lớp 5 thì có 

Câu a:

2.(3\(x\) - \(\frac12\)) - 2\(x\) = \(\frac12\).(2\(x\) - 3)

6\(x\) - 1 - 2\(x\) = \(x\) - \(\frac32\)

6\(x\) - 2\(x\) - \(x\) = 1 - \(\frac32\)

4\(x\) - \(x\) = - \(\frac12\)

3\(x\) = - \(\frac12\)

\(x\) = - \(\frac12\) : 3

\(x=-\frac16\)

Vậy \(x=-\frac16\)

Câu b:

(2\(x\) - \(\frac35\))\(^2\) = \(\frac{4}{25}\)

(2\(x-\frac35\))\(^2\) = \(\left(\frac{2}{25}\right)\)\(^2\)

2\(x\) - \(\frac35\) = \(\frac25\) hoặc 2\(x\) - \(\frac35\) = - \(\frac25\)

TH: 2\(x\) - \(\frac35\) = \(\frac25\)

2\(x\) = \(\frac25+\frac35\)

2\(x\) = 1

\(x=\frac12\)

2\(x\) - \(\frac35\) = - \(\frac25\)

2\(x\) = - \(\frac25\) + \(\frac35\)

2\(x\) = \(\frac15\)

\(x\) = \(\frac{13}{25}\) : 2

\(x\) = \(\frac15\)

Vậy \(x\) ∈ {1/5; 1/2}

a, \(-\frac{5}{7}-\left(\frac{1}{2}-x\right)=-\frac{11}{4}\)

\(\frac{1}{2}-x=\frac{57}{28}\)

\(x=-\frac{43}{28}\)

b, \(\left(2x-1\right)^2-5=20\)

\(\Rightarrow\left(2x-1\right)^2=25\)

\(\Rightarrow2x-1=\pm5\)

\(\Rightarrow\left[{}\begin{matrix}2x-1=5\\2x-1=-5\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x=6\\2x=-4\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)

Câd

\(\frac{x-6}{4}=\frac{4}{x-6}\)

(\(x-6\))(\(x-6\)) =4.4

(\(x-6\))\(^2\) = 4\(^2\)

\(x-6=-4\) hoặc \(x\) - 6 = 4

\(x-6\) = -4

\(x=-4+6\)

\(x=2\)

\(x-6=4\)

\(x=4+6\)

\(x=10\)

Vậy \(x\) ∈ {2; 10}

b, \(\left(2x-1\right)^2-5=20\)

\(\Rightarrow\left(2x-1\right)^2=25\)

\(\Rightarrow\left(2x-1\right)^2=5^2\)

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

\(\Rightarrow\left[{}\begin{matrix}2x=7\\2x=-5\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\frac{7}{2}\\x=-\frac{5}{2}\end{matrix}\right.\)

Vậy ...

a) \(-\frac{5}{7}-\left(\frac{1}{2}-x\right)=\frac{-11}{4}\)

\(\Rightarrow\left(\frac{1}{2}-x\right)=\left(-\frac{5}{7}\right)+\frac{11}{4}\)

\(\Rightarrow\frac{1}{2}-x=\frac{57}{28}\)

\(\Rightarrow x=\frac{1}{2}-\frac{57}{28}\)

\(\Rightarrow x=-\frac{43}{28}\)

Vậy \(x=-\frac{43}{28}.\)

b) \(\left(2x-1\right)^2-5=20\)

\(\Rightarrow\left(2x-1\right)^2=20+5\)

\(\Rightarrow\left(2x-1\right)^2=25\)

\(\Rightarrow2x-1=\pm5\)

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

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

Vậy \(x\in\left\{3;-2\right\}.\)

d) \(\frac{x-6}{4}=\frac{4}{x-6}\)

\(\Rightarrow\left(x-6\right).\left(x-6\right)=4.4\)

\(\Rightarrow\left(x-6\right).\left(x-6\right)=16\)

\(\Rightarrow\left(x-6\right)^2=16\)

\(\Rightarrow x-6=\pm4\)

\(\Rightarrow\left[{}\begin{matrix}x-6=4\\x-6=-4\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=4+6\\x=\left(-4\right)+6\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=10\\x=2\end{matrix}\right.\)

Vậy \(x\in\left\{10;2\right\}.\)

Chúc bạn học tốt!

Câu d:

-1\(\frac23\) - (|2\(x\)| + \(\frac56\)) = - 2

-\(\frac53\) - |2\(x\)| - \(\frac56\) = - 2

|2\(x\)| = - \(\frac53\) - \(\frac56\) + 2

|2\(x\)| = - \(\frac52\) + 2

|2\(x\)| = - \(\frac12\) (vô lí vì trị tuyệt đối của một số luôn là một số không âm)

Không có giá trị nào của x thỏa mãn đề bài.

x ∈ ∅

Câu a:

|\(x\) - 3| = \(x\) + 4

Vì |\(x\) - 3| ≥ 0 ∀ \(x\) nên \(x\) + 4 ≥ 0 ⇒ \(x\) ≥ - 4

Với -4 ≤ \(x\) ≤ 3 ta có:

-\(x\) + 3 = \(x\) + 4

\(x\) + \(x\) = -4 + 3

2\(x\) = -1

\(x=\frac{-1}{2}\)

Với x > 3 ta có:

x - 3 = x + 4

x - x = 3 + 4

0 = 7 (vô lí)

Vậy x = -1/2 là nghiện duy nhất của phương trình.

