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a) \(5.2^{x+1}.2^{-2}-2^x=384\Leftrightarrow2^x\left(5.2^{-2}.2-1\right)=384\)\(\Leftrightarrow2^x.1,5=384\Leftrightarrow2^x=384:1,5=256=2^8\)
\(\Rightarrow x=8\)
b) \(3^{x+2}.5^y=45^x\Leftrightarrow3^{x+2}.5^y=3^{2x}.5^x\Leftrightarrow\frac{3^{2x}}{3^{x+2}}=\frac{5^y}{5^x}\)\(\Leftrightarrow3^{2x-x+2}=5^{y-x}\Leftrightarrow3^{x+2}=5^{y-x}\)
\(\Rightarrow x+2=y-x=0\Rightarrow x=y=-2\)
Ta có bất đẳng thức giá trị tuyệt đối:
\(\left|A\right|+\left|B\right|\ge\left|A+B\right|\)
Dấu \(=\)khi \(AB\ge0\).
d) \(\left|x+1\right|+\left|x+2\right|+\left|2x-3\right|\)
\(\ge\left|x+1+x+2\right|+\left|2x-3\right|\)
\(=\left|2x+3\right|+\left|3-2x\right|\)
\(\ge\left|2x+3+3-2x\right|=6\)
Dấu \(=\)khi \(\hept{\begin{cases}\left(x+1\right)\left(x+2\right)\ge0\\\left(2x+3\right)\left(3-2x\right)\ge0\end{cases}}\Leftrightarrow-1\le x\le\frac{3}{2}\).
e) \(\left|x+1\right|+\left|x+2\right|+\left|x-3\right|+\left|x-5\right|\)
\(=\left(\left|x+1\right|+\left|3-x\right|\right)+\left(\left|x+2\right|+\left|5-x\right|\right)\)
\(\ge\left|x+1+3-x\right|+\left|x+2+5-x\right|\)
\(=4+7=11\)
Dấu \(=\)khi \(\hept{\begin{cases}\left(x+1\right)\left(3-x\right)\ge0\\\left(x+2\right)\left(5-x\right)\ge0\end{cases}}\Leftrightarrow-1\le x\le3\).
Do đó phương trình đã cho vô nghiệm.
Bài 1 :
\(A=x^2-2xy^2+y^4=\left(x-y^2\right)^2=-\left(y^2-x\right)^2\)
Mà \(B=-\left(y^2-x\right)^2\)
Nên ta có : đpcm
Bài 2
Đặt \(\left(x+1\right)\left(x-2\right)\left(2x-1\right)=0\)
TH1 : x = -1
TH2 : x = 2
TH3 : x = 1/2
Bài 4 :
a, \(\left(2x+3\right)\left(5-x\right)=0\Leftrightarrow x=-\frac{3}{2};5\)
b, \(\left(x-\frac{1}{2}\right)\left(3x+1\right)\left(2-x\right)=0\Leftrightarrow x=\frac{1}{2};-\frac{1}{3};2\)
c, \(x^2+2x=0\Leftrightarrow x\left(x+2\right)=0\Leftrightarrow x=0;-2\)
d, \(x^2-x=0\Leftrightarrow x\left(x-1\right)=0\Leftrightarrow x=0;1\)
\(\left(2x-5\right)^2=0,81\)
\(\left(2x-5\right)^2=0,9^2\)
\(\Rightarrow2x-5=0,9\)
\(2x=0,9+5\)
\(2x=5,9\)
\(x=5,9:2\)
\(x=2,95\)
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\(\left(x-\frac{1}{3}\right)^3=0,027\)
\(\left(x-\frac{1}{3}\right)^3=0,3^3\)
\(\Rightarrow x-\frac{1}{3}=0,3\)
\(x=0,3+\frac{1}{3}\)
\(x=\frac{19}{30}\)
\(\left(2x-5\right)^2=0,81\)
\(\Rightarrow2x-5=0,9\)
\(\Rightarrow2x=5,9\)
\(\Rightarrow x=2,95\)
\(\left(x-\frac{1}{3}\right)^3=0,027\)
\(\Rightarrow x-\frac{1}{3}=0,3\)
\(\Rightarrow x=\frac{19}{30}\)
a) \(5.2^{x+1}.2^{-2}-2^x=384\)
\(\Leftrightarrow2^x.2.\frac{5}{4}-2^x=384\)
\(\Leftrightarrow2^x.\left(\frac{5}{2}-1\right)=384\)
\(\Leftrightarrow2^x.\frac{3}{2}=384\)
\(\Leftrightarrow2^x=256\)
\(\Leftrightarrow2^x=2^8\)
\(\Leftrightarrow x=8\)
c) \(\left(x+1\right)^{x+1}=\left(x+1\right)^{x+3}\)
\(\Leftrightarrow\left(x+1\right)^{x+3}-\left(x+1\right)^{x+1}=0\)
\(\Leftrightarrow\left(x+1\right)^{x+1}\left[\left(x+1\right)^2-1\right]=0\)
\(\Leftrightarrow\orbr{\begin{cases}\left(x+1\right)^{x+1}=0\\\left(x+1\right)^2-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x+1=0\\\left(x+1\right)^2=1\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-1\\x\in\left\{0;-2\right\}\end{cases}}}\)
Vậy \(x\in\left\{0;-1;-2\right\}\)