12 x 5 + 12 x 3 + (-12) x (-2) = ?
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a, \(\left(\dfrac{12}{25}\right)^x=\left(\dfrac{3}{5}\right)^2-\left(-\dfrac{3}{5}\right)^4\)
\(\Rightarrow\left(\dfrac{12}{25}\right)^x=\left(\dfrac{3}{5}\right)^2.\left[1-\left(\dfrac{3}{5}\right)^2\right]\)
\(\Rightarrow\left(\dfrac{12}{25}\right)^x=\dfrac{9}{25}.\dfrac{16}{25}\Rightarrow\left(\dfrac{12}{25}\right)^x=\dfrac{144}{625}=\left(\dfrac{12}{25}\right)^2\)
Vì \(\dfrac{12}{25}\ne-1;\dfrac{12}{25}\ne0;\dfrac{12}{25}\ne1\) nên \(x=2\)
b, Sửa đề:
\(\left(-\dfrac{3}{4}\right)^{3x-1}=\dfrac{81}{256}\)
\(\Rightarrow\left(-\dfrac{3}{4}\right)^{3x-1}=\left(-\dfrac{3}{4}\right)^4\)
Vì \(\dfrac{-3}{4}\ne-1;\dfrac{-3}{4}\ne0;\dfrac{-3}{4}\ne1\) nên \(3x-1=4\)
\(\Rightarrow3x=5\Rightarrow x=\dfrac{5}{3}\)
Vậy......
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e) \(\dfrac{x-5}{7}=\dfrac{x+2}{3}\Leftrightarrow3\left(x-5\right)=7\left(x+2\right)\Leftrightarrow3x-15=7x+14\)
\(\Leftrightarrow7x-3x=-15-14\Leftrightarrow4x=-29\Leftrightarrow x=\dfrac{-29}{4}\) vậy \(x=\dfrac{-29}{4}\)
f) \(\left(x-5\right)\left(x+2\right)=\left(x-3\right)\left(x+4\right)\Leftrightarrow x^2+2x-5x-10=x^2+4x-3x-12\)
\(\Leftrightarrow x^2+2x-5x-10-x^2-4x+3x+12=0\Leftrightarrow-4x+2=0\Leftrightarrow4x=2\Leftrightarrow x=\dfrac{2}{4}=\dfrac{1}{2}\)
vậy \(x=\dfrac{1}{2}\)
a) \(\left|\dfrac{1}{2}x\right|=3-2x\)
+) TH1: \(\dfrac{1}{2}x\ge0\Rightarrow x\ge0\)
Khi đó: \(\dfrac{1}{2}x=3-2x\)
\(\Rightarrow x=2\left(3-2x\right)\)
\(\Rightarrow x=6-4x\)
\(\Rightarrow x+4x=6\)
\(\Rightarrow5x=6\Rightarrow x=\dfrac{6}{5}\) (thỏa mãn)
+) TH2: \(-\dfrac{1}{2}x< 0\Rightarrow x>0\)
\(-\dfrac{1}{2}x=3-2x\)
\(\Rightarrow-x=6-4x\)
\(\Rightarrow-x+4x=6\)
\(\Rightarrow3x=6\Rightarrow x=2\) (t/m)
Vậy \(\left\{{}\begin{matrix}x=\dfrac{6}{5}\\x=2\end{matrix}\right.\).
b) \(\left|x-1\right|=3x+2\)
+) TH1: \(x-1\ge0\Rightarrow x\ge1\)
\(x-1=3x+2\)
\(\Rightarrow x-3x=1+2\)
\(\Rightarrow-2x=3\)
\(\Rightarrow x=\dfrac{-3}{2}\) (loại)
+) TH2: \(x-1< 0\Rightarrow x< 1\)
\(-x+1=3x+ 2\)
\(\Rightarrow-x-3x=-1+2\)
\(\Rightarrow-4x=1\)
\(\Rightarrow x=\dfrac{-1}{4}\) (t/m)
Vậy \(x=-\dfrac{1}{4}.\)
c) Tương tự.
a) Vì \(\left|\dfrac{1}{2}x\right|\ge0\)
\(\Leftrightarrow3-2x\ge0\)
Mà \(\left|\dfrac{1}{2}x\right|=3-2x\)
\(\Leftrightarrow\dfrac{1}{2}x=3-2x\)
\(\Leftrightarrow0,5x=3-2x\)
\(\Leftrightarrow0,5x+2x=3\)
\(\Leftrightarrow2,5x=3\)
\(\Leftrightarrow x=\dfrac{3}{2,5}\)
Vậy ................
b) Vì \(\left|x-1\right|\ge0\)
Mà \(\left|x-1\right|=3x+2\)
\(\Leftrightarrow3x+2\ge0\)
\(\Leftrightarrow x-1=3x+2\)
\(\Leftrightarrow x-3x=2+1\)
\(\Leftrightarrow-2x=3\)
\(\Leftrightarrow x=-\dfrac{3}{2}\left(tm\right)\)
Vậy ....................
c) Vì \(\left|5x\right|\ge0\)
Mà \(\left|5x\right|=x+12\)
\(\Leftrightarrow x+12\ge0\)
\(\Leftrightarrow5x=x+12\)
\(\Leftrightarrow5x-x=12\)
\(\Leftrightarrow4x=12\)
\(\Leftrightarrow x=3\left(tm\right)\)
Vậy ............
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