a) 2^x - 15 = 17

2^x = 2⁵

b) ( 7x - 11 )³ = 2⁵. 5² + 200

(7x - 11)³ = 32 . 25 + 200

(7x -11)³ = 1000

(7x-11)³ = 10³

7x - 11 = 10

7x = 10+11

7x = 21

x = 21 : 7

x = 3

like nha

e) \(\left(2x-15\right)^5=\left(2x-15\right)^3\)

\(\Rightarrow\left(2x-15\right)^5-\left(2x-15\right)^3=0\)

\(\Rightarrow\left(2x-15\right)^3\cdot\left(2x-15\right)^2-\left(2x-15\right)^3=0\)

\(\Rightarrow\left(2x-15\right)^3\cdot\left[\left(2x-15\right)^2-1\right]=0\)

\(\Rightarrow\left(2x-15\right)^3=0\) hoặc \(\left(2x-15\right)^2-1=0\)

+)TH1: \(\left(2x-15\right)^3=0\)

\(\Rightarrow2x-15=0\)

\(\Rightarrow2x=15\)

\(\Rightarrow x=\frac{15}{2}\)

+)TH2: \(\left(2x-15\right)^2-1=0\)

\(\Rightarrow\left(2x-15\right)^2=1\)

\(\Rightarrow\left[{}\begin{matrix}2x-15=1\\2x-15=-1\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}2x=16\\2x=14\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=8\\x=7\end{matrix}\right.\)

Vậy \(x=\frac{15}{2}\) hoặc \(x=8\) hoặc \(x=7\)

a) \(2^x-17=15\Rightarrow2^x=32\)

Mà \(2^5=32\Rightarrow x=5\)

Vậy x = 5

b)\(\left(7x-11\right)^3=2^5\cdot5^2+200\)

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

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

\(\Rightarrow7x-11=10\)

\(\Rightarrow7x=21\)

\(\Rightarrow x=3\)

Vậy x = 3

c)\(x^{10}=1^x\Rightarrow x^{10}=1\)(số 1 có luỹ thừa là bao nhiêu thì vẫn là 1 thui)\(\Rightarrow x=1\)

Vậy x = 1

d) \(x^{10}=x\Rightarrow x^{10}-x=0\)

\(\Rightarrow x\left(x^9-1\right)=0\)

\(\Rightarrow x=0\) hoặc \(x^9-1=0\)

+)TH1: \(x=0\)

+)TH2: \(x^9-1=0\Rightarrow x^9=1\Rightarrow x=1\)

Vậy x = 0 hoặc x = 1

tìm x:

a)

2^x= 2⁵

=> x = 5

b)

(7x−11)³= 32 . 25 + 200

(7x−11)³= 800+200

(7x−11)³= 1000

10³= 1000

=> (7x−11)^{3 =} 10³

7x-11 = 10

7x= 11+10

7x = 21

x = 21: 7 = 3

=> x = 3

c)

=> x = 1

Vì cơ số 1 với số mũ nào thì cũng bằng 1

d)

k;

k;ụkh

jk

hk

k

gh

khk

\(a.x^{10}=x\)

\(=>x=1\)

a) \(x^{10}=x\)

\(\Rightarrow x^{10}-x=0\)

\(\Rightarrow x\left(x^9-1\right)=0\)

=> x=0 HOẶC   \(x^9-1=0\)

=>x=0 HOẶC     x=1

1) \(\frac{7}{8}x-5\left(x-9\right)=\frac{20x+1,5}{6}\)

<=> \(\frac{21x}{24}-\frac{100\left(x-9\right)}{24}=\frac{80x+6}{24}\)

<=> 21x - 100x + 900 = 80x + 6

<=> -79x - 80x = 6 - 900

<=> -159x = -894

<=> x = 258/53

Vậy S = {258/53}

2) \(\frac{\left(2x+1\right)^2}{5}-\frac{\left(x+1\right)^2}{3}=\frac{7x^2-14x-5}{15}\)

<=> \(\frac{3\left(4x^2+4x+1\right)}{15}-\frac{5\left(x^2+2x+1\right)}{15}=\frac{7x^2-14x-5}{15}\)

<=> 12x² + 12x + 3 - 5x² - 10x - 5 = 7x² - 14x - 5

<=> 7x² + 2x - 7x² + 14x = -5 + 2

<=> 16x = 3

<=> x = 3/16

Vậy S  = {3/16}

3) 4(3x - 2) - 3(x - 4) = 7x+  10

<=> 12x - 8 - 3x + 12 = 7x + 10

<=> 9x - 7x = 10 - 4

<=> 2x = 6

<=> x = 3

Vậy S = {3}

4) \(\frac{\left(x+10\right)\left(x+4\right)}{12}-\frac{\left(x+4\right)\left(2-x\right)}{4}=\frac{\left(x+10\right)\left(x-2\right)}{3}\)

<=> \(\frac{x^2+14x+40}{12}+\frac{3\left(x^2+2x-8\right)}{12}=\frac{4\left(x^2+8x-20\right)}{12}\)

<=> x² + 14x + 40 + 3x² + 6x - 24 = 4x² + 32x - 80

<=> 4x² + 20x - 4x² - 32x = -80 - 16

<=> -12x = -96

<=> x = 8

Vậy S = {8}

\(a,x^{10}=1\Leftrightarrow x=1\)

b, 2^x = 256 <=> 2^x = 2⁸ <=> x = 8

c, x¹⁰ = x

<=> \(x^{10}-x=0\)

<=> \(x\left[x^9-1\right]=0\)

<=> x = 0 hoặc x = 1

d, \((2x-15)^5=(2x-15)^3\)

<=> \((2x-15)^5-(2x-15)^3=0\)

<=> \((2x-15)^2.\left[1-(2x-15)^3\right]=0\)

<=> \(\orbr{\begin{cases}2x-15=0\\1-(2x-15)^3=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{15}{2}\\2x-15=\pm1\end{cases}}\)

Tìm nốt x đi .

