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Giải
Tìm x:
a)\(\left(x-2\right)^2=1\Leftrightarrow\left(x-2\right)^2=1^2.\)
\(\Rightarrow\orbr{\begin{cases}x-2=1\Rightarrow x=1+2=3\\x-2=-1\Rightarrow x=-1+2=1\end{cases}}\)
=> Vậy \(x=\orbr{\begin{cases}3\\1\end{cases}}\)
b) \(\left(2x-1\right)^3=-8\Leftrightarrow\left(2x-1\right)^3=\left(-2\right)^3\)
\(\Rightarrow\left(2x-1\right)=-2\Rightarrow2x=-2+1=-1\)
\(\Rightarrow x=-1:2=-\frac{1}{2}\)
Vậy \(x=-\frac{1}{2}\)
c) \(\left(x+\frac{1}{2}\right)^2=\frac{1}{16}\Leftrightarrow\left(x+\frac{1}{2}\right)^2=\orbr{\begin{cases}\left(-\frac{1}{4}\right)^2\\\left(\frac{1}{4}\right)^2\end{cases}}\)
\(\Rightarrow\left(x+\frac{1}{2}\right)=\orbr{\begin{cases}-\frac{1}{4}\\\frac{1}{4}\end{cases}}\Rightarrow\orbr{\begin{cases}x+\frac{1}{2}=-\frac{1}{4}\Rightarrow x=-\frac{1}{4}-\frac{1}{2}=-\frac{3}{4}\\x+\frac{1}{2}=\frac{1}{4}\Rightarrow x=\frac{1}{4}-\frac{1}{2}=-\frac{1}{4}\end{cases}}\)
Vậy \(x=-\frac{3}{4};-\frac{1}{4}\)
BT2:
Giải
a) \(9.3^3.\frac{1}{81}.3^2=3^2.3^3.\left(\frac{1}{3}\right)^4.3^2=\left(3^2.3^3.3^2\right).\left(\frac{1}{3}\right)^4\)
\(=3^{2+3+2}.\left(\frac{1}{3}\right)^4=3^7.\left(\frac{1}{3}\right)^4=\frac{3^7.1^4}{1.3^4}=3^3\)
b) \(4.2^5:\left(2^3.\frac{1}{16}\right)=2^2.2^5:\left(2^3.\left(\frac{1}{2}\right)^4\right)=2^{2+5}:\left(\frac{2^3.1^4}{2^4}\right)\)
\(=2^7:\left(\frac{1}{2}\right)=2^7.\frac{2}{1}=2^8\)
c) Chị đang nghĩ...
Câu 1 a)x+1/3=1/4
x= -1/12
b) x-3=1
x=4
c)3x-1=-4
3x=-3
x=-1
Câu 2:
a) \(9.3^3.\frac{1}{81}.3^2=3^2.3^3.\frac{1}{3^4}.3^2=3^7.\frac{1}{3^4}=3^3\)
b) \(4.2^5:\left(2^3.\frac{1}{16}\right)=2^2.2^5:2^3:\frac{1}{16}=2^7:2^3.16=2^4.2^4=2^8\)
c) \(3^2.2^5.\left(\frac{2}{3}\right)^2=3^2.2^5.\frac{2^2}{3^2}=2^5.2^2=2^7\)
d) \(\left(\frac{1}{3}\right)^2.\frac{1}{3}.9^2=\left(\frac{1}{3}\right)^3.\left(3^2\right)^2=\frac{1^3}{3^3}.3^4=1^3.3=3^1\)
Câu 1:
a) 2225 và 3150
Ta có:2225=(29)25=51225
3150=(36)25=72925
Vì 51225<72925
Suy ra: 2225<3150
Câu 2:
a)\(25^3:5^2=\left(5^2\right)^3:5^2=5^6:5^2=5^4\)
b)\(\left(\frac{3}{7}\right)^{21}:\left(\frac{9}{49}\right)^6=\left(\frac{3}{7}\right)^{21}:\left[\left(\frac{3}{7}\right)^2\right]^6=\left(\frac{3}{7}\right)^{21}:\left(\frac{3}{7}\right)^{12}=\left(\frac{3}{7}\right)^9\)
