Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) \(\frac{1}{2}\sqrt{x-1}-\frac{3}{2}\sqrt{9x-9}+24\sqrt{\frac{x-1}{64}}=-17\)
<=> \(\frac{1}{2}\sqrt{x-1}-\frac{3}{2}\sqrt{9\left(x-1\right)}+24\frac{\sqrt{x-1}}{\sqrt{64}}=-17\)
<=>\(\frac{1}{2}\sqrt{x-1}-\frac{9}{2}\sqrt{x-1}+3\sqrt{x-1}=-17\)
<=>\(\sqrt{x-1}\left(\frac{1}{2}-\frac{9}{2}+\frac{6}{2}\right)=-17\)
<=>\(\sqrt{x-1}=-17\)
<=>x-1=17
<=>x=18
Vậy pt có nghiệm là x=18
\(a.ĐK:x-1\ge0\Leftrightarrow x\ge1\)
\(\frac{1}{2}\sqrt{x-1}-\frac{3}{2}\sqrt{9x-9}+24\sqrt{\frac{x-1}{64}}=-17\)
\(\Leftrightarrow\frac{1}{2}\sqrt{x-1}-\frac{27}{2}\sqrt{x-1}+24\sqrt{\frac{x-1}{64}}=-17\)
\(\Leftrightarrow\sqrt{x-1}\left(\frac{1}{2}-\frac{27}{2}+24\sqrt{\frac{1}{64}}\right)=-17\)
\(\Leftrightarrow\sqrt{x-1}.\left(-10\right)=-17\)
\(\Leftrightarrow\sqrt{x-1}=\frac{-17}{-10}=\frac{17}{10}\)
\(\Leftrightarrow x-1=\left(\frac{17}{10}\right)^2\)
\(\Leftrightarrow x=\frac{289}{100}+1=3,89\left(TM\right)\)
Vậy \(S=\left\{3,89\right\}\)
\(b.ĐK:x^2+2\ge0\)
\(\sqrt{9x^2+18}+2\sqrt{x^2+2}-\sqrt{25x^2+50}+3=0\)
\(\Leftrightarrow9\sqrt{x^2+2}+2\sqrt{x^2+2}-25\sqrt{x^2+2}=-3\)
\(\Leftrightarrow\sqrt{x^2+2}\left(9+2-25\right)=-3\)
\(\Leftrightarrow\sqrt{x^2+2}=\frac{-3}{-14}=\frac{3}{14}\)
\(\Leftrightarrow x^2+2=\left(\frac{3}{14}\right)^2\)
\(\Leftrightarrow x=\sqrt{\frac{9}{196}-2}=\sqrt{-\frac{383}{196}}\left(vl\right)\)
Vậy \(S=\varnothing\)
Mấy câu kia làm tương tự
Bài 1:
Ta có: \(\sqrt{16x-32}+\sqrt{25x-50}=18+\sqrt{9x-18}\)
\(\Leftrightarrow\sqrt{16\left(x-2\right)}+\sqrt{25\left(x-2\right)}=18+\sqrt{9\left(x-2\right)}\)
\(\Leftrightarrow4\sqrt{x-2}+5\sqrt{x-2}=18+3\sqrt{x-2}\)
\(\Leftrightarrow6\sqrt{x-2}=18\)
\(\Leftrightarrow\sqrt{x-2}=3\)
\(\Leftrightarrow\left(\sqrt{x-2}\right)^2=3^2\)
\(\Leftrightarrow x-2=9\)
\(\Leftrightarrow x=11\)
Vậy tập nghiệm của PT \(S=\left\{11\right\}\)
a. \(\sqrt{x^2}=7\Leftrightarrow\left|x\right|=7\Leftrightarrow\left[{}\begin{matrix}x=7\\x=-7\end{matrix}\right..\)
b. \(\sqrt{x^2}=\left|-8\right|\Leftrightarrow\left|x\right|=8\Leftrightarrow\left[{}\begin{matrix}x=8\\x=-8\end{matrix}\right..\)
c. \(\sqrt{4x^2}=6\Leftrightarrow\left|2x\right|=6\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right..\)
d. \(\sqrt{9x^2}=\left|-12\right|\Leftrightarrow\left|3x\right|=12\Leftrightarrow\left[{}\begin{matrix}3x=12\\3x=-12\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-4\end{matrix}\right..\)
B1:
\(\sqrt{\left(\sqrt{2}-\sqrt{3}\right)^2}+\sqrt{18}\)
\(=\left|\sqrt{2}-\sqrt{3}\right|+3\sqrt{2}\)
\(=\sqrt{3}-\sqrt{2}+3\sqrt{2}\)
\(=\sqrt{3}+2\sqrt{2}\)
\(\sqrt{7-4\sqrt{3}}+\sqrt{\left(1+\sqrt{3}\right)^2}\)
\(=\sqrt{4-4\sqrt{3}+3}+\left|1+\sqrt{3}\right|\)
\(=\sqrt{\left(2-\sqrt{3}\right)^2}+1+\sqrt{3}\)
\(=2-\sqrt{3}+1+\sqrt{3}\)
\(=3\)
B2:
