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\(\sqrt{x+3-4\sqrt{x-1}}+\sqrt{x+8-6\sqrt{x-1}}=1\) ( SỬA ĐỀ)
\(\sqrt{x-1-2.2.\sqrt{x-1}+4}+\sqrt{x-1-2.3.\sqrt{x-1}+9}=1\)
\(|x-1-2|+|x-1-3|=1\)
\(|x-3|+|x-4|=1\)
Với \(x\le3\)thì PT thành \(3-x+4-x=1\) \(\Rightarrow-2x=-6\Rightarrow x=3\)(thõa mãn)
Với \(3\le x< 4\)thì PT thành \(x-3+4-x=1\Leftrightarrow0x=0\Rightarrow\)Đúng với mọi x từ \(3\le x< 4\)
Với \(x\ge4\)thì PT thành \(x-3+x-4=1\Leftrightarrow2x=8\Leftrightarrow x=4\)(thõa mãn)
Vậy \(3\le x\le4\)
\(\sqrt{x^2-2x+1}+\sqrt{x^2-4x+4}=3\Leftrightarrow\sqrt{\left(x-1\right)^2}+\sqrt{\left(x-2\right)^2}=3\Leftrightarrow\left|x-1\right|+\left|x-2\right|=3\) \(+,x\ge2\Rightarrow\left\{{}\begin{matrix}x-2\ge0\\x-1\ge1\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\left|x-2\right|=x-2\\\left|x-1\right|=x-1\end{matrix}\right.\Rightarrow\left|x-2\right|+\left|x-1\right|=x-2+x-1=3\Leftrightarrow2x-3=3\Leftrightarrow x=3\left(\text{t/m}\right)\) \(+,1\le x< 2\Rightarrow\left\{{}\begin{matrix}x-1\ge0\\x-2< 0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\left|x-1\right|=x-1\\\left|x-2\right|=-\left(x-2\right)=2-x\end{matrix}\right.\Rightarrow\left|x-1\right|+\left|x-2\right|=x-1+2-x=1\left(l\right)\) \(+,x< 1\Rightarrow\left\{{}\begin{matrix}x-1< 0\\x-2< 0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\left|x-1\right|=-\left(x-1\right)=1-x\\\left|x-2\right|=-\left(x-2\right)=2-x\end{matrix}\right.\Rightarrow\left|x-1\right|+\left|x-2\right|=1-x+2-x=3\Leftrightarrow3-2x=3\Leftrightarrow x=0\left(\text{t/m}\right)\) \(f,\left\{{}\begin{matrix}\sqrt{x^2-9}\ge0\\\sqrt{x^2-6x+9}\ge0\end{matrix}\right.mà:\sqrt{x^2-9}+\sqrt{x^2-6x+9}=0\Rightarrow\left\{{}\begin{matrix}\sqrt{x^2-9}=0\\\sqrt{x^2-6x+9}=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x^2-9=0\\\sqrt{\left(x-3\right)^2}=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x^2-9=0\\\left|x-3\right|=0\end{matrix}\right.\Leftrightarrow x=3\)\thay vào ta thấy thoa man => x=3
\(ĐK:x\ge4\)\(\sqrt{x^2+x-20}=\sqrt{x^2+5x-4x-20}=\sqrt{x\left(x+5\right)-4\left(x+5\right)}=\sqrt{\left(x-4\right)\left(x+5\right)}=\sqrt{x-4}.\sqrt{x+5}=\sqrt{x-4}\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-4}=0\\\sqrt{x+5}=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x+5=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\left(tm\right)\\x=-4\left(l\right)\end{matrix}\right.