X : 0,125 : 8 = X : ....
A . 0 B . 1 C . 4 D . 64
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Bài 2:
a: \(\left(0.25\right)^3\cdot32=\dfrac{1}{4^3}\cdot32=\dfrac{32}{64}=\dfrac{1}{2}\)
b: \(\left(-0.125\right)^3\cdot80^4=\left(-0.125\cdot80\right)^3\cdot80=-80\)
c: \(\dfrac{8^2\cdot4^5}{2^{20}}=\dfrac{2^6\cdot2^{10}}{2^{20}}=\dfrac{1}{2^4}=\dfrac{1}{16}\)
d: \(\dfrac{81^{11}\cdot3^{17}}{27^{10}\cdot9^{15}}=\dfrac{3^{44}\cdot3^{17}}{3^{30}\cdot3^{30}}=\dfrac{3^{61}}{3^{60}}=3\)
a) \(\left|x\right|-\frac{7}{6}=\frac{9}{15}\)
=> \(\left|x\right|=\frac{9}{15}+\frac{7}{6}=\frac{53}{30}\)
=> \(\orbr{\begin{cases}x=\frac{53}{30}\\x=-\frac{53}{30}\end{cases}}\)
b) \(\left|x-\frac{4}{3}\right|=\frac{1}{6}\)
=> \(\orbr{\begin{cases}x-\frac{4}{3}=\frac{1}{6}\\x-\frac{4}{3}=-\frac{1}{6}\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{3}{2}\\x=\frac{7}{6}\end{cases}}\)
c) \(\left|x-\frac{4}{3}\right|-\frac{1}{3}=\frac{1}{2}\)
=> \(\left|x-\frac{4}{3}\right|=\frac{1}{2}+\frac{1}{3}\)
=> \(\left|x-\frac{4}{3}\right|=\frac{5}{6}\)
=> \(\orbr{\begin{cases}x-\frac{4}{3}=\frac{5}{6}\\x-\frac{4}{3}=-\frac{5}{6}\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{13}{6}\\x=\frac{1}{2}\end{cases}}\)
d) \(\frac{8}{3}-\left|\frac{7}{9}-x\right|=-\frac{1}{5}\)
=> \(\left|\frac{7}{9}-x\right|=\frac{43}{15}\)
=> \(\orbr{\begin{cases}\frac{7}{9}-x=\frac{43}{15}\\\frac{7}{9}-x=-\frac{43}{15}\end{cases}}\Rightarrow\orbr{\begin{cases}x=-\frac{94}{45}\\x=\frac{164}{45}\end{cases}}\)
e) \(\left|x-\left(\frac{1}{4}\right)^2\right|-\frac{25}{64}=0\)
=> \(\left|x-\frac{1}{16}\right|=\frac{25}{64}\)
=> \(\orbr{\begin{cases}x-\frac{1}{16}=\frac{25}{64}\\x-\frac{1}{16}=-\frac{25}{64}\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{29}{64}\\x=-\frac{21}{64}\end{cases}}\)
f) \(\left(x-\frac{1}{4}\right)^2+\frac{17}{64}=\frac{21}{32}\)
=> \(\left(x-\frac{1}{4}\right)^2=\frac{25}{64}\)
=> \(\left(x-\frac{1}{4}\right)^2=\left(\frac{5}{8}\right)^2\)
=> \(\orbr{\begin{cases}x-\frac{1}{4}=\frac{5}{8}\\x-\frac{1}{4}=-\frac{5}{8}\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{7}{8}\\x=-\frac{3}{8}\end{cases}}\)
Tính:
a) \(8^3.\left(0,125\right)^3\)
\(=\left(8.0,125\right)^3\)
\(=1^3\)
\(=1.\)
b) \(7^{200}.\left(\frac{1}{7}\right)^{200}\)
\(=\left(7.\frac{1}{7}\right)^{200}\)
\(=1^{200}\)
\(=1.\)
c) \(\left(0,25\right)^3.64\)
\(=\left(0,25\right)^3.4^3\)
\(=\left(0,25.4\right)^3\)
\(=1^3\)
\(=1.\)
a) x^4 - 2x^2 + 1 = 0
=> ( x^2 - 1 )^2 = 0
=> x^2 - 1 = 0
=> x^2 = 1
=> x = 1 hoặc x = -1
a) x4-2x2+1=0
(thang Tran giải rồi nhé)
b) x4-2x2-8=0
<=> x^4 - 2x^2 +1 -9 =0
<=> (x^2 -1)^2 -9 =0
\(\Leftrightarrow\orbr{\begin{cases}x^2-1=-3\\x^2-1=3\end{cases}\Leftrightarrow\orbr{\begin{cases}x^2=-2\left(VN\right)\\x=+_-\sqrt{2}\end{cases}}}\)
Vậy x=+- căn 2
c) x4-4x2-60=0
\(\Leftrightarrow x^4-4x^2+4-64=0\)
\(\Leftrightarrow\left(x^2-2\right)-64=0\)
\(\Leftrightarrow\left(x^2+62\right)\left(x^2-66\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x^2+62=0\\x^2-66=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x^2=-62\left(VN\right)\\x^2=+_-\sqrt{66}\end{cases}}}\)
Vậy x=+- căn 66
d) x6-16x2+64=0
B
Nhớ bấm L I K E cho mk nhá :))))