Tìm x: x(x+2)-(x-2)^2=8
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Đặt \(A=2^{x+9}-2^{x+8}-2^{x+7}-...-2^{x+1}-2^x\)
\(\Rightarrow2A=2\left(2^{x+9}-2^{x+8}-...-2^x\right)\)
\(\Rightarrow2A=2^{x+9}.2^1-2^{x+8}.2^1-...-2^x.2^1\)
\(\Rightarrow2A=2^{x+10}-2^{x+9}-...-2^{x+1}\)
\(\Rightarrow A=2A-A=2^{x+10}-2^{x+9}-...-2^{x+1}-\left(2^{x+9}-2^{x+8}-...-2^{x+1}-2^x\right)=2^{x+10}-2^{x+9}-2^{x+9}+2^x\)
\(\Rightarrow A=2^{x+10}-2.2^{x+9}+2^x=2^{x+10}-2^{x+10}+2^x=2^x\)
\(\Rightarrow2^x=1024\Rightarrow x=10\)
\(\left|x+\frac{1}{x}\right|=3x-1\)
\(\orbr{\begin{cases}x+\frac{1}{x}=3x-1\\-x-\frac{1}{x}=3x-1\end{cases}}\)
\(\orbr{\begin{cases}x+\frac{1}{x}-3x+1=0\\-x-\frac{1}{x}-3x+1=0\end{cases}}\)
\(\orbr{\begin{cases}-2x+\frac{1}{x}+1=0\\-4x-\frac{1}{x}+1=0\end{cases}}\)
\(\orbr{\begin{cases}-2x^2+1+x=0\\-4x^2-1+x=0\end{cases}}\)
\(\orbr{\begin{cases}x=-\frac{1}{2};x=1\\x=\frac{1-\sqrt{15t}}{8}\end{cases}}\)
| x + \(\frac{1}{3}\)| = 3x - 1
\(\Rightarrow\)x + \(\frac{1}{3}\)= \(\pm\)( 3x - 1 )
TH1 : x + \(\frac{1}{3}\)= 3x - 1
\(\Rightarrow\)2x = \(\frac{4}{3}\)
\(\Rightarrow\)x = \(\frac{2}{3}\)
TH2 : x + \(\frac{1}{3}\)= - 3x + 1
\(\Rightarrow\)4x = \(\frac{2}{3}\)
\(\Rightarrow\)x = \(\frac{1}{6}\)
\(5^x-2-3^2=2^4-\left(2^8x2^4-2^{10}x2^2\right)\)
\(5^x-2-3^2=2^4-\left(2^{8+4}-2^{10+2}\right)\)
\(5^x-2-3^2=2^4-\left(2^{12}-2^{12}\right)\)
\(5^x-2-3^2=2^4-0\)
\(5^x-2-3^2=2^4\)
\(5^x-2-9=16\)
\(5^x-2=16+9\)
\(5^x-2=25\)
\(5^x=25+2\)
\(5^x=27\)
Bởi vì 27 không phân tích được 1 số có số mũ là 2
\(\Rightarrow\) Không tồn tại x
\(5^{x-2}-9=16-\left(256.16-1024.4\right)\)
\(\Rightarrow5^{x-2}-9=16-\left(4096-4096\right)\)
\(\Rightarrow5^{x-2}-9=16-0\)
\(\Rightarrow5^{x-2}-9=16\)
\(\Rightarrow5^{x-2}=25\)
\(\Rightarrow x-2=25:5\)
\(\Rightarrow x-2=3\)
\(\Rightarrow x=5\)
a) ĐKXĐ: \(x\ne2\)
\(\Rightarrow\left(x+2\right)\left(x-2\right)=5.1\)
\(\Rightarrow x^2-4=5\Rightarrow x^2=9\)
\(\Rightarrow\left[{}\begin{matrix}x=3\left(tm\right)\\x=-3\left(tm\right)\end{matrix}\right.\)
b) ĐKXĐ: \(x\ne-1\)
\(\Rightarrow\left(x+1\right)^2=2.8=16\)
\(\Rightarrow\left[{}\begin{matrix}x+1=4\\x+1=-4\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=3\left(tm\right)\\x=-5\left(tm\right)\end{matrix}\right.\)
c) giống câu a
d) ĐKXĐ: \(x\ne5,x\ne-1\)
\(\Rightarrow\left(x+1\right)\left(x+2\right)=\left(x-3\right)\left(x-5\right)\)
\(\Rightarrow x^2+3x+2=x^2-8x+15\)
\(\Rightarrow11x=13\)
\(\Rightarrow x=\dfrac{13}{11}\left(tm\right)\)
a) \(\dfrac{1}{7}< \dfrac{x}{35}< \dfrac{2}{5}\)
\(\Rightarrow\dfrac{5}{35}< \dfrac{x}{35}< \dfrac{14}{35}\)
\(\Rightarrow5< x< 14\)
b) \(\dfrac{5}{13}< 2-x< \dfrac{5}{8}\)
\(\Rightarrow2-\dfrac{5}{8}< x< 2-\dfrac{5}{13}\)
\(\Rightarrow\dfrac{11}{8}< x< \dfrac{21}{13}\)
\(y^2=x\left(x+1\right)\left(x+7\right)\left(x+8\right)\)
\(=\left(x^2+8x\right)\left(x^2+8x+7\right)\)
\(\Rightarrow4y^2=\left(2x^2+16x\right)\left(2x^2+16x+14\right)\)
\(=\left(2x^2+16x+7-7\right)\left(2x^2+16x+7+7\right)\)
\(=\left(2x^2+16x+7\right)^2-49\)
\(\Leftrightarrow\left(2x^2+16x+7\right)^2-4y^2=49\)
\(\Leftrightarrow\left(2x^2+16x+7-2y\right)\left(2x^2+16x+7+2y\right)=49=1.49=7.7\)
Xét các trường hợp và thu được các nghiệm là: \(\left(-3,0\right),\left(0,0\right)\).
\(\frac{x+5}{3}=\frac{x-1}{4}\)
\(\Rightarrow\left(x+5\right).4=\left(x-1\right).3\)
\(\Rightarrow4x+20=3x-3\)
\(\Rightarrow4x-3x=-3-20\Rightarrow x=-23\)
\(\frac{x+5}{3}=\frac{x-1}{4}\)
\(\Rightarrow\left(x+5\right)\cdot4=\left(x-1\right)\cdot3\)
\(4x+20=3x-3\)
\(4x-3x=-3-20\)
\(x=-23\)
Vậy \(x=-23\)
\(x\left(x+2\right)-\left(x-2\right)^2=8\)
\(x^2+2x-x^2+4x-4=8\)
\(6x-4=8\)
\(x=2\)