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\(\sqrt{4x^2}=3\left(ĐK:4x^2\ge0\forall x\in R\right)\\ \Leftrightarrow\sqrt{\left(2x\right)^2}=3\\ \Leftrightarrow\left|2x\right|=3\\ \Leftrightarrow\left[{}\begin{matrix}2x=-3\\2x=3\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\left(tm\right)\\x=\dfrac{3}{2}\left(tm\right)\end{matrix}\right.\)
Vậy \(S=\left\{-\dfrac{3}{2};\dfrac{3}{2}\right\}\)
\(\sqrt{x^2-6x+9}=2\\ \Leftrightarrow\sqrt{\left(x-3\right)^2}=2\left(ĐK:\left(x-3\right)^2\ge0\forall x\in R\right)\\ \Leftrightarrow\left|x-3\right|=2\\ \Leftrightarrow\left[{}\begin{matrix}x-3=2\\x-3=-2\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=2+3\\x=-2-3\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=5\left(tm\right)\\x=-5\left(tm\right)\end{matrix}\right.\)
Vậy \(S=\left(\pm5\right)\)
\(\sqrt{\left(2x-3\right)^2}=6\left(ĐK:\left(2x-3\right)^2\ge0\forall x\in R\right)\\ \Leftrightarrow\left|2x-3\right|=6\\ \Leftrightarrow\left[{}\begin{matrix}2x-3=6\\2x-3=-6\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}2x=3+6\\2x=-6+3\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}2x=9\\2x=-3\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=4,5\left(tm\right)\\x=-1,5\left(tm\right)\end{matrix}\right.\)
Vậy \(S=\left\{4,5;-1,5\right\}\)
\(\sqrt{25x^2}=100\\ \sqrt{\left(5x\right)^2}=100\left(ĐK:\left(5x\right)^2\ge0\forall x\in R\right)\\\Leftrightarrow \left|5x\right|=100\\ \Leftrightarrow\left[{}\begin{matrix}5x=100\\5x=-100\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=20\left(tm\right)\\x=-20\left(tm\right)\end{matrix}\right.\)
Vậy \(S=\left\{\pm20\right\}\)
\(\sqrt{x+6-4\sqrt{x+2}}-\sqrt{9-4\sqrt{5}}=0\left(đk:x\ge-2\right)\)
\(\Leftrightarrow\sqrt{\left(\sqrt{x+2}-2\right)^2}=\sqrt{\left(\sqrt{5}-2\right)^2}\)
\(\Leftrightarrow\left|\sqrt{x+2}-2\right|=\left|\sqrt{5}-2\right|\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x+2}-2=\sqrt{5}-2\\\sqrt{x+2}-2=2-\sqrt{5}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=5\\x+2=21-8\sqrt{5}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=3\left(tm\right)\\x=19-8\sqrt{5}\left(tm\right)\end{matrix}\right.\)
Vậy \(S=\left\{3;19-8\sqrt{5}\right\}\)
1. đk: pt luôn xác định với mọi x
\(\sqrt{x^2-2x+1}-\sqrt{x^2-6x+9}=10\)
\(\Leftrightarrow\sqrt{\left(x-1\right)^2}-\sqrt{\left(x-3\right)^2}=10\)
\(\Leftrightarrow\left|x-1\right|-\left|x-3\right|=10\)
Bạn mở dấu giá trị tuyệt đối như lớp 7 là ok rồi!
2. đk: \(x\geq 1\)
\(\sqrt{x+2\sqrt{x-1}}=3\sqrt{x-1}-5\)
\(\Leftrightarrow\sqrt{x-1+2\sqrt{x-1}+1}=3\sqrt{x-1}-5\)
\(\Leftrightarrow\sqrt{\left(\sqrt{x-1}-1\right)^2}-3\sqrt{x-1}+5=0\)
\(\Leftrightarrow\left|\sqrt{x-1}-1\right|-3\sqrt{x-1}+5=0\)
Đến đây thì ổn rồi! bạn cứ xét khoảng rồi mở trị và bình phương 1 chút là ok cái bài!
