right)=-4\)
=>\(-\sqrt{x-2}=-4\)
=>x-2=16
=>x=18(nhận)
d: ĐKXĐ: x+3>=0
=>x>=-3
\(\sqrt{9x+27}+4\sqrt{x+3}-\dfrac{3}{4}\cdot\sqrt{16x+48}=0\)
=>\(3\sqrt{x+3}+4\sqrt{x+3}-\dfrac{3}{4}\cdot4\sqrt{x+3}=0\)
=>\(4\sqrt{x+3}=0\)
=>x+3=0
=>x=-3(nhận)
NV
15 tháng 10 2023
a) \(\sqrt{4x-20}+\sqrt{x-5}-\dfrac{1}{3}\sqrt{9x-45}=4\)
= \(2\sqrt{x-5}+\sqrt{x-5}-\dfrac{1}{3}\sqrt{9\left(x-5\right)}=4\)
= \(2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}=4\)
= \(2\sqrt{x-5}=4\)
= \(\sqrt{x-5}=2\)
= \(\left|x-5\right|=4\)
=> \(x-5=\pm4\)
\(x=\pm4+5\)
\(x=9;x=1\)
Vậy x=9; x=1
H9
HT.Phong (9A5)
CTVHS
5 tháng 11 2023
6) \(\sqrt{x^2-4x+1}=x\left(x\ge0\right)\)
\(\Leftrightarrow x^2-4x+1=x^2\)
\(\Leftrightarrow x^2-x^2=4x-1\)
\(\Leftrightarrow4x=1\)
\(\Leftrightarrow x=\dfrac{1}{4}\left(tm\right)\)
8) \(\sqrt{x^2-x-6}=\sqrt{x-3}\left(x\ge3\right)\)
\(\Leftrightarrow x^2-x-6=x-3\)
\(\Leftrightarrow x^2-2x-3=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\left(tm\right)\\x=-1\left(ktm\right)\end{matrix}\right.\)
9) \(\sqrt{x-1}+\sqrt{4x-4}-\sqrt{25x-25}+2=0\left(x\ge1\right)\)
\(\Leftrightarrow\sqrt{x-1}+2\sqrt{x-1}-5\sqrt{x-1}+2=0\)
\(\Leftrightarrow-2\sqrt{x-1}+2=0\)
\(\Leftrightarrow-2\sqrt{x-1}=-2\)
\(\Leftrightarrow\sqrt{x-1}=1\)
\(\Leftrightarrow x-1=1\)
\(\Leftrightarrow x=1+1\)
\(\Leftrightarrow x=2\left(tm\right)\)
H
18 tháng 8 2023
\(a,đk:x\ge5\\ \Leftrightarrow\sqrt{x-5}+\sqrt{4\left(x-5\right)}-\dfrac{1}{5}\sqrt{9\left(x-5\right)}=3\\ \Leftrightarrow\sqrt{x-5}+2\sqrt{x-5}-\dfrac{1}{5}.3\sqrt{x-5}=3\\ \Leftrightarrow\dfrac{12}{5}\sqrt{x-5}=3\\ \Rightarrow\sqrt{x-5}=\dfrac{5}{4}\\ \Leftrightarrow\left(\sqrt{x-5}\right)^2=\left(\dfrac{5}{4}\right)^2\\ \Leftrightarrow x-5=\dfrac{25}{16}\\ \Rightarrow x=\dfrac{25}{16}+5\\ \Rightarrow x=\dfrac{105}{16}\left(t|m\right)\)
\(b,đk:x\ge1\\ \Leftrightarrow\sqrt{x-1}+\sqrt{4\left(x-1\right)}-\sqrt{25\left(x-1\right)}=-2\\ \Leftrightarrow\sqrt{x-1}+2\sqrt{x-1}-5\sqrt{x-1}=-2\\ \Leftrightarrow-2\sqrt{x-1}=-2\\ \Leftrightarrow\sqrt{x-1}=1\\ \Leftrightarrow x-1=1\\ \Leftrightarrow x=2\left(t|m\right)\)
H9
HT.Phong (9A5)
CTVHS
24 tháng 9 2023
a) \(\sqrt{x-1}+\sqrt{4x-4}-\sqrt{25x-25}+2=0\) (ĐK: \(x\ge1\))
\(\Leftrightarrow\sqrt{x-1}+\sqrt{4\left(x-1\right)}-\sqrt{25\left(x-1\right)}+2=0\)
\(\Leftrightarrow\sqrt{x-1}+2\sqrt{x-1}-5\sqrt{x-1}+2=0\)
\(\Leftrightarrow-2\sqrt{x-1}=-2\)
\(\Leftrightarrow\sqrt{x-1}=\dfrac{2}{2}\)
\(\Leftrightarrow\sqrt{x-1}=1\)
\(\Leftrightarrow x-1=1\)
\(\Leftrightarrow x=2\left(tm\right)\)
b) \(\sqrt{16x+16}-\sqrt{9x+9}+\sqrt{4x+4}+\sqrt{x+1}=16\) (ĐK: \(x\ge-1\))
