Tim x biết:
a) \(8\sqrt{x}-3\sqrt{\frac{4}{81}}=5,2\)
b)\(12-3x^2=10+\sqrt{\frac{25}{16}}\)
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c) \(\sqrt{\left(x-2\right)^2}=10\)
\(x-2=10\)
\(x=12\)
d) \(\sqrt{9x^2-6x+1}=15\)
\(\sqrt{\left(3x\right)^2-2.3x.1+1^2}=15\)
\(\sqrt{\left(3x-1\right)^2}=15\)
\(3x-1=15\)
\(3x=16\)
\(x=\dfrac{16}{3}\)
a) \(đk:x\ge0\)
\(pt\Leftrightarrow3\sqrt{2x}+4\sqrt{2x}-3\sqrt{2x}=12\)
\(\Leftrightarrow4\sqrt{2x}=12\Leftrightarrow\sqrt{2x}=3\Leftrightarrow2x=9\Leftrightarrow x=\dfrac{9}{2}\left(tm\right)\)
b) \(đk:x\ge-2\)
\(pt\Leftrightarrow3\sqrt{x+2}+12\sqrt{x+2}-2\sqrt{x+2}=26\)
\(\Leftrightarrow13\sqrt{x+2}=26\)
\(\Leftrightarrow\sqrt{x+2}=2\Leftrightarrow x+2=4\Leftrightarrow x=2\left(tm\right)\)
c) \(pt\Leftrightarrow\left|x-2\right|=10\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=10\\x-2=-10\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=12\\x=-8\end{matrix}\right.\)
d) \(pt\Leftrightarrow\sqrt{\left(3x-1\right)^2}=15\)
\(\Leftrightarrow\left|3x-1\right|=15\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-1=15\\3x-1=-15\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{16}{3}\\x=-\dfrac{14}{3}\end{matrix}\right.\)
e) \(đk:x\ge\dfrac{8}{3}\)
\(pt\Leftrightarrow3x+4=9x^2-48x+64\)
\(\Leftrightarrow9x^2-51x+60=0\)
\(\Leftrightarrow3\left(x-4\right)\left(5x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=4\left(tm\right)\\x=\dfrac{5}{3}\left(ktm\right)\end{matrix}\right.\)
\(a,\frac{-5}{9}.\left(\frac{3}{10}-\frac{2}{5}\right)\)
\(=\frac{-5}{9}.\frac{-1}{10}\)
\(=\frac{1}{18}\)
\(b,2^8:2^5+3^3.2-12\)
\(=2^3+9.2-12\)
\(=8+18-12\)
\(=26-12\)
\(=14\)
Câu c,d em chưa học nên không biết làm ạ, mong mọi người thông cảm!!!
Sửa lại câu b
\(=2^3+27.2-12\)
\(=8+54-12\)
\(=62-12\)
\(=50\)
a, \(-\frac{187}{70}\)
b,\(\frac{27}{70}\)
c,\(\frac{53}{14}\)
d,\(\frac{27}{4}\)
e,1
f,\(\frac{23}{4}\)
g,-1
i,6
k,315
l,\(\frac{9}{2}\)
a)\(8\sqrt{x}-3\sqrt{\frac{4}{81}}=5,2\)
\(\Rightarrow8\sqrt{x}-3.\frac{2}{9}=5,2\)
\(\Rightarrow8\sqrt{x}-\frac{2}{3}=5,2\)
\(\Rightarrow8\sqrt{x}=5,2+\frac{2}{3}\)
\(\Rightarrow8\sqrt{x}=\frac{40}{3}\)
\(\Rightarrow\sqrt{x}=\frac{40}{3}:8\)
\(\Rightarrow\sqrt{x}=\frac{5}{3}\)
\(\Rightarrow x=\frac{25}{9}\)
b)\(12-3x^2=10+\sqrt{\frac{25}{16}}\)
\(\Rightarrow12-3x^2=10+\frac{5}{4}\)
\(\Rightarrow12-3x^2=11,25\)
\(\Rightarrow3x^2=12-11,25\)
\(\Rightarrow3x^2=0,75\)
\(\Rightarrow x^2=0,25\)
\(\Rightarrow x=\sqrt{0,25}\)
\(\Rightarrow x=0,5\)