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1) \(\Leftrightarrow\sqrt{\left(x+5\right)^2}=4\)
\(\Leftrightarrow\left|x+5\right|=4\)
\(\Leftrightarrow\left[{}\begin{matrix}x+5=4\\x+5=-4\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-9\end{matrix}\right.\)
2) \(ĐK:x\ge2\)
\(\Leftrightarrow\sqrt{x-2}=2\)
\(\Leftrightarrow x-2=4\Leftrightarrow x=6\left(tm\right)\)
3) \(\Leftrightarrow\left(x^2-x+4\right)-\sqrt{x^2-x+4}+\dfrac{1}{4}=\dfrac{9}{4}\)
\(\Leftrightarrow\left(\sqrt{x^2-x+4}-\dfrac{1}{2}\right)^2=\dfrac{9}{4}\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x^2-x+4}-\dfrac{1}{2}=\dfrac{3}{2}\\\sqrt{x^2-x+4}-\dfrac{1}{2}=-\dfrac{3}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x^2-x+4}=2\\\sqrt{x^2-x+4}=-1\left(VLý\right)\end{matrix}\right.\)
\(\Leftrightarrow x^2-x+4=4\Leftrightarrow x\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
4) \(ĐK:x\ge0\)
\(\Leftrightarrow3\sqrt{x}-3=\sqrt{x}+2\)
\(\Leftrightarrow\sqrt{x}=\dfrac{5}{2}\Leftrightarrow x=\dfrac{25}{4}\left(tm\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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) \(\sqrt{x^2-10x+25}=x-2\left(đk:x\ge2\right)\)
\(\Leftrightarrow\sqrt{\left(x-5\right)^2}=x-2\)
\(\Leftrightarrow\left|x-5\right|=x-2\)
\(\Leftrightarrow\left[{}\begin{matrix}x-5=x-2\left(x\ge5\right)\\x-5=2-x\left(2\le x< 5\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}5=2\left(VLý\right)\\x=\dfrac{7}{2}\left(tm\right)\end{matrix}\right.\)
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a)ĐKXĐ: \(x\ge0\)
Ta có: \(\sqrt{x}=2\)
\(\Leftrightarrow x=2^2=4\)(nhận)
Vậy: S={4}
b) ĐKXĐ: \(x\ge1\)
Ta có: \(\sqrt{25x-25}-10=0\)
\(\Leftrightarrow\sqrt{25}\cdot\sqrt{x-1}=10\)
\(\Leftrightarrow\sqrt{x-1}=2\)
\(\Leftrightarrow x-1=2^2=4\)
hay x=5(nhận)
Vậy: S={5}
c)ĐKXĐ: \(x\in Z\)
Ta có: \(\sqrt{25-10x+x^2}=7\)
\(\Leftrightarrow\sqrt{\left(x-5\right)^2}=7\)
\(\Leftrightarrow\left|x-5\right|=7\)
\(\Leftrightarrow\left[{}\begin{matrix}x-5=7\\x-5=-7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=12\left(nhận\right)\\x=-2\left(nhận\right)\end{matrix}\right.\)
Vậy: S={12;-2}
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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!
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\(\left(\sqrt{x+5}-\sqrt{x+2}\right)\left(1+\sqrt{x^2+7x+10}\right)=3\left(đk:x\ge-2\right)\)
Đặt \(a=\sqrt{x+5},b=\sqrt{x+2}\left(đk:a,b\ge0,a\ne b\right)\)
\(\Rightarrow\left\{{}\begin{matrix}ab=\sqrt{\left(x+5\right)\left(x+2\right)}=\sqrt{x^2+7x+10}\\a^2-b^2=x+5-x-2=3\end{matrix}\right.\)
PT trở thành: \(\left(a-b\right)\left(1+ab\right)=a^2-b^2\)
\(\Leftrightarrow\left(a-b\right)\left(ab+1\right)=\left(a-b\right)\left(a+b\right)\)
\(\Leftrightarrow\left(a-b\right)\left(ab+1-a-b\right)=0\)
\(\Leftrightarrow\left(a-b\right)\left(b-1\right)\left(a-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=b\left(loại\right)\\a=1\\b=1\end{matrix}\right.\)
+ Với a=1
\(\Rightarrow\sqrt{x+5}=1\Leftrightarrow x+5=1\Leftrightarrow x=-4\left(ktm\right)\)
+ Với b=1
\(\Rightarrow\sqrt{x+2}=1\Leftrightarrow x+2=1\Leftrightarrow x=-1\left(tm\right)\)
Vậy \(S=\left\{-1\right\}\)
Đặt \(\left\{{}\begin{matrix}\sqrt{x+5}=a\\\sqrt{x+2=b}\end{matrix}\right.\)
Thì được:
\(\left(a-b\right)\left(1+ab\right)=a^2-b^2\)
\(\Leftrightarrow\left(a-1\right)\left(b-1\right)\left(a-b\right)=0\)
Làm tiếp
![](https://rs.olm.vn/images/avt/0.png?1311)
=> \(\sqrt{2-x}.\sqrt{3-x}+\sqrt{3-x}.\sqrt{5-x}+\sqrt{5-x}.\sqrt{2-x}+5-x=5\)
=> \(\sqrt{3-x}\left(\sqrt{2-x}+\sqrt{5-x}\right)+\sqrt{5-x}\left(\sqrt{2-x}+\sqrt{5-x}\right)=5\)
=> \(\left(\sqrt{5-x}+\sqrt{2-x}\right)\left(\sqrt{5-x}+\sqrt{3-x}\right)=5\)
=> giải tiếp nhé , mình biết lớp 10
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ĐKXĐ: \(\dfrac{74}{9}\le x\le10\)
Đặt \(\sqrt{10-x}=t\Rightarrow0\le t\le\dfrac{4}{3}\) \(\Rightarrow x=10-t^2\)
Ta được:
\(2+\sqrt{4-3t}=\dfrac{10-t^2}{3}\)
\(\Leftrightarrow\sqrt{4-3t}-1=\dfrac{10-t^2}{3}-3\)
\(\Leftrightarrow\dfrac{3\left(1-t\right)}{\sqrt{4-3t}+1}=\dfrac{\left(1-t\right)\left(1+t\right)}{3}\)
\(\Rightarrow\left[{}\begin{matrix}t=1\Rightarrow x=9\\\dfrac{3}{\sqrt{4-3t}+1}=\dfrac{t+1}{3}\left(1\right)\end{matrix}\right.\)
Xét (1), do \(0\le t\le\dfrac{4}{3}\Rightarrow\left\{{}\begin{matrix}\dfrac{3}{\sqrt{4-3t}+1}\ge1\\\dfrac{t+1}{3}\le\dfrac{\dfrac{4}{3}+1}{3}=\dfrac{7}{9}< 1\end{matrix}\right.\)
\(\Rightarrow\left(1\right)\) vô nghiệm
Vậy pt có nghiệm duy nhất \(x=9\)
ĐK : tụ làm nha
Đặt 10 - x^2 = a
pt <=> \(\sqrt{15+a}-\sqrt{a}=3\)
=> \(\sqrt{15+a}=3+\sqrt{a}\)
=> \(15+a=9+6\sqrt{a}+a\)
=> \(6\sqrt{a}=6\)
=> a = 1
=> 10 - x^2 = 1
=> x^2 = 9
=> x = 3 hoặc x = -3