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 = \(\frac{1}{6}\)

X=2 ,Y=3 nha!

x=2,y=3

\(5^x-2-3^2=2^4-\left(2^8x2^4-2^{10}x2^2\right)\)

\(5^x-2-3^2=2^4-\left(2^{8+4}-2^{10+2}\right)\)

\(5^x-2-3^2=2^4-\left(2^{12}-2^{12}\right)\)

\(5^x-2-3^2=2^4-0\)

\(5^x-2-3^2=2^4\)

\(5^x-2-9=16\)

\(5^x-2=16+9\)

\(5^x-2=25\)

\(5^x=25+2\)

\(5^x=27\)

Bởi vì 27 không phân tích được 1 số có số mũ là 2

\(\Rightarrow\) Không tồn tại x

\(5^{x-2}-9=16-\left(256.16-1024.4\right)\)

\(\Rightarrow5^{x-2}-9=16-\left(4096-4096\right)\)

\(\Rightarrow5^{x-2}-9=16-0\)

\(\Rightarrow5^{x-2}-9=16\)

\(\Rightarrow5^{x-2}=25\)

\(\Rightarrow x-2=25:5\)

\(\Rightarrow x-2=3\)

\(\Rightarrow x=5\)

\(5^{x-2}-3^2=2^4-\left(2^8\cdot2^4-2^{10}\cdot2^2\right)\)

\(\Rightarrow5^{x-2}-3^2=2^4-\left(2^{12}-2^{12}\right)\)

\(\Rightarrow5^{x-2}-9=16-0\)

\(\Rightarrow5^{x-2}=16+9\)

\(\Rightarrow5^{x-2}=25\)

\(\Rightarrow5^{x-2}=5^2\)

\(\Rightarrow x-2=2\)

\(\Rightarrow x=2+2\)

\(\Rightarrow x=4\)

Vậy: x=4

a: Ta có: \(\dfrac{x+2}{5}=\dfrac{1}{x-2}\)

\(\Leftrightarrow x^2-4=5\)

hay \(x\in\left\{3;-3\right\}\)

d: Ta có: \(\dfrac{x-3}{x+1}=\dfrac{x+2}{x-5}\)

\(\Leftrightarrow x^2-8x+15=x^2+3x+2\)

\(\Leftrightarrow-11x=-13\)

hay \(x=\dfrac{13}{11}\)

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) \(\dfrac{5}{13}< 2-x< \dfrac{5}{8}\)

\(\Rightarrow2-\dfrac{5}{8}< x< 2-\dfrac{5}{13}\)

\(\Rightarrow\dfrac{11}{8}< x< \dfrac{21}{13}\)

ai mần tr mình cho tích

a) ta có:\(\dfrac{1}{7}=\dfrac{1.5}{7.5}=\dfrac{5}{35}\);\(\dfrac{2}{5}=\dfrac{2.7}{5.7}=\dfrac{14}{35}\)

\(\Rightarrow x\in\left\{6;7;8;9;10;11;12;13\right\}\)

b)ta có:\(2-x=\dfrac{5}{9};\dfrac{5}{10};\dfrac{5}{11};\dfrac{5}{12}\)

\(\Rightarrow x=2-\dfrac{5}{9};2-\dfrac{5}{10};2-\dfrac{5}{11};2-\dfrac{5}{12}\)

\(\Rightarrow x=\dfrac{13}{9};\dfrac{3}{2};\dfrac{17}{11};\dfrac{19}{12}\)

\(y^2=x\left(x+1\right)\left(x+7\right)\left(x+8\right)\)

\(=\left(x^2+8x\right)\left(x^2+8x+7\right)\)

\(\Rightarrow4y^2=\left(2x^2+16x\right)\left(2x^2+16x+14\right)\)

\(=\left(2x^2+16x+7-7\right)\left(2x^2+16x+7+7\right)\)

\(=\left(2x^2+16x+7\right)^2-49\)

\(\Leftrightarrow\left(2x^2+16x+7\right)^2-4y^2=49\)

\(\Leftrightarrow\left(2x^2+16x+7-2y\right)\left(2x^2+16x+7+2y\right)=49=1.49=7.7\)

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 \(x=-23\)