\(\sqrt{16x-32}+\sqrt{25x-50}=18+\sqrt{9x-18}\)
Cho \(A=\left(\frac{2\sqrt{x}}{x-\sqrt{x}}-\frac{2x-2}{x\sqrt{x-2x+\sqrt{x}}}\right):\left(\frac{1}{\sqrt{x}}\right)^2\)
a) Rút gọn A
B) Tìm x để A>0
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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
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 \(S=\left\{11\right\}\)