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ĐKXĐ: \(x\ge-5\)
\(\Leftrightarrow\sqrt{4\left(x+5\right)}-3\sqrt{x+5}+\frac{4}{3}\sqrt{9\left(x+5\right)}=6\)
\(\Leftrightarrow2\sqrt{x+5}-3\sqrt{x+5}+4\sqrt{x+5}=6\)
\(\Leftrightarrow3\sqrt{x+5}=6\)
\(\Leftrightarrow\sqrt{x+5}=2\)
\(\Leftrightarrow x+5=4\Rightarrow x=-1\)
1) \(\sqrt{5-2x}=6\left(đk:x\le\dfrac{5}{2}\right)\)
\(\Leftrightarrow5-2x=36\)
\(\Leftrightarrow2x=-31\Leftrightarrow x=-\dfrac{31}{2}\left(tm\right)\)
2) \(\sqrt{2-x}=\sqrt{x+1}\left(đk:2\ge x\ge-1\right)\)
\(\Leftrightarrow2-x=x+1\)
\(\Leftrightarrow2x=1\Leftrightarrow x=\dfrac{1}{2}\left(tm\right)\)
3) \(\Leftrightarrow\sqrt{\left(2x+1\right)^2}=6\)
\(\Leftrightarrow\left|2x+1\right|=6\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+1=6\\2x+1=-6\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-\dfrac{7}{2}\end{matrix}\right.\)
4) \(\sqrt{x^2-10x+25}=x-2\left(đk:x\ge2\right)\)
\(\Leftrightarrow\sqrt{\left(x-5\right)^2}=x-2\)
\(\Leftrightarrow\left|x-5\right|=x-2\)
\(\Leftrightarrow\left[{}\begin{matrix}x-5=x-2\left(x\ge5\right)\\x-5=2-x\left(2\le x< 5\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}5=2\left(VLý\right)\\x=\dfrac{7}{2}\left(tm\right)\end{matrix}\right.\)
\(a,\Leftrightarrow x-1=4\Leftrightarrow x=5\\ b,\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{3}{4}\\3x+1=4x-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{3}{4}\\x=4\left(tm\right)\end{matrix}\right.\Leftrightarrow x=4\\ c,ĐK:x\ge-5\\ PT\Leftrightarrow2\sqrt{x+5}-3\sqrt{x+5}+4\sqrt{x+5}=6\\ \Leftrightarrow3\sqrt{x+5}=6\\ \Leftrightarrow\sqrt{x+5}=3\\ \Leftrightarrow x+5=9\\ \Leftrightarrow x=4\left(tm\right)\)
\(d,\Leftrightarrow\sqrt{\left(x-2\right)^2}=\sqrt{\left(\sqrt{5}+1\right)^2}\\ \Leftrightarrow\left|x-2\right|=\sqrt{5}+1\\ \Leftrightarrow\left[{}\begin{matrix}x-2=\sqrt{5}+1\\2-x=\sqrt{5}+1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{5}+3\\x=1-\sqrt{5}\end{matrix}\right.\)
\(d,ĐK:x\ge0\\ PT\Leftrightarrow\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=2\\\sqrt{x}=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\left(tm\right)\\x=9\left(tm\right)\end{matrix}\right.\\ e,ĐK:x\ge1\\ PT\Leftrightarrow\sqrt{x-1}+\dfrac{3}{2}\cdot2\sqrt{x-1}-\dfrac{2}{5}\cdot5\sqrt{x-1}=4\\ \Leftrightarrow2\sqrt{x-1}=4\Leftrightarrow\sqrt{x-1}=2\\ \Leftrightarrow x-1=4\Leftrightarrow x=5\left(tm\right)\\ f,ĐK:x\ge5\\ PT\Leftrightarrow\sqrt{x-5}+2\sqrt{x-5}-\dfrac{1}{3}\cdot3\sqrt{x-5}=6\\ \Leftrightarrow2\sqrt{x-5}=6\Leftrightarrow\sqrt{x-5}=3\\ \Leftrightarrow x-5=9\Leftrightarrow x=14\left(tm\right)\)
a) \(\dfrac{9x-7}{\sqrt{7x+5}}=\sqrt{7x+5}\) (1)
\(\Leftrightarrow9x-7=\sqrt{\left(7x+5\right)\left(7x+5\right)}\)
\(\Leftrightarrow9x-\sqrt{\left(7x+5\right)\left(7x+5\right)}=7\)
\(\Leftrightarrow9x-\sqrt{\left(7x+5\right)^2}=7\)
\(\Leftrightarrow9x-\left|7x+5\right|=7\)
\(\Leftrightarrow\left[{}\begin{matrix}9x-\left(7x+5\right)=7\left(đk:7x+5\ge0\right)\\9x-\left[-\left(7x+5\right)\right]=7\left(đk:7x+5< 0\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=6\left(đk:x\ge-\dfrac{5}{7}\right)\\x=\dfrac{1}{8}\left(đk:x< -\dfrac{5}{7}\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=6\\x\in\varnothing\end{matrix}\right.\)
