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\(A=\frac{1}{\sqrt{1}-\sqrt{2}}-\frac{1}{\sqrt{2}-\sqrt{3}}+\frac{1}{\sqrt{3}-\sqrt{4}}-....-\frac{1}{\sqrt{24}-\sqrt{25}}\)
\(=\frac{\sqrt{1}+\sqrt{2}}{(\sqrt{1}-\sqrt{2})(\sqrt{1}+\sqrt{2})}-\frac{\sqrt{2}+\sqrt{3}}{(\sqrt{2}-\sqrt{3})(\sqrt{2}+\sqrt{3})}+\frac{\sqrt{3}+\sqrt{4}}{(\sqrt{3}-\sqrt{4})(\sqrt{3}+\sqrt{4})}-...-\frac{\sqrt{24}+\sqrt{25}}{(\sqrt{24}-\sqrt{25})(\sqrt{24}+\sqrt{25})}\)
\(=\frac{\sqrt{1}+\sqrt{2}}{-1}-\frac{\sqrt{2}+\sqrt{3}}{-1}+\frac{\sqrt{3}+\sqrt{4}}{-1}-...-\frac{\sqrt{24}+\sqrt{25}}{-1}\)
\(=\frac{(1+\sqrt{2})-(\sqrt{2}+\sqrt{3})+(\sqrt{3}+\sqrt{4})-...-(\sqrt{24}+\sqrt{25})}{-1}\)
\(=\frac{1-\sqrt{25}}{-1}=4\)
\(B=\frac{5}{4+\sqrt{11}}+\frac{11-3\sqrt{11}}{\sqrt{11}-3}-\frac{4}{\sqrt{5}-1}+\sqrt{(\sqrt{5}-2)^2}\)
\(=\frac{5(4-\sqrt{11})}{(4+\sqrt{11})(4-\sqrt{11})}+\frac{\sqrt{11}(\sqrt{11}-3)}{\sqrt{11}-3}-\frac{4(\sqrt{5}+1)}{(\sqrt{5}-1)(\sqrt{5}+1)}+\sqrt{5}-2\)
\(=\frac{5(4-\sqrt{11})}{5}+\sqrt{11}-\frac{4(\sqrt{5}+1)}{4}+\sqrt{5}-2\)
\(=4-\sqrt{11}+\sqrt{11}-(\sqrt{5}+1)+\sqrt{5}-2\)
\(=1\)
1)
ĐK: \(x\geq 5\)
PT \(\Leftrightarrow \sqrt{4(x-5)}+3\sqrt{\frac{x-5}{9}}-\frac{1}{3}\sqrt{9(x-5)}=6\)
\(\Leftrightarrow \sqrt{4}.\sqrt{x-5}+3\sqrt{\frac{1}{9}}.\sqrt{x-5}-\frac{1}{3}.\sqrt{9}.\sqrt{x-5}=6\)
\(\Leftrightarrow 2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}=6\)
\(\Leftrightarrow 2\sqrt{x-5}=6\Rightarrow \sqrt{x-5}=3\Rightarrow x=3^2+5=14\)
2)
ĐK: \(x\geq -1\)
\(\sqrt{x+1}+\sqrt{x+6}=5\)
\(\Leftrightarrow (\sqrt{x+1}-2)+(\sqrt{x+6}-3)=0\)
\(\Leftrightarrow \frac{x+1-2^2}{\sqrt{x+1}+2}+\frac{x+6-3^2}{\sqrt{x+6}+3}=0\)
\(\Leftrightarrow \frac{x-3}{\sqrt{x+1}+2}+\frac{x-3}{\sqrt{x+6}+3}=0\)
\(\Leftrightarrow (x-3)\left(\frac{1}{\sqrt{x+1}+2}+\frac{1}{\sqrt{x+6}+3}\right)=0\)
Vì \(\frac{1}{\sqrt{x+1}+2}+\frac{1}{\sqrt{x+6}+3}>0, \forall x\geq -1\) nên $x-3=0$
\(\Rightarrow x=3\) (thỏa mãn)
Vậy .............
