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đề =
\(2\sqrt{9\left(x-3\right)}-\frac{1}{5}\sqrt{25\left(x-3\right)}-\frac{1}{7}\sqrt{49\left(x-3\right)}=20\)
=>\(6\sqrt{x-3}-\sqrt{x-3}-\sqrt{x-3}=20\)
=>\(4\sqrt{x-3}=20\)
=>\(\sqrt{x-3}=5\)
=>\(x-3=25\)
=>\(x=28\)
1. ĐKXĐ: \(x\ge3\)
\(2\sqrt{9x-27}-\frac{1}{5}\sqrt{25x-75}-\frac{1}{7}\sqrt{49x-147}=20\)
⇔ \(6\sqrt{x-3}-\sqrt{x-3}-\sqrt{x-3}=20\)
⇔ \(4\sqrt{x-3}=20\)
⇔ \(\sqrt{x-3}=5\)
⇔ \(x-3=25\)
⇔ \(x=28\left(TMĐKXĐ\right)\)
Vậy....
2. ĐKXĐ: \(x\ge0\)
\(\frac{3}{2}\sqrt{5x}+\sqrt{5x}-7=\frac{1}{2}\sqrt{5x}\)
⇔ \(\frac{3}{2}\sqrt{5x}+\sqrt{5x}-\frac{1}{2}\sqrt{5x}=7\)
⇔ \(2\sqrt{5x}=7\)
⇔ \(\sqrt{5x}=\frac{7}{2}\)
⇔ \(5x=\frac{49}{4}\)
⇔ \(x=\frac{49}{20}\left(TMĐKXĐ\right)\)
Vậy...
\(2\sqrt{9x-27}-\frac{1}{5}\sqrt{25x-75}-\frac{1}{7}\sqrt{49x-147}=20\)
\(< =>2\sqrt{9\left(x-3\right)}-\frac{1}{5}\sqrt{25\left(x-3\right)}-\frac{1}{7}\sqrt{49\left(x-3\right)}=20\)
\(< =>2\cdot3\sqrt{\left(x-3\right)}-\frac{1}{5}.5\sqrt{\left(x-3\right)}-\frac{1}{7}.7\sqrt{\left(x-3\right)}=20\) \(đk:x\ge0\)
\(< =>6\sqrt{\left(x-3\right)}-\sqrt{\left(x-3\right)}-\sqrt{\left(x-3\right)}=20\)
\(< =>\sqrt{\left(x-3\right)}\left(6-1-1\right)=20\)
\(< =>4\sqrt{\left(x-3\right)}=20\)
\(< =>\sqrt{\left(x-3\right)}=5\)
\(< =>x-3=25\)
\(< =>x=28\left(tm\right)\)
a/ \(\sqrt{x-1}+\sqrt{4x-4}-\sqrt{25x-25}+2=0\) (ĐKXĐ : \(x\ge1\))
\(\Leftrightarrow\sqrt{x-1}+2\sqrt{x-1}-5\sqrt{x-1}+2=0\)
\(\Leftrightarrow2\sqrt{x-1}=2\Leftrightarrow x-1=1\Leftrightarrow x=2\)
b/ \(\sqrt{9x^2+18}+2\sqrt{x^2+2}-\sqrt{25x^2+50}+3=0\)
\(\Leftrightarrow3\sqrt{x^2+2}+2\sqrt{x^2+2}-5\sqrt{x^2+2}+3=0\)
<=> 3 = 0 (vô lý)
=> pt vô nghiệm.
c/ \(\frac{9x-7}{\sqrt{7x+5}}=\sqrt{7x+5}\) (ĐKXĐ : x>-5/7)
\(\Leftrightarrow9x-7=7x+5\Leftrightarrow2x=12\Leftrightarrow x=6\)
d/ \(\frac{\sqrt{2x-3}}{\sqrt{x-1}}=2\) (ĐKXĐ : \(x\ge\frac{3}{2}\))
\(\Leftrightarrow2x-3=4\left(x-1\Leftrightarrow\right)2x=1\Leftrightarrow x=\frac{1}{2}\) (loại)
Vậy pt vô nghiệm.