\(\sqrt{25x-25}-\dfrac{15}{2}\sqrt{\dfrac{x-1}{9}}=6+\sqrt{x-1}\left(x\ge1\right)\)
\(< =>5\sqrt{x-1}-\dfrac{15}{2}\cdot\dfrac{\sqrt{x-1}}{3}=6+\sqrt{x-1}\)
\(< =>30\sqrt{x-1}-15\sqrt{x-1}=36+6\sqrt{x-1}\)
\(< =>9\sqrt{x-1}=36\\ < =>\sqrt{x-1}=4\\ < =>x-1=16\\ < =>x=17\left(tm\right)\)
\(\Leftrightarrow5\sqrt{x-1}-\dfrac{15}{2}\cdot\dfrac{1}{3}\sqrt{x-1}-\sqrt{x-1}=6\)
=>\(1.5\cdot\sqrt{x-1}=6\)
=>\(\sqrt{x-1}=4\)
=>x-1=16
=>x=17
a:
\(\Leftrightarrow\sqrt{\left(x-3\right)^2}=3\)
=>|x-3|=3
=>x-3=3 hoặc x-3=-3
=>x=0 hoặc x=6
b: \(\Leftrightarrow\sqrt{x-1+2\sqrt{x-1}+1}=2\)
=>\(\sqrt{\left(\sqrt{x-1}+1\right)^2}=2\)
=>\(\left|\sqrt{x-1}+1\right|=2\)
=>\(\left[{}\begin{matrix}\sqrt{x-1}+1=2\\\sqrt{x-1}+1=-2\left(loại\right)\end{matrix}\right.\Leftrightarrow\sqrt{x-1}=1\)
=>x-1=1
=>x=2
c:
ĐKXĐ: x>4/5
PT \(\Leftrightarrow\sqrt{\dfrac{5x-4}{x+2}}=2\)
=>\(\dfrac{5x-4}{x+2}=4\)
=>5x-4=4x+8
=>x=12(nhận)
d: ĐKXĐ: x-4>=0 và x+1>=0
=>x>=4
PT =>\(\left(\sqrt{x-4}+\sqrt{x+1}\right)^2=5^2=25\)
=>\(x-4+x+1+2\sqrt{\left(x-4\right)\left(x+1\right)}=25\)
=>\(\sqrt{4\left(x^2-3x-4\right)}=25-2x+3=28-2x\)
=>\(\sqrt{x^2-3x-4}=14-x\)
=>x<=14 và x^2-3x-4=(14-x)^2=x^2-28x+196
=>x<=14 và -3x-4=-28x+196
=>x<=14 và 25x=200
=>x=8(nhận)
a) \(\sqrt{x^2-6x+9}=3\)
\(\Leftrightarrow\sqrt{\left(x-3\right)^2}=3\)
\(\Leftrightarrow\left|x-3\right|=3 \)
TH1: \(\left|x-3\right|=x-3\) với \(x\ge3\)
Pt trở thành:
\(x-3=3\) (ĐK: \(x\ge3\))
\(\Leftrightarrow x=3+3\)
\(\Leftrightarrow x=6\left(tm\right)\)
TH2: \(\left|x-3\right|=-\left(x-3\right)\) với \(x< 3\)
Pt trở thành:
\(-\left(x-3\right)=3\) (ĐK: \(x< 3\))
\(\Leftrightarrow x-3=-3\)
\(\Leftrightarrow x=-3+3\)
\(\Leftrightarrow x=0\left(tm\right)\)
b) \(\sqrt{x+2\sqrt{x-1}}=2\) (ĐK: \(x\ge1\))
\(\Leftrightarrow x+2\sqrt{x-1}=4\)
\(\Leftrightarrow2\sqrt{x-1}=4-x\)
\(\Leftrightarrow4\left(x-1\right)=16-8x+x^2\)
\(\Leftrightarrow4x-4=16-8x+x^2\)
\(\Leftrightarrow x^2-12x+20=0\)
\(\Leftrightarrow\left(x-10\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=10\left(tm\right)\\x=2\left(tm\right)\end{matrix}\right.\)
c) \(\dfrac{\sqrt{5x-4}}{\sqrt{x+2}}=2\) (ĐK: \(x\ge\dfrac{4}{5}\))
\(\Leftrightarrow\dfrac{5x-4}{x+2}=4\)
\(\Leftrightarrow5x-4=4x+8\)
\(\Leftrightarrow x=12\left(tm\right)\)
\(x+\sqrt{x}+\sqrt{x+3}+\sqrt{x^2+3x}=6\left(đk:x\ge0\right)\)
\(\Leftrightarrow x+\sqrt{x}+\sqrt{x+3}+\sqrt{x\left(x+3\right)}=6\)
\(\Leftrightarrow\sqrt{x}\left(\sqrt{x}+1\right)+\sqrt{x+3}\left(\sqrt{x}+1\right)=6\)
\(\Leftrightarrow\left(\sqrt{x}+1\right)\left(\sqrt{x}+\sqrt{x+3}\right)=6\)
Do \(x\ge0\Leftrightarrow\sqrt{x}\ge0\Leftrightarrow\sqrt{x}+\sqrt{x+3}\ge\sqrt{x}+\sqrt{3}\ge\sqrt{x}+1>0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}\sqrt{x}+1=2\\\sqrt{x}+\sqrt{x+3}=3\end{matrix}\right.\\\left\{{}\begin{matrix}\sqrt{x}+1=1\\\sqrt{x}+\sqrt{x+3}=6\end{matrix}\right.\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=1\left(tm\right)\\\left\{{}\begin{matrix}x=0\\\sqrt{x}+\sqrt{x+3}=6\left(VLý\right)\end{matrix}\right.\end{matrix}\right.\)
Vậy \(S=\left\{1\right\}\)
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) \(\sqrt{x^2-10x+25}=x-2\left(đk:x\ge2\right)\)
\(\Leftrightarrow\sqrt{\left(x-5\right)^2}=x-2\)
\(\Leftrightarrow\left|x-5\right|=x-2\)
\(\Leftrightarrow\left[{}\begin{matrix}x-5=x-2\left(x\ge5\right)\\x-5=2-x\left(2\le x< 5\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}5=2\left(VLý\right)\\x=\dfrac{7}{2}\left(tm\right)\end{matrix}\right.\)
ĐKXĐ : \(x^3+x^2+6\ge0\)
\(pt\Leftrightarrow x^2+x+9=6+x^2+x^3\)
\(\Leftrightarrow x^3-x-3=0\)
Đến đây có lẽ dùng công thức Cardano là ra , nhưng mà không biết bạn học Cardano chưa nhỉ ?
Incursion_03Cardano mình học rồi