\(1,x:\left(-\dfrac{1}{3}\right)^3=\left(-\dfrac{1}{3}\right)\\ \Leftrightarrow x=\left(-\dfrac{1}{3}\right)\times\left(-\dfrac{1}{3}\right)^3\\ \Leftrightarrow x=\left(-\dfrac{1}{3}\right)^4=\dfrac{1}{81}\\ 2,\left(\dfrac{4}{5}\right)^5.x=\left(\dfrac{4}{5}\right)^7\\ \Leftrightarrow x=\left(\dfrac{4}{5}\right)^7:\left(\dfrac{4}{5}\right)^5=\left(\dfrac{4}{5}\right)^{7-5}=\left(\dfrac{4}{5}\right)^2=\dfrac{16}{25}\)
\(3,\left(x+\dfrac{1}{2}\right)^2=\dfrac{1}{16}\\ \Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=\dfrac{1}{4}\\x+\dfrac{1}{2}=-\dfrac{1}{4}\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{4}\\x=-\dfrac{3}{4}\end{matrix}\right.\)
\(4,\left(3x+1\right)^3=-27\\ \Leftrightarrow\left(3x+1\right)^3=\left(-3\right)^3\\ \Leftrightarrow3x+1=-3\\ \Leftrightarrow3x=-4\\ \Leftrightarrow x=-\dfrac{4}{3}\)
\(5,\left(\dfrac{1}{2}\right)^2.x=\left(\dfrac{1}{2}\right)^5\\ \Leftrightarrow x=\left(\dfrac{1}{2}\right)^5:\left(\dfrac{1}{2}\right)^2\\ \Leftrightarrow x=\left(\dfrac{1}{2}\right)^{5-2}=\left(\dfrac{1}{2}\right)^3=\dfrac{1}{8}\)
\(6,\left(-\dfrac{1}{3}\right)^3.x=\dfrac{1}{81}\\ \Leftrightarrow\left(-\dfrac{1}{3}\right)^3.x=\left(-\dfrac{1}{3}\right)^4\\ \Leftrightarrow x=\left(-\dfrac{1}{3}\right)^4:\left(-\dfrac{1}{3}\right)^3=-\dfrac{1}{3}\)
\(7,\left(2x-3\right)^2=16\\ \Leftrightarrow\left[{}\begin{matrix}2x-3=4\\2x-3=-4\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}\\x=-\dfrac{1}{2}\end{matrix}\right.\)
\(8,\left(x-\dfrac{2}{3}\right)^3=\dfrac{1}{27}\\ \Leftrightarrow\left(x-\dfrac{2}{3}\right)^3=\left(\dfrac{1}{3}\right)^3\\ \Leftrightarrow x-\dfrac{2}{3}=\dfrac{1}{3}\\ \Leftrightarrow x=\dfrac{1}{3}+\dfrac{2}{3}=\dfrac{3}{3}=1\)
`@` `\text {Ans}`
`\downarrow`
(Vế 1)
`1.`
`x \div(-1/3)^3 =-1/3`
`=> x= (-1/3) \times (-1/3)^3`
`=> x= (-1/3)^4`
`2.`
`(4/5)^5 *x = (4/5)^7`
`=> x = (4/5)^7 \div (4/5)^5`
`=> x=(4/5)^2`
`3.`
`(x+1/2)^2 =1/16`
`=> (x+1/2)^2 = (+-1/4)^2`
`=>`\(\left[{}\begin{matrix}x+\dfrac{1}{2}=\dfrac{1}{4}\\x+\dfrac{1}{2}=-\dfrac{1}{4}\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=\dfrac{1}{4}-\dfrac{1}{2}\\x=-\dfrac{1}{4}-\dfrac{1}{2}\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=-\dfrac{1}{4}\\x=-\dfrac{3}{4}\end{matrix}\right.\)
`4.`
`(3x+1)^3 = -27`
`=> (3x+1)^3 = (-3)^3`
`=> 3x+1=-3`
`=> 3x=-3-1`
`=> 3x =-4`
`=> x=-4/3`
`5.`
`(1/2)^2*x=(1/2)^5`
`=> x=(1/2)^5 \div (1/2)^2`
`=> x=(1/2)^3`
`6.`
`(-1/3)^3*x=1/81`
`=> (-1/3)^3*x = (1/3)^4`
`=> x= (1/3)^4 \div (-1/3)^3`
`=> x=(-1/3)`
`7.`
`(2x-3)^2 = 16`
`=> (2x-3)^2 = (+-4)^2`
`=>`\(\left[{}\begin{matrix}2x-3=4\\2x-3=-4\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}2x=7\\2x=-1\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=\dfrac{7}{2}\\x=-\dfrac{1}{2}\end{matrix}\right.\)
`8.`
`(x-2/3)^3 = 1/27`
`=> (x-2/3)^3 = (1/3)^3`
`=> x-2/3=1/3`
`=> x=1/3 + 2/3`
`=> x=1`
