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a) \(2\sqrt{2x}-5\sqrt{8x}+7\sqrt{18x}=28\) (*)
đk: x >/ 0
(*) \(\Leftrightarrow2\sqrt{2x}-10\sqrt{2x}+21\sqrt{2x}=28\)
\(\Leftrightarrow13\sqrt{2x}=28\) \(\Leftrightarrow\sqrt{2x}=\dfrac{28}{13}\Leftrightarrow2x=\left(\dfrac{28}{13}\right)^2\Leftrightarrow x=\dfrac{392}{169}\left(N\right)\)
Kl: \(x=\dfrac{392}{169}\)
b) \(\sqrt{4x-20}+\sqrt{x-5}-\dfrac{1}{3}\sqrt{9x-45}=4\) (*)
đk: x >/ 5
(*) \(\Leftrightarrow2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}=4\)
\(\Leftrightarrow2\sqrt{x-5}=4\Leftrightarrow\sqrt{x-5}=2\Leftrightarrow x-5=4\Leftrightarrow x=9\left(N\right)\)
Kl: x=9
c) \(\sqrt{\dfrac{3x-2}{x+1}}=2\) (*)
Đk: \(\left[{}\begin{matrix}x< -1\\x\ge\dfrac{2}{3}\end{matrix}\right.\)
(*) \(\Leftrightarrow\dfrac{3x-2}{x+1}=4\Leftrightarrow3x-2=4x+4\Leftrightarrow x=-6\left(N\right)\)
Kl: x=-6
d) \(\dfrac{\sqrt{5x-4}}{\sqrt{x+2}}=2\) (*)
Đk: \(x\ge\dfrac{4}{5}\)
(*) \(\Leftrightarrow\sqrt{5x-4}=2\sqrt{x+2}\Leftrightarrow5x-4=4x+8\Leftrightarrow x=12\left(N\right)\)
Kl: x=12
a) Đk: \(\left[{}\begin{matrix}x\le-1\\x\ge1\end{matrix}\right.\)
\(\sqrt{x^2-1}-x^2+1=0\)
\(\Leftrightarrow x^2-1-\sqrt{x^2-1}= 0\)
\(\Leftrightarrow\left(\sqrt{x^2-1}-1\right)\sqrt{x^2-1}=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x^2-1}-1=0\\\sqrt{x^2-1}=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x^2-1}=1\\x^2-1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2=2\left(1\right)\\x^2=1\left(2\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow x=\pm\sqrt{2}\left(N\right)\)
\(\left(2\right)\Leftrightarrow x=\pm1\left(N\right)\)
Kl: \(x=\pm\sqrt{2}\), \(x=\pm1\)
b) Đk: \(\left[{}\begin{matrix}x\le-2\\x\ge2\end{matrix}\right.\)
\(\sqrt{x^2-4}-x+2=0\)
\(\Leftrightarrow\sqrt{x^2-4}=x-2\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2-4=x^2-4x+4\\x\ge2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}4x=8\\x\ge2\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=2\left(N\right)\\x\ge2\end{matrix}\right.\)
kl: x=2
c) \(\sqrt{x^4-8x^2+16}=2-x\)
\(\Leftrightarrow\sqrt{\left(x^2-4\right)^2}=2-x\)
\(\Leftrightarrow\left|x^2-4\right|=2-x\) (*)
Th1: \(x^2-4< 0\Leftrightarrow-2< x< 2\)
(*) \(\Leftrightarrow x^2-4=x-2\Leftrightarrow x^2-x-2=0\Leftrightarrow\left[{}\begin{matrix}x=2\left(L\right)\\x=-1\left(N\right)\end{matrix}\right.\)
Th2: \(x^2-4\ge0\Leftrightarrow\left[{}\begin{matrix}x\le-2\\x\ge2\end{matrix}\right.\)
(*)\(\Leftrightarrow x^2-4=2-x\Leftrightarrow x^2+x-6=0\Leftrightarrow\left[{}\begin{matrix}x=2\left(N\right)\\x=-3\left(N\right)\end{matrix}\right.\)