1.
a) \(\sqrt{3-2\sqrt{2}}+\sqrt{6-4\sqrt{2}}+\sqrt{9-4\sqrt{2}}=\sqrt{2-2\sqrt{2}+1}+\sqrt{4-2.2.\sqrt{2}+2}+\sqrt{8-2.2\sqrt{2}.1+1}=\sqrt{\left(\sqrt{2}\right)^2-2.\sqrt{2}.1+1^2}+\sqrt{2^2-2.2.\sqrt{2}+\left(\sqrt{2}\right)^2}+\sqrt{\left(2\sqrt{2}\right)^2-2.2\sqrt{2}.1+1^2}=\sqrt{\left(\sqrt{2}-1\right)^2}+\sqrt{\left(2-\sqrt{2}\right)^2}+\sqrt{\left(2\sqrt{2}-1\right)^2}=\left|\sqrt{2}-1\right|+\left|2-\sqrt{2}\right|+\left|2\sqrt{2}-1\right|=\sqrt{2}-1+2-\sqrt{2}+2\sqrt{2}-1=2\sqrt{2}\)
b) \(\sqrt{\left(4+\sqrt{10}\right)^2}-\sqrt{\left(4-\sqrt{10}\right)^2}=\left|4+\sqrt{10}\right|-\left|4-\sqrt{10}\right|=4+\sqrt{10}-4+\sqrt{10}=2\sqrt{10}\)
c) \(\dfrac{1}{\sqrt{2013}-\sqrt{2014}}-\dfrac{1}{\sqrt{2014}-\sqrt{2015}}=\dfrac{\sqrt{2013}+\sqrt{2014}}{\left(\sqrt{2013}-\sqrt{2014}\right)\left(\sqrt{2013}+\sqrt{2014}\right)}-\dfrac{\sqrt{2014}+\sqrt{2015}}{\left(\sqrt{2014}-\sqrt{2015}\right)\left(\sqrt{2014}+\sqrt{2015}\right)}=\dfrac{\sqrt{2013}+\sqrt{2014}}{2013-2014}-\dfrac{\sqrt{2014}+\sqrt{2015}}{2014-2015}=-\left(\sqrt{2013}+\sqrt{2014}\right)+\sqrt{2014}+\sqrt{2015}=-\sqrt{2013}-\sqrt{2014}+\sqrt{2014}+\sqrt{2015}=\sqrt{2015}-\sqrt{2013}\)
2.
a) \(x^2-2\sqrt{5}x+5=0\Leftrightarrow x^2-2.x.\sqrt{5}+\left(\sqrt{5}\right)^2=0\Leftrightarrow\left(x-\sqrt{5}\right)^2=0\Leftrightarrow x-\sqrt{5}=0\Leftrightarrow x=\sqrt{5}\)Vậy S={\(\sqrt{5}\)}
b) ĐK:x\(\ge-3\)
\(\sqrt{x+3}=1\Leftrightarrow\left(\sqrt{x+3}\right)^2=1^2\Leftrightarrow x+3=1\Leftrightarrow x=-2\left(tm\right)\)
Vậy S={-2}
3.
a) \(A=\dfrac{x-\sqrt{x}}{x+\sqrt{x}+1}-\dfrac{2x+\sqrt{x}}{\sqrt{x}}+\dfrac{2\left(x-1\right)}{\sqrt{x}-1}=\dfrac{\sqrt{x}\left(x\sqrt{x}-1\right)}{x+\sqrt{x}+1}-\dfrac{\sqrt{x}\left(2\sqrt{x}+1\right)}{\sqrt{x}}+\dfrac{2\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}-1}=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{x+\sqrt{x}+1}-\left(2\sqrt{x}+1\right)+2\left(\sqrt{x}+1\right)=\sqrt{x}\left(\sqrt{x}-1\right)-2\sqrt{x}-1+2\sqrt{x}+2=x-\sqrt{x}+1\)
b) Ta có \(A=x-\sqrt{x}+1=x-2\sqrt{x}.\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{3}{4}=\left(\sqrt{x}-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\)
Ta có \(\left(\sqrt{x}-\dfrac{1}{2}\right)^2\ge0\Leftrightarrow\left(\sqrt{x}-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\Leftrightarrow A\ge\dfrac{3}{4}\)
Dấu bằng xảy ra khi x=\(\dfrac{1}{4}\)
Vậy GTNN của A=\(\dfrac{3}{4}\)
\(a,\sqrt{3-x}+\sqrt{2-x}=1\)
\(\Rightarrow\sqrt{3+x}=1-\sqrt{2-x}\)
\(\Rightarrow3+x=1-2\sqrt{2-x}+2-x\)
\(\Rightarrow2x+2\sqrt{2-x}=0\)
\(\Rightarrow x+\sqrt{2-x}=0\)
\(\Rightarrow2-x=\left(-x\right)^2\)
\(\Rightarrow2-x=x^2\)
\(\Rightarrow2-x^2-x=0\)
\(\Rightarrow x^2+x-2=0\)
\(\Rightarrow\orbr{\begin{cases}x+2=0\\x-1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-2\\x=1\end{cases}}}\)
Vậy....