Lâu lâu chưa dạng gặp dạng này

e) \(\frac{11.3^{22}.9.35-9.15}{\left(2.3^{14}\right)^2}\)

\(=\frac{11.3^{22}.3^2.5.7-3^2.3.5}{2^2.3^{28}}\)

\(=\frac{3^3.5.\left(11.3^{20}.7-1\right)}{2^2.3^{28}}\)

\(=\frac{5.\left(11.3^{20}.7-1\right)}{2^2.3^{25}}\)

Đề bài sai ko vậy ?? kết quả ko có ra phân số hoặc số nguyên mà là số mà bạn chưa học đâu

a)

\(\begin{array}{l}x + \left( { - \frac{1}{5}} \right) = \frac{{ - 4}}{{15}}\\x = \frac{{ - 4}}{{15}} + \frac{1}{5}\\x = \frac{{ - 4}}{{15}} + \frac{3}{{15}}\\x = \frac{{ - 1}}{{15}}\end{array}\)                 

Vậy \(x = \frac{{ - 1}}{{15}}\).

b)

\(\begin{array}{l}3,7 - x = \frac{7}{{10}}\\x = 3,7 - \frac{7}{{10}}\\x = \frac{{37}}{{10}} - \frac{7}{{10}}\\x=\frac{30}{10}\\x = 3\end{array}\)

Vậy \(x = 3\).

c)

\(\begin{array}{l}x.\frac{3}{2} = 2,4\\x.\frac{3}{2} = \frac{{12}}{5}\\x = \frac{{12}}{5}:\frac{3}{2}\\x = \frac{{12}}{5}.\frac{2}{3}\\x = \frac{8}{5}\end{array}\)

Vậy \(x = \frac{8}{5}\)       

d)

\(\begin{array}{l}3,2:x =  - \frac{6}{{11}}\\\frac{{16}}{5}:x =  - \frac{6}{{11}}\\x = \frac{{16}}{5}:\left( { - \frac{6}{{11}}} \right)\\x = \frac{{16}}{5}.\frac{{ - 11}}{6}\\x = \frac{{ - 88}}{{15}}\end{array}\)

Vậy \(x = \frac{{ - 88}}{{15}}\).

Mấy ý này bản chất ko khác nhau nhé, mình làm mẫu, bạn làm tương tự mấy ý kia nhé 

a, \(\left|5x\right|=x+2\)

Với \(x\ge0\)thì \(5x=x+2\Leftrightarrow x=\dfrac{1}{2}\)

Với \(x< 0\)thì \(5x=-x-2\Leftrightarrow6x=-2\Leftrightarrow x=-\dfrac{1}{3}\)

b, \(\left|7x-3\right|-2x+6=0\Leftrightarrow\left|7x-3\right|=2x-6\)

Với \(x\ge\dfrac{3}{7}\)thì \(7x-3=2x-6\Leftrightarrow5x=-3\Leftrightarrow x=-\dfrac{3}{5}\)( ktm )

Với \(x< \dfrac{3}{7}\)thì \(7x-3=-2x+6\Leftrightarrow9x=9\Leftrightarrow x=1\)( ktm )

Vậy phương trình vô nghiệm 

1.

\(\frac{2x+3}{4}-\frac{5x+3}{6}=\frac{3-4x}{12}\)

\(MC:12\)

Quy đồng :

\(\Rightarrow\frac{3.\left(2x+3\right)}{12}-\left(\frac{2.\left(5x+3\right)}{12}\right)=\frac{3x-4}{12}\)

\(\frac{6x+9}{12}-\left(\frac{10x+6}{12}\right)=\frac{3x-4}{12}\)

\(\Leftrightarrow6x+9-\left(10x+6\right)=3x-4\)

\(\Leftrightarrow6x+9-3x=-4-9+16\)

\(\Leftrightarrow-7x=3\)

\(\Leftrightarrow x=\frac{-3}{7}\)

2.\(\frac{3.\left(2x+1\right)}{4}-1=\frac{15x-1}{10}\)

\(MC:20\)

Quy đồng :

\(\frac{15.\left(2x+1\right)}{20}-\frac{20}{20}=\frac{2.\left(15x-1\right)}{20}\)

\(\Leftrightarrow15\left(2x+1\right)-20=2\left(15x-1\right)\)

\(\Leftrightarrow30x+15-20=15x-2\)

\(\Leftrightarrow15x=3\)

\(\Leftrightarrow x=\frac{3}{15}=\frac{1}{5}\)