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đề =
\(2\sqrt{9\left(x-3\right)}-\frac{1}{5}\sqrt{25\left(x-3\right)}-\frac{1}{7}\sqrt{49\left(x-3\right)}=20\)
=>\(6\sqrt{x-3}-\sqrt{x-3}-\sqrt{x-3}=20\)
=>\(4\sqrt{x-3}=20\)
=>\(\sqrt{x-3}=5\)
=>\(x-3=25\)
=>\(x=28\)
\(2\sqrt{9x-27}-\frac{1}{5}\sqrt{25x-75}-\frac{1}{7}\sqrt{49x-147}=20\)
\(< =>2\sqrt{9\left(x-3\right)}-\frac{1}{5}\sqrt{25\left(x-3\right)}-\frac{1}{7}\sqrt{49\left(x-3\right)}=20\)
\(< =>2\cdot3\sqrt{\left(x-3\right)}-\frac{1}{5}.5\sqrt{\left(x-3\right)}-\frac{1}{7}.7\sqrt{\left(x-3\right)}=20\) \(đk:x\ge0\)
\(< =>6\sqrt{\left(x-3\right)}-\sqrt{\left(x-3\right)}-\sqrt{\left(x-3\right)}=20\)
\(< =>\sqrt{\left(x-3\right)}\left(6-1-1\right)=20\)
\(< =>4\sqrt{\left(x-3\right)}=20\)
\(< =>\sqrt{\left(x-3\right)}=5\)
\(< =>x-3=25\)
\(< =>x=28\left(tm\right)\)
a) ĐK: \(x\ge5\)
\(\sqrt{4x-20}+\frac{1}{3}\sqrt{9x-45}-\frac{1}{5}\sqrt{16x-80}=0\)
\(\Leftrightarrow\)\(\sqrt{4\left(x-5\right)}+\frac{1}{3}\sqrt{9\left(x-5\right)}-\frac{1}{5}\sqrt{16\left(x-5\right)}=0\)
\(\Leftrightarrow\)\(2\sqrt{x-5}+\sqrt{x-5}-\frac{4}{5}\sqrt{x-5}=0\)
\(\Leftrightarrow\)\(\frac{11}{5}\sqrt{x-5}=0\)
\(\Leftrightarrow\)\(x-5=0\)
\(\Leftrightarrow\)\(x=5\) (t/m)
Vậy
b) \(-5x+7\sqrt{x}=-12\)
\(\Leftrightarrow\)\(5x-7\sqrt{x}-12=0\)
\(\Leftrightarrow\)\(\left(\sqrt{x}+1\right)\left(5\sqrt{x}-12\right)=0\)
đến đây tự làm
c) d) e) bạn bình phương lên
f) \(VT=\sqrt{3\left(x^2+2x+1\right)+9}+\sqrt{5\left(x^4-2x^2+1\right)+25}\)
\(=\sqrt{3\left(x+1\right)^2+9}+\sqrt{5\left(x^2-1\right)^2}\)
\(\ge\sqrt{9}+\sqrt{25}=8\)
Dấu "=" xảy ra \(\Leftrightarrow\)\(\hept{\begin{cases}x+1=0\\x^2-1=0\end{cases}}\)\(\Leftrightarrow\)\(x=-1\)
Vậy...
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
a/ Giải rồi
b/ ĐKXĐ: \(x\ge-1\)
Đặt \(\sqrt{2x+3}+\sqrt{x+1}=t>0\)
\(\Rightarrow t^2=3x+4+2\sqrt{2x^2+5x+3}\) (1)
Pt trở thành:
\(t=t^2-6\Leftrightarrow t^2-t-6=0\Rightarrow\left[{}\begin{matrix}t=3\\t=-2\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\sqrt{2x+3}+\sqrt{x+1}=3\)
\(\Leftrightarrow3x+4+2\sqrt{2x^2+5x+3}=9\)
\(\Leftrightarrow2\sqrt{2x^2+5x+3}=5-3x\left(x\le\frac{5}{3}\right)\)
\(\Leftrightarrow4\left(2x^2+5x+3\right)=\left(5-3x\right)^2\)
\(\Leftrightarrow...\)
e/ ĐKXD: \(x>0\)
\(5\left(\sqrt{x}+\frac{1}{2\sqrt{x}}\right)=2\left(x+\frac{1}{4x}\right)+4\)
Đặt \(\sqrt{x}+\frac{1}{2\sqrt{x}}=t\ge\sqrt{2}\)
\(\Rightarrow t^2=x+\frac{1}{4x}+1\)
Pt trở thành:
\(5t=2\left(t^2-1\right)+4\)
\(\Leftrightarrow2t^2-5t+2=0\Rightarrow\left[{}\begin{matrix}t=2\\t=\frac{1}{2}\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\sqrt{x}+\frac{1}{2\sqrt{x}}=2\)
\(\Leftrightarrow2x-4\sqrt{x}+1=0\)
\(\Rightarrow\sqrt{x}=\frac{2\pm\sqrt{2}}{2}\)
\(\Rightarrow x=\frac{3\pm2\sqrt{2}}{2}\)
\(3x-7\sqrt{x}+4=0\)
\(3x-3\sqrt{x}-4\sqrt{x}+4=0\)
\(3\sqrt{x}\left(\sqrt{x}-1\right)-4\left(\sqrt{x}-1\right)=0\)
\(\left(\sqrt{x}-1\right)\left(3\sqrt{x}-4\right)=0\)
\(\orbr{\begin{cases}\sqrt{x}-1=0\\3\sqrt{x}-4=0\end{cases}}\Rightarrow\orbr{\begin{cases}\sqrt{x}=1\\3\sqrt{x}=4\end{cases}}\Rightarrow\orbr{\begin{cases}x=1\\x=\frac{16}{9}\end{cases}}\)
ĐK: \(x\ge1\)
\(\frac{1}{2}\sqrt{x-1}-\frac{3}{2}\sqrt{9x-9}+24\sqrt{\frac{x-1}{64}}=-17\)
<=> \(\frac{1}{2}\sqrt{x-1}-\frac{3}{2}\sqrt{9\left(x-1\right)}+24\sqrt{\frac{1}{64}\left(x-1\right)}=-17\)
<=> \(\frac{1}{2}\sqrt{x-1}-\frac{9}{2}\sqrt{x-1}+3\sqrt{x-1}=-17\)
<=> \(-\sqrt{x-1}=-17\)
<=> \(x-1=17^2\)
<=> \(x=290\)
Vậy....
