a: \(\Leftrightarrow-x^2-3x+x+3+x^2-6x=11\)

=>-8x+3=11

=>-8x=8

hay x=-1

b: \(\Leftrightarrow3x^2-15x+x-5-3x^2+3x=5\)

=>-11x=10

hay x=-10/11

đề sai thì phải

\(a,3^{x+1}=27\)

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

\(x+1=3\)

\(x=3-1=2\)

\(b,6^{x-1}=36\)

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

\(x-1=2\)

\(x=2+1=3\)

\(c,3^{2x+1}=243\)

\(3^{2x+1}=3^5\)

\(2x+1=5\)

\(2x=5+1=6\)

\(x=6:2=3\)

Mình hong hiểu câu d :<

\(x^{50}=x\\ \Rightarrow x^{50}-x=0\\ \Rightarrow x\left(x^{49}-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=0\\x^{49}-1=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=0\\x^{49}=1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)

Giúp mình nhanh với ạ! Thank,s

a) `1/4 x + 7/4 x = -6`

`=> x(1/4 + 7/4) = -6`

`=> x. 2 = -6`

`=> x= -6/2`

`=> x= -3`

Vậy `x= -3`

b) `1/4 x +2x = 9/2`

`=> x(1/4 +2) =9/2`

`=> x(1/4 + 8/4) = 9/2`

`=>x. 9/4 =9/2`

`=> x= 9/2 : 9/4`

`=> x= 2`

Vậy `x=2`

`a) 14,58 - x= -3,72`

`=> x= 14,58 - ( -3,72)`

`=> x= 14,58+ 3,72`

`=>x=18,3`

`b)3/4 + 1/2.x= -5/6`

`=>  1/2.x= -5/6-3/4`

\(a)14,58-x=-3,72\\ x=14,58-\left(-3,72\right)\\ x=14,58+3,72\\ x=18,3\\ b)\dfrac{3}{4}+\dfrac{1}{2}.x=-\dfrac{5}{6}\\ \dfrac{1}{2}.x=-\dfrac{5}{6}-\dfrac{3}{4}\\ \dfrac{1}{2}.x=-\dfrac{19}{12}\\ x=-\dfrac{19}{12}:\dfrac{1}{2}\\ x=-\dfrac{19}{6}.\)

a) x³-1-(x²+2x)(x-2)=5

⇔ x³-1-x³+4x=5

⇔ 4x=6

⇔ \(x=\dfrac{3}{2}\)

\(a,\Rightarrow1-6x+9x^2-9x-9x^2=-29\\ \Rightarrow-15x=-30\Rightarrow x=2\\ b,\Rightarrow8x^3-12x^2+6x-1-x^2+4x-4=4x-25x^2-6\\ \Rightarrow8x^3+12x^2+6x+1=0\\ \Rightarrow\left(2x+1\right)^3=0\\ \Rightarrow2x+1=0\Rightarrow x=-\dfrac{1}{2}\)

\(a,\Rightarrow5x+3x^2-3x^2-x+2=6\\ \Rightarrow4x=4\Rightarrow x=1\\ b,\Rightarrow\left(2x+\dfrac{1}{2}-1+2x\right)\left(2x+\dfrac{1}{2}+1-2x\right)=2\\ \Rightarrow\dfrac{3}{2}\left(4x-\dfrac{1}{2}\right)=2\\ \Rightarrow6x-\dfrac{3}{4}=2\\ \Rightarrow6x=\dfrac{11}{4}\\ \Rightarrow x=\dfrac{11}{24}\\ c,\Rightarrow\left(x+3\right)\left(x-2\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=2\\x=-3\end{matrix}\right.\)

Lời giải:
a. Vì $x,y$ thuộc $Z$ nên $x-3, y+5\in\mathbb{Z}$. Tích của chúng $=11$ nên ta có bảng sau:

b. Vì $x,y\in\mathbb{Z}$ nên $2x+1, 6-y\in\mathbb{Z}$.

Với $x$ nguyên thì $2x+1$ là số nguyên lẻ nên ta có bảng sau:

\(a.\)

\(x-\dfrac{5}{6}=\dfrac{1}{2}\)

\(\Leftrightarrow x=\dfrac{1}{2}+\dfrac{5}{6}=\dfrac{3+5}{6}=\dfrac{8}{6}=\dfrac{4}{3}\)

\(b.\)

\(-\dfrac{3}{4}-x=-\dfrac{7}{12}\)

\(\Leftrightarrow x=-\dfrac{3}{4}--\dfrac{7}{12}=-\dfrac{3}{4}+\dfrac{7}{12}=\dfrac{\left(-3\right)\cdot3+7}{12}=\dfrac{-2}{12}=-\dfrac{1}{6}\)

a)x=1/2+5/6

x=/3/6+5/6

x=8/6=4/3

b)x=-3/4- (-7/12)

x=-9/12-(-7/12)

x=-1/6

a) \(x-\dfrac{5}{6}=\dfrac{1}{2}\) 

           \(x=\dfrac{1}{2}+\dfrac{5}{6}\) 

          \(x=\dfrac{4}{3}\) 

b) \(\dfrac{-3}{4}-x=\dfrac{-7}{12}\) 

               \(x=\dfrac{-3}{4}-\dfrac{-7}{12}\) 

               \(x=\dfrac{-1}{6}\)