a: \(4x^3+12=120\)

=>\(4x^3=108\)

=>\(x^3=27=3^3\)

=>x=3

b: \(\left(x-4\right)^2=64\)

=>\(\left[{}\begin{matrix}x-4=8\\x-4=-8\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=12\\x=-4\end{matrix}\right.\)

c: (x+1)^3-2=5^2

=>\(\left(x+1\right)^3=25+2=27\)

=>x+1=3

=>x=2

d: 136-(x+5)^2=100

=>(x+5)^2=36

=>\(\left[{}\begin{matrix}x+5=6\\x+5=-6\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=-11\end{matrix}\right.\)

e: \(4^x=16\)

=>\(4^x=4^2\)

=>x=2

f: \(7^x\cdot3-147=0\)

=>\(3\cdot7^x=147\)

=>\(7^x=49\)

=>x=2

g: \(2^{x+3}-15=17\)

=>\(2^{x+3}=32\)

=>x+3=5

=>x=2

h: \(5^{2x-4}\cdot4=10^2\)

=>\(5^{2x-4}=\dfrac{100}{4}=25\)

=>2x-4=2

=>2x=6

=>x=3

i: (32-4x)(7-x)=0

=>(4x-32)(x-7)=0

=>4(x-8)*(x-7)=0

=>(x-8)(x-7)=0

=>\(\left[{}\begin{matrix}x-8=0\\x-7=0\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=8\\x=7\end{matrix}\right.\)

k: (8-x)(10-2x)=0

=>(x-8)(x-5)=0

=>\(\left[{}\begin{matrix}x-8=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=8\\x=5\end{matrix}\right.\)

m: \(3^x+3^{x+1}=108\)

=>\(3^x+3^x\cdot3=108\)

=>\(4\cdot3^x=108\)

=>\(3^x=27\)

=>x=3

n: \(5^{x+2}+5^{x+1}=750\)

=>\(5^x\cdot25+5^x\cdot5=750\)

=>\(5^x\cdot30=750\)

=>\(5^x=25\)

=>x=2

\(\frac{7^{x+2}+7^{x+1}+7x}{57}=\frac{5^{2x}+5^{2x+1}+5^{2x+3}}{131}\)

\(\Rightarrow\frac{7x\left(7^2+7^1+1\right)}{57}=\frac{5^{2x}\left(1+5^1+5^3\right)}{131}\)

\(\Rightarrow\frac{7x\left(49+7+1\right)}{57}=\frac{5^{2x}\left(1+5+125\right)}{131}\)

\(\Rightarrow\frac{7x.57}{57}=\frac{5^{2x}.131}{131}\)

\(\Rightarrow7x=25x\)

\(\Rightarrow x=0\)

\(\left(4x-3\right)^4=\left(4x-3\right)^2\)

\(\Rightarrow\left(4x-3\right)^4-\left(4x-3\right)^2=0\)

\(\Rightarrow\left(4x-3\right)^2\left[\left(4x-3\right)^2-1\right]=0\)

\(\Leftrightarrow\hept{\begin{cases}\left(4x-3\right)^2=0\\\left(4x-3\right)^2=1\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}4x-3=0\\4x-3=-1\\4x-3=1\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x=\frac{3}{4}\\x=\frac{1}{2}\\x=1\end{cases}}\)

Bài mình làm sai à hay sao mà t]ick sai thế???

16. 7 x − 16. 5 2 x  = 0

⇔  7 x  =  5 2 x  ⇔  7 / 25 x  =  7 / 25 0  ⇔ x = 0

a: 7x+58=100

nên 7x=42

hay x=6

c: x-56:x=16

nên x-14=16

hay x=30

c)x - 56 : 4  = 16

x - 56       = 16 : 4 

x- 56        = 4

x               =4 + 56

x               = 60 

d)101 + (36 - 4x) = 105 

          (36- 4x ) = 105 - 101 

          36 - 4x = 4

                4x = 36 - 4 

                4x = 32

                  x = 32:4

                  x = 8

a) 58 + 7x = 100

            7x = 100 - 58

            7x = 42

              x = 42 : 7

              x = 6

b) (x - 12) : 12 = 12

              x - 12 = 12 . 12

             x - 12 = 144

                    x = 144 + 12

                    x = 156

d, \(=>\left(\frac{4}{5}\right)^{2x+7}=\left(\frac{4}{5}\right)^4.\)

=> \(2x+7=4\) 

=> 2x= -3

=> x=-3/2     . Vậy x=-3/2

e, => \(\frac{7^x.7^2+7^x.7+7^x}{57}=\frac{5^{2x}+5^{2x}.5+5^{2x}.5^2}{131}.\)