1 slot xíu nữa làm :)))))
8h lên giúp bạn trước rồi giúp mấy bạn khác sau :v
a, nhóm can x vào một nhóm cái trong ngoặc còn lại thì tính ra
\(11\sqrt{5x}=33\)
chia cả hai vế cho 11 căn 5 rồi bình phương hai vế do x>=0
b,sai đề
1,
\(D=\frac{1}{\sqrt{h+2\sqrt{h-1}}}+\frac{1}{\sqrt{h-2\sqrt{h-1}}}\)
\(=\frac{1}{\sqrt{h-1+2\sqrt{h-1}+1}}+\frac{1}{\sqrt{h-1-2\sqrt{h-1}+1}}\)
\(=\frac{1}{\sqrt{h-1}+1}+\frac{1}{\sqrt{h-1}-1}\)
\(=\frac{\sqrt{h-1}-1+\sqrt{h-1}+1}{h-1-1}\)
\(=\frac{2\sqrt{h-1}}{h-2}\)
Thay \(h=3\)vào D ta có:
\(D=\frac{2\sqrt{3-1}}{3-2}=2\sqrt{2}\)
Vậy với \(h=3\)thì \(D=2\sqrt{2}\)
2,
a, \(\sqrt{x-1}+\sqrt{4x-4}-\sqrt{25x-25}+2=0\)(ĐK: \(x\ge1\))
\(\Leftrightarrow\sqrt{x-1}+2\sqrt{x-1}-5\sqrt{x-1}+2=0\)
\(\Leftrightarrow-2\sqrt{x-1}=-2\)
\(\Leftrightarrow\sqrt{x-1}=1\Leftrightarrow x=2\left(TM\right)\)
Vậy PT có nghiệm là \(x=2\)
b, \(\sqrt{9x^2+18}+2\sqrt{x^2+2}-\sqrt{25x^2+50}+3=0\)(ĐK: \(-\sqrt{2}\le x\le\sqrt{2}\))
\(\Leftrightarrow3\sqrt{x^2+2}+2\sqrt{x^2+2}-5\sqrt{x^2+2}=-3\)
\(\Leftrightarrow0=-3\)(vô lí)
Vậy PT đã cho vô nghiệm.
Ta có: \(2\sqrt{9x-27}-\dfrac{1}{5}\sqrt{25x-75}-\dfrac{1}{7}\sqrt{49x-147}=20\)
\(\Leftrightarrow2\cdot3\sqrt{x-3}-\dfrac{1}{5}\cdot5\sqrt{x-3}-\dfrac{1}{7}\cdot7\cdot\sqrt{x-3}=20\)
\(\Leftrightarrow4\sqrt{x-3}=20\)
\(\Leftrightarrow x-3=25\)
hay x=28
\(a)ĐK:x\ge-1\\ \Leftrightarrow x+1=2\sqrt{x+1}\\ \Leftrightarrow x^2+2x+1=4x+4\\ \Leftrightarrow x^2+2x-4x+1-4=0\\ \Leftrightarrow x^2-2x-3=0\\ \Leftrightarrow\left[{}\begin{matrix}x=3\left(tm\right)\\x=-1\left(tm\right)\end{matrix}\right.\)
Vậy \(S=\left\{3;-1\right\}\)
\(b)ĐK:x\ge2\\ \Leftrightarrow2x-4=\sqrt{x-2}\\ \Leftrightarrow4x^2-16x+16=x-2\\ \Leftrightarrow4x^2-16x-x+16+2=0\\ \Leftrightarrow4x^2-17x+18=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{9}{4}\left(tm\right)\\x=2\left(tm\right)\end{matrix}\right.\)
Vậy \(S=\left\{\dfrac{9}{4};2\right\}\)
\(c)ĐK:x\ge3\\ \Leftrightarrow2\sqrt{9\left(x-3\right)}-\dfrac{1}{5}\sqrt{25\left(x-3\right)}-\dfrac{1}{7}\sqrt{49\left(x-3\right)}=20\\ \Leftrightarrow2.3\sqrt{x-3}-\dfrac{1}{5}\cdot5\sqrt{x-3}-\dfrac{1}{7}\cdot7\sqrt{x-3}=20\\ \Leftrightarrow6\sqrt{x-3}-\sqrt{x-3}-\sqrt{x-3}=20\\ \Leftrightarrow4\sqrt{x-3}=20\\ \Leftrightarrow\sqrt{x-3}=5\\ \Leftrightarrow x-3=25\\ \Leftrightarrow x=25+3\\ \Leftrightarrow x=28\left(tm\right)\)
Vậy \(S=\left\{28\right\}\)
a: \(\Leftrightarrow\left\{{}\begin{matrix}\left(2x+6\right)^2=\left(1-x\right)^2\\-3< =x< =1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left(2x+6+x-1\right)\left(2x+6+1-x\right)=0\\-3< =x< =1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(3x+5\right)\left(x+7\right)=0\\-3< =x< =1\end{matrix}\right.\Leftrightarrow x=-\dfrac{5}{3}\)
b: \(\Leftrightarrow2\cdot3\sqrt{x-3}-\dfrac{1}{5}\cdot5\sqrt{x-3}-\dfrac{1}{7}\cdot7\sqrt{x-3}=2x\)
\(\Leftrightarrow4\sqrt{x-3}=2x\)
\(\Leftrightarrow2\sqrt{x-3}=x\)
\(\Leftrightarrow\sqrt{4x-12}=x\)
\(\Leftrightarrow\left\{{}\begin{matrix}x>=3\\x^2=4x-12\end{matrix}\right.\Leftrightarrow x\in\varnothing\)
\(ĐKXĐ:x\ge3\)
\(2\sqrt{9x-27}-\frac{1}{5}\sqrt{25x-75}-\frac{1}{7}\sqrt{49x-147}=0\)
\(2\sqrt{9\left(x-3\right)}-\frac{1}{5}\sqrt{25\left(x-3\right)}-\frac{1}{7}\sqrt{49\left(x-3\right)}=0\)
\(6\sqrt{x-3}-\sqrt{x-3}-\sqrt{x-3}=0\)
\(4\sqrt{x-3}=0\)
\(x-3=0\)
\(x=3\left(TM\right)\)