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a: =>5x+1=6/7 hoặc 5x+1=-6/7
=>5x=-1/7 hoặc 5x=-13/7
=>x=-1/35 hoặc x=-13/35
b: =>x-2/9=4/9
=>x=6/9=2/3
c: =>8x+1=5
=>8x=4
hay x=1/2
Bài 1:
\(\dfrac{\left(\dfrac{2}{5}\right)^7\cdot5^7+\left(\dfrac{9}{4}\right)^3:\left(\dfrac{3}{16}\right)^3}{2^7\cdot5^2+512}\)
\(=\dfrac{\left(\dfrac{2}{5}\cdot5\right)^7+\left(\dfrac{9}{4}:\dfrac{3}{16}\right)^3}{2^7\cdot5^2+512}\)
\(=\dfrac{2^7+12^3}{2^7\cdot5^2+512}\)
\(=\dfrac{1856}{3712}\)
\(=0,5\)
Bài 2:
\(\left(5x+1\right)^2=\dfrac{36}{49}\)
\(\Rightarrow5x+1=\dfrac{6}{7}\)
\(\Rightarrow5x=\dfrac{-1}{7}\)
\(\Rightarrow x=\dfrac{-1}{35}\)
1: \(\left(\dfrac{1}{16}\right)^x=\left(\dfrac{1}{8}\right)^6\)
\(\Leftrightarrow\left(\dfrac{1}{2}\right)^{4x}=\left(\dfrac{1}{2}\right)^{18}\)
=>4x=18
hay x=9/2
2: \(\left(\dfrac{1}{16}\right)^x=\left(\dfrac{1}{8}\right)^{36}\)
\(\Leftrightarrow\left(\dfrac{1}{2}\right)^{4x}=\left(\dfrac{1}{2}\right)^{108}\)
=>4x=108
hay x=27
3: \(\left(\dfrac{1}{81}\right)^x=\left(\dfrac{1}{27}\right)^4\)
\(\Leftrightarrow\left(\dfrac{1}{3}\right)^{4x}=\left(\dfrac{1}{3}\right)^{12}\)
=>4x=12
hay x=3
a) \(\left(5x+1\right)^2=\dfrac{36}{49}\)
\(\left(5x+1\right)^2=\left(\pm\dfrac{6}{9}\right)\)\(^2\)
\(5x+1=\pm\dfrac{6}{9}\)
+) \(5x+1=\dfrac{6}{9}\)
\(5x=\dfrac{6}{9}-1=\dfrac{6}{9}-\dfrac{9}{9}\)
\(5x=\dfrac{-5}{9}\)
\(x=\dfrac{-5}{9}:5=\dfrac{-1}{45}\)
+) \(5x+1=\dfrac{-6}{9}\)
\(5x=\dfrac{-6}{9}-1=\dfrac{-6}{9}-\dfrac{9}{9}\)
\(5x=\dfrac{-5}{3}\)
\(x=\dfrac{-5}{3}:5=\dfrac{-5}{15}\)
vậy \(x\in\left\{\dfrac{-5}{15};\dfrac{-1}{45}\right\}\)
a: \(=2\cdot\dfrac{5}{4}-3\cdot\dfrac{7}{6}+4\cdot\dfrac{9}{8}=\dfrac{5}{2}-\dfrac{7}{2}+\dfrac{9}{2}=\dfrac{7}{2}\)
b: \(=18-16\cdot\dfrac{1}{2}+\dfrac{1}{16}\cdot\dfrac{3}{4}\)
=10+3/64
=643/64
c: \(=\dfrac{2}{3}\cdot\dfrac{9}{4}-\dfrac{3}{4}\cdot\dfrac{8}{3}+\dfrac{7}{5}\cdot\dfrac{5}{14}=\dfrac{3}{2}-2+\dfrac{1}{2}=2-2=0\)
b, \(\left(5x+1\right)^2=\dfrac{36}{49}\)
\(\left(5x+1\right)^2=\left(\pm\dfrac{6}{7}\right)^2\)
\(\Rightarrow\left[{}\begin{matrix}5x+1=\dfrac{6}{7}\\5x+1=-\dfrac{6}{7}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}5x=-\dfrac{1}{7}\\5x=\dfrac{-13}{7}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-\dfrac{1}{35}\\x=\dfrac{-13}{35}\end{matrix}\right.\)
Vậy .....
Nguyễn Thanh Hằng ;Hồng Phúc Nguyễn ;Mới vô; ... các bn giúp mik vs mik đang cần gấp !
Lớp 7 chưa học phân thức qua bên lớp 8 đi lộn địa chỉ ròi
a,\(\left(5x-1\right)^6=729\)
\(\left(5x-1\right)^6=3^6\)
\(5x-1=3\)
\(5x=4\)
\(x=\dfrac{4}{5}\)
b Ta có \(8=2^3\),\(25=5^2\)
Mà \(\dfrac{8}{25}=\dfrac{2^x}{5^{x-1}}\)
=> \(2^3=2^x,5^2=5^{x-1}\)
=> x=3
Câu 1 :
a) \(\left(\dfrac{-1}{3}\right)^3.x=\dfrac{1}{81}\)
\(x=\left(\dfrac{-1}{3}\right)^4:\left(\dfrac{-1}{3}\right)^3\Rightarrow x=\dfrac{-1}{3}\)
b) \(\left(5.x\right)^3=-64\)
\(\left(5.x\right)^3=\left(-4\right)^3\Rightarrow5x=-4\Rightarrow x=\dfrac{-4}{5}\)
c) \(\left(2x-3\right)^2-9=0\)
\(\left(2x-3\right)^2=9=\left(\pm3\right)^2\)
\(\Rightarrow\left[{}\begin{matrix}2x-3=3\\2x-3=-3\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3\\x=0\end{matrix}\right.\)
d) \(\left(5X+1\right)^2=\dfrac{36}{49}=\left(\pm\dfrac{6}{7}\right)^2\)
\(\Rightarrow\left[{}\begin{matrix}5X+1=\dfrac{6}{7}\\5x+1=\dfrac{-6}{7}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}\dfrac{-1}{35}\\\dfrac{-13}{35}\end{matrix}\right.\)
Câu 2: mik chỉ nêu đáp án thôi nhé :
a) \(x=0\)
\(y=\dfrac{1}{2}\) hoặc \(y=\dfrac{-1}{2}\)
b) x =10 còn y giống câu a