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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) \(\left(x^4\right)^2=\dfrac{x^{12}}{x^5}\\ x^8=x^7\\ \Rightarrow x=1;x=-1\)
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 \(x=\frac{1}{9}\)
\(\left(3x-1\right)^3=\left(\frac{2}{3}\right)^3\)
=> 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 ^^
\(a.\frac{x-1}{x+2}=\frac{4}{5}\)
\(\Rightarrow\frac{x+2-3}{x+2}=\frac{4}{5}\)
\(\Rightarrow1-\frac{3}{x+2}=\frac{4}{5}\)
\(\Rightarrow\frac{3}{x+2}=1-\frac{4}{5}\)
\(\Rightarrow\frac{3}{x+2}=\frac{1}{5}\)
\(\Rightarrow\frac{3}{x+2}=\frac{3}{15}\Rightarrow x+2=15\)
\(\Rightarrow x=13\)( thỏa mãn )
a)Ta có:
\(\frac{x-1}{x+2}=\frac{4}{5}\Leftrightarrow5\left(x-1\right)=4\left(x+2\right)\)
\(\Leftrightarrow5x-5=4x+8\)
\(\Leftrightarrow5x-4x=8+5\)
\(\Leftrightarrow x=13\)
b)Ta có:
\(2^{2x+1}+4^{x+3}=2^{2x+1}+2^{2x+6}=2^{2x+1}\left(1+2^5\right)=2^{2x+1}.33=264\Leftrightarrow2^{2x+1}=8=2^3\)\(\Rightarrow2x+1=3\Leftrightarrow2x=2\Leftrightarrow x=1\)
c)Ta có:
\(\frac{x^2}{-8}=\frac{27}{x}\Leftrightarrow x^3=-8.27=-216\Leftrightarrow x=-6\)
d)Ta có:
\(\frac{x+7}{-20}=\frac{-5}{x+7}\Leftrightarrow\left(x+7\right)^2=\left(-20\right)\left(-5\right)=100\Leftrightarrow\left[{}\begin{matrix}x+7=10\\x+7=-10\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-17\end{matrix}\right.\)e)Ta có:
\(\frac{x}{-8}=\frac{2}{-x^3}\Leftrightarrow x.\left(-x^3\right)=-8.2\)
\(\Leftrightarrow-x^4=-16\Leftrightarrow x^4=16\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)
X bằng căn bậc 3 của 25
X^8=(X^4)^2 và (X^4)^2=X^12/X^5 x#0
Nên X^10=25.(X^4)^2=25.X^12/X^5
=> X^10.X^5/X^12=25
X^3=25
X bằng căn bậc 3 của 25