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Bài a,b,c,e,g,i thì đặt điều kiện rồi bình phương 2 vế rồi giải, bài j chuyển vế rồi bình phương
Chỉ trình bày lời giải, tự tìm điều kiện nha :v
d) \(\sqrt{x+2\sqrt{x-1}}=2\)
\(\Leftrightarrow\sqrt{x-1+2\sqrt{x-1}+1}=2\)
\(\Leftrightarrow\sqrt{\left(\sqrt{x-1}+1\right)^2}=2\)
\(\Leftrightarrow\sqrt{x-1}+1=2\)
\(\Leftrightarrow\sqrt{x-1}=1\)
\(\Rightarrow x-1=1\Leftrightarrow x=2\)
f) \(\sqrt{x+4\sqrt{x-4}}=2\)
\(\Leftrightarrow\sqrt{x-4+2.2\sqrt{x-4}+4}=2\)
\(\Leftrightarrow\sqrt{\left(\sqrt{x-4}+2\right)^2}=2\)
\(\Leftrightarrow\sqrt{x-4}+2=2\)
\(\Leftrightarrow\sqrt{x-4}=0\)
\(\Rightarrow x-4=0\Leftrightarrow x=4\)
\(đk:x\ge1\)
\(pt\Leftrightarrow3\sqrt{x-1}-\sqrt{x-1}+4\sqrt{x-1}=12\)
\(\Leftrightarrow6\sqrt{x-1}=12\Leftrightarrow\sqrt{x-1}=2\)
\(\Leftrightarrow x-1=4\Leftrightarrow x=1+4=5\left(N\right)\)
\(\sqrt{36x-36}-\sqrt{9x-9}-\sqrt{4x-4}=16-\sqrt{x-1}\)
\(\Rightarrow\sqrt{36\left(x-1\right)}-\sqrt{9\left(x-1\right)}-\sqrt{4\left(x-1\right)}=16-\sqrt{x-1}\)
\(\Leftrightarrow\sqrt{36}\sqrt{x-1}-\sqrt{9}\sqrt{x-1}-\sqrt{4}\sqrt{x-1}=16-\sqrt{x-1}\)
\(\Leftrightarrow6\sqrt{x-1}-3\sqrt{x-1}-2\sqrt{x-1}=16-\sqrt{x-1}\)
\(\Leftrightarrow\sqrt{x-1}=16-\sqrt{x-1}\)
\(\Leftrightarrow\sqrt{x-1}+\sqrt{x-1}=16\)
\(\Leftrightarrow2\sqrt{x-1}=16\)
\(\Leftrightarrow\sqrt{x-1}=8\)
\(\Leftrightarrow x-1=64\)
\(\Leftrightarrow x=64+1\)
\(\Leftrightarrow x=65\)
Vậy \(x=65\)
\(\sqrt{36x-36}-\sqrt{9x-9}-\sqrt{4x-4}=16-\sqrt{x-1}\)
<=> \(6\sqrt{x-1}-3\sqrt{x-1}-2\sqrt{x-1}+\sqrt{x-1}=16\)
<=> \(\sqrt{x-1}\left(6-3-2+1\right)=16\)
<=> \(\sqrt{x-1}=8\)
<=> \(x-1=64\)
<=> \(x=65\)
Vậy nghiệm của PT: S= \(\left\{65\right\}\)
P/s: Sai đừng trách mk nha!
đk: x > = 0
\(\left(\sqrt{x}-1\right)^2+\sqrt{x}\left(4-\sqrt{x}\right)=11\)
<=> \(x-2\sqrt{x}+1-x+4\sqrt{x}=11\)
<=> \(2\sqrt{x}=11\)
<=> \(\sqrt{x}=\frac{11}{2}\)
<=> x = 121/4
b) 4x2 - 4 = 0
<=> 4(x - 1)(x + 1) = 0
<=> x = 1 hoặc x = -1
Trả lời:
a, \(\left(\sqrt{x}-1\right)^2+\sqrt{x}\left(4-\sqrt{x}\right)=11\)
\(\Leftrightarrow\left(\sqrt{x}\right)^2-2\sqrt{x}+1+4\sqrt{x}-\left(\sqrt{x}\right)^2=11\)
\(\Leftrightarrow2\sqrt{x}+1=11\)
\(\Leftrightarrow2\sqrt{x}=10\)
\(\Leftrightarrow\sqrt{x}=5\)
\(\Leftrightarrow\sqrt{x}=\sqrt{25}\)
\(\Rightarrow x=25\)
Vậy x = 25
b, \(4x^2-4=0\)
\(\Leftrightarrow\)\(4\left(x^2-1\right)=0\)
\(\Leftrightarrow4\left(x-1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\x+1=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=1\\x=-1\end{cases}}\)
Vậy x = 1; x = -1
a.
\(\sqrt{x^2-4}=\sqrt{x-2}\)
\(\Leftrightarrow\sqrt{\left(x-2\right)\left(x+2\right)}=\sqrt{x-2}\)
\(\Leftrightarrow\sqrt{\left(x-2\right)}.\sqrt{\left(x+2\right)}-\sqrt{x-2}=0\)
\(\Leftrightarrow\sqrt{x-2}\left(\sqrt{x+2}-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-2}=0\\\sqrt{x+2}-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)
Vậy x=2 hoặc x=-1
b)
\(\Leftrightarrow\sqrt{x-1}+5\sqrt{4.\left(x-1\right)}-\sqrt{9.\left(x-1\right)}< 4\)
\(\Leftrightarrow\sqrt{x-1}+10\sqrt{x-1}-3\sqrt{x-1}< 4\)
\(\Leftrightarrow\left(1+10-3\right)\sqrt{x-1}< 4\)
\(\Leftrightarrow8\sqrt{x-1}< 4\)
\(\Leftrightarrow\sqrt{x-1}< \frac{1}{2}\)
\(\Leftrightarrow x-1< \frac{1}{4}\)
\(\Leftrightarrow x< \frac{5}{4}\)
Vậy...