\Rightarrow x=4\) \(b,ĐK:x\le2;\sqrt{x+1}+\sqrt{2-x}=\sqrt{6}\Leftrightarrow x+1+2-x+2\sqrt{\left(x+1\right)\left(2-x\right)}=6\Leftrightarrow3+2\sqrt{\left(x+1\right)\left(2-x\right)}=6\Leftrightarrow2\sqrt{\left(x+1\right)\left(2-x\right)}=3\Leftrightarrow\sqrt{\left(x-1\right)\left(2-x\right)}=1,5\Leftrightarrow\left(x-1\right)\left(2-x\right)=\frac{9}{4}\Leftrightarrow\left(x-1\right)\left(x-2\right)=-\frac{9}{4}\Leftrightarrow x^2-3x+2=-\frac{9}{4}\Leftrightarrow x^2-3x+\frac{9}{4}=-2\Leftrightarrow\left(x-\frac{3}{2}\right)^2=-2\Rightarrow vonghiem\)
Bài 1:
Để căn thức có nghĩa thì:
a)
\(-5x-10\geq 0\Leftrightarrow 5x+10\leq 0\Leftrightarrow x\leq -2\)
b)
\(x^2-3x+2\geq 0\Leftrightarrow (x-1)(x-2)\geq 0\)
\(\Leftrightarrow \left[\begin{matrix} x-1\geq 0; x-2\geq 0\\ x-1\leq 0; x-2\leq 0\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x\geq 2\\ x\leq 1\end{matrix}\right.\)
c) \(\frac{x+3}{5-x}\geq 0\)
\(\Leftrightarrow \left[\begin{matrix} x+3\geq 0; 5-x>0\\ x+3\leq 0; 5-x< 0\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} -3\leq x< 5\\ -3\geq x>5 (\text{vô lý})\end{matrix}\right.\)
\(\Rightarrow -3\leq x< 5\)
d) \(-x^2+4x-4\geq 0\)
\(\Leftrightarrow -(x^2-4x+4)\geq 0\Leftrightarrow -(x-2)^2\geq 0\)
Vì \((x-2)^2\geq 0, \forall x\in\mathbb{R}\)
\(\Rightarrow x=2\)
a) Do VT >=0 nên VP >=0 nên \(x\ge4\)
\(PT\Leftrightarrow\left(x-2\right)-\sqrt{x-2}-2=0\)
Đặt \(\sqrt{x-2}=t\ge\sqrt{4-2}=\sqrt{2}\) thì \(t^2-t-2=0\)
\(\Leftrightarrow t=2\left(loại t = -1 vì nó không thỏa mãn đk\right)\Leftrightarrow x-2=4\Leftrightarrow x=6\)
Bài 1:
a, Sai đề
b, \(\sqrt{x^2-4x+4}=x-2\)
\(\Leftrightarrow\sqrt{\left(x-2\right)^2}=x-2\)
\(\Leftrightarrow\left|x-2\right|=x-2\)(*)
TH1: \(x\ge2\Rightarrow\left|x-2\right|=x-2\)
(*)\(\Leftrightarrow x-2=x-2\)
\(\Leftrightarrow0x=0\)\(\Rightarrow\)PT có vô số nghiệm
TH2: \(x< 2\Rightarrow\left|x-2\right|=2-x\)
(*)\(\Leftrightarrow2-x=x-2\)
\(\Leftrightarrow-2x=-4\)
\(\Leftrightarrow x=2\)
Bài 2:
a, \(A=\sqrt{13+4\sqrt{10}}+\sqrt{13-4\sqrt{10}}\)
\(=\sqrt{\left(2\sqrt{2}+\sqrt{5}\right)^2}+\sqrt{\left(2\sqrt{2}-\sqrt{5}\right)^2}\)
\(=2\sqrt{2}+\sqrt{5}+2\sqrt{2}-\sqrt{5}\)
\(=2\sqrt{2}+2\sqrt{2}=4\sqrt{2}\)
b, \(B=\sqrt{2x+4+6\sqrt{2x-5}}+\sqrt{2x-4-2\sqrt{2x-5}}\)\(\left(x\ge\dfrac{5}{2}\right)\)
\(=\sqrt{2x-5+6\sqrt{2x-5}+9}+\sqrt{2x-5-2\sqrt{2x-5}+1}\)
\(=\sqrt{\left(\sqrt{2x-5}+3\right)^2}+\sqrt{\left(\sqrt{2x-5}-1\right)^2}\)
\(=\left|\sqrt{2x-5}+3\right|+\left|\sqrt{2x-5}-1\right|\)
\(=\sqrt{2x-5}+3+\sqrt{2x-5}-1\)
\(=2\sqrt{2x-5}+2\)
\(=2\left(\sqrt{2x-5}+1\right)\)
Sai thì nhớ báo nhé bạn.
câu a là \(\sqrt{x-2}+x=3\)