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THEO ĐỀ BÀI TA CÓ
1^2+2^2+3^2+...+10^2=385
MÀ 2^2+4^2+....+20^2=2(1^2+2^2+....+10^2)=2.385=770
VẬY 2^2+2^4+....+20^2=770
a) \(\frac{4}{3}-\frac{2}{5}\)
\(=\frac{20}{15}-\frac{6}{15}=\frac{14}{15}\)
b) \(\left|-\frac{1}{10}\right|-\left(-\frac{1}{3}\right)^2\div\frac{5}{9}\)
\(=\frac{1}{10}-\frac{1}{9}\cdot\frac{9}{5}\)
\(=\frac{1}{10}-\frac{1}{5}=\frac{1}{10}-\frac{2}{10}\)
\(=-\frac{1}{10}\)
c) Đề bài có vấn đề!!!
d) \(\left(-0,2\right)^2\cdot5-8^2\cdot\frac{9^4}{3^7}\cdot4^3\)
\(=0,04\cdot5-64\cdot\frac{\left(3^2\right)^4}{3^7}\cdot64\)
\(=0,2-4096\cdot\frac{3^8}{3^7}=0,2-4096\cdot3\)
\(=0,2-12288=-128878\)
a) \(\frac{0,5}{0,2}=\frac{1,25}{0,1x}\Leftrightarrow0,1x.0,5=0,2.1,25\)
\(\Leftrightarrow0,1x.0,5=0,25\Leftrightarrow0,1x=0,5\Leftrightarrow x=5\)
b) \(x-\frac{3}{2}=2x-\frac{4}{3}\Leftrightarrow x-2x=\frac{-4}{3}+\frac{3}{2}\)
\(\Leftrightarrow x-2x=\frac{1}{6}\Leftrightarrow-x=\frac{1}{6}\Leftrightarrow x=\frac{-1}{6}\)
c) \(x+\frac{13}{14}=\frac{4}{7}\Rightarrow x=\frac{4}{7}-\frac{13}{14}\Rightarrow x=\frac{-5}{14}\)
d)\(-3\left(x-2\right)=2x+1\)
\(\Leftrightarrow-3x+6=2x+1\Leftrightarrow-3x-2x=1-6\)
\(\Leftrightarrow-5x=-5\Leftrightarrow x=1\)
e) \(\left(x-1\right)^2-4=0\Leftrightarrow\left(x-1\right)^2=4\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=2\\x-1=\left(-2\right)\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=3\\x=-1\end{cases}}\)
cậu có thể tham khảo bài trên ạ, nếu thấy đúng thì cho mk 1 t.i.c.k ạ, thank nhiều
\(d,-3\left(x-2\right)=2x+1\)
\(< =>-3x+6=2x+1\)
\(< =>-3x-2x+6-1=0\)
\(< =>5-5x=0\)
\(< =>5\left(1-x\right)=0< =>x=1\)
\(e,\left(x-1\right)^2-4=0\)
\(< =>\left(x-1+2\right)\left(x-1-2\right)=\left(x+1\right)\left(x-3\right)=0\)
\(< =>\orbr{\begin{cases}x+1=0\\x-3=0\end{cases}< =>\orbr{\begin{cases}x=-1\\x=3\end{cases}}}\)
1)Tính:
a)\(4^2\cdot2=\left(2^2\right)^2\cdot2=2^4\cdot2=2^5=32\)
b)\(36^2:6^2=\left(36:6\right)^2=6^2=48\)
c)\(\left(\frac{2}{5}\right)^{10}:\left(\frac{4}{25}\right)^2=\left(\frac{2}{5}\right)^{10}\cdot\left(\frac{25}{4}\right)^2=\)\(\left(1\right)^{10}\cdot\left(\frac{5}{2}\right)^2=1\cdot\frac{5^2}{2^2}=1\cdot\frac{25}{4}=\frac{25}{4}\)
a
\(4^2.2=16.2=32\)
b\(36^2:6^2=36.36:6.6=36.36:36=36\)
c
a) \(\left(\frac{2}{3}\right)^3=\frac{8}{27}\)
b) \(\left(-\frac{2}{3}\right)^3=-\frac{8}{27}\)
c) \(\left(-1\frac{3}{4}\right)^2=\left(-\frac{7}{4}\right)^2=\frac{49}{16}\)
Bài 8:
a: \(\left(\dfrac{2}{5}+\dfrac{3}{4}\right)^2=\left(\dfrac{8+15}{20}\right)^2=\left(\dfrac{23}{20}\right)^2=\dfrac{529}{400}\)
b: \(\left(\dfrac{5}{4}-\dfrac{1}{6}\right)^2=\left(\dfrac{15}{12}-\dfrac{2}{12}\right)^2=\left(\dfrac{13}{12}\right)^2=\dfrac{169}{144}\)
a) \(\left(\frac{-1}{3}\right)^4=\frac{\left(-1\right)^4}{3^4}=\frac{1}{81}\)
b) \(\left(-2\frac{1}{4}\right)^3=\left(\frac{-9}{4}\right)^3=\frac{\left(-9\right)^3}{4^3}=\frac{-729}{64}\)
c) \(\left(-0,2\right)^2=\left(\frac{-1}{5}\right)^2=\frac{\left(-1\right)^2}{5^2}=\frac{1}{25}\)
d) \(\left(-5,3\right)^0=1\)
a)\(\left(\frac{-1}{3}\right)^4=\frac{1}{81}\)
b) \(\left(-2\frac{1}{4}\right)^3=\frac{-729}{64}\)
c) \(\left(-0,2\right)^2=\frac{1}{25}\)
d) \(\left(-5,3\right)^0=1\)
Cbht