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\(3,5+\left|x+\dfrac{3}{2}\right|=-1,5\cdot\left(-\sqrt{9}\right)\)
\(3,5+\left|x+\dfrac{3}{2}\right|=-1,5\cdot\left(-3\right)\)
\(3,5+\left|x+\dfrac{3}{2}\right|=4,5\)
\(\left|x+\dfrac{3}{2}\right|=4,5-3,5\)
\(\left|x+\dfrac{3}{2}\right|=1\)
\(\Rightarrow x+\dfrac{3}{2}=1\) hoặc \(x+\dfrac{3}{2}=-1\)
\(x=1-\dfrac{3}{2}\) \(x=-1-\dfrac{3}{2}\)
\(x=\dfrac{-1}{2}\) \(x=\dfrac{-5}{2}\)
Vậy \(x=\dfrac{-1}{2}\)hoặc \(x=\dfrac{-5}{2}\)
\(3,5+\left|x+\dfrac{3}{2}\right|=-1,5.\left(-\sqrt{9}\right)\)
\(\Rightarrow3,5+\left|x+\dfrac{3}{2}\right|=4,5\)
\(\Rightarrow\left|x+\dfrac{3}{2}\right|=4,5-3,5=1\)
\(\Rightarrow\left[{}\begin{matrix}x+\dfrac{3}{2}=1\\x+\dfrac{3}{2}=-1\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=1-\dfrac{3}{2}\\x=-1-\dfrac{3}{2}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\dfrac{-1}{2}\\x=\dfrac{-5}{2}\end{matrix}\right.\)
Vậy..................
a)TH1 x>=3 \(\left|x-3\right|\)=x-3
pttt: x-3-2x=1 suy ra x=-4 <3 -> loại
TH2 x=< 3 pttt 3-x-2x=1 suy ra x =2/3 thỏa mãn
b) VT=\(\dfrac{4^{x+2}+4^{x+1}+4^x}{21}=\dfrac{4^x\left(4^2+4+1\right)}{21}=4^x\)
VP= \(\dfrac{3^{2x}+3^{2x+1}+3^{2x+3}}{31}=\dfrac{9^x\left(1+3+27\right)}{31}=9^x\)
vậy pt đã cho tương đương với 4^x=9^x \(\Leftrightarrow\left(\dfrac{4}{9}\right)\)^x =1 suy ra x =0
3,5 + /x + \(\frac{3}{2}\) / = -1,5(-\(\sqrt{9}\))
=> 3,5 +/ x +\(\frac{3}{2}\) / = -1,5 ( -3 )
=> 3,5 + / x + \(\frac{3}{2}\) / =4,5
=> / x + \(\frac{3}{2}\) / = 4,5 - 3,5
=> / x + \(\frac{3}{2}\) / = 1
=> \(\hept{\begin{cases}x+\frac{3}{2}=1\\x+\frac{3}{2}=-1\end{cases}}\)
=> \(\hept{\begin{cases}x=1-\frac{3}{2}\\x=-1-\frac{3}{2}\end{cases}}\)
=> \(\hept{\begin{cases}x=\frac{-1}{2}\\x=\frac{-5}{2}\end{cases}}\)
vậy x = \(\frac{-1}{2}\)hay x = \(\frac{-5}{2}\)
\(3,5+\left|x+\frac{3}{2}\right|=-1,5.\left(-\sqrt{9}\right)\) \(3,5+\left|x+\frac{3}{2}\right|=-1,5.\left(-3\right)\) \(3,5+\left|x+\frac{3}{2}\right|=4,5\) \(\left|x+\frac{3}{2}\right|=4,5-3,5\) \(\left|x+\frac{3}{2}\right|=1\) \(\Rightarrow\orbr{\begin{cases}x+\frac{3}{2}=1\\x+\frac{3}{2}=-1\end{cases}}\) \(\Rightarrow\orbr{\begin{cases}x=-\frac{1}{2}\\x=-\frac{5}{2}\end{cases}}\) Vậy x=\(-\frac{1}{2}\) hoặc x=\(-\frac{5}{2}\)
a)\(\left|x-3,5\right|=0,25\)
\(\Leftrightarrow\orbr{\begin{cases}x-3,5=0,25\\x-3,5=-0,25\end{cases}\Leftrightarrow\orbr{\begin{cases}x=3,75\\x=3,25\end{cases}}}\)
