Giải:

a) Ta có:

\(\left(x^4\right)^2=\frac{x^{12}}{x^5}\left(x\ne0\right)\Leftrightarrow x^8=x^7\)

\(\Leftrightarrow x^8-x^7=0\Leftrightarrow x^7\left(x-1\right)=0\)

\(\Leftrightarrow x-1=0\left(x^7\ne0\right)\Leftrightarrow x=1\)

Vậy \(x=1\)

b) Ta có:

\(x^{10}=25x^8\Leftrightarrow x^{10}-25x^8=0\)

\(\Leftrightarrow x^8\left(x^2-25\right)=0\Leftrightarrow\) \(\left[\begin{array}{}x^8=0\\x^2-25=0\end{array}\right.\)

\(\Leftrightarrow\) \(\left[\begin{array}{}x=0\\x=5\\x=-5\end{array}\right.\) Vậy...

a) \(\left(x^4\right)^2=\frac{x^{12}}{x^5}\left(x\ne0\right)\)

\(\Rightarrow x^8=x^7\)

\(\Rightarrow x^8-x^7=0\)

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

\(\Rightarrow x-1=0\) ( vì \(x^7\ne0\) )

Vậy \(x=1\)

b ) \(x^{10}=25x^8\)

\(\Rightarrow x^{10}-25x^8=0\)

\(\Rightarrow x^8.\left(x^2-25\right)=0\)

\(\Leftrightarrow x^8=0\) hoặc \(x^2-25=0\)

Do đó \(x=0\) hoặc \(x=5\) hoặc \(x=-5\)

Vậy \(x\in\left\{0;5;-5\right\}\)

a.

\(\left(x^4\right)^2=\frac{x^{12}}{x^5}\)

\(x^8=x^7\)

\(x\ne0\)

\(\Rightarrow x=1\)

b.

\(x^{10}=25\times x^8\)

\(\frac{x^{10}}{x^8}=25\)

\(x^2=\left(\pm5\right)^2\)

\(x=\pm5\)

Vậy x = 5 hoặc x = -5

Chúc bạn học tốt

a) \(\left(x^4\right)^2=\dfrac{x^{12}}{x^5}\\ x^8=x^7\\ \Rightarrow x=1;x=-1\)

b)\(x^{10}=25.x^8\\ x^2=25\\ \Rightarrow\left\{{}\begin{matrix}x=5\\x=-5\end{matrix}\right.\)

a) \(\left(x^4\right)^2=\dfrac{x^{12}}{x^5}\)

\(\Rightarrow x^8=x^7\)

\(\Rightarrow x^8-x^7=0\)

\(\Rightarrow x^7.x-x^7=0\)

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

\(\Rightarrow x-1=0\) (vì x^7 \(\ne\)0)

\(\Rightarrow\) x=1

b) x^10=25x^8

\(\Rightarrow x^8.x^2-25x^8=0\)

\(\Rightarrow x^8\left(x^2-25\right)=0\)

\(\Rightarrow x^8=0\) hoặc \(x^2-25=0\)

1) x^8=0

\(\Rightarrow\) x=0(1)

2) x^2 -25=0

x^2=0+25

x^2=25

x^2=5^2 hay x^2=(-5)^2

Suy ra x=5 hoặc x=-5 (2)

Từ (1) và (2)\(\Rightarrow\)x\(\in\left\{0;5;-5\right\}\)

EM KO CHÉP ĐÁP ÁN NHÉ

Bài 1 : \(M=\frac{8^{20}+4^{20}}{4^{25}+64^5}=\frac{\left(2^3\right)^{20}+\left(2^2\right)^{20}}{\left(2^2\right)^{25}+\left(2^6\right)^5}=\frac{2^{60}+2^{40}}{2^{50}+2^{30}}=\frac{2^{40}\left(2^{20}+1\right)}{2^{30}\left(2^{20}+1\right)}=2^{10}=1024\)

Bài 2 : a) \(\left(x^4\right)^2=\frac{x^{12}}{x^5}\)=> \(x^8=x^7\)

=> \(x^8-x^7=0\)

=> \(x^7\left(x-1\right)=0\)

=> \(x-1=0\Rightarrow x=1\)(vì x⁷ = 0 => x = 0 mà x \(\ne\)0 nên loại)

b) \(x^{10}-25x^8=0\)

=> \(x^8\left(x^2-25\right)=0\)

=> x⁸ = 0 hoặc x² - 25 = 0

=> x = 0 hoặc x² = 25

=> x = 0 hoặc x = \(\pm\)5

Bài 3 : a) \(\left(2x+3\right)^2=\frac{9}{121}=\left(\pm\frac{3}{11}\right)^2\)

