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b) (5/2-3x)=25/9
3x = 5/2-25/9
3x =-5/18
x =-5/18:3
x=-5/54
\(e.\left(x-1\right)^5=-32\)
\(\left(x-1\right)^5=\left(-2\right)^5\)
\(x-1=-2\)
\(x\) \(=-2+1\)
\(x\) \(=-1\)
Vậy \(x=-1\)
\(a)x+\left(-5\right)=-14\)
\(\Leftrightarrow x=-14-\left(-5\right)\)
\(\Leftrightarrow x=-14+5\)
\(\Leftrightarrow x=-9\)
\(b)-x+7=-23\)
\(\Leftrightarrow-x=-23+ \left(-7\right)\)
\(\Leftrightarrow-x=-30\)
\(\Leftrightarrow x=30\)
\(c)112-x=\left(-3\right).\left(-15\right)\)
\(\Leftrightarrow112-x=45\)
\(\Leftrightarrow x=112-45\)
\(\Leftrightarrow x=67\)
\(d)\left(x-15\right)-27=5^5:5^3\)
\(\Leftrightarrow\left(x-15\right)-27=5^2\)
\(\Leftrightarrow\left(x-15\right)-27=25\)
\(\Leftrightarrow x-15=52\)
\(\Leftrightarrow x=67\)
\(e)\left(2x+1\right)^2=81\)
\(\Leftrightarrow\left(2x+1\right)^2=9^2\)
\(\Leftrightarrow2x+1=9\)
\(\Leftrightarrow2x=8\)
\(\Leftrightarrow x=4\)
\(f)(x-5^3)=-27\)
\(f)(x-5^3)=-9^3\)
\(\Leftrightarrow x-5=-9\)
\(\Leftrightarrow x=-4\)
P/s: Bạn tự kết luận.
3.
a) \(\left(x-1\right)^3=125\)
=> \(\left(x-1\right)^3=5^3\)
=> \(x-1=5\)
=> \(x=5+1\)
=> \(x=6\)
Vậy \(x=6.\)
b) \(2^{x+2}-2^x=96\)
=> \(2^x.\left(2^2-1\right)=96\)
=> \(2^x.3=96\)
=> \(2^x=96:3\)
=> \(2^x=32\)
=> \(2^x=2^5\)
=> \(x=5\)
Vậy \(x=5.\)
c) \(\left(2x+1\right)^3=343\)
=> \(\left(2x+1\right)^3=7^3\)
=> \(2x+1=7\)
=> \(2x=7-1\)
=> \(2x=6\)
=> \(x=6:2\)
=> \(x=3\)
Vậy \(x=3.\)
Chúc bạn học tốt!
3^x*5^x-1=224
3^x*5^x/5=224
15^x=224*5
15^x=1120
=>ko tồn tại x thỏa mãn đề bài vị 15^x luôn có tận cùng bằng 5 (x khác 0 ) hoặc 1 ( x=0) ma 1120 co tận cùng bằng 0
Bài 6 :
a) \(\dfrac{625}{5^n}=5\Rightarrow\dfrac{5^4}{5^n}=5\Rightarrow5^{4-n}=5^1\Rightarrow4-n=1\Rightarrow n=3\)
b) \(\dfrac{\left(-3\right)^n}{27}=-9\Rightarrow\dfrac{\left(-3\right)^n}{\left(-3\right)^3}=\left(-3\right)^2\Rightarrow\left(-3\right)^{n-3}=\left(-3\right)^2\Rightarrow n-3=2\Rightarrow n=5\)
c) \(3^n.2^n=36\Rightarrow\left(2.3\right)^n=6^2\Rightarrow\left(6\right)^n=6^2\Rightarrow n=6\)
d) \(25^{2n}:5^n=125^2\Rightarrow\left(5^2\right)^{2n}:5^n=\left(5^3\right)^2\Rightarrow5^{4n}:5^n=5^6\Rightarrow\Rightarrow5^{3n}=5^6\Rightarrow3n=6\Rightarrow n=3\)
Bài 7 :
a) \(3^x+3^{x+2}=9^{17}+27^{12}\)
\(\Rightarrow3^x\left(1+3^2\right)=\left(3^2\right)^{17}+\left(3^3\right)^{12}\)
\(\Rightarrow10.3^x=3^{34}+3^{36}\)
\(\Rightarrow10.3^x=3^{34}\left(1+3^2\right)=10.3^{34}\)
\(\Rightarrow3^x=3^{34}\Rightarrow x=34\)
b) \(5^{x+1}-5^x=100.25^{29}\Rightarrow5^x\left(5-1\right)=4.5^2.\left(5^2\right)^{29}\)
\(\Rightarrow4.5^x=4.25^{2.29+2}=4.5^{60}\)
\(\Rightarrow5^x=5^{60}\Rightarrow x=60\)