\(a,\frac{9x-7}{\sqrt{7x+5}}=\sqrt{7x+5}\)\(ĐKXĐ:x\ge-\frac{5}{7}\)
\(\Leftrightarrow9x-7=7x+5\)
\(\Leftrightarrow9x-7x=5+7\)
\(\Leftrightarrow2x=12\)
\(\Leftrightarrow x=6\)
\(b,\sqrt{4x-20}+3\sqrt{\frac{x-5}{9}}-\frac{1}{3}\sqrt{9x-45}=4\)
\(\Leftrightarrow\sqrt{4\left(x-5\right)}+3.\frac{\sqrt{x-5}}{\sqrt{9}}-\frac{1}{3}\sqrt{9\left(x-5\right)}=4\)
\(\Leftrightarrow2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}=4\)
\(\Leftrightarrow\sqrt{x-5}\left(2+1-1\right)=4\)
\(\Leftrightarrow2\sqrt{x-5}=4\)
\(\Leftrightarrow\sqrt{x-5}=2\)
\(\Leftrightarrow x-5=4\)
\(\Leftrightarrow x=9\)
ĐKXĐ: bạn tự tìm
a/ Có vẻ bạn ghi nhầm đề, nhưng nói chung vẫn giải được, nghiệm xấu
\(\Leftrightarrow2\sqrt{x}+\frac{1}{2}\sqrt{x}-\frac{3}{4}\sqrt{5x}=5\)
\(\Leftrightarrow\sqrt{x}\left(\frac{5}{2}-\frac{3\sqrt{5}}{4}\right)=5\)
\(\Rightarrow\sqrt{x}=\frac{40+12\sqrt{5}}{11}\Rightarrow x=\left(\frac{40+12\sqrt{5}}{11}\right)^2\)
b/ \(\sqrt{3-x}-3\sqrt{3-x}+5\sqrt{3-x}=6\)
\(\Leftrightarrow3\sqrt{3-x}=6\)
\(\Leftrightarrow\sqrt{3-x}=2\Rightarrow3-x=4\Rightarrow x=-1\)
c/ \(7\left(5\sqrt{x}-2\right)=2\left(8\sqrt{x}+\frac{5}{2}\right)\)
\(\Leftrightarrow35\sqrt{x}-14=16\sqrt{x}+5\)
\(\Leftrightarrow19\sqrt{x}=19\)
\(\Rightarrow\sqrt{x}=1\Rightarrow x=1\)
d/ \(\sqrt{3x^2+12x+4}=4\)
\(\Leftrightarrow3x^2+12x+4=16\)
\(\Leftrightarrow3x^2+12x-12=0\)
\(\Rightarrow x=-2\pm2\sqrt{2}\)
1. ĐKXĐ: \(x\ge3\)
\(2\sqrt{9x-27}-\frac{1}{5}\sqrt{25x-75}-\frac{1}{7}\sqrt{49x-147}=20\)
⇔ \(6\sqrt{x-3}-\sqrt{x-3}-\sqrt{x-3}=20\)
⇔ \(4\sqrt{x-3}=20\)
⇔ \(\sqrt{x-3}=5\)
⇔ \(x-3=25\)
⇔ \(x=28\left(TMĐKXĐ\right)\)
Vậy....
2. ĐKXĐ: \(x\ge0\)
\(\frac{3}{2}\sqrt{5x}+\sqrt{5x}-7=\frac{1}{2}\sqrt{5x}\)
⇔ \(\frac{3}{2}\sqrt{5x}+\sqrt{5x}-\frac{1}{2}\sqrt{5x}=7\)
⇔ \(2\sqrt{5x}=7\)
⇔ \(\sqrt{5x}=\frac{7}{2}\)
⇔ \(5x=\frac{49}{4}\)
⇔ \(x=\frac{49}{20}\left(TMĐKXĐ\right)\)
Vậy...