giúp me được ko

\(a,2^x.4=128\\2^x.2^2=2^7\\ 2^x=\dfrac{2^7}{2^2}=2^{7-2}=2^5\\ Vậy:x=5\\ ----\\ b,\left(2x+1\right)^3=125=5^3\\ \Rightarrow 2x+1=5\\ 2x=5-1=4\\ x=\dfrac{4}{2}=2\\ ----\\ c,2x-2^6=6\\ 2x=6+2^6=6+64\\ 2x=70\\ x=\dfrac{70}{2}=35\\ ----\\ d,49.7^x=2401\\ 7^x=\dfrac{2401}{49}=49=7^2\\ Vậy:x=2\)

b,   <=> (x-1)(3x+7x^2)=0

<=> x(x-1)(7x+3)=0

<=> x=0;x=1;x=-3/7

a,  2x^3+3x^2-2x-3=0

<=> 2x^3-2x^2+5x^2-5x+3x-3=0

<=> 2x^2(x-1)+5x(x-1)+3(x-1)=0

<=> (x-1)(2x^2+5x+3)=0

<=> (x-1)(2x^2+2x+3x+3)=0

<=> (x-1)[2x(x+1)+3(x+1)]=0

<=> (x-1)(x+1)(2x+3)=0

<=>x=1;x=-1;x=-3/2

<=>

3:

a: 3^x*3=243

=>3^x=81

=>x=4

b; 2^x*16^2=1024

=>2^x=4

=>x=2

c: 64*4^x=16^8

=>4^x=4^16/4^3=4^13

=>x=13

d: 2^x=16

=>2^x=2^4

=>x=4

a) \(\left(x-1\right)^3=8=2^3\)

\(x-1=2\)

\(x=2+1=3\)

b) \(7^{2x-6}=49=7^2\)

\(2x-6=2\)

\(2x=6+2=8\)

\(x=8:2=4\)

c) \(\left(2x-14\right)^7=128=2^7\)

\(2x-14=2\)

\(2x=14+2=16\)

\(x=16:2=8\)

d) \(x^4\cdot x^5=5^3\cdot5^6=5^4\cdot5^5\)

\(x=5\)

e) \(3\cdot\left(x+2\right):7\cdot4=120\)

\(x+2=120:3\cdot7:4\)

\(x+2=70\)

\(x=70-2=68\)

Lời giải:

a. $(x-1)^3=8=2^3$
$\Rightarrow x-1=2$

$\Rightarrow x=3$

b. $7^{2x-6}=49=7^2$
$\Rightarrow 2x-6=2$

$\Rightarrow 2x=8$

$\Rightarrow x=4$

c. $(2x-14)^7=128=2^7$

$\Rightarrow 2x-14=2$

$\Rightarrow 2x=16$

$\Rightarrow x=18$

d.

$x^4.x^5=5^3.5^6$

$x^9=5^9$

$\Rightarrow x=5$

e. 

$3(x+2):7=120:4=30$

$3(x+2)=30.7=210$

$x+2=210:3=70$

$x=70-2=68$

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)

a) \(2^x.4=128\Rightarrow2^x=32=2^5\Rightarrow x=5\)

b) \(x^{17}=x\Rightarrow x^{17}-x=0\Rightarrow x\left(x^{16}-1\right)=0\Rightarrow x=0\) hay \(x=1\)

c) \(\left(2x-2\right)^3=8\Rightarrow\left(2x-2\right)^3=2^3\Rightarrow2x-2=2\Rightarrow2x=4\Rightarrow x=2\)

d) \(\left(x-6\right)^3=\left(x-6\right)^2\Rightarrow\left(x-6\right)^3-\left(x-6\right)^2=0\)

\(\Rightarrow\left(x-6\right)^2\left(x-6-1\right)=0\Rightarrow\Rightarrow\left(x-6\right)^2\left(x-7\right)=0\)

\(\Rightarrow x-6=0\) hay \(x-7=0\Rightarrow x=6\) hay \(x=7\)

e) \(\left(7x-11\right)^3=2^5.5^2+200\Rightarrow\left(7x-11\right)^3=32.25+200\)

\(\Rightarrow\left(7x-11\right)^3=1000=10^3\Rightarrow7x-11=10\Rightarrow7x=21\Rightarrow x=3\)

f) \(3+2^{x-1}=24-\left[4^2-\left(2^2-1\right)\right]\Rightarrow2^{x-1}=24-\left[16-3\right]-3\)

\(\Rightarrow2^{x-1}=24-13-3\Rightarrow2^{x-1}=8=2^3\Rightarrow2x-1=3\Rightarrow2x=4\Rightarrow x=2\)

Cảm ơn bạn Nguyễn Đức Trí rất nhìu nha!

a) 2^x . 4 = 128

<=> 2^x = 32 

<=> 2^x = 2⁵

<=> x = 5

b) x¹⁵ = x¹

<=> x¹⁵ - x = 0

<=> x(x¹⁴ - 1) = 0

<=> \(\orbr{\begin{cases}x=0\\x^{14}-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x^{14}=1^{14}\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=\pm1\end{cases}}\)

c) (2x + 1)³ = 125

<=> (2x + 1)³ = 5³

<=> 2x + 1 = 5

<=> 2x = 4

<=> x = 2

d) (x - 5)⁴ = (x - 5)⁶

<=> (x - 5)⁶ - (x - 5)⁴ = 0

<=> (x - 5)⁴[(x - 5)² - 1] = 0

<=> \(\orbr{\begin{cases}\left(x-5\right)^4=0\\\left(x-5\right)^2-1=0\end{cases}}\)

Khi (x - 5)⁴ = 0 => x - 5 = 0 => x = 5

Khi (x - 5)² - 1 = 0 <=> (x - 5)² = 1² <=> \(\orbr{\begin{cases}x-5=1\\x-5=-1\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=6\\x=4\end{cases}}\)

a, 2x . 4 = 128

=> 2x = 128 : 4 = 32

=> x = 32 : 2 = 16

Vậy x = 16

b, x15 = x 1 => Sai đề

c, (2x + 1)³ = 125

=> ( 2x + 1 ) = 5³

=> 2x + 1 = 5 

=> 2x = 5 - 1

=> 2x = 4

=> x = 4 : 2

=> x = 2