\(2,\left(\dfrac{2}{3}x-\dfrac{1}{3}\right)^5=\dfrac{1}{243}\\ \Leftrightarrow\left(\dfrac{2}{3}x-\dfrac{1}{3}\right)^5=\left(\dfrac{1}{3}\right)^5\\ \Leftrightarrow\dfrac{2}{3}x-\dfrac{1}{3}=\dfrac{1}{3}\\ \Leftrightarrow\dfrac{2}{3}x=\dfrac{1}{3}+\dfrac{1}{3}=\dfrac{2}{3}\\ \Leftrightarrow x=\dfrac{2}{3}:\dfrac{2}{3}=1\)
\(4,\left(2x-1\right)^{10}=49^5\\ \Leftrightarrow\left(2x-1\right)^{10}=\left(7^2\right)^5\\ \Leftrightarrow\left(2x-1\right)^{10}=7^{10}\\ \Leftrightarrow2x-1=7\\ \Leftrightarrow2x=8\\ \Leftrightarrow x=4\)
\(5,x:5^2=\left(\dfrac{3}{5}\right)^2:3^2\\ \Leftrightarrow x:5^2=\dfrac{9}{25}:9\\ \Leftrightarrow x:25-\dfrac{1}{25}\\ \Leftrightarrow x=\dfrac{1}{25}\times25=1\)
\(8,\left(2x-5\right)^4=-81\)
Xem lại đề nhé số mũ chẵn thì phải ra chẵn chứ
Sửa đề \(\left(2x-5\right)^4=81\\ \Leftrightarrow\left(2x-5\right)^4=3^4\\ \Leftrightarrow2x-5=3\\ \Leftrightarrow2x=8\\ \Leftrightarrow x=4\)
Phần bên này mình chỉ làm những câu khác và khó hơn 1 tý thôi còn các câu còn lại bạn có thể tự làm được nhé
Vế 2:
`1.`
\(\left(x+0,7\right)^3=-27\)
`=> (x+0,7)^3 = (-3)^3`
`=> x+0,7 = -3`
`=> x= -3 - 0,7`
`=> x=-3,7`
`2.`
\(\left(\dfrac{2}{3}x-\dfrac{1}{3}\right)^5=\dfrac{1}{243}\)
`=>`\(\left(\dfrac{2}{3}x-\dfrac{1}{3}\right)^5=\left(\dfrac{1}{3}\right)^5\)
`=>`\(\dfrac{2}{3}x-\dfrac{1}{3}=\dfrac{1}{3}\)
`=>`\(\dfrac{2}{3}x=\dfrac{1}{3}+\dfrac{1}{3}\)
`=>`\(\dfrac{2}{3}x=\dfrac{2}{3}\)
`=> x=1`
`3.`
\(\left(\dfrac{2}{5}-3x\right)^2=\dfrac{9}{25}\)
`=>`\(\left(\dfrac{2}{5}-3x\right)^2=\left(\pm\dfrac{3}{5}\right)^2\)
`=>`\(\left[{}\begin{matrix}\dfrac{2}{5}-3x=\dfrac{3}{5}\\\dfrac{2}{5}-3x=-\dfrac{3}{5}\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}3x=-\dfrac{1}{5}\\3x=1\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=-\dfrac{1}{15}\\x=\dfrac{1}{3}\end{matrix}\right.\)
`4.`
\(\left(2x-1\right)^{10}=49^5\)
`=>`\(\left(2x-1\right)^{10}=7^{10}\)
`=>` `2x-1=7`
`=> 2x=8`
`=> x = 8 \div 2`
`=> x=4`
`5.`
\(x\div5^2=\left(\dfrac{3}{5}\right)^2\div3^2\)
`=>`\(x\div5^2=\left(\dfrac{3}{5}\div3\right)^2\)
`=>`\(x\div5^2=\dfrac{1}{25}\)
`=>`\(x=\left(\dfrac{1}{5}\right)^2\cdot5^2\)
`=>`\(x=\left(\dfrac{1}{5}\cdot5\right)^2=1\)
`6.`
\(\left(x-\dfrac{3}{5}\right)^2=4\)
`=>`\(\left(x-\dfrac{3}{5}\right)^2=\left(\pm2\right)^2\)
`=>`\(\left[{}\begin{matrix}x-\dfrac{3}{5}=2\\x-\dfrac{3}{5}=-2\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=2+\dfrac{3}{5}\\x=-2+\dfrac{3}{5}\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=\dfrac{13}{5}\\x=-\dfrac{7}{5}\end{matrix}\right.\)
`7.`
\(\left(x-\dfrac{1}{4}\right)^2=\dfrac{4}{9}\)
`=>`\(\left(x-\dfrac{1}{4}\right)^2=\left(\pm\dfrac{2}{3}\right)^2\)
`=>`\(\left[{}\begin{matrix}x-\dfrac{1}{4}=\dfrac{2}{3}\\x-\dfrac{1}{4}=-\dfrac{2}{3}\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=\dfrac{2}{3}+\dfrac{1}{4}\\x=-\dfrac{2}{3}+\dfrac{1}{4}\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=\dfrac{11}{12}\\x=-\dfrac{5}{12}\end{matrix}\right.\)
`8.`
\(\left(2x-5\right)^4=-81\)
`=>`\(\left(2x-5\right)=-\left(3\right)^4\)
`=>` `2x-5=-3`
`=> 2x=-3+5`
`=> 2x=2`
`=> x=1`
