[ 0,( 3 )]\(^2\) - \(\dfrac{81}{82}\)+ 2
3- \(\dfrac{1}{49}\)+ [0,( 142857)]\(^2\)
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Những câu hỏi liên quan
7 tháng 9 2023
a) \(5^6:5^5+\left(\dfrac{4}{9}\right)^0=5^{6-5}+1=5+1=6\)
b) \(\left(\dfrac{3}{7}\right)^{21}:\left(1-\dfrac{40}{49}\right)^3\)
\(=\left(\dfrac{3}{7}\right)^{21}:\left(\dfrac{9}{49}\right)^3\)
\(=\left(\dfrac{3}{7}\right)^{21}:\left[\left(\dfrac{3}{7}\right)^2\right]^3\)
\(=\left(\dfrac{3}{7}\right)^{21}:\left(\dfrac{3}{7}\right)^6\)
\(=\left(\dfrac{3}{7}\right)^{21-6}=\left(\dfrac{3}{7}\right)^{15}\)
c) \(\left(\dfrac{2}{3}\right)^3-\left(\dfrac{-52}{3}\right)^0+\dfrac{4}{9}\)
\(=\dfrac{8}{27}-1+\dfrac{4}{9}\)
\(=\dfrac{8-27+12}{27}=-\dfrac{7}{27}\)
7 tháng 9 2023
\(a)5^6:5^5+\left(\dfrac{4}{9}\right)^0=5^1+1=6\)
\(b,\left(\dfrac{3}{7}\right)^{21}:\left(1-\dfrac{40}{49}\right)^3\)
\(=\left(\dfrac{3}{7}\right)^{21}:\left(\dfrac{49-40}{49}\right)^3\)
\(=\left(\dfrac{3}{7}\right)^{21}:\left(\dfrac{9}{49}\right)^3=\left(\dfrac{3}{7}\right)^{21}:[\left(\dfrac{3}{7}\right)^2]^3\)
\(=\left(\dfrac{3}{7}\right)^{21}:\left(\dfrac{3}{7}\right)^6=\left(\dfrac{3}{7}\right)^{21-6}\)
\(=\left(\dfrac{3}{7}\right)^{15}\)
\(c,3.\left(\dfrac{2}{3}\right)^3-\left(\dfrac{-52}{3}\right)^0+\dfrac{4}{9}\)
\(=3.\dfrac{8}{27}-1+\dfrac{4}{9}\)
\(=\dfrac{8}{9}-1+\dfrac{4}{9}\)
\(=\dfrac{8-9+4}{9}=\dfrac{1}{3}\)
H
1 tháng 3 2023
`(2/3 x +1/2) (-2x+3)=0`
\(\Rightarrow\left[{}\begin{matrix}\dfrac{2}{3}x+\dfrac{1}{2}=0\\-2x+3=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\dfrac{2}{3}x=-\dfrac{1}{2}\\-2x=-3\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}.\dfrac{3}{2}\\x=\dfrac{3}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{3}{4}\\x=\dfrac{3}{2}\end{matrix}\right.\)
NH
1 tháng 3 2023
\(\left(\dfrac{2}{3}x+\dfrac{1}{2}\right)\cdot\left(-2x+3\right)=0\\ =>\left[{}\begin{matrix}\dfrac{2}{3}x+\dfrac{1}{2}=0\\-2x+3=0\end{matrix}\right.\\ =>\left[{}\begin{matrix}\dfrac{2}{3}x=-\dfrac{1}{2}\\-2x=-3\end{matrix}\right.\\ =>\left[{}\begin{matrix}x=-\dfrac{3}{4}\\x=\dfrac{3}{2}\end{matrix}\right.\)
AT
14 tháng 7 2021
\(25\sqrt{\dfrac{x-3}{25}}-7\sqrt{\dfrac{4x-12}{9}}-7\sqrt{x^2-9}+18\sqrt{\dfrac{9x^2-81}{81}}=0\left(x\ge3\right)\)
\(=25\sqrt{\dfrac{1}{25}.\left(x-3\right)}-7\sqrt{\dfrac{4}{9}.\left(x-3\right)}-7\sqrt{x^2-9}+18\sqrt{\dfrac{1}{9}.\left(x^2-9\right)}=0\)
\(=5\sqrt{x-3}-\dfrac{14}{3}\sqrt{x-3}-7\sqrt{x^2-9}+6\sqrt{x^2-9}=0\)
\(\Rightarrow\dfrac{1}{3}\sqrt{x-3}-\sqrt{\left(x-3\right)\left(x+3\right)}=0\Rightarrow\sqrt{x-3}-3\sqrt{\left(x-3\right)\left(x+3\right)}=0\)