a) Đk: \(\left[{}\begin{matrix}x\le-1\\x\ge1\end{matrix}\right.\)
\(\sqrt{x^2-1}-x^2+1=0\)
\(\Leftrightarrow x^2-1-\sqrt{x^2-1}= 0\)
\(\Leftrightarrow\left(\sqrt{x^2-1}-1\right)\sqrt{x^2-1}=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x^2-1}-1=0\\\sqrt{x^2-1}=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x^2-1}=1\\x^2-1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2=2\left(1\right)\\x^2=1\left(2\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow x=\pm\sqrt{2}\left(N\right)\)
\(\left(2\right)\Leftrightarrow x=\pm1\left(N\right)\)
Kl: \(x=\pm\sqrt{2}\), \(x=\pm1\)
b) Đk: \(\left[{}\begin{matrix}x\le-2\\x\ge2\end{matrix}\right.\)
\(\sqrt{x^2-4}-x+2=0\)
\(\Leftrightarrow\sqrt{x^2-4}=x-2\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2-4=x^2-4x+4\\x\ge2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}4x=8\\x\ge2\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=2\left(N\right)\\x\ge2\end{matrix}\right.\)
kl: x=2
c) \(\sqrt{x^4-8x^2+16}=2-x\)
\(\Leftrightarrow\sqrt{\left(x^2-4\right)^2}=2-x\)
\(\Leftrightarrow\left|x^2-4\right|=2-x\) (*)
Th1: \(x^2-4< 0\Leftrightarrow-2< x< 2\)
(*) \(\Leftrightarrow x^2-4=x-2\Leftrightarrow x^2-x-2=0\Leftrightarrow\left[{}\begin{matrix}x=2\left(L\right)\\x=-1\left(N\right)\end{matrix}\right.\)
Th2: \(x^2-4\ge0\Leftrightarrow\left[{}\begin{matrix}x\le-2\\x\ge2\end{matrix}\right.\)
(*)\(\Leftrightarrow x^2-4=2-x\Leftrightarrow x^2+x-6=0\Leftrightarrow\left[{}\begin{matrix}x=2\left(N\right)\\x=-3\left(N\right)\end{matrix}\right.\)
Kl: x=-3, x=-1,x=2
d) \(\sqrt{9x^2+6x+1}=\sqrt{11-6\sqrt{2}}\)
\(\Leftrightarrow\sqrt{\left(3x+1\right)^2}=\sqrt{\left(3-\sqrt{2}\right)^2}\)
\(\Leftrightarrow\left|3x+1\right|=3-\sqrt{2}\) (*)
Th1: \(3x+1\ge0\Leftrightarrow x\ge-\dfrac{1}{3}\)
(*) \(\Leftrightarrow3x+1=3-\sqrt{2}\Leftrightarrow x=\dfrac{2-\sqrt{2}}{3}\left(N\right)\)
Th2: \(3x+1< 0\Leftrightarrow x< -\dfrac{1}{3}\)
(*) \(\Leftrightarrow3x+1=-3+\sqrt{2}\Leftrightarrow x=\dfrac{-4+\sqrt{2}}{3}\left(N\right)\)
Kl: \(x=\dfrac{2-\sqrt{2}}{3}\), \(x=\dfrac{-4+\sqrt{2}}{3}\)
e) Đk: \(x\ge-\dfrac{3}{2}\)
\(\sqrt{4^2-9}=2\sqrt{2x+3}\) \(\Leftrightarrow\sqrt{7}=2\sqrt{2x+3}\) \(\Leftrightarrow7=8x+12\)
\(\Leftrightarrow8x=-5\Leftrightarrow x=-\dfrac{5}{8}\left(N\right)\)
kl: \(x=-\dfrac{5}{8}\)
f) Đk: x >/ 5
\(\sqrt{4x-20}+3\sqrt{\dfrac{x-5}{9}}-\dfrac{1}{3}\sqrt{9x-45}=4\)
\(\Leftrightarrow2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}=4\)
\(\Leftrightarrow2\sqrt{x-5}=4\)
\(\Leftrightarrow\sqrt{x-5}=2\)
\(\Leftrightarrow x-5=4\)
\(\Leftrightarrow x=9\left(N\right)\)
kl: x=9
