Thanks !!!!!!!!!!!!!!!

câu E

\(\left\{{}\begin{matrix}x\ne\dfrac{5}{2}\\\left(2x-5\right)\left(5-2x\right)=-\left(\dfrac{3}{2}\right)^4\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\ne\dfrac{5}{2}\\\left|2x-5\right|=\left(\dfrac{3}{2}\right)^2\end{matrix}\right.\)

\(\left[{}\begin{matrix}\left\{{}\begin{matrix}x< \dfrac{5}{2}\\2x-5=-\left(\dfrac{3}{2}\right)^2\Rightarrow x=\dfrac{11}{8}< \dfrac{5}{2}\left(n\right)\end{matrix}\right.\\\left\{{}\begin{matrix}x>\dfrac{5}{2}\\2x-5=\left(\dfrac{3}{2}\right)^2\Rightarrow x=\dfrac{29}{8}>\dfrac{5}{2}\left(n\right)\end{matrix}\right.\end{matrix}\right.\)

câu F (bạn cho vào lớp 7.2=lớp 14 nhé. )

a) Ta có 

x⁸=(x⁴)²=>n=4

lol

1)

\(2^{x-1}=16\\ 2^{x-1}=2^4\\ \Rightarrow x-1=4\\ x=4+1\\ x=5\)

5)

\(\left(x-1\right)^2=25\Rightarrow\left[{}\begin{matrix}x-1=5\\x-1=-5\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=6\\x=-4\end{matrix}\right.\)

6)

\(\left|2x-1\right|=5\Rightarrow\left[{}\begin{matrix}2x-1=5\\2x-1=-5\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)

5) (x-1)² = 25

(x-1)² = 5²

x-1 = 5

x = 5+1

x = 6

6) \(\left|2x-1\right|=5 \)

\(TH1:\) \(2x-1=5\)

\(\Leftrightarrow2x=5+1\)

\(\Leftrightarrow2x=6\)

\(\Leftrightarrow x=6:2\)

\(\Leftrightarrow x=3\)

\(TH2:2x-1=-5\)

\(\Leftrightarrow2x=-5+1\)

\(\Leftrightarrow2x=-4\)

\(\Leftrightarrow x=-4:2\)

\(\Leftrightarrow x=-2\)

Vậy x = 3 hoặc x = -2.

Tick nha!

1.a) Sửa lại đề: \(\frac{11}{17}\)ở mẫu chuyển thành \(\frac{11}{7}\)

\(\frac{0,75+0,6-\frac{3}{7}-\frac{3}{13}}{2,75+2,2-\frac{11}{7}-\frac{11}{13}}=\frac{\frac{3}{4}+\frac{3}{5}-\frac{3}{7}-\frac{3}{13}}{\frac{11}{4}+\frac{11}{5}-\frac{11}{7}-\frac{11}{13}}\)\(=\frac{3\left(\frac{1}{4}+\frac{1}{5}-\frac{1}{7}-\frac{1}{13}\right)}{11\left(\frac{1}{4}+\frac{1}{5}-\frac{1}{7}-\frac{1}{13}\right)}=\frac{3}{11}\)

( vì \(\frac{1}{4}+\frac{1}{5}-\frac{1}{7}-\frac{1}{13}\ne0\))

2.a) \(\frac{3}{5}+\frac{3}{2}.x=\frac{-5}{7}\)\(\Leftrightarrow\frac{3}{2}.x=\frac{-5}{7}-\frac{3}{5}\)

\(\Leftrightarrow\frac{3}{2}.x=\frac{-46}{35}\)\(\Leftrightarrow x=\frac{-46}{35}:\frac{3}{2}\)\(\Leftrightarrow x=\frac{-92}{105}\)

Vậy \(x=\frac{-92}{105}\)

b) \(\left(4x-\frac{1}{3}\right).\left(\frac{3}{2}x+\frac{5}{6}\right)=0\)\(\Leftrightarrow\orbr{\begin{cases}4x-\frac{1}{3}=0\\\frac{3}{2}x+\frac{5}{6}=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}4x=\frac{1}{3}\\\frac{3}{2}x=\frac{-5}{6}\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{12}\\x=\frac{-5}{9}\end{cases}}\)

Vậy \(x=\frac{-5}{9}\)hoặc \(x=\frac{1}{12}\)

a) \(\frac{x-2}{5}=\frac{3}{8}\)

(x-2).8=5.3

(x-2).8=15

x-2=15:8

x-2=\(\frac{15}{8}\)

x=\(\frac{15}{8}+2\)

x=\(\frac{31}{8}\)

b)\(\frac{x-1}{x+5}=\frac{6}{7}\)

(x-1).7=(x+5).6

7x-7=6x+30

7x=6x+30+7

7x=6x+37

7x-6x=37

x=37

c)\(\frac{x^2}{6}=\frac{24}{25}\)

\(x^2.25=6.24\)

\(x^2.25=144\)

\(x^2=144:25\)

\(x^2=\frac{144}{25}\)

\(x^2=\left(\frac{12}{5}\right)^2\)

\(x=\frac{12}{5}\)

\(\frac{x-2}{5}=\frac{3}{8}\)=>\(x-2=\frac{15}{8}\)

                           =>\(x=\frac{31}{8}\)

\(\frac{x^2}{6}=\frac{24}{25}\Rightarrow\)x²= \(\frac{144}{25}\)

                      => x = \(\frac{12}{5}\)

a,

\(\dfrac{5^x}{125}=5^4\\ 5^x:5^3=5^4\\ 5^x=5^4\cdot5^3\\ 5^x=5^7\\ \Rightarrow x=7\)

b,

\(\dfrac{3^x}{3}+3^{x-2}=4\\ 3^{x-1}+3^{x-2}=3^1+3^0\\ \Rightarrow x=2\)

c,

\(\left(x+\dfrac{2006}{2007}\right)^6=0\\ \Rightarrow x+\dfrac{2006}{2007}=0\\ x=0-\dfrac{2006}{2007}\\ x=\dfrac{-2006}{2007}\)

d,

\(\left(x-\dfrac{1}{5}\right)^3=\dfrac{8}{125}\\ \left(x-\dfrac{1}{5}\right)^3=\left(\dfrac{2}{5}\right)^3\\ \Rightarrow x-\dfrac{1}{5}=\dfrac{2}{5}\\ x=\dfrac{2}{5}+\dfrac{1}{5}\\ x=\dfrac{3}{5}\)

e,

\(3^x+3^{x-2}=810\\ 3^x\left(1+3^2\right)=810\\ 3^x\cdot10=810\\ 3^x=810:10\\ 3^x=81\\ 3^x=3^4\\ \Rightarrow x=4\)

g,

\(5^{x+2}+5^{x+1}+5^x=19375\\ 5^x\left(5^2+5+1\right)=19375\\ 5^x\cdot31=19375\\ 5^x=19375:31\\ 5^x=625\\ 5^x=5^4\\ \Rightarrow x=4\)

cảm ơn nha bn. mk kết bn vs nhau nhé