=> \(\orbr{\begin{cases}2x+3=\frac{3}{11}\\2x+3=-\frac{3}{11}\end{cases}}\Rightarrow\orbr{\begin{cases}x=-\frac{15}{11}\\x=-\frac{18}{11}\end{cases}}\)

b) \(\left(3x-1\right)^3=-\frac{8}{27}=\left(-\frac{2}{3}\right)^3\)

=> 3x - 1 = -2/3

=> 3x = 1/3

=> x = 1/3 : 3 = 1/9

1) Ta có \(M=\frac{8^{20}+4^{20}}{4^{25}+64^5}=\frac{\left(2^3\right)^{20}+\left(2^2\right)^{20}}{\left(2^2\right)^{25}+\left(2^6\right)^5}=\frac{2^{60}+2^{40}}{2^{50}+2^{30}}=\frac{2^{40}\left(2^{20}+1\right)}{2^{30}\left(2^{30}+1\right)}=2^{10}=1024\)

2) a) \(\left(x^4\right)^2=\frac{x^{12}}{x^5}\)

=> x⁸ = x⁷

=> x⁸ - x⁷ = 0

=> x⁷(x - 1) = 0

=> \(\orbr{\begin{cases}x^7=0\\x-1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=1\end{cases}}\)

Vậy x \(\in\left\{0;1\right\}\)

b) x¹⁰ = 25x⁸

=> x¹⁰ - 25x⁸ = 0

=> x⁸(x² - 25) = 0

=> \(\orbr{\begin{cases}x^8=0\\x^2-25=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=\pm5\end{cases}}\)

Vậy \(x\in\left\{0;5;-5\right\}\)

3) \(\left(2x+3\right)^2=\frac{9}{121}\)

=> \(\left(2x+3\right)^2=\left(\frac{3}{11}\right)^2\)

=> \(\orbr{\begin{cases}2x+3=\frac{3}{11}\\2x+3=-\frac{3}{11}\end{cases}}\Rightarrow\orbr{\begin{cases}2x=\frac{-30}{11}\\2x=-\frac{36}{11}\end{cases}}\Rightarrow\orbr{\begin{cases}x=-\frac{15}{11}\\x=-\frac{18}{11}\end{cases}}\)

Vậy \(x\in\left\{-\frac{15}{11};-\frac{18}{11}\right\}\)

b) \(\left(3x-1\right)^3=-\frac{8}{27}\)

=> \(\left(3x-1\right)^3=\left(-\frac{2}{3}\right)^3\)

=> \(3x-1=-\frac{2}{3}\)

=> \(3x=\frac{1}{3}\)

=> \(x=\frac{1}{9}\)

Vậy \(x=\frac{1}{9}\)

cảm ơn các bạn nhiều lắm nuôn.I love u

Tìm x, biết:

3(x+2)(x+5) +5(x+5)(x+10) +7(x+10)(x+17) =x(x+2)(x+17) (x∉−2;−5;−10;−17)

2(x−1)(x−3) +5(x−3)(x−8) +12(x−8)(x−20) −1x−20 =−34 (x∉1;3;8;20)

x+110 +2+111 x+112 =x+113 +x+114 

x−1030 +x−1443 +x−595 +x−1488 =0

Trả lời luôn à bạn

Phần c khó để tớ giải cho

\(\left(3x-1\right)^3=\left(\frac{2}{3}\right)^3\)

=> 3x -1 = 2/3

3x = 5/3

x = 5/9

học tốt ^^

\(\left(x^4\right)^2=x^{12-5}\)

\(x^8-x^7=0\)

\(x^7\cdot x-x^7=0\)

\(x^7\cdot\left(x-1\right)=0\)

+) x^7 = 0 => x = 0

+) x -1 = 0 => x = 1

Vậy,...........

học tốt ^^

\(\left(x^4\right)^2=\frac{x^{12}}{x^5}\Rightarrow\left(x^4\right)^2x^5=\frac{x^{12}}{x^5}x^5\Rightarrow x^{13}=x^{12}\)

\(\Rightarrow x=1\)

a) \(\frac{x+5}{4}-\frac{2x-3}{3}=\frac{6x-1}{8}+\frac{2x-1}{12}\)

<=> \(\frac{x}{4}+\frac{5}{4}-\frac{2x}{3}+1=\frac{6x}{8}-\frac{1}{8}+\frac{2x}{12}-\frac{1}{12}\)

<=> \(-\frac{4}{3}x=-\frac{59}{24}\)

<=> \(x=\frac{59}{32}\)

Vậy S = { 59/32}

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

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

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

<=> \(-x=-8\)

<=> x = 8 

Vậy S = { 8 }