\(\Leftrightarrow x=6\)
Vậy tập nghiệm phương trình (1) là \(S=\left\{6\right\}\)
b) \(\sqrt{4x-20}+3\sqrt{\dfrac{x+5}{9}}-\dfrac{1}{3}\sqrt{9x-45}=4\) (2)
\(\Leftrightarrow\sqrt{4\left(x-5\right)}+3\cdot\dfrac{\sqrt{x+5}}{3}-\dfrac{1}{3}\cdot\sqrt{9\left(x-5\right)}=4\)
\(\Leftrightarrow\sqrt{4}\sqrt{x-5}+\sqrt{x+5}-\dfrac{1}{3}\cdot\sqrt{9}\sqrt{x-5}=4\)
\(\Leftrightarrow2\sqrt{x-5}+\sqrt{x+5}-\dfrac{1}{3}\cdot3\sqrt{x-5}=4\)
\(\Leftrightarrow2\sqrt{x-5}+\sqrt{x+5}-\sqrt{x-5}=4\)
\(\Leftrightarrow\sqrt{x-5}+\sqrt{x+5}=4\)
\(\Leftrightarrow\sqrt{x-5}=4-\sqrt{x+5}\)
\(\Leftrightarrow x-5=\left(4-\sqrt{x+5}\right)^2\)
\(\Leftrightarrow x-5=16-8\sqrt{x+5}+x+5\)
\(\Leftrightarrow-5=16-8\sqrt{x+5}+5\)
\(\Leftrightarrow-5=21-8\sqrt{x+5}\)
\(\Leftrightarrow8\sqrt{x+5}=21+5\)
\(\Leftrightarrow8\sqrt{x+5}=26\)
\(\Leftrightarrow\sqrt{x+5}=\dfrac{13}{4}\)
\(\Leftrightarrow x+5=\dfrac{169}{16}\)
\(\Leftrightarrow x=\dfrac{169}{16}-5\)
\(\Leftrightarrow x=\dfrac{89}{16}\)
Vậy tập nghiệm phương trình (2) là \(S=\left\{\dfrac{89}{16}\right\}\)
Nick cũ không đi giải lấy nick mới giải làm gì vậy Tuấn Anh Phan Nguyễn ? :D
a) \(\sqrt{x-1}+\sqrt{4x-4}-\sqrt{25x-25}+2=0\) (ĐK: \(x\ge1\))
\(\Leftrightarrow\sqrt{x-1}+\sqrt{4\left(x-1\right)}-\sqrt{25\left(x-1\right)}+2=0\)
\(\Leftrightarrow\sqrt{x-1}+2\sqrt{x-1}-5\sqrt{x-1}+2=0\)
\(\Leftrightarrow-2\sqrt{x-1}=-2\)
\(\Leftrightarrow\sqrt{x-1}=\dfrac{2}{2}\)
\(\Leftrightarrow\sqrt{x-1}=1\)
\(\Leftrightarrow x-1=1\)
\(\Leftrightarrow x=2\left(tm\right)\)
b) \(\sqrt{16x+16}-\sqrt{9x+9}+\sqrt{4x+4}+\sqrt{x+1}=16\) (ĐK: \(x\ge-1\))
\(\Leftrightarrow\sqrt{16\left(x+1\right)}-\sqrt{9\left(x+1\right)}+\sqrt{4\left(x+1\right)}+\sqrt{x+1}=16\)
\(\Leftrightarrow4\sqrt{x+1}-3\sqrt{x+1}+2\sqrt{x+1}+\sqrt{x+1}=16\)
\(\Leftrightarrow4\sqrt{x+1}=16\)
\(\Leftrightarrow\sqrt{x+1}=4\)
\(\Leftrightarrow x+1=16\)
\(\Leftrightarrow x=15\left(tm\right)\)
1 bài thôi nhé, tui còn phải xem World Cup :vv
\(\sqrt{x^4-4x+4}=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-6\sqrt{20}}}}\)
\(\Leftrightarrow x^4-4x+4=\sqrt{5}-\sqrt{3-\sqrt{20-2\cdot\sqrt{20}\cdot3+9}}\)
\(\Leftrightarrow x^4-4x+4=\sqrt{5}-\sqrt{3-\left(\sqrt{20}-3\right)}\)
\(\Leftrightarrow x^4-4x+4=\sqrt{5}-\sqrt{6-2\sqrt{5}}\)
\(\Leftrightarrow x^4-4x+4=\sqrt{5}-\sqrt{\left(\sqrt{5}-1\right)^2}\)
\(\Leftrightarrow x^4-4x+4=\sqrt{5}-\sqrt{5}+1\)
\(\Leftrightarrow x^4-4x+3=0\)
\(\Leftrightarrow x^4+2x^3+3x^2-2x^3-4x^2-6x+x^2+2x+3=0\)
\(\Leftrightarrow x^2\left(x^2+2x+3\right)-2x\left(x^2+2x+3\right)+\left(x^2+2x+3\right)=0\)
\(\Leftrightarrow\left(x^2+2x+3\right)\left(x^2-2x+1\right)=0\)
\(\Leftrightarrow\left(x-1\right)^2\left(x^2+2x+3\right)=0\)
Vì: \(x^2+2x+3=\left(x^2+2x+1\right)+2=\left(x+1\right)^2+2\ge2>0\)
=> \(\left(x-1\right)^2=0\Leftrightarrow x-1=0\Leftrightarrow x=1\) (thỏa mãn)
Vậy pt có nghiệm x = 1
p/s: đkxđ là x thuộc R nên tui k ghi vào :v
cảm ơn nhiều