Kl: x=-3, x=-1,x=2
d) \(\sqrt{9x^2+6x+1}=\sqrt{11-6\sqrt{2}}\)
\(\Leftrightarrow\sqrt{\left(3x+1\right)^2}=\sqrt{\left(3-\sqrt{2}\right)^2}\)
\(\Leftrightarrow\left|3x+1\right|=3-\sqrt{2}\) (*)
Th1: \(3x+1\ge0\Leftrightarrow x\ge-\dfrac{1}{3}\)
(*) \(\Leftrightarrow3x+1=3-\sqrt{2}\Leftrightarrow x=\dfrac{2-\sqrt{2}}{3}\left(N\right)\)
Th2: \(3x+1< 0\Leftrightarrow x< -\dfrac{1}{3}\)
(*) \(\Leftrightarrow3x+1=-3+\sqrt{2}\Leftrightarrow x=\dfrac{-4+\sqrt{2}}{3}\left(N\right)\)
Kl: \(x=\dfrac{2-\sqrt{2}}{3}\), \(x=\dfrac{-4+\sqrt{2}}{3}\)
e) Đk: \(x\ge-\dfrac{3}{2}\)
\(\sqrt{4^2-9}=2\sqrt{2x+3}\) \(\Leftrightarrow\sqrt{7}=2\sqrt{2x+3}\) \(\Leftrightarrow7=8x+12\)
\(\Leftrightarrow8x=-5\Leftrightarrow x=-\dfrac{5}{8}\left(N\right)\)
kl: \(x=-\dfrac{5}{8}\)
f) Đk: x >/ 5
\(\sqrt{4x-20}+3\sqrt{\dfrac{x-5}{9}}-\dfrac{1}{3}\sqrt{9x-45}=4\)
\(\Leftrightarrow2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}=4\)
\(\Leftrightarrow2\sqrt{x-5}=4\)
\(\Leftrightarrow\sqrt{x-5}=2\)
\(\Leftrightarrow x-5=4\)
\(\Leftrightarrow x=9\left(N\right)\)
kl: x=9
ĐKXĐ:
a/ \(x-2020>0\Rightarrow x>2020\)
b/ \(x\ne0\)
c/ \(3x+5< 0\Rightarrow x< -\frac{5}{3}\)
d/ \(\frac{x-3}{1-x}\ge0\Rightarrow1< x\le3\)
Bài 2: ĐKXĐ tự tìm
a/ \(2\sqrt{2x}-10\sqrt{2x}+21\sqrt{2x}=28\)
\(\Leftrightarrow13\sqrt{2x}=28\Rightarrow\sqrt{2x}=\frac{28}{13}\)
\(\Rightarrow x=\frac{392}{169}\)
b/ \(2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}=4\)
\(\Leftrightarrow\sqrt{x-5}=2\Rightarrow x=9\)
c/ \(3\sqrt{2x+1}>15\Rightarrow\sqrt{2x+1}>5\)
\(\Rightarrow2x+1>25\Rightarrow x>12\)
d/ \(\sqrt{x}+1>12\Rightarrow\sqrt{x}>11\Rightarrow x>121\)
\(\sqrt{16x+16}-\sqrt{9x+9}+\sqrt{4x+4}+\sqrt{x+1}=16\)
\(\Leftrightarrow4\sqrt{x+1}-3\sqrt{x+1}+2\sqrt{x+1}+\sqrt{x+1}=16\)
\(\Leftrightarrow4\sqrt{x+1}=16\)
\(\Leftrightarrow\sqrt{x+1}=4\)
<=> 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)
a: \(\Leftrightarrow\dfrac{2x-3}{x-1}=4\)
=>4x-4=2x-3
=>2x=1
hay x=1/2
b: \(\Leftrightarrow\sqrt{\dfrac{2x-3}{x-1}}=2\)
=>(2x-3)=4x-4
=>4x-4=2x-3
=>2x=1
hay x=1/2(nhận)
c: \(\Leftrightarrow\sqrt{2x+3}\left(\sqrt{2x-3}-2\right)=0\)
=>2x+3=0 hoặc 2x-3=4
=>x=-3/2 hoặc x=7/2
e: \(\Leftrightarrow2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}=4\)
=>căn (x-5)=2
=>x-5=4
hay x=9
Bài 2:
a: =>25x=35^2=1225
=>x=49
b: \(\Leftrightarrow2\sqrt{x+5}-3\sqrt{x+5}+\dfrac{4}{3}\cdot3\sqrt{x+5}=6\)
\(\Leftrightarrow3\sqrt{x+5}=6\)
=>x+5=4
=>x=-1
Mình làm một vài câu thôi nhé, các câu còn lại tương tự.