đk: \(x\ge-2\)
Ta có: \(\sqrt{9x+18}-5\sqrt{x+2}+\frac{4}{5}\sqrt{25x+50}=6\)
\(\Leftrightarrow3\sqrt{x+2}-5\sqrt{x+2}+4\sqrt{x+2}=6\)
\(\Leftrightarrow2\sqrt{x+2}=6\)
\(\Leftrightarrow\sqrt{x+2}=3\)
\(\Leftrightarrow x+2=9\)
\(\Rightarrow x=7\)
Vậy x = 7
2) \(\frac{1}{5}\sqrt{25x+50}-5\sqrt{x+2}+\sqrt{9x+18}+9=0\)
\(\frac{1}{5}\sqrt{25\left(x+2\right)}-5\sqrt{x+2}+\sqrt{9x+18}+9=0\)
\(\frac{1}{5}.\sqrt{25}.\sqrt{x+2}-5\sqrt{x+2}+\sqrt{9x+18}+9=0\)
\(\frac{1}{5}.5\sqrt{x+2}-5\sqrt{x+2}+\sqrt{9x+18}+9=0\)
\(\frac{1}{5}.5\sqrt{x+2}-5\sqrt{x+2}+\sqrt{9\left(x+2\right)}+9=0\)
\(\frac{1}{5}.5\sqrt{x+2}-5\sqrt{x+2}+\sqrt{9}.\sqrt{x+2}+9=0\)
\(\frac{1}{5}.5\sqrt{x+2}-5\sqrt{x+2}+3\sqrt{x+2}+9=0\)
\(\sqrt{x+2}-5\sqrt{x+2}+3\sqrt{x+2}+9=0\)
\(-\sqrt{x+2}=-9\)
\(x+2=81\)
\(\Rightarrow x=79\)
3) \(\sqrt{x^2-4x+4}=7x-1\)
\(\sqrt{x^2-2.x.2+2^2}=7x-1\)
\(\sqrt{\left(x-2\right)^2}=7x-1\)
\(x-2=7x-1\)
\(-2=7x-1-x\)
\(-2+1=7x-x\)
\(-1=6x\)
\(-\frac{1}{6}=x\)
\(\Rightarrow x=-\frac{1}{6}\)
a/ \(\sqrt{x-1}+\sqrt{4x-4}-\sqrt{25x-25}+2=0\) (ĐKXĐ : \(x\ge1\))
\(\Leftrightarrow\sqrt{x-1}+2\sqrt{x-1}-5\sqrt{x-1}+2=0\)
\(\Leftrightarrow2\sqrt{x-1}=2\Leftrightarrow x-1=1\Leftrightarrow x=2\)
b/ \(\sqrt{9x^2+18}+2\sqrt{x^2+2}-\sqrt{25x^2+50}+3=0\)
\(\Leftrightarrow3\sqrt{x^2+2}+2\sqrt{x^2+2}-5\sqrt{x^2+2}+3=0\)
<=> 3 = 0 (vô lý)
=> pt vô nghiệm.
c/ \(\frac{9x-7}{\sqrt{7x+5}}=\sqrt{7x+5}\) (ĐKXĐ : x>-5/7)
\(\Leftrightarrow9x-7=7x+5\Leftrightarrow2x=12\Leftrightarrow x=6\)
d/ \(\frac{\sqrt{2x-3}}{\sqrt{x-1}}=2\) (ĐKXĐ : \(x\ge\frac{3}{2}\))
\(\Leftrightarrow2x-3=4\left(x-1\Leftrightarrow\right)2x=1\Leftrightarrow x=\frac{1}{2}\) (loại)
Vậy pt vô nghiệm.
Bài 1:
a: \(=\left|5-\sqrt{3}\right|-\left|\sqrt{3}-2\right|\)
\(=5-\sqrt{3}-2+\sqrt{3}=3\)
b; \(B=\dfrac{\left(2-\sqrt{3}\right)\cdot\sqrt{52+30\sqrt{3}}-\left(2+\sqrt{3}\right)\cdot\sqrt{52-30\sqrt{3}}}{\sqrt{2}}\)
\(=\dfrac{\left(2-\sqrt{3}\right)\cdot\left(3\sqrt{3}+5\right)-\left(2+\sqrt{3}\right)\left(3\sqrt{3}-5\right)}{\sqrt{2}}\)
\(=\dfrac{6\sqrt{3}+10-9-5\sqrt{3}-6\sqrt{3}+10-9+5\sqrt{3}}{\sqrt{2}}\)
\(=\dfrac{20-18}{\sqrt{2}}=\sqrt{2}\)
c: \(C=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{\left(2\sqrt{5}-3\right)^2}}}\)
\(=\sqrt{\sqrt{5}-\sqrt{3+3-2\sqrt{5}}}\)
\(=\sqrt{\sqrt{5}-\left(\sqrt{5}-1\right)}=1\)
d: \(A=\left(\sqrt{5}-1\right)\cdot\sqrt{6+2\sqrt{5}}\)
\(=\left(\sqrt{5}-1\right)\left(\sqrt{5}+1\right)=5-1=4\)
a)
\(\sqrt{9x^2}=12\\ < =>\left(\sqrt{9x^2}\right)^2=12^2\\ < =>9x^2=144\\ < =>x^2=16\\ < =>\left[{}\begin{matrix}x=4\\x=-4\end{matrix}\right.\)
b)
\(\sqrt{25x^2}=\left|-50\right|\\ < =>\sqrt{25x^2}=50\left(vì-50< 0\right)\)
\(< =>\left(\sqrt{25x^2}\right)^2=50^2\\ =>25x^2=2500\\ < =>x^2=100\\ < =>\left[{}\begin{matrix}x=10\\x=-10\end{matrix}\right.\)
a, x = 4
b, x = 10