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a) ĐKXĐ: \(x\geq -3\)
Ta có: \(\sqrt{x+3}=1+\sqrt{2}\)
\(\Rightarrow x+3=(1+\sqrt{2})^2\)
\(\Leftrightarrow x+3=1+2+2\sqrt{2}=3+2\sqrt{2}\)
\(\Leftrightarrow x=2\sqrt{2}\) (thỏa mãn)
Vậy \(x=2\sqrt{2}\)
b) ĐK: \(x\geq 0\)
Có: \(\sqrt{10+\sqrt{5x}}=\sqrt{6}+2\)
\(\Rightarrow 10+\sqrt{5x}=(\sqrt{6}+2)^2=6+4+4\sqrt{6}\)
\(\Leftrightarrow \sqrt{5x}=4\sqrt{6}=\sqrt{96}\)
\(\Leftrightarrow x=\frac{96}{5}\) (thỏa mãn)
Vậy.....
c) ĐK: \(x\geq 4\)
Ta có: \(\sqrt{x^2-16}-\sqrt{x-4}=0\)
\(\Leftrightarrow \sqrt{(x-4)(x+4)}-\sqrt{x-4}=0\)
\(\Leftrightarrow \sqrt{x-4}(\sqrt{x+4}-1)=0\)
\(\Leftrightarrow \left[\begin{matrix} \sqrt{x-4}=0\\ \sqrt{x+4}=1\end{matrix}\right. \Leftrightarrow \left[\begin{matrix} x=4\\ x=-3\end{matrix}\right.\) (loại $x=-3$ vì $x\geq 4$)
Vậy \(x=4\)
d) ĐK: \(x\ge 0\)
Ta có: \(x-6\sqrt{x}+5=0\)
\(\Leftrightarrow (x-\sqrt{x})-5(\sqrt{x}-1)=0\)
\(\Leftrightarrow \sqrt{x}(\sqrt{x}-1)-5(\sqrt{x}-1)=0\)
\(\Leftrightarrow (\sqrt{x}-5)(\sqrt{x}-1)=0\)
\(\Leftrightarrow \left[\begin{matrix} \sqrt{x}-5=0\\ \sqrt{x}-1=0\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=25\\ x=1\end{matrix}\right.\) (đều t/m)
e) ĐK: \(x\geq 3\)
\(\sqrt{x-3}\geq 7\)
\(\Leftrightarrow x-3\geq 49\)
\(\Leftrightarrow x\geq 52\). Kết hợp với ĐK suy ra \(x\geq 52\)
f) ĐK: \(x\geq -1\)
Ta có: \(\sqrt{x+1}\leq 3\)
\(\Leftrightarrow x+1\leq 9\)
\(\Leftrightarrow x\leq 8\)
Kết hợp với ĐK suy ra \(-1\leq x\leq 8\)
Bài 1: Giải phương trình
a) ĐKXĐ: \(x\ge3\)
Ta có: \(\sqrt{100\cdot\left(x-3\right)}=\sqrt{20}\)
\(\Leftrightarrow\left|100\cdot\left(x-3\right)\right|=\left|20\right|\)
\(\Leftrightarrow100\cdot\left|x-3\right|=20\)
\(\Leftrightarrow\left|x-3\right|=\frac{1}{5}\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=\frac{1}{5}\\x-3=-\frac{1}{5}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{16}{5}\left(nhận\right)\\x=\frac{14}{5}\left(loại\right)\end{matrix}\right.\)
Vậy: \(S=\left\{\frac{16}{5}\right\}\)
b) Ta có: \(\sqrt{\left(x-3\right)^2}=7\)
\(\Leftrightarrow\left|x-3\right|=7\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=7\\x-3=-7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=10\\x=-4\end{matrix}\right.\)
Vậy: S={10;-4}
c) Ta có: \(\sqrt{4x^2+4x+1}=6\)
\(\Leftrightarrow\sqrt{\left(2x+1\right)^2}=6\)
\(\Leftrightarrow\left|2x+1\right|=6\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+1=6\\2x+1=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=5\\2x=-7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{5}{2}\\x=\frac{-7}{2}\end{matrix}\right.\)
Vậy: \(S=\left\{\frac{5}{2};\frac{-7}{2}\right\}\)
Bài 1:
b: \(\Leftrightarrow2+\sqrt{3x-5}=x+1\)
\(\Leftrightarrow\sqrt{3x-5}=x-1\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2-2x+1=3x-5\\x>=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x^2-5x+6=0\\x>=1\end{matrix}\right.\Leftrightarrow x\in\left\{2;3\right\}\)
c: \(\Leftrightarrow5x+7=16\left(x+3\right)\)
=>16x+48=5x+7
=>11x=-41
hay x=-41/11
Làm hơi tắt xíu, có gì ko hiểu cmt nha :>