\(\Rightarrow\sqrt{x-3}\left(1-3\sqrt{x+3}\right)=0\Rightarrow\left[{}\begin{matrix}\sqrt{x-3}=0\\1=3\sqrt{x+3}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{26}{9}\left(l\right)\end{matrix}\right.\)
13 tháng 10 2021
a) \(\Rightarrow\left(\dfrac{1}{2}x-\dfrac{1}{3}\right)^2=\dfrac{4}{25}\)
\(\Rightarrow\left[{}\begin{matrix}\dfrac{1}{2}x-\dfrac{1}{3}=\dfrac{2}{5}\\\dfrac{1}{2}x-\dfrac{1}{3}=-\dfrac{2}{5}\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{22}{15}\\x=-\dfrac{2}{15}\end{matrix}\right.\)
b) \(\Rightarrow\left(1-\dfrac{1}{4}x\right)^2=\dfrac{121}{49}\)
\(\Rightarrow\left[{}\begin{matrix}1-\dfrac{1}{4}x=\dfrac{11}{7}\\1-\dfrac{1}{4}x=-\dfrac{11}{7}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{16}{7}\\x=\dfrac{72}{7}\end{matrix}\right.\)
NV
Nguyễn Việt Lâm
Giáo viên
7 tháng 2 2022
\(VT=\dfrac{a^2}{b+ab^2c}+\dfrac{b^2}{b+abc^2}+\dfrac{c^2}{c+a^2bc}\ge\dfrac{\left(a+b+c\right)^2}{a+b+c+abc\left(a+b+c\right)}=\dfrac{9}{3+3abc}\)
\(VT\ge\dfrac{9}{3+\dfrac{\left(a+b+c\right)^3}{9}}=\dfrac{3}{2}\)
Dấu "=" xảy ra khi \(a=b=c=1\)
T
9 tháng 8 2023
a) \(\dfrac{49}{81}=\dfrac{7^x}{9^x}\)(sửa đề)
\(\Leftrightarrow\left(\dfrac{7}{9}\right)^2=\left(\dfrac{7}{9}\right)^x\)\(\Rightarrow x=2\)
b) \(\dfrac{-64}{343}=\left(-\dfrac{4^x}{7^x}\right)\)(sửa đề)
\(\Leftrightarrow\left(-\dfrac{4}{7}\right)^3=\left(-\dfrac{4}{7}\right)^x\) \(\Rightarrow x=3\)
c) \(\dfrac{9}{144}=\dfrac{3^x}{12^x}\)(sửa đề)
\(\Leftrightarrow\left(\dfrac{3}{12}\right)^2=\left(\dfrac{3}{12}\right)^x\Rightarrow x=2\)
d) \(-\dfrac{1}{32}=\left(-\dfrac{1^x}{2^x}\right)\)(sửa đề)
\(\Leftrightarrow\left(-\dfrac{1}{2}\right)^5=\left(-\dfrac{1}{2}\right)^x\Rightarrow x=5\)
Mong bạn xem lại đề bài.
đùa à bất kì số thập phân hữu hạn nào luỹ thừa lên cũng = 0 nên nó rất dễ giải:
a) \(\left[0,\left(3\right)\right]^2-\dfrac{81}{82}+2\)
= \(0-\dfrac{81}{82}+2\)
= \(\left(0+2\right)-\dfrac{81}{82}\)
= \(2-\dfrac{81}{82}\)
= \(\dfrac{83}{82}\)
b) \(3-\dfrac{1}{49}+\left[0,\left(142857\right)\right]^2\)
= \(3-\dfrac{1}{49}+0\)
= \(\left(3+0\right)-\dfrac{1}{49}\)
= \(3-\dfrac{1}{49}\)
= \(\dfrac{146}{49}\)
mình rút gọn luôn đó nhe
a)
\(\left[0,\left(3\right)\right]^2-\dfrac{81}{82}+2\\ =\dfrac{1}{9}-\dfrac{81}{82}+2\\ =\dfrac{82}{738}-\dfrac{729}{738}+\dfrac{1479}{738}\\ =\dfrac{82-729+1479}{738}\\ =\dfrac{832}{738}\\ \approx1,13\)
b)
\(3-\dfrac{1}{49}+\dfrac{1}{7}\\ =\dfrac{147}{49}-\dfrac{1}{49}+\dfrac{7}{49}\\ =\dfrac{147-1+7}{49}\\ =\dfrac{153}{49}\\ \approx3,12\)