c) Bài C bạn xem lại đề
d) \(\dfrac{3}{2.4^x}+\dfrac{5}{3.4^{x+2}}=\dfrac{3}{2.4^8}+\dfrac{5}{3.4^{10}}\)
\(\Rightarrow\dfrac{3}{2.4^x}-\dfrac{3}{2.4^8}+\dfrac{5}{3.4^{x+2}}-\dfrac{5}{3.4^{10}}=0\)
\(\Rightarrow\dfrac{3}{2}\left(\dfrac{1}{4^x}-\dfrac{1}{4^8}\right)+\dfrac{5}{3.4^2}\left(\dfrac{1}{4^x}-\dfrac{1}{4^8}\right)=0\)
\(\Rightarrow\left(\dfrac{1}{4^x}-\dfrac{1}{4^8}\right)\left(\dfrac{3}{2}+\dfrac{5}{3.4^2}\right)=0\)
\(\Rightarrow\dfrac{1}{4^x}-\dfrac{1}{4^8}=0\)
\(\Rightarrow\dfrac{4^8-4^x}{4^{x+8}}=0\Rightarrow4^8-4^x=0\left(4^{x+8}>0\right)\Rightarrow4^x=4^8\Rightarrow x=8\)
Bài 1:
a: \(\dfrac{14}{15}+4\dfrac{5}{17}+1\dfrac{1}{15}-3\dfrac{5}{17}\)
\(=\dfrac{14}{15}+1+\dfrac{1}{15}+4+\dfrac{5}{17}-3-\dfrac{5}{17}\)
\(=1+1+1=3\)
b: \(\dfrac{5}{9}:\left(\dfrac{1}{11}-\dfrac{5}{22}\right)+\dfrac{5}{9}:\left(\dfrac{1}{15}-\dfrac{2}{3}\right)\)
\(=\dfrac{5}{9}:\left(\dfrac{2}{22}-\dfrac{5}{22}\right)+\dfrac{5}{9}:\left(\dfrac{1}{15}-\dfrac{10}{15}\right)\)
\(=\dfrac{5}{9}:\dfrac{-3}{22}+\dfrac{5}{9}:\dfrac{-9}{15}\)
\(=\dfrac{5}{9}\cdot\dfrac{-22}{3}+\dfrac{5}{9}\cdot\dfrac{-5}{3}=\dfrac{5}{9}\left(-\dfrac{22}{3}+\dfrac{-5}{3}\right)\)
\(=\dfrac{5}{9}\cdot\left(-9\right)=-5\)
c: \(27\left(-\dfrac{1}{3}\right)^2+\dfrac{1}{3}-25\cdot\left(-\dfrac{1}{5}\right)^3-\dfrac{1}{5}\)
\(=27\cdot\dfrac{1}{9}+\dfrac{1}{3}-25\cdot\dfrac{-1}{125}-\dfrac{1}{5}\)
\(=3+\dfrac{1}{3}+\dfrac{1}{5}-\dfrac{1}{5}=3+\dfrac{1}{3}=\dfrac{10}{3}\)
Bài 2:
a: \(\dfrac{5}{2}-\dfrac{3}{2}\left(x-2\right)=2\)
=>\(\dfrac{3}{2}\left(x-2\right)=\dfrac{5}{2}-2=\dfrac{1}{2}\)
=>\(x-2=\dfrac{1}{2}:\dfrac{3}{2}=\dfrac{1}{3}\)
=>\(x=2+\dfrac{1}{3}=\dfrac{7}{3}\)
b: \(\left(2x-2,5\right)\cdot\left(2,5+\dfrac{-7}{5}:x\right)=0\)
=>\(\left[{}\begin{matrix}2x-2,5=0\\-\dfrac{7}{5}:x+2,5=0\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}2x=2,5\\-\dfrac{7}{5}:x=-2,5=-\dfrac{5}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2.5}{2}=1.25=\dfrac{5}{4}\\x=\dfrac{5}{2}:\dfrac{7}{5}=\dfrac{25}{14}\end{matrix}\right.\)
c: \(\left(1,5-\dfrac{3}{5}x\right)^3=27^{-1}\)
=>\(\left(-\dfrac{3}{5}x+\dfrac{3}{2}\right)^3=\dfrac{1}{27}=\left(\dfrac{1}{3}\right)^3\)
=>\(-\dfrac{3}{5}x+\dfrac{3}{2}=\dfrac{1}{3}\)
=>\(-\dfrac{3}{5}x=\dfrac{1}{3}-\dfrac{3}{2}=\dfrac{2-9}{6}=\dfrac{-7}{6}\)
=>\(x=\dfrac{7}{6}:\dfrac{3}{5}=\dfrac{7}{6}\cdot\dfrac{5}{3}=\dfrac{35}{18}\)
d: \(\dfrac{\left(2x-5\right)^2}{-4}=\dfrac{16}{5-2x}\)
=>\(\dfrac{\left(2x-5\right)^2}{-4}=\dfrac{-16}{2x-5}\)
=>\(\left(2x-5\right)^3=\left(-4\right)\cdot\left(-16\right)=64\)
=>2x-5=4
=>2x=9
=>\(x=\dfrac{9}{2}\)
và là sao vậy bn??