b) \(B=\dfrac{x-\sqrt{x}}{1-\sqrt{x}}-\dfrac{x\sqrt{x}}{\sqrt{x}}=\dfrac{\sqrt{x}\left(x-\sqrt{x}\right)-x\sqrt{x}\left(1-\sqrt{x}\right)}{\sqrt{x}\left(1-\sqrt{x}\right)}\) = \(\dfrac{x\sqrt{x}-x-x\sqrt{x}+x^2}{\sqrt{x}-x}=\dfrac{x^2-x}{\sqrt{x}-x}\)
c) \(C=\dfrac{x+2\sqrt{x}}{\sqrt{x}-x}-\dfrac{x\sqrt{x}}{\sqrt{x}+1}=\dfrac{\left(\sqrt{x}+1\right)\left(x+2\sqrt{x}\right)-x\sqrt{x}\left(\sqrt{x}-x\right)}{\left(\sqrt{x}-x\right)\left(\sqrt{x}+1\right)}=x+2\sqrt{x}-x\sqrt{x}\)
\(d,D=\dfrac{x+2\sqrt{x}}{\sqrt{x}+2}+\dfrac{5\sqrt{x}-2}{x-4}=\dfrac{x+2\sqrt{x}}{\sqrt{x}+2}+\dfrac{5\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}=\) \(\dfrac{\left(x+2\sqrt{x}\right)\left(\sqrt{x}-2\right)+5\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}=\dfrac{x+7\sqrt{x}-2}{\sqrt{x}+2}\)
e) \(E=\dfrac{\sqrt{x}}{\sqrt{x}-3}+\dfrac{\sqrt{x}-24}{x-9}=\dfrac{\sqrt{x}}{\sqrt{x}-3}+\dfrac{\sqrt{x}-24}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\dfrac{\sqrt{x}\left(\sqrt{x}+3\right)+\sqrt{x}-24}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\) = \(\dfrac{2\sqrt{x}-24}{\sqrt{x}+3}\)
F) F = \(\dfrac{3}{\sqrt{x}+5}+\dfrac{20-2\sqrt{x}}{x-25}=\dfrac{3\left(\sqrt{x}-5\right)+20-2\sqrt{x}}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}=\dfrac{23-2\sqrt{x}}{\sqrt{x}+5}\)
2. \(\dfrac{\sqrt{x^2}-16}{\sqrt{x-3}}+\sqrt{x+3}=\dfrac{7}{\sqrt{x-3}}\) (2)
\(\Leftrightarrow\dfrac{\sqrt{x^2}-16}{\sqrt{x-3}}+\sqrt{x+3}-\dfrac{7}{\sqrt{x-3}}=0\)
\(\Leftrightarrow\dfrac{\sqrt{x^2}-16+\sqrt{\left(x-3\right)\left(x+3\right)}-7}{\sqrt{x-3}}=0\)
\(\Leftrightarrow\sqrt{x^2}-16+\sqrt{\left(x-3\right)\left(x+3\right)}-7=0\)
\(\Leftrightarrow\left|x\right|-16+\sqrt{x^2-9}-7=0\)
\(\Leftrightarrow\left|x\right|-23+\sqrt{x^2-9}=0\)
\(\Leftrightarrow\sqrt{x^2-9}=-\left|x\right|+23\)
\(\Leftrightarrow x^2-9=-\left(-\left|x\right|+23\right)^2\)
\(\Leftrightarrow x^2-9=-\left(-\left|x\right|\right)^2-46\cdot\left|x\right|+529\)
\(\Leftrightarrow x^2-9=\left|x\right|^2-46+\left|x\right|+529\)
\(\Leftrightarrow x^2-9=x^2-46\cdot\left|x\right|+529\)
\(\Leftrightarrow-9=-46\cdot\left|x\right|+529\)
\(\Leftrightarrow46\cdot\left|x\right|=529+9\)
\(\Leftrightarrow49\cdot\left|x\right|=538\)
\(\Leftrightarrow\left|x\right|=\dfrac{269}{23}\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{269}{23}\\x=-\dfrac{269}{23}\end{matrix}\right.\)
Sau khi dùng phép thử ta nhận thấy \(x\ne-\dfrac{269}{23}\)
Vậy tập nghiệm phương trình (1) là \(S=\left\{\dfrac{269}{23}\right\}\)
3. sửa đề: \(\sqrt{14-x}=\sqrt{x-4}\sqrt{x-1}\) (3)
\(\Leftrightarrow\sqrt{14-x}=\sqrt{\left(x-4\right)\left(x-1\right)}\)
