a) \(\frac{5}{6}\)x - \(\frac{3}{8}\)x - 10 = 12

=> \(\left(\frac{5}{6}-\frac{3}{8}\right)\)x = 12 + 10

=> \(\frac{11}{24}\)x = 22

=> x = 22 : \(\frac{11}{24}\)

=> x = 48

Vậy x = 48.

b) (\(\left|x\right|\) - \(\frac{1}{8}\)) . \(\left(-\frac{1}{8}\right)^5\) = \(\left(-\frac{1}{8}\right)^7\)

=> \(\left|x\right|\) - \(\frac{1}{8}\) = \(\left(-\frac{1}{8}\right)^7\) : \(\left(-\frac{1}{8}\right)^5\)

=> \(\left|x\right|\) - \(\frac{1}{8}\) = \(\left(-\frac{1}{8}\right)^{7-5}\)

=> \(\left|x\right|\) - \(\frac{1}{8}\) = \(\frac{1}{64}\)

=> \(\left|x\right|\) = \(\frac{1}{64}\) + \(\frac{1}{8}\)

=> \(\left|x\right|\) = \(\frac{9}{64}\)

=> x = \(\frac{9}{64}\) hoặc x = \(\frac{-9}{64}\)

Vậy x = \(\frac{9}{64}\) hoặc x = \(\frac{-9}{64}\)

hên quá làm đúng hì hì, cảm ơn nhen, hết sợ bị sai ồi

Bài 1:

1. \(\left(\frac{1^5}{2}\right):\left(\frac{1^8}{2}\right)\)\(=\frac{1^5:1^8}{2:2}=\frac{1}{1}=1.\)

Mình chỉ làm bài 1 thôi nhé.

Chúc bạn học tốt!

a,(=)\(3^{x+1}.\left(3+4\right)=7.3^6\)

(=)\(3^{x+1}=3^6\)

=>x+1=6(=)x=5

b

1) a) \(x^2=2x\Leftrightarrow x^2-2x=0\Leftrightarrow x\left(x-2\right)=0\) \(\Leftrightarrow\left\{{}\begin{matrix}x=0\\x-2=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\x=2\end{matrix}\right.\) vậy \(x=0;x=2\)

b) \(x^3=x\Leftrightarrow x^3-x=0\Leftrightarrow x\left(x^2-1\right)=0\) \(\Leftrightarrow x\left(x+1\right)\left(x-1\right)=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\x+1=0\\x-1=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=0\\x=-1\\x=1\end{matrix}\right.\) vậy \(x=0;x=-1;x=1\)

\(x^2=2x\Rightarrow x^2-2x=0\Rightarrow x\left(x-2\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x-2=0\Rightarrow x=2\end{matrix}\right.\)

\(x^3=x\Rightarrow x^3-x=0\Rightarrow x\left(x^2-1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x^2-1=0\Rightarrow x^2=1\Rightarrow x=\pm1\end{matrix}\right.\)

\(A=\left(\dfrac{1}{4}-1\right)\left(\dfrac{1}{9}-1\right)\left(\dfrac{1}{16}-1\right)\left(\dfrac{1}{25}-1\right)...\left(\dfrac{1}{121}-1\right)\)

\(A=\dfrac{-3}{4}.\dfrac{-8}{9}.\dfrac{-15}{16}.\dfrac{-24}{25}...\dfrac{-120}{121}\)

\(A=\dfrac{3.8.15.24....120}{4.9.16.25...121}\)

\(A=\dfrac{1.3.2.4.3.5.4.6....10.12}{2.2.3.3.4.4.5.5....11.11}\)

\(A=\dfrac{1.2.4....10}{2.3.4.5...11}.\dfrac{3.4.5....12}{2.3.4.5....11}\)

\(A=\dfrac{1}{11}.6=\dfrac{6}{11}\)

3) Áp dụng tính chất:

\(\dfrac{a}{b}< 1\Rightarrow\dfrac{a+m}{b+m}< 1\left(m\in N\right)\)

\(B=\dfrac{8^{2017}+1}{8^{2018}+1}< 1\)

\(B< \dfrac{8^{2017}+1+8}{8^{2018}+1+8}\)

\(B< \dfrac{8^{2017}+8}{8^{2018}+8}\)

\(B< \dfrac{8\left(8^{2016}+1\right)}{8\left(8^{2017}+1\right)}\)

\(B< \dfrac{8^{2016}+1}{8^{2017}+1}=A\)

\(B< A\)

a) \(3^{7x-1}=815\)

\(\Leftrightarrow3^{7x}:3=815\)

\(\Leftrightarrow\left(3^7\right)^x:3=815\)

\(\Leftrightarrow2187^x=815.3\)

\(\Leftrightarrow2187^x=2445\)

\(\Rightarrow\) Không có x thoả mãn.

b) \(\left(-0,6\right)^5.x=\left(\dfrac{-3}{5}\right)^8\)

\(\Leftrightarrow x=\left(\dfrac{-3}{5}\right)^8:\left(-0,6\right)^5\)

\(\Leftrightarrow x=\left(-0,6\right)^8:\left(-0,6\right)^5\)

\(\Leftrightarrow x=\left(-0,6\right)^3\)

c) \(\left(0,5-x\right)^3=-8\)

\(\Leftrightarrow\left(0,5-x\right)^3=\left(-2\right)^3\)

\(\Rightarrow0,5-x=-2\)

\(\Leftrightarrow x=0,5+2\)

\(\Leftrightarrow x=2,5\).

Câu b hình như bạn làm sai rồi đó !

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: \(\Leftrightarrow4^x\left(\dfrac{3}{2}+\dfrac{5}{3}\cdot4^2\right)=4^8\left(\dfrac{3}{2}+\dfrac{5}{3}\cdot4^2\right)\)

=>4^x=4^8

=>x=8

b: \(\Leftrightarrow2^x\cdot\dfrac{1}{2}+2^x\cdot2=2^{10}\left(2^2+1\right)\)

=>2^x=2^11

=>x=11

c: =>1/6*6^x+6^x*36=6^15(1+6^3)

=>6^x=6*6^15

=>x=16

d: \(\Leftrightarrow8^x\left(\dfrac{5}{3}\cdot8^2-\dfrac{3}{5}\right)=8^9\left(\dfrac{5}{3}\cdot8^2-\dfrac{3}{5}\right)\)

=>x=9

a. \(\left(\frac{-1}{3}\right)^3.x=\frac{1}{81}\)

\(\Leftrightarrow\frac{1}{81}:\left(-\frac{1}{27}\right)\)

\(\Leftrightarrow x=\frac{-1}{3}\)

b. x⁸ = 16 . x⁶

  <=>  x⁸ : x⁶ = 16

 <=> x²          = 4²

<=> x            = 4

c. (2x - 1)⁶ = (2x - 1)⁸

    <=> x = \(\orbr{\begin{cases}x=1\\x=0\end{cases}}\)

Vậy x = 1 hoặc 0

dùng fx gõ cho hẳn hoi