=> \(\frac{7^x\left(7^2+7+1\right)}{57}=\frac{5^{2x}\left(1+5+5^2\right)}{131}\)

= > \(\frac{7^x.57}{57}=\frac{5^{2x}.131}{131}\)

=> \(7^x=5^{2x}\)

Đến đoạn này là mik nghĩ không ra nhé

Cô làm tiếp giúp Linh Đan:

\(7^x=5^{2x}\Rightarrow7^x=25^x\Rightarrow\frac{7^x}{25^x}=1\Rightarrow\left(\frac{7}{25}\right)^x=1\Rightarrow x=0\)

a) \(\left(x-4\right)^2=\left(x-4\right)^4\)

\(\Rightarrow\left(x-4\right)^2-\left(x-4^4\right)=0\)

\(\Rightarrow\left(x-4\right)^2.\left[1-\left(x-4\right)^2\right]=0\)

\(\Rightarrow\left[{}\begin{matrix}\left(x-4\right)^2=0\\1-\left(x-4\right)^2=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x-4=0\\\left(x-4\right)^2=1^2\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x-4=0\\x-4=1\\x-4=-1\end{matrix}\right.\)

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

mk ghi kết quả thôi nhé, nếu từ kết quả mak k biết biến đổi thì ib cho mk

\(x^5-7x^4-x^3+43x^2-36=\left(x-6\right)\left(x-3\right)\left(x-1\right)\left(x+1\right)\left(x+2\right)\)

câu thứ 2 bạn ktra lại đề

\(x^4+2x^3-15x^2-18x+64=\left(x-2\right)\left(x^3+4x^2-7x-32\right)\)

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

\(x^3-3x^2-4x+12=\left(x-3\right)\left(x-2\right)\left(x+2\right)\)

a)  \(x^5-7x^4-x^3+43x^2-36\)

\(=x^3\left(x^2-1\right)-7x^2\left(x^2-1\right)+36\left(x^2-1\right)\)

\(=\left(x^2-1\right)\left(x^3-7x^2+36\right)=\left(x-1\right)\left(x+1\right)\left(x^3+2x^2-9x^2-18x+18x+36\right)\)

\(=\left(x-1\right)\left(x+1\right)\left(x+2\right)\left(x^9-9x+18\right)\)

\(=\left(x-1\right)\left(x+1\right)\left(x+2\right)\left(x-3\right)\left(x-6\right)\)

c)  \(x^4+2x^3-15x^2-18x+64\)

\(=x^3\left(x-2\right)+4x^2\left(x-2\right)-7x\left(x-2\right)-32\left(x-2\right)\)

\(=\left(x-2\right)\left(x^3+4x^2-7x-32\right)\)

d)  \(x^3-x^2-4=x^2\left(x-2\right)+x\left(x-2\right)+2\left(x-2\right)\)

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

e)  \(x^3-3x^2-4x+12=x\left(x^2-4\right)-3\left(x^2-4\right)\)

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

a) 2x + 23x = 625

   x (2 + 23) = 625

   x. 25        = 625

             x    = 625 : 25

             x    = 25

b) (2^2+4^2) x +2^4 x 5 * x = 100 (dấu* là dấu nhân vì mik không muốn trùng nhau nên viết vậy)

    ( 4 + 16 ) x + 2^4 x 5 * x = 100

         20 * x +   16 x 5 * x =100

            x ( 20 + 16 x 5 )  = 100

           x ( 20 + 80 ) = 100

           x * 100 =100

                   x=100 : 100

                   x= 1

c) ( 3x +1)^2 = 6^2+8^2

    (3x +1)^2  = 36 +64

    (3x +1)^2   =100

     (3x +1)^2  = 10^2

           3x + 1 =10

                  3x=10 - 1=9

                    x = 9 :3

                    x = 3

( d và e làm tương tự )

k mik nha!

Dể mà,k mik mik giải cho

Giai giup minh di ban

\(1,\\ a,=7x^3-49x^2+21x\\ b,=x^2-x-42\\ c,=x^2-16x+64\\ d,=9x^2+12x+4\\ e,=x^2-16-25+10x-x^2=10x-41\\ 2,\\ a,\Rightarrow2\left(x-7\right)=19\\ \Rightarrow x-7=\dfrac{19}{2}\Rightarrow x=\dfrac{33}{2}\\ b,\Rightarrow4x^2-20x+25-4x^2+3x-2x=50\\ \Rightarrow-19x=25\Rightarrow x=-\dfrac{25}{19}\)