(2x² + 1)(x-3)=0

\(\Rightarrow\orbr{\begin{cases}2x^2+1=0\\x-3=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}2x^2=-1\Rightarrow x^2=-\frac{1}{2}\left(vl\right)\\x=3\end{cases}}\)

Vậy x=3

48-(15-x)⁵=48

     (15-x)⁵=48-48

     (15-x)⁵=0

=>  15-x   =0

           x    =15-0

           x    =15

Vậy x=15

\(\Leftrightarrow\dfrac{2}{2.4}+\dfrac{2}{4.6}+...+\dfrac{2}{\left(2x-2\right).2x}=\dfrac{11}{24}\)

\(\Leftrightarrow\dfrac{4-2}{2.4}+\dfrac{6-4}{4.6}+...+\dfrac{2x-\left(2x-2\right)}{\left(2x-2\right).2x}=\dfrac{11}{24}\)

\(\Leftrightarrow\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{2x-2}-\dfrac{1}{2x}=\dfrac{11}{24}\)

\(\Leftrightarrow\dfrac{1}{2}-\dfrac{1}{2x}=\dfrac{11}{24}\)

\(\Leftrightarrow\dfrac{1}{2x}=\dfrac{1}{2}-\dfrac{11}{24}\)

\(\Leftrightarrow\dfrac{1}{2x}=\dfrac{1}{24}\)

\(\Rightarrow2x=24\)

\(\Rightarrow x=12\)

=>(2x-1)^2=24^2

=>2x-1=24 hoặc 2x-1=-24

=>x=-23/2 hoặc x=25/2

\(\dfrac{2x-1}{-12}=\dfrac{48}{1-2x}\) (ĐK: \(x\ne\dfrac{1}{2}\))

\(\Leftrightarrow\dfrac{2x-1}{12}=\dfrac{48}{2x-1}\)

\(\Leftrightarrow-12\cdot48=\left(2x-1\right)\left(2x-1\right)\)

\(\Leftrightarrow567=\left(2x-1\right)^2\)

\(\Leftrightarrow24^2=\left(2x-1\right)^2\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-1=-24\\2x-1=24\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2x=-23\\2x=25\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{23}{2}\left(tm\right)\\x=\dfrac{25}{2}\left(tm\right)\end{matrix}\right.\)

\(x=3y=2z\)

\(\Rightarrow\frac{x}{1}=\frac{y}{2}=\frac{z}{3}\)

\(\Rightarrow\frac{2x}{2}=\frac{3y}{6}=\frac{4z}{12}=\frac{2x-3y+4z}{2-6+12}=\frac{48}{8}=6\)

Rồi thế vào là ra thôi :

 \(\frac{2x}{2}=6\Rightarrow x=..........\)

Rồi tương tự thôi

6)

\(x=3y=2z\)

\(\Rightarrow\frac{x}{6}=\frac{y}{2}=\frac{z}{3}\)

\(\Rightarrow\frac{2x}{12}=\frac{3y}{6}=\frac{4z}{12}\)

Áp dụng tc của dãy tỉ số bằng nhau ta có

\(\frac{2x}{12}=\frac{3y}{6}=\frac{4z}{12}=\frac{2x-3y+4z}{12-6+12}=\frac{48}{18}=\frac{24}{9}\)

\(\Rightarrow\begin{cases}x=16\\y=\frac{16}{3}\\z=8\end{cases}\)

7)

\(2x=3y=-2z\)

\(\Rightarrow\frac{2x}{1}=\frac{3y}{1}=\frac{-4z}{2}\)

Áp dụng tc của dãy tỉ số bằng nhau ta có

\(\frac{2x}{1}=\frac{3y}{1}=\frac{-4z}{2}=\frac{2x-3y-\left(-4z\right)}{1-1-2}=\frac{48}{-2}=-24\)

\(\Rightarrow\begin{cases}x=-12\\y=-8\\z=12\end{cases}\)

6) *2x - 3y + 4z = 48

<=> 4z -2z +4z = 48

=> ( 4-2+4)z = 48

=> z=8  => 2z= 16

* 2x -3y + 4z =48

<=> 6y - 3y +6y =48

=> (6 - 3+ 6)y = 48

=> y= \(\frac{16}{3}\) => 3y = 16

* 2x - 3y + 4z =48

<=> 2x -x + 2x = 48

=> ( 2 -1 +2)x =48

=>x= 16

a) \(2^x=8.64=2^3.2^6=2^9\Rightarrow x=9\)

b) \(3.2^x=48\Rightarrow2^x=16=2^4\Rightarrow x=4\)

Ta có : \(\frac{1+2x}{36}=\frac{1+4x}{48}=\frac{1+6x}{6y}\Rightarrow\frac{2+4x}{72}=\frac{1+4x}{48}=\frac{1+6x}{6y}\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có :

\(\frac{1+2x}{36}=\frac{2+4x}{72}=\frac{1+4x}{48}=\frac{1+6x}{6y}=\frac{2+4x-1-4x}{72-48}=\frac{1}{24}\)

=> \(\frac{1+4x}{48}=\frac{1}{24}\Rightarrow\frac{1+4x}{48}=\frac{2}{48}\Rightarrow1+4x=2\Rightarrow x=0,25\)

\(\frac{1+2x}{36}=\frac{1+4x}{48}=\frac{1+6x}{6x}\Rightarrow\frac{2+4x}{72}=\frac{1+4x}{48}=\frac{1+6x}{6y}\)

Áp dụng t/c dãy tỉ số bằng nhau ta có :

\(\frac{1+2x}{36}=\frac{2+4x}{72}=\frac{1+4x}{48}=\frac{1+6x}{6y}=\frac{2+4x-1-4x}{72-48}=\frac{1}{24}\)

\(\Rightarrow\frac{1+4x}{48}=\frac{1}{24}\Rightarrow\frac{1+4x}{48}=\frac{2}{48}\Rightarrow1+4x=2\Rightarrow x=0,25\)

Ta có : 2^x - 2^y = 48

=> x > y => x = y + n với n ∈ N*

=> 2^x - 2^y = 2^{y + n} - 2^y = 48

=> 2^y . 2ⁿ - 2^y = 48

=> 2^y . (2ⁿ - 1) = 48

=> 2^y ; 2ⁿ - 1 ∈ Ư(48) ∈ {1;2;3;4;6;8;12;16;24;48}

Mà 2ⁿ - 1 luôn lẻ với mọi n ∈ N*

=> 2ⁿ - 1 = 3

=> 2^y . 3 = 48 

=> 2^y = 16 = 2⁴

=> y = 4 

=> 2^x - 2⁴ = 48

=> 2^x = 48 + 16 = 64 = 2⁶

=> x = 6

Vậy x = 6 ; y = 4