`-5(x+4) -3(x-2)=-114`

`=>-5x-20 - (3x-6)=-114`

`=> -5x-20 - 3x+6=-114`

`=> -8x -14=-114`

`=>-8x= -100`

`=>x=-100 : (-8)`

`=>x= 12,5`

giúp mik nha mik cần gấp

\(-5\left(x+4\right)-3\left(x-2\right)=-114\)

\(\Leftrightarrow-5x-20-3x+6=-114\)

\(\Leftrightarrow-5x-3x=-114+20-6\)

\(\Leftrightarrow-8x=-100\)

\(\Leftrightarrow x=\dfrac{-100}{-8}=\dfrac{100}{8}\)

\(\left(x-2\right)^3=-27\)

\(\Rightarrow x-2=-3\)

\(\Rightarrow x=-3+2\)

\(\Rightarrow x=-1\)

Vậy \(x=-1\)

Phần còn lại , đề sai thì phải 

a) \(\left(x-2\right)^3=-27\)

\(\left(x-2\right)^3=\left(-3\right)^3\)

\(\Rightarrow x-2=-3\)

\(x=\left(-3\right)+2\)

\(x=-1\)

b) \(2^x+2=114\)

\(2^x=114-2\)

\(2^x=112\)

...

1, Ta có:

\(\left(x-2\right)^3=-27=\left(-3\right)^3\)

\(\Rightarrow x-2=-3\)

\(\Rightarrow x=-3+2=-1\)

Vậy x = -1

2, \(2^x+2=114\)

\(\Rightarrow2^x=114-2=112\approx2^{6.81}\)

\(\Rightarrow x\approx6.81\)

Vậy x \(\approx6.81\)

1) Ta có: \(2x\left(x-3\right)+5\left(x-3\right)=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{5}{2}\end{matrix}\right.\)

2) Ta có: \(\left(x^2-4\right)-\left(x-2\right)\left(3-2x\right)=0\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{1}{3}\end{matrix}\right.\)

3) Ta có: \(\left(2x-1\right)^2-\left(2x+5\right)^2=11\)

\(\Leftrightarrow4x^2-4x-1-4x^2-20x-25=11\)

\(\Leftrightarrow-24x=11+1+25=37\)

hay \(x=-\dfrac{37}{24}\)

5) Ta có: \(3x^2-5x-8=0\)

\(\Leftrightarrow3x^2+3x-8x-8=0\)

\(\Leftrightarrow3x\left(x+1\right)-8\left(x+1\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(3x-8\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=\dfrac{8}{3}\end{matrix}\right.\)

8) Ta có: \(\left|x-5\right|=3\)

\(\Leftrightarrow\left[{}\begin{matrix}x-5=3\\x-5=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=8\\x=2\end{matrix}\right.\)

10) Ta có: \(\left|2x+1\right|=\left|x-1\right|\)

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

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

câu 1: 11

câu 2 : a/ 1               b/4                                 c/ -2

câu 3: a/ 4655482                                                      b/ 790

\(\dfrac{24}{x^2-2x+4}=\dfrac{3x}{x+2}+\dfrac{72}{x^2+8}\) (1)

đkxđ: \(x\ne-2\)

\(\Leftrightarrow\dfrac{24}{x^2-2x+4}-\dfrac{3}{x+2}-\dfrac{72}{\left(x+2\right)\left(x^2-2x+4\right)}=0\)\(\Leftrightarrow\dfrac{24\left(x+2\right)}{x^3+8}-\dfrac{3\left(x^2-2x+4\right)}{x^3+8}-\dfrac{72}{x^3+8}=0\)\(\Rightarrow24x+48-3x^2+6x-12-72=0\)

\(\Leftrightarrow-3x^2+30x-36=0\)

\(\Leftrightarrow-3\left(x^2-10x+12\right)=0\)

\(\Leftrightarrow x^2-10x+12=0\)

\(\Leftrightarrow\left(x^2-10x+25\right)=13\)

\(\Leftrightarrow\left(x-5\right)^2=13\)

\(\Leftrightarrow\left[{}\begin{matrix}x-5=\sqrt{13}\\x-5=-\sqrt{13}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\sqrt{13}+5\\x=-\sqrt{13}+5\end{matrix}\right.\)

\(\dfrac{24}{x^2-2x+4}=\dfrac{3x}{x+2}+\dfrac{72}{x^3+8}\) ĐK: x>0

\(\Leftrightarrow\dfrac{24\left(x+2\right)}{\left(x+2\right)\left(x^2-2x+4\right)}=\dfrac{3x\left(x^2-2x+4\right)}{\left(x+2\right)\left(x^2-2x+4\right)}+\dfrac{72}{\left(x+2\right)\left(x^2-2x+4\right)}\)

\(\Rightarrow24\left(x+2\right)=3x\left(x^2-2x+4\right)+72\\ \Leftrightarrow24x+48-3x^3+6x^2-12x-216x-72=0\)

\(\Leftrightarrow-3x^3+6x^2-204x-24=0\)

Tới đây là cách giải phương trình bậc 3 thì mk xin chịu :3

Cách giải Phương trình bậc 3 bạn xem tham khải tại đây: https://vi.wikipedia.org/wiki/Ph%C6%B0%C6%A1ng_tr%C3%ACnh_b%E1%BA%ADc_ba

_​a) 3^x+2 + 5.3^x+1 = 648

=> 3^x+1.3+ 5.3^x+1= 648

=> 3^x+1(3+ 5) = 648

=> 3^x+1.8 = 648

=> 3^x+1 = 81

=>3^x+1= 3⁴

=> x+1 = 4

=> x = 3

vậy x= 3

\(3^x+10=13.7\)

\(3^x=91-10\)

\(3^x=81=3^4\)

Vậy x = 4

\(3^x+10=13\times7\)

\(\Rightarrow3^x=91-10\)

\(\Rightarrow3^x=81=3^4\)

\(\Rightarrow x=4\)

2: =(2x+1)^2-y^2

=(2x+1+y)(2x+1-y)

3: =x^2(x^2+2x+1)

=x^2(x+1)^2

4: =x^2+6x-x-6

=(x+6)(x-1)

5: =-6x^2+3x+4x-2

=-3x(2x-1)+2(2x-1)

=(2x-1)(-3x+2)

6: =5x(x+y)-(x+y)

=(x+y)(5x-1)

7: =2x^2+5x-2x-5

=(2x+5)(x-1)

8: =(x^2-1)*(x^2-4)

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

9: =x^2(x-5)-9(x-5)

=(x-5)(x-3)(x+3)