Giải:
a) ??? Đề thiếu
b) \(\sqrt{-3x+4}=12\)
\(\Leftrightarrow-3x+4=144\)
\(\Leftrightarrow-3x=140\)
\(\Leftrightarrow x=\dfrac{-140}{3}\)
Vậy ...
c), d), g), h), i), p), q), v), a') Tương tự b)
w), x) Mình đã làm ở đây:
Câu hỏi của Ami Yên - Toán lớp 9 | Học trực tuyến
z) \(\sqrt{16\left(x+1\right)^2}-\sqrt{9\left(x+1\right)^2}=4\)
\(\Leftrightarrow4\left(x+1\right)-3\left(x+1\right)=4\)
\(\Leftrightarrow x+1=4\)
\(\Leftrightarrow x=3\)
Vậy ...
b') \(\sqrt{9x+9}+\sqrt{4x+4}=\sqrt{x+1}\)
\(\Leftrightarrow3\sqrt{x+1}+2\sqrt{x+1}=\sqrt{x+1}\)
\(\Leftrightarrow3\sqrt{x+1}+2\sqrt{x+1}-\sqrt{x+1}=0\)
\(\Leftrightarrow4\sqrt{x+1}=0\)
\(\Leftrightarrow x+1=0\)
\(\Leftrightarrow x=-1\)
Vậy ...
- Câu a có chút thiếu sót, mong thông cảm :)
\(\sqrt{3x-1}\) = 4
1.
a) \(\sqrt{3-2\sqrt{2}}+\sqrt{6-4\sqrt{2}}+\sqrt{9-4\sqrt{2}}=\sqrt{2-2\sqrt{2}+1}+\sqrt{4-2.2.\sqrt{2}+2}+\sqrt{8-2.2\sqrt{2}.1+1}=\sqrt{\left(\sqrt{2}\right)^2-2.\sqrt{2}.1+1^2}+\sqrt{2^2-2.2.\sqrt{2}+\left(\sqrt{2}\right)^2}+\sqrt{\left(2\sqrt{2}\right)^2-2.2\sqrt{2}.1+1^2}=\sqrt{\left(\sqrt{2}-1\right)^2}+\sqrt{\left(2-\sqrt{2}\right)^2}+\sqrt{\left(2\sqrt{2}-1\right)^2}=\left|\sqrt{2}-1\right|+\left|2-\sqrt{2}\right|+\left|2\sqrt{2}-1\right|=\sqrt{2}-1+2-\sqrt{2}+2\sqrt{2}-1=2\sqrt{2}\)
b) \(\sqrt{\left(4+\sqrt{10}\right)^2}-\sqrt{\left(4-\sqrt{10}\right)^2}=\left|4+\sqrt{10}\right|-\left|4-\sqrt{10}\right|=4+\sqrt{10}-4+\sqrt{10}=2\sqrt{10}\)
c) \(\dfrac{1}{\sqrt{2013}-\sqrt{2014}}-\dfrac{1}{\sqrt{2014}-\sqrt{2015}}=\dfrac{\sqrt{2013}+\sqrt{2014}}{\left(\sqrt{2013}-\sqrt{2014}\right)\left(\sqrt{2013}+\sqrt{2014}\right)}-\dfrac{\sqrt{2014}+\sqrt{2015}}{\left(\sqrt{2014}-\sqrt{2015}\right)\left(\sqrt{2014}+\sqrt{2015}\right)}=\dfrac{\sqrt{2013}+\sqrt{2014}}{2013-2014}-\dfrac{\sqrt{2014}+\sqrt{2015}}{2014-2015}=-\left(\sqrt{2013}+\sqrt{2014}\right)+\sqrt{2014}+\sqrt{2015}=-\sqrt{2013}-\sqrt{2014}+\sqrt{2014}+\sqrt{2015}=\sqrt{2015}-\sqrt{2013}\)
2.
