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\(\left(0,4x-2\right)-\left(1,5x+1\right)\) \(\left(-4x-0,8\right)\)\(=3,6\)
<=>\(\left(0,4x-2\right)-\)\(\left(-6x^2-1,2x-4x-0,8\right)\)\(=3,6\)
<=> \(0,4x-2+6x^2+1,2x\)\(+4x+0,8=3,6\)
<=> \(6x^2+5,6x-1,2=3,6\)
<=> \(6x^2+5,6x-4,8=0\)
<=>
\(\left(-2,5\right).\left(-4\right)=10\)
\(\left(-2,5\right).0,5.\left(-2\right).2=\left[\left(-2,5\right).\left(-2\right).\left(0,5.2\right)\right]=5.1=5\)
\(\left(-0,5\right).0,5.\left(-2\right).2=\left[\left(-0,5\right).\left(-2\right).\left(0,5.2\right)\right]=1.1=1\)
\(25.\left(-5\right).\left(-0,4\right).\left(-0,2\right)=\left[25.\left(-0,4\right).\left(-5\right).\left(-0,2\right)\right]=-10.1=-10\)
x:0,2=0,8:x
\(\dfrac{x}{0,2}=\dfrac{0,8}{x}\\ \Leftrightarrow x.x=0,8.0,2\\ \Rightarrow x^2=0,16\\ \Rightarrow x^2=\left(\pm0,4\right)^2\\ \Rightarrow x=\pm0,4\)
Vậy...
\(\dfrac{x}{0,2}=\dfrac{0,8}{x}\)
\(\Leftrightarrow x.x=0,2.0,8\)
\(\Leftrightarrow x^2=0,16\)
\(\Leftrightarrow x^2=\left(0,4\right)^2\)
\(\Leftrightarrow x=0,4\)
a,\(\Leftrightarrow\frac{x}{0,2}=\frac{x}{0,8}\Leftrightarrow0,8x=0,2x\Leftrightarrow0,6x=0\Leftrightarrow x=0\)
b,\(\Leftrightarrow\frac{x+11}{14-x}=\frac{2}{3}\)
\(\Leftrightarrow3\left(x+11\right)=2\left(14-x\right)\)
\(\Leftrightarrow3x+33=28-2x\)
\(\Leftrightarrow3x+2x=28-33\)
\(\Leftrightarrow5x=-5\Leftrightarrow x=-1\)
Bài 1:
a)\(\frac{\left(0,8\right)^5}{\left(0,4\right)^6}=\frac{\left(0,2\cdot4\right)^5}{\left(0,2\cdot2\right)^6}=\frac{\left(0,2\right)^5\cdot\left(2^2\right)^5}{\left(0,2\right)^6\cdot2^6}=\frac{\left(0,2\right)^5\cdot2^{10}}{\left(0,2\right)^6\cdot2^6}=\frac{2^4}{0,2}=\frac{16}{\frac{2}{10}}=80\)
b)\(\frac{8^{10}+4^{10}}{8^4+4^{11}}=\frac{\left(2^3\right)^{10}+\left(2^2\right)^{10}}{\left(2^3\right)^4+\left(2^2\right)^{11}}=\frac{2^{30}+2^{20}}{2^{12}+2^{22}}=\frac{2^{20}\left(2^{10}+1\right)}{2^{12}\left(1+2^{10}\right)}=\frac{2^{20}}{2^{12}}=256\)
Bài 2:
a)\(2^{x-1}=16\)
\(\Rightarrow2^{x-1}=2^4\)
\(\Rightarrow x-1=4\Rightarrow x=5\)
b)\(\left(x-1\right)^2=25\)
\(\Rightarrow\left(x-1\right)^2=5^2=\left(-5\right)^2\)
\(\Rightarrow x-1=5\) hoặc \(x-1=-5\)
\(\Rightarrow x=6\) hoặc \(x=-4\)
Vậy \(x=6\) hoặc \(x=-4\)
c)\(\left(x-1\right)^{x+2}=\left(x-1\right)^{x+6}\)
\(\Rightarrow\left(x-1\right)^{x+2}-\left(x-1\right)^{x+6}=0\)
\(\Leftrightarrow\left(x-1\right)^{x+2}\left[1-\left(x-1\right)^4\right]\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}\left(x-1\right)^{x+2}=0\\1-\left(x-1\right)^4=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x-1=0\\1=\left(x-1\right)^4\end{array}\right.\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=1\\\left(x-1\right)^4=\left(-1\right)^4=1^4\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=1\\x-1=1\\x-1=-1\end{array}\right.\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=1\\x=2\\x=0\end{array}\right.\)
d)\(\left(x+20\right)^{100}+\left|y+4\right|=0\left(1\right)\)
Ta thấy: \(\begin{cases}\left(x+20\right)^{100}\ge0\\\left|y+4\right|\ge0\end{cases}\)
\(\Rightarrow\left(x+20\right)^{100}+\left|y+4\right|\ge0\left(2\right)\)
Từ (1) và (2) suy ra \(\begin{cases}\left(x+20\right)^{100}=0\\\left|y+4\right|=0\end{cases}\)
\(\Rightarrow\begin{cases}x+20=0\\y+4=0\end{cases}\)\(\Rightarrow\begin{cases}x=-20\\y=-4\end{cases}\)