\(\Leftrightarrow\sqrt{16\left(x+1\right)}-\sqrt{9\left(x+1\right)}+\sqrt{4\left(x+1\right)}+\sqrt{x+1}=16\)
\(\Leftrightarrow4\sqrt{x+1}-3\sqrt{x+1}+2\sqrt{x+1}+\sqrt{x+1}=16\)
\(\Leftrightarrow4\sqrt{x+1}=16\)
\(\Leftrightarrow\sqrt{x+1}=4\)
\(\Leftrightarrow x+1=16\)
\(\Leftrightarrow x=15\left(tm\right)\)
14 tháng 10 2021
\(a,ĐK:x\ge1\\ PT\Leftrightarrow\sqrt{x-1}+2\sqrt{x-1}-5\sqrt{x-1}=-2\\ \Leftrightarrow-2\sqrt{x-1}=-2\Leftrightarrow\sqrt{x-1}=1\\ \Leftrightarrow x-1=1\Leftrightarrow x=2\left(tm\right)\\ b,ĐK:x\ge0\\ PT\Leftrightarrow\dfrac{1}{3}\sqrt{2x}-2\sqrt{2x}+3\sqrt{2x}=12\\ \Leftrightarrow\dfrac{4}{3}\sqrt{2x}=12\Leftrightarrow\sqrt{2x}=9\\ \Leftrightarrow2x=81\Leftrightarrow x=\dfrac{81}{2}\left(tm\right)\)
25 tháng 9 2021
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: \(x^2+4\ge0\left(đúng\forall x\right)\)
6 tháng 9 2021
a: ta có: \(\sqrt{x-1}+\sqrt{4x-4}-\sqrt{25x-25}+2=0\)
\(\Leftrightarrow\sqrt{x-1}+2\sqrt{x-1}-5\sqrt{x-1}+2=0\)
\(\Leftrightarrow\sqrt{x-1}=1\)
hay x=2
c: Ta có: \(\sqrt{1-2x^2}=x-1\)
\(\Leftrightarrow1-2x^2=x^2-2x+1\)
\(\Leftrightarrow-3x^2+2x=0\)
\(\Leftrightarrow-x\left(3x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(loại\right)\\x=\dfrac{2}{3}\left(loại\right)\end{matrix}\right.\)
KH
1
H
23 tháng 8 2023
mình nghĩ căn đầu tiên phải là `x+1` mới đúng kiểu đề á, còn không phải thì bạn cmt nói mình nha=))
ĐK: \(x\ge-1\)
PT trở thành:
\(\sqrt{x+1}+\sqrt{4}.\sqrt{x+1}-\sqrt{25}.\sqrt{x+1}=-8\\ \Leftrightarrow\sqrt{x+1}+2\sqrt{x+1}-5\sqrt{x+1}=-8\\ \Leftrightarrow\left(1+2-5\right)\sqrt{x+1}=-8\\ \Leftrightarrow-2\sqrt{x+1}=-8\\ \Leftrightarrow\sqrt{x+1}=-\dfrac{8}{-2}=4\\ \Leftrightarrow x+1=4^2=16\\ \Leftrightarrow x=16-1=15\left(tm\right)\)
HT
1
CD
15 tháng 8 2020
DK: \(x\ge1\)
\(PT\Leftrightarrow\sqrt{x-1}+\sqrt{4x-4}-\sqrt{25x-25}+2=0\\ \Leftrightarrow\sqrt{x-1}+2\sqrt{x-1}-5\sqrt{x-1}+2=0\\ \Leftrightarrow2-2\sqrt{x-1}=0\\ \Leftrightarrow1-\sqrt{x-1}=0\\\Leftrightarrow \sqrt{x-1}=1\Leftrightarrow x-1=1\Leftrightarrow x=2\left(TM\right)\)
Vậy phương trình đã cho có 1 nghiệm là x = 2
ĐK: \(x\ge1\)
Ta có:
\(\sqrt{4x-4}+\sqrt{25x-25}+\sqrt{81x-81}=1\)
\(\Rightarrow\sqrt{4\left(x-1\right)}+\sqrt{25\left(x-1\right)}+\sqrt{81\left(x-1\right)}=1\)
\(\Rightarrow2\sqrt{x-1}+5\sqrt{x-1}+9\sqrt{x-1}=1\)
\(\Rightarrow16\sqrt{x-1}=1\)
\(\Rightarrow\sqrt{x-1}=\dfrac{1}{16}\)
\(\Rightarrow\left[{}\begin{matrix}x-1=\dfrac{1}{4}\\x-1=-\dfrac{1}{4}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{5}{4}\\x=\dfrac{3}{4}\end{matrix}\right.\)
bằng \(-\)1 mà bạn