\(\Leftrightarrow\sqrt{14-x}=\sqrt{x^2-x-4x+4}\)
\(\Leftrightarrow\sqrt{14-x}=\sqrt{x^2-5x+4}\)
\(\Leftrightarrow14-x=x^2-5x+4\)
\(\Leftrightarrow14-x-x^2+5x-4=0\)
\(\Leftrightarrow10+4x-x^2=0\)
\(\Leftrightarrow-x^2+4x+10=0\)
\(\Leftrightarrow x^2-4x-10=0\)
\(\Leftrightarrow x=\dfrac{-\left(-4\right)\pm\sqrt{\left(-4\right)^2-4\cdot1\cdot\left(-10\right)}}{2\cdot1}\)
\(\Leftrightarrow x=\dfrac{4\pm\sqrt{16+40}}{2}\)
\(\Leftrightarrow x=\dfrac{4\pm\sqrt{56}}{2}\)
\(\Leftrightarrow x=\dfrac{4\pm2\sqrt{14}}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{4-2\sqrt{14}}{2}\\x=\dfrac{4+2\sqrt{14}}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2+\sqrt{14}\\x=2-\sqrt{14}\end{matrix}\right.\)
sau khi dùng phép thử ta nhận thấy \(x\ne2-\sqrt{14}\)
Vậy tập nghiệm phương trình (3) là \(S=\left\{2+\sqrt{14}\right\}\)
Câu 1:
PT \(\Leftrightarrow x^2+3x+8=(x+5)\sqrt{x^2+x+2}\)
\(\Leftrightarrow (x^2+x+2)+2(x+5)-4=(x+5)\sqrt{x^2+x+2}\)
Đặt \(\sqrt{x^2+x+2}=a; x+5=b(a\geq 0)\)
\(PT\Leftrightarrow a^2+2b-4=ba\)
\(\Leftrightarrow (a^2-4)-b(a-2)=0\)
\(\Leftrightarrow (a-2)(a+2-b)=0\Rightarrow \left[\begin{matrix} a=2\\ a+2=b\end{matrix}\right.\)
Nếu \(a=2\Rightarrow x^2+x+2=a^2=4\)
\(\Leftrightarrow x^2+x-2=0\Leftrightarrow (x-1)(x+2)=0\Rightarrow x=1; x=-2\) (đều thỏa mãn)
Nếu \(a+2=b\Leftrightarrow \sqrt{x^2+x+2}+2=x+5\)
\(\Leftrightarrow \sqrt{x^2+x+2}=x+3\)
\(\Rightarrow \left\{\begin{matrix} x+3\geq 0\\ x^2+x+2=(x+3)^2\end{matrix}\right.\Rightarrow \left\{\begin{matrix} x+3\geq 0\\ 5x+7=0\end{matrix}\right.\Rightarrow x=\frac{-7}{5}\) (thỏa mãn)
Vậy..........
Câu 2:
ĐKXĐ: \(x\geq 1\) hoặc \(x\leq \frac{1}{2}\)
\(10x^2-9x-8x\sqrt{2x^2-3x+1}+3=0\)
\(\Leftrightarrow 3(2x^2-3x+1)-8x\sqrt{2x^2-3x+1}+4x^2=0\)
Đặt \(\sqrt{2x^2-3x+1}=a(a\geq 0)\)
Khi đó PT \(\Leftrightarrow 3a^2-8xa+4x^2=0\)
\(\Leftrightarrow (a-2x)(3a-2x)=0\) \(\Rightarrow \left[\begin{matrix} a=2x\\ 3a=2x\end{matrix}\right.\)
Nếu \(a=\sqrt{2x^2-3x+1}=2x\Rightarrow \left\{\begin{matrix} x\geq 0\\ 2x^2-3x+1=4x^2\end{matrix}\right.\)
\(\Rightarrow \left\{\begin{matrix} x\geq 0\\ 2x^2+3x-1=0\end{matrix}\right.\Rightarrow x=\frac{-3+\sqrt{17}}{4}\) (t/m)
Nếu \(3a=3\sqrt{2x^2-3x+1}=2x\Rightarrow \left\{\begin{matrix} x\geq 0\\ 9(2x^2-3x+1)=4x^2\end{matrix}\right.\)
\(\Rightarrow \left\{\begin{matrix} x\geq 0\\ 14x^2-27x+9=0\end{matrix}\right.\Rightarrow x=\frac{3}{2}; x=\frac{3}{7}\) (t/m)
Vậy...........
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) \(ĐK:x\ge0\)
\(\Leftrightarrow3\sqrt{x}-3=\sqrt{x}+2\)
\(\Leftrightarrow\sqrt{x}=\dfrac{5}{2}\Leftrightarrow x=\dfrac{25}{4}\left(tm\right)\)