Vậy \(x\) = -1/2

a)\(\left(-3\right)^{x+3}=-\frac{1}{27}\)

\(\left(-3\right)^{x+3}=\left(-\frac{1}{3}\right)^3\)

\(\left(-3\right)^{x+3}=\left(-\frac{3^0}{3^1}\right)^3\)

\(\left(-3\right)^{x+3}=\left(-3^{-1}\right)^3\)

\(\left(-3\right)^{x+3}=\left(-3\right)^{-3}\)

\(\Rightarrow x+3=-3\)

\(\Rightarrow x=-6\)

b)\(\left(-6\right)^{2x+2}=\frac{1}{36}\)

\(\left(-6\right)^{2x+2}=\left(-\frac{1}{6}\right)^2\)

\(\left(-6\right)^{2x+2}=\left(-\frac{6^0}{6^1}\right)^2\)

\(\left(-6\right)^{2x+2}=\left(-6^{-1}\right)^2\)

\(\left(-6\right)^{2x+2}=\left(-6\right)^{-2}\)

\(\Rightarrow2x+2=-2\)

\(\Rightarrow2x=-4\)

\(\Rightarrow x=-2\)

c)\(\left(-3\right)^{x+5}=\frac{1}{81}\)

\(\left(-3\right)^{x+5}=\left(-\frac{1}{3}\right)^4\)

\(\left(-3\right)^{x+5}=\left(-\frac{3^0}{3^1}\right)^4\)

\(\left(-3\right)^{x+5}=\left(-3^{-1}\right)^4\)

\(\left(-3\right)^{x+5}=\left(-3\right)^{-4}\)

\(\Rightarrow x+5=-4\)

\(\Rightarrow x=-9\)

d)\(\left(\frac{1}{9}\right)^x=\left(\frac{1}{27}\right)^6\)

\(\left[\left(\frac{1}{3}\right)^2\right]^x=\left[\left(\frac{1}{3}\right)^3\right]^6\)

\(\left(\frac{1}{3}\right)^{2x}=\left(\frac{1}{3}\right)^{18}\)

\(\Rightarrow2x=18\)

\(\Rightarrow x=9\)

e)\(\left(\frac{4}{9}\right)^x=\left(\frac{8}{27}\right)^6\)

\(\left[\left(\frac{2}{3}\right)^2\right]^x=\left[\left(\frac{2}{3}\right)^3\right]^6\)

\(\left(\frac{2}{3}\right)^{2x}=\left(\frac{2}{3}\right)^{18}\)

\(\Rightarrow2x=18\)

\(\Rightarrow x=9\)

\(\frac{x}{\left(-\frac{1}{3}\right)^3}=-\frac{1}{3}\Rightarrow x=\left(-\frac{1}{3}\right)\left(-\frac{1}{3}\right)^3=\left(-\frac{1}{3}\right)^4\)

\(\left(\frac{4}{5}\right)^5\cdot x=\left(\frac{4}{5}\right)^7\)

=> \(x=\frac{\left(\frac{4}{5}\right)^7}{\left(\frac{4}{5}\right)^5}=\left(\frac{4}{5}\right)^2=\frac{16}{25}\)

\(\left(x+\frac{1}{2}\right)^2=\frac{1}{16}=\left(\pm\frac{1}{4}\right)^2\)

=> \(\orbr{\begin{cases}x+\frac{1}{2}=\frac{1}{4}\\x+\frac{1}{2}=-\frac{1}{4}\end{cases}}\Rightarrow\orbr{\begin{cases}x=-\frac{1}{4}\\x=-\frac{3}{4}\end{cases}}\)

(3x + 1)³ = -27 => (3x + 1)³ = (-3)³ => 3x + 1 = -3 => 3x = -4 => x = -4/3

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

\(=>x:\frac{-1}{27}=\frac{-1}{3}\)

\(=>x=\frac{-1}{3}.\frac{-1}{27}=>x=\frac{1}{81}\)

b) \(\left(\frac{4}{5}\right)^5.x=\left(\frac{4}{5}\right)^7\)

\(=>x=\left(\frac{4}{5}\right)^7:\left(\frac{4}{5}\right)^5=>x=\left(\frac{4}{5}\right)^2=\frac{16}{25}\)

c)\(\left(x+\frac{1}{2}\right)^2=\frac{1}{16}\)

\(=>\orbr{\begin{cases}\left(x+\frac{1}{2}\right)^2=\left(\frac{1}{4}\right)^2\\\left(x+\frac{1}{2}\right)^2=\left(\frac{-1}{4}\right)^2\end{cases}}\)

\(=>\orbr{\begin{cases}x+\frac{1}{2}=\frac{1}{4}\\x+\frac{1}{2}=\frac{-1}{4}\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{-1}{4}\\x=-1\end{cases}}}\)

d|) \(\left(3x+1\right)^3=-27\)

\(=>\left(3x+1\right)^3=\left(-3\right)^3\)

\(=>3x+1=-3\)

\(=>3x=-4=>x=\frac{-4}{3}\)

cậu có thể tham khảo bài làm trên đây ạ, chúc cậu học tốt:>