c)\(3-\left(-\frac{6}{7}\right)^0+\left(\frac{1}{2}\right)^2:2=3+\frac{1}{4}:2=3+\frac{1}{8}=\frac{25}{8}\)
Câu 3:
a)\(9.3^3.\frac{1}{81}.3^2=3^2.3^3.3^2.\left(\frac{1}{3^4}\right)=3^7:3^4=3^3\)
b)\(4.2^5:\left(2^3.\frac{1}{16}\right)=2^2.2^5:\left(2^3.\frac{1}{2^4}\right)=2^7:\frac{1}{2}=2^8\)
c)\(3^2.2^5.\left(\frac{2}{3}\right)^2=288.\frac{4}{9}=2^7\)
d)\(\left(\frac{1}{3}\right)^3.\frac{1}{3}.9^2=\left(\frac{1}{3}\right)^4.\left(3^2\right)^2=3^4.\left(\frac{1}{3}\right)^4=3^4:3^4=1\)
a: \(=3^2\cdot3^3\cdot3^{-4}\cdot3^2=3^{2+3-4+2}=3^3\)
b: \(=2^2\cdot2^5:\left(2^3\cdot\dfrac{1}{2^4}\right)=2^7:\dfrac{1}{2}=2^8\)
c: \(=9\cdot32\cdot\dfrac{4}{9}=128=2^7\)
d: \(=\dfrac{1}{27}\cdot3^4=3^1\)
a) 9 . 33 . 1/81 . 32
=32.33.1/34.32
=33
b) 4 . 25 : (23 . 1/16)
=22.25:(23.1/24)
=22.25:1/2
=22.24
=22+4
=26
c) 32 . 25 . (2/3)2
=32.25.22/32
=25.22
=25+2
=27
d) (1/3)2 . 1/3 . 92
=1/32.1/3.(32)2
=1/32.1/3.34
=1/32.33
=3
=31
Bài làm:
Bài 1
a) \(\left(x-\frac{1}{2}\right)^2=0\)
\(\rightarrow\left(x-\frac{1}{2}\right)^2=0^2\)
\(\rightarrow x-\frac{1}{2}=0\)
\(\Rightarrow x=\frac{1}{2}\)
Bài 2
a) \(25^3\div5^2=\left(5^2\right)^3\div5^2=5^6\div5^2=5^4\)
b) \(\left(\frac{3}{7}\right)^{21}\div\left(\frac{9}{49}\right)^6=\left(\frac{3}{7}\right)^{21}\div\left[\left(\frac{3}{7}\right)^2\right]^6=\left(\frac{3}{7}\right)^{21}\div\left(\frac{3}{7}\right)^{12}=\left(\frac{3}{7}\right)^9\)
c) \(3-\left(\frac{-6}{7}\right)^0+\left(\frac{1}{2}\right)^2\div2=3-1+\frac{1}{4}\times\frac{1}{2}=2+\frac{1}{8}=\frac{17}{8}\)
Bài 3
a) \(9\times3^3\times\frac{1}{81}\times3^2=3^2\times3^3\times\frac{1}{3^4}\times3^2=3^3\)
b) \(4\times2^5\div\left(2^3\times\frac{1}{16}\right)=2^2\times2^5\div\left(2^3\times\frac{1}{2^4}\right)=2^7\div\frac{1}{2}=2^6\)
c) \(3^2\times2^5\times\left(\frac{2}{3}\right)^2=3^2\times2^5\times\frac{2^2}{3^2}=3^2\times\frac{2^7}{3^2}=2^7\)
d) \(\left(\frac{1}{3}\right)^2\times\frac{1}{3}\times9^2=\left(\frac{1}{3}\right)^3\times3^4=\frac{1}{3^3}\times3^4=3^1\)
Các bạn giải từng bước ra cho mình nhé, cảm ơn các bạn