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a: (x-3)2=49
=>x-3=7 hoặc x-3=-7
=>x=10 hoặc x=-4
b: \(\left(x^4\right)^2=\dfrac{x^{12}}{x^5}\)
\(\Leftrightarrow x^8-x^7=0\)
\(\Leftrightarrow x^7\left(x-1\right)=0\)
=>x=0 hoặc x=1
c: \(\Leftrightarrow x^{10}-25x^8=0\)
\(\Leftrightarrow x^8\left(x^2-25\right)=0\)
\(\Leftrightarrow x^8\left(x-5\right)\left(x+5\right)=0\)
hay \(x\in\left\{0;5;-5\right\}\)
1.
\(-3x^5y^4+3x^2y^3-7x^2y^3+5x^5y^4\)
\(=(-3x^5y^4+5x^5y^4)+(3x^2y^3-7x^2y^3)\)
\(=2x^5y^4-4x^2y^3\)
2.
\(\frac{1}{2}x^4y-\frac{3}{2}x^3y^4+\frac{5}{3}x^4y-x^3y^4\)
\(=(\frac{1}{2}x^4y+\frac{5}{3}x^4y)-(\frac{3}{2}x^3y^4+x^3y^4)\)
\(=\frac{13}{6}x^4y-\frac{5}{2}x^3y^4\)
3.
\(5x-7xy^2+3x-\frac{1}{2}xy^2\)
\(=(5x+3x)-(7xy^2+\frac{1}{2}xy^2)\)
\(=8x-\frac{15}{2}xy^2\)
4.
\(\frac{-1}{5}x^4y^3+\frac{3}{4}x^2y-\frac{1}{2}x^2y+x^4y^3\)
\(=(\frac{-1}{5}x^4y^3+x^4y^3)+(\frac{3}{4}x^2y-\frac{1}{2}x^2y)\)
\(=\frac{4}{5}x^4y^3+\frac{1}{4}x^2y\)
5.
\(\frac{7}{4}x^5y^7-\frac{3}{2}x^2y^6+\frac{1}{5}x^5y^7+\frac{2}{3}x^2y^6\)
\(=(\frac{7}{4}x^5y^7+\frac{1}{5}x^5y^7)+(-\frac{3}{2}x^2y^6+\frac{2}{3}x^2y^6)\)
\(=\frac{39}{20}x^5y^7-\frac{5}{6}x^2y^6\)
6.
\(\frac{1}{3}x^2y^5(-\frac{3}{5}x^3y)+x^5y^6=(\frac{1}{3}.\frac{-3}{5})(x^2.x^3)(y^5.y)+x^5y^6\)
\(=\frac{-1}{5}x^5y^6+x^5y^6=\frac{4}{5}x^5y^6\)
1, \(\left(xy\right)^2-\frac{1}{2}x^2y^2+3xy^2.\left(-\frac{1}{3}x\right)\)
\(=x^2y^2-\frac{1}{2}x^2y^2-x^2y^2\)
\(=-\frac{1}{2}x^2y^2\)
2, \(4.\left(-\frac{1}{2}x\right)^2-\frac{3}{2}x.\left(-x\right)+\frac{1}{3}x^2\)
\(=x^2+\frac{3}{2}x^2+\frac{1}{3}x^2\)
\(=\frac{17}{6}x^2\)
3, \(-4.\left(2x\right)^2y^3+\frac{1}{2}xy.\left(-2xy^2\right)+\frac{1}{4}x^2y^3\)
\(=-16x^2y^3-x^2y^3+\frac{1}{4}x^2y^3\)
\(=-\frac{67}{4}x^2y^3\)
4, \(\frac{1}{3}x^4y-\frac{5}{3}x^3.\left(\frac{5}{2}xy\right)+\frac{3}{4}x^4y\)
\(=\frac{1}{3}x^4y-\frac{25}{6}x^4y+\frac{3}{5}x^4y\)
\(=-\frac{97}{30}x^4y\)
5, \(\left(-2x^3y^4\right)^2-5x^2y.\left(\frac{3}{4}x^4y^7\right)-\frac{2}{3}x^6y^8\)
\(=4x^6y^8-\frac{15}{4}x^6y^8-\frac{2}{3}x^6y^8\)
\(=-\frac{5}{12}x^6y^8\)
a)
\((3x-7)^5=0\Rightarrow 3x-7=0\Rightarrow x=\frac{7}{3}\)
b)
\(\frac{1}{4}-(2x-1)^2=0\)
\(\Leftrightarrow (2x-1)^2=\frac{1}{4}=(\frac{1}{2})^2=(-\frac{1}{2})^2\)
\(\Rightarrow \left[\begin{matrix} 2x-1=\frac{1}{2}\\ 2x-1=\frac{-1}{2}\end{matrix}\right.\Rightarrow \Rightarrow \left[\begin{matrix} x=\frac{3}{4}\\ x=\frac{1}{4}\end{matrix}\right.\)
c)
\(\frac{1}{16}-(5-x)^3=\frac{31}{64}\)
\(\Leftrightarrow (5-x)^3=\frac{1}{16}-\frac{31}{64}=\frac{-27}{64}=(\frac{-3}{4})^3\)
\(\Leftrightarrow 5-x=\frac{-3}{4}\)
\(\Leftrightarrow x=\frac{23}{4}\)
d)
\(2x=(3,8)^3:(-3,8)^2=(3,8)^3:(3,8)^2=3,8\)
\(\Rightarrow x=3,8:2=1,9\)
e)
\((\frac{27}{64})^9.x=(\frac{-3}{4})^{32}\)
\(\Leftrightarrow [(\frac{3}{4})^3]^9.x=(\frac{3}{4})^{32}\)
\(\Leftrightarrow (\frac{3}{4})^{27}.x=(\frac{3}{4})^{32}\)
\(\Leftrightarrow x=(\frac{3}{4})^{32}:(\frac{3}{4})^{27}=(\frac{3}{4})^5\)
f)
\(5^{(x+5)(x^2-4)}=1\)
\(\Leftrightarrow (x+5)(x^2-4)=0\)
\(\Leftrightarrow \left[\begin{matrix} x+5=0\\ x^2-4=0\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x+5=0\\ x^2=4=2^2=(-2)^2\end{matrix}\right.\)
\(\Rightarrow \left[\begin{matrix} x=-5\\ x=\pm 2\end{matrix}\right.\)
g)
\((x-2,5)^2=\frac{4}{9}=(\frac{2}{3})^2=(\frac{-2}{3})^2\)
\(\Rightarrow \left[\begin{matrix} x-2,5=\frac{2}{3}\\ x-2,5=\frac{-2}{3}\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=\frac{19}{6}\\ x=\frac{11}{6}\end{matrix}\right.\)
h)
\((2x+\frac{1}{3})^3=\frac{8}{27}=(\frac{2}{3})^3\)
\(\Rightarrow 2x+\frac{1}{3}=\frac{2}{3}\Rightarrow x=\frac{1}{6}\)
a/ 3^x+3.3^x=2.3^5 ( QUY ĐỒNG)
4.3^x=2.3^5=> 3^x=3^5/2 lớp 7 đề ko phù hợp
B. (16/32)^X=1024=2^10
2^(-x)=2^10=> x=-10