\(a.\sqrt{x-1}=3\left(ĐK:x\ge1\right)\Leftrightarrow x-1=9\Leftrightarrow x=10\)
\(b.\sqrt{x^2-4x+4}=2\\ \Leftrightarrow\sqrt{\left(x-2\right)^2}=2\\ \Leftrightarrow\left|x-2\right|=2\\ \Leftrightarrow\left[{}\begin{matrix}x-2=2\left(x\ge2\right)\\2-x=2\left(x< 2\right)\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=4\\x=0\end{matrix}\right.\)
\(c.\sqrt{25x^2-10x+1}=4x-9\\ \Leftrightarrow\sqrt{\left(5x-1\right)^2}=4x-9\\ \Leftrightarrow\left|5x-1\right|=4x-9\\\Leftrightarrow \left[{}\begin{matrix}5x-1=4x-9\left(x\ge\frac{1}{5}\right)\\1-5x=4x-9\left(x< \frac{1}{5}\right)\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-8\left(ktm\right)\\x=\frac{10}{9}\left(ktm\right)\end{matrix}\right.\)
\(d.\sqrt{x^2+2x+1}=\sqrt{x+1}\left(ĐK:x\ge-1\right)\\ \Leftrightarrow x^2+2x+1=x+1\\ \Leftrightarrow x^2+x=0\Leftrightarrow x\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)
e. ĐK: \(\left[{}\begin{matrix}x\ge3\\x\le-3\end{matrix}\right.\)
\(\sqrt{x^2-9}+\sqrt{x^2-6x+9}=0\\ \Leftrightarrow\sqrt{\left(x-3\right)\left(x+3\right)}+\sqrt{\left(x-3\right)^2}=0\\ \Leftrightarrow\sqrt{x-3}\left(\sqrt{x+3}+\sqrt{x-3}\right)=0\\ \Leftrightarrow\sqrt{x-3}=0\\ \Leftrightarrow x-3=0\Leftrightarrow x=3\)
Câu cuối chưa nghĩ ra, sorry :<
a) \(\sqrt{x+3}-\sqrt{x-1}=\sqrt{2x+2}\)
Điều kiện: \(\hept{\begin{cases}x+3\ge0\\x-1\ge0\\2x+2\ge0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ge-3\\x\ge1\\x\ge-1\end{cases}\Leftrightarrow x\ge1}\)
\(\Leftrightarrow\left(\sqrt{x+3}-\sqrt{x-1}\right)^2=\left(\sqrt{2x+2}\right)^2\)
\(\Leftrightarrow x+3-2\sqrt{\left(x+3\right)\left(x-1\right)}+x-1=2x+2\)
\(\Leftrightarrow2x+2-2\sqrt{\left(x+3\right)\left(x-1\right)}=2x+2\)
\(\Leftrightarrow-2\sqrt{\left(x+3\right)\left(x-1\right)}=0\)
\(\Leftrightarrow\left(x+3\right)\left(x-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+3=0\\x-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-3\left(l\right)\\x=1\left(n\right)\end{cases}}\)
Vậy \(S=\left\{1\right\}\)
b,
+ Với \(x=0\) \(\Rightarrow PTVN\)
+ Với \(x\ne0\), chia cả 2 vế cho \(x^2\) :
\(PT\Leftrightarrow x^2-16x+46+\frac{144}{x}+\frac{81}{x^2}=0\)
\(\Leftrightarrow\left(x^2+\frac{81}{x^2}\right)-16\left(x-\frac{9}{x}\right)+46=0\)
Đặt \(x-\frac{9}{x}=t\Rightarrow t^2=x^2+\frac{81}{x^2}-18\)
\(\Leftrightarrow t^2+18-16t+46=0\)
\(\Leftrightarrow t^2-16t+64=0\Rightarrow t=8\)
\(\Leftrightarrow x-\frac{9}{x}=8\Leftrightarrow x^2-8x-9=0\) \(\Rightarrow\left[{}\begin{matrix}x=-1\\x=9\end{matrix}\right.\) (t/m)
cậu xem làm được mấy bài kia không làm giùm với (đang gấp) :))
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=\(\dfrac{3}{4}\)
1.
\(x-6\sqrt{x}-\sqrt{x}+6=0\)
\(\Leftrightarrow\left(\sqrt{x}-6\right)\left(\sqrt{x}-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=36\\x=1\end{cases}}\)
2.
\(\sqrt{\left(x-3\right)\left(x+3\right)}-3\sqrt{x-3}=0\)
\(\Leftrightarrow\sqrt{x-3}\left(\sqrt{x+3}-3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=3\\\sqrt{x-3}=3\end{cases}\Leftrightarrow\orbr{\begin{cases}x=3\\x=12\end{cases}}}\)