b)\(\left|1,5-x\right|=1,75\)
\(\Leftrightarrow\orbr{\begin{cases}1,5-x=1,75\\1,5-x=-1,75\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-0,25\\x=3,25\end{cases}}}\)
c)\(\left|x-1,5\right|=1,5\)
\(\Leftrightarrow\orbr{\begin{cases}x-1,5=1,5\\x-1,5=-1,5\end{cases}\Leftrightarrow\orbr{\begin{cases}x=3\\x=0\end{cases}}}\)
\(K=5\left(\dfrac{1}{2}+\dfrac{1}{6}+...+\dfrac{1}{45\cdot46}\right)\)
\(=5\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{45}-\dfrac{1}{46}\right)\)
=5*45/46=225/46
\(T=\dfrac{1}{5}\cdot\sqrt{6\cdot\dfrac{2}{3}}-\dfrac{3}{2}\cdot\sqrt{\dfrac{2}{3}\cdot\dfrac{8}{75}}+\dfrac{1}{2}\cdot\sqrt{6\cdot\dfrac{8}{75}}\)
\(=\dfrac{1}{5}\cdot2-\dfrac{3}{2}\cdot\dfrac{4}{15}+\dfrac{1}{2}\cdot\dfrac{4}{5}\)
=2/5-12/30+4/10
=2/5
Chọn B. Thay \(\dfrac{1}{3}\)vào x và \(\dfrac{1}{2}\)vào y
giải để ra được m
a: \(\Leftrightarrow\left|x-3\right|=12-5x-8=-5x+4\)
\(\Leftrightarrow\left\{{}\begin{matrix}x< =\dfrac{4}{5}\\\left(-5x+4\right)^2=\left(x-3\right)^2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x< =\dfrac{4}{5}\\\left(5x-4-x+3\right)\left(5x-4+x-3\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x< =\dfrac{4}{5}\\\left(4x-1\right)\left(6x-7\right)=0\end{matrix}\right.\Leftrightarrow x=\dfrac{1}{4}\)
b: \(\left(\sqrt{x}+3\right)^{10}=1024\cdot125^2\cdot25^2\)
\(\Leftrightarrow\left(\sqrt{x}+3\right)^{10}=2^{10}\cdot5^6\cdot5^4=10^{10}\)
\(\Leftrightarrow\sqrt{x}+3=10\)
hay x=49
c: \(\dfrac{3-0.2x}{5}=\dfrac{7}{15}+1.4x\)
\(\Leftrightarrow\dfrac{9-0.6x}{15}=\dfrac{7}{15}+\dfrac{21x}{15}\)
=>21x+7=9-0,6x
=>21,6x=-2
hay x=-5/54
d: \(\Leftrightarrow\left(\dfrac{4}{3}\right)^{3x}=\dfrac{5^9\cdot7^9\left(4\cdot7-5^2\right)}{5^9\cdot7^9\cdot4}\)
\(\Leftrightarrow\left(\dfrac{4}{3}\right)^{3x}=\dfrac{28-25}{4}=\dfrac{3}{4}\)
=>3x=-1
hay x=-1/3
\(\left(2x-3\right)\left(x-\dfrac{1}{2}\right)=0\\ \Rightarrow\left[{}\begin{matrix}2x-3=0\\x-\dfrac{1}{2}=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}2x=3\\x=\dfrac{1}{2}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=\dfrac{1}{2}\end{matrix}\right.\) (Thêm KL cuối dòng: Vậy \(x\in\left\{\dfrac{3}{2};\dfrac{1}{2}\right\}\))
\(\Rightarrow x+\dfrac{2}{3}x=1,5+3,5\Rightarrow\dfrac{5}{3}x=5\Rightarrow x=5:\dfrac{5}{3}=3\)
⇒x+23x=1,5+3,5⇒53x=5⇒x=5:53=3