\(\sqrt{x-5}+\dfrac{1}{3}\sqrt{9x-45}=\dfrac{1}{5}\sqrt{25x-125+6}\)
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PK
1
12 tháng 8 2021
Ta có: \(\sqrt{25x-125}-3\cdot\sqrt{\dfrac{x-5}{9}}-\dfrac{1}{3}\sqrt{9x-45}=6\)
\(\Leftrightarrow5\sqrt{x-5}-3\cdot\dfrac{\sqrt{x-5}}{3}-\dfrac{1}{3}\cdot3\sqrt{x-5}=6\)
\(\Leftrightarrow3\sqrt{x-5}=6\)
\(\Leftrightarrow x-5=4\)
hay x=9
NH
giai phuong trinh
\(\sqrt{x}-5+\dfrac{1}{3}\sqrt{9x}-45=\dfrac{1}{5}\sqrt{25x}-125=6\)
giup minh voi
1
25 tháng 5 2022
Sửa đề: \(\sqrt{x-5}+\dfrac{1}{3}\sqrt{9x-45}=\dfrac{1}{5}\sqrt{25x-125}+6\)
\(\Leftrightarrow\sqrt{x-5}+\dfrac{1}{3}\cdot3\cdot\sqrt{x-5}-\dfrac{1}{5}\cdot5\sqrt{x-5}=6\)
\(\Leftrightarrow\sqrt{x-5}=6\)
=>x-5=36
hay x=41
18 tháng 10 2021
\(d,ĐK:x\ge0\\ PT\Leftrightarrow\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=2\\\sqrt{x}=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\left(tm\right)\\x=9\left(tm\right)\end{matrix}\right.\\ e,ĐK:x\ge1\\ PT\Leftrightarrow\sqrt{x-1}+\dfrac{3}{2}\cdot2\sqrt{x-1}-\dfrac{2}{5}\cdot5\sqrt{x-1}=4\\ \Leftrightarrow2\sqrt{x-1}=4\Leftrightarrow\sqrt{x-1}=2\\ \Leftrightarrow x-1=4\Leftrightarrow x=5\left(tm\right)\\ f,ĐK:x\ge5\\ PT\Leftrightarrow\sqrt{x-5}+2\sqrt{x-5}-\dfrac{1}{3}\cdot3\sqrt{x-5}=6\\ \Leftrightarrow2\sqrt{x-5}=6\Leftrightarrow\sqrt{x-5}=3\\ \Leftrightarrow x-5=9\Leftrightarrow x=14\left(tm\right)\)
5 tháng 8 2021
a) Ta có: \(2\sqrt{9x-27}-\dfrac{1}{5}\sqrt{25x-75}-\dfrac{1}{7}\sqrt{49x-147}=20\)
\(\Leftrightarrow6\sqrt{x-3}-\sqrt{x-3}-\sqrt{x-3}=20\)
\(\Leftrightarrow4\sqrt{x-3}=20\)
\(\Leftrightarrow x-3=25\)
hay x=28
b) Ta có: \(\sqrt{9x+18}-5\sqrt{x+2}+\dfrac{4}{5}\sqrt{25x+50}=6\)
\(\Leftrightarrow3\sqrt{x+2}-5\sqrt{x+2}+4\sqrt{x+2}=6\)
\(\Leftrightarrow2\sqrt{x+2}=6\)
\(\Leftrightarrow x+2=9\)
hay x=7
NH
2
PA
31 tháng 7 2017
\(\sqrt{16x+16}-\sqrt{9x+9}+\sqrt{4x+4}+\sqrt{x+1}=16\)
\(\Leftrightarrow4\sqrt{x+1}-3\sqrt{x+1}+2\sqrt{x+1}+\sqrt{x+1}=16\)
\(\Leftrightarrow4\sqrt{x+1}=16\)
\(\Leftrightarrow\sqrt{x+1}=4\)
<=> x + 1 = 16
<=> x = 15 (nhận)
~ ~ ~
\(\sqrt{4x+20}-3\sqrt{5+x}+\dfrac{4}{3}\sqrt{9x+45}=6\)
\(\Leftrightarrow2\sqrt{x+5}-3\sqrt{x+5}+4\sqrt{x+5}=6\)
\(\Leftrightarrow3\sqrt{x+5}=6\)
\(\Leftrightarrow\sqrt{x+5}=2\)
<=> x + 5 = 4
<=> x = - 1 (nhận)
\(ĐK:x\ge5\\ \Leftrightarrow\sqrt{x-5}+\dfrac{1}{3}\cdot3\sqrt{x-5}=\dfrac{1}{5}\sqrt{25x-119}\\ \Leftrightarrow2\sqrt{x-5}=\dfrac{1}{5}\sqrt{25x-119}\\ \Leftrightarrow4\left(x-5\right)=\dfrac{1}{25}\left(25x-119\right)\\ \Leftrightarrow4x-20=x-\dfrac{119}{25}\\ \Leftrightarrow3x=\dfrac{381}{25}\Leftrightarrow x=\dfrac{127}{25}\)
Ta có: \(\sqrt{x-5}+\dfrac{1}{3}\sqrt{9x-45}=\dfrac{1}{5}\sqrt{25x-125}+6\)
\(\Leftrightarrow x-5=36\)
hay x=41