Tính: \(\sqrt{9x-27}\)+\(\sqrt{x-3}\)-\(\frac{1}{2}\)\(\sqrt{4x-12}\)=7
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
22 tháng 7 2023
\(a) \sqrt{4x^2− 9} = 2\sqrt{x + 3}\)
\(ĐK:x\ge\dfrac{3}{2}\)
\(pt\Leftrightarrow4x^2-9=4\left(x+3\right)\)
\(\Leftrightarrow4x^2-9=4x+12\)
\(\Leftrightarrow4x^2-4x-21=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1-\sqrt{22}}{2}\left(l\right)\\x=\dfrac{1+\sqrt{22}}{2}\left(tm\right)\end{matrix}\right.\)
\(b)\sqrt{4x-20}+3.\sqrt{\dfrac{x-5}{9}}-\dfrac{1}{3}\sqrt{9x-45}=4\)
\(ĐK:x\ge5\)
\(pt\Leftrightarrow2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}=4\)
\(\Leftrightarrow2\sqrt{x-5}=4\Leftrightarrow\sqrt{x-5}=2\)
\(\Leftrightarrow x-5=4\Leftrightarrow x=9\left(tm\right)\)
22 tháng 7 2023
\(c)\dfrac{2}{3}\sqrt{9x-9}-\dfrac{1}{4}\sqrt{16x-16}+27.\sqrt{\dfrac{x-1}{81}}=4\)
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 \(\dfrac{5}{3}+\sqrt{x+3}>0\))
\(\Leftrightarrow x-3=0\Leftrightarrow x=3\left(nhận\right)\)
24 tháng 7 2023
=>\(5\cdot\dfrac{3\sqrt{x-3}}{5}-7\cdot\dfrac{2\sqrt{x-3}}{3}-7\cdot\sqrt{x^2-9}+18\cdot\sqrt{\dfrac{9}{81}\left(x^2-9\right)}=0\)
=>\(3\cdot\sqrt{x-3}-\dfrac{14}{3}\sqrt{x-3}=7\cdot\sqrt{x^2-9}-18\cdot\dfrac{3}{9}\cdot\sqrt{x^2-9}\)
=>\(-\dfrac{5}{3}\sqrt{x-3}=\sqrt{x^2-9}\)
=>\(\sqrt{x-3}\left(\sqrt{x+3}+\dfrac{5}{3}\right)=0\)
=>x-3=0
=>x=3
I
1 tháng 4 2020
đề bài như trên
\(\Leftrightarrow\sqrt{9\left(x-3\right)}+\sqrt{x-3}-\frac{1}{2}\sqrt{4\left(x-3\right)}=7\)
\(\Leftrightarrow3\sqrt{x-3}+\sqrt{x-3}-\frac{1}{2}.2\sqrt{x-3}=7\)
\(\Leftrightarrow3\sqrt{x-3}=7\)
\(\Leftrightarrow\sqrt{x-3}=\frac{7}{3}\left(đk:x\ge3\right)\)
\(\Leftrightarrow x-3=\frac{49}{9}=>x=\frac{76}{9}\left(thoảman\right)\)
CM
2
24 tháng 8 2020
\(\sqrt{4x^2-4x+1}=\sqrt{x^2+10x+25}\)
\(\Leftrightarrow\sqrt{\left(2x-1\right)^2}=\sqrt{\left(x+5\right)^2}\)
\(\Leftrightarrow\left|2x-1\right|=\left|x+5\right|\)
\(\Leftrightarrow\orbr{\begin{cases}2x-1=x+5\\2x-1=-\left(x+5\right)\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}2x-1=x+5\\2x-1=-x-5\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=6\\x=-\frac{4}{3}\end{cases}}\)
24 tháng 8 2020
a)
\(\sqrt{x+3}+2\sqrt{4\left(x+3\right)}-\frac{1}{3}\sqrt{9\left(x+3\right)}=8\)
\(\sqrt{x+3}+2\cdot2\sqrt{x+3}-\frac{1}{3}\cdot3\sqrt{x+3}=8\)
\(\sqrt{x+3}+4\sqrt{x+3}-\sqrt{x+3}=8\)
\(4\sqrt{x+3}=8\)
\(\sqrt{x+3}=2\)
\(\orbr{\begin{cases}2\ge0\left(llđ\right)\\x+3=2^2\end{cases}}\)
\(x+3=4\)
\(x=1\)
b)
\(\orbr{\begin{cases}x^2+10x+25\ge0\\4x^2-4x+1=x^2+10x+25\end{cases}}\)
\(\orbr{\begin{cases}\left(x+5\right)^2\ge0\left(lld\right)\\3x^2-6x-24=0\end{cases}}\)
\(\orbr{\begin{cases}x=6\\x=-\frac{4}{3}\end{cases}}\)
19 tháng 10 2020
ghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghghgh
by
duc
30 tháng 10 2023
a: ĐKXĐ: \(x\in R\)
\(\sqrt{x^2-4x+4}=7\)
=>\(\sqrt{\left(x-2\right)^2}=7\)
=>|x-2|=7
=>\(\left[{}\begin{matrix}x-2=7\\x-2=-7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=9\\x=-5\end{matrix}\right.\)
b: ĐKXĐ: x>=-3
\(\sqrt{4x+12}-3\sqrt{x+3}+\dfrac{4}{3}\cdot\sqrt{9x+27}=6\)
=>\(2\sqrt{x+3}-3\sqrt{x+3}+\dfrac{4}{3}\cdot3\sqrt{x+3}=6\)
=>\(3\sqrt{x+3}=6\)
=>\(\sqrt{x+3}=2\)
=>x+3=4
=>x=1(nhận)
PN
1
7 tháng 11 2021
\(<=>15\sqrt{x+3} - 3\sqrt{x+3}=4\sqrt{x+3} <=> 15\sqrt{x+3} - 3\sqrt{x+3}-4\sqrt{x+3}=0 <=> 8\sqrt{x+3}=0 <=>\sqrt{x+3}=0 => x+3=0 =>x=-3\)