a) \(x^2-2\sqrt{5}x+5=0\Leftrightarrow x^2-2.x.\sqrt{5}+\left(\sqrt{5}\right)^2=0\Leftrightarrow\left(x-\sqrt{5}\right)^2=0\Leftrightarrow x-\sqrt{5}=0\Leftrightarrow x=\sqrt{5}\)Vậy S={\(\sqrt{5}\)}
b) ĐK:x\(\ge-3\)
\(\sqrt{x+3}=1\Leftrightarrow\left(\sqrt{x+3}\right)^2=1^2\Leftrightarrow x+3=1\Leftrightarrow x=-2\left(tm\right)\)
Vậy S={-2}
3.
a) \(A=\dfrac{x-\sqrt{x}}{x+\sqrt{x}+1}-\dfrac{2x+\sqrt{x}}{\sqrt{x}}+\dfrac{2\left(x-1\right)}{\sqrt{x}-1}=\dfrac{\sqrt{x}\left(x\sqrt{x}-1\right)}{x+\sqrt{x}+1}-\dfrac{\sqrt{x}\left(2\sqrt{x}+1\right)}{\sqrt{x}}+\dfrac{2\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}-1}=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{x+\sqrt{x}+1}-\left(2\sqrt{x}+1\right)+2\left(\sqrt{x}+1\right)=\sqrt{x}\left(\sqrt{x}-1\right)-2\sqrt{x}-1+2\sqrt{x}+2=x-\sqrt{x}+1\)
b) Ta có \(A=x-\sqrt{x}+1=x-2\sqrt{x}.\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{3}{4}=\left(\sqrt{x}-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\)
Ta có \(\left(\sqrt{x}-\dfrac{1}{2}\right)^2\ge0\Leftrightarrow\left(\sqrt{x}-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\Leftrightarrow A\ge\dfrac{3}{4}\)
Dấu bằng xảy ra khi x=\(\dfrac{1}{4}\)
Vậy GTNN của A=\(\dfrac{3}{4}\)
\(a,3\sqrt{2x}+\sqrt{8x}-\sqrt{18x}=16\left(dk:x\ge0\right)\\ \Leftrightarrow3\sqrt{2x}+2\sqrt{2x}-3\sqrt{2x}=16\\ \Leftrightarrow\sqrt{2x}\left(3+2-3\right)=16\\ \Leftrightarrow2\sqrt{2x}=16\\ \Leftrightarrow\sqrt{2x}=8\\ \Leftrightarrow\left|2x\right|=64\\ \Leftrightarrow2x=64\\ \Leftrightarrow x=32\left(tm\right)\)
Vậy \(S=\left\{32\right\}\)
\(b,\sqrt{4x+20}-3\sqrt{x+5}+\dfrac{4}{3}\sqrt{9x+45}=6\left(dk:x\ge-5\right)\)
\(\Leftrightarrow\sqrt{4\left(x+5\right)}-3\sqrt{x+5}+\dfrac{4}{3}\sqrt{9\left(x+5\right)}=6\\ \Leftrightarrow2\sqrt{x+5}-3\sqrt{x+5}+4\sqrt{x+5}=6\\ \Leftrightarrow\sqrt{x+5}\left(2-3+4\right)=6\\ \Leftrightarrow3\sqrt{x+5}=6\\ \Leftrightarrow\sqrt{x+5}=2\\ \Leftrightarrow\left|x+5\right|=4\\ \Leftrightarrow x+5=4\\ \Leftrightarrow x=-1\left(tm\right)\)
Vậy \(S=\left\{-1\right\}\)