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a: \(=2^2\cdot9\cdot\dfrac{1}{6\cdot9}\cdot\dfrac{4^2}{9^2}=\dfrac{2^2}{6}\cdot\dfrac{2^4}{3^4}=\dfrac{2^6}{2\cdot3\cdot3^4}=\dfrac{2^5}{3^5}=\left(\dfrac{2}{3}\right)^5\)
b: \(=2^8\cdot\dfrac{3^4}{2^4}=3^4\cdot2^4=6^4\)
c: \(=\dfrac{\left(\dfrac{1}{2}\right)^3\cdot2^3\cdot\left(\dfrac{1}{2}\right)^2}{\left(-8\right)^2\cdot16}\cdot2^6=\dfrac{\dfrac{1}{2^2}}{64\cdot16}\cdot64=\dfrac{1}{4}:16=\dfrac{1}{64}=\left(\dfrac{1}{8}\right)^2\)
a: \(=2^2\cdot9\cdot\dfrac{1}{3^3\cdot2}\cdot\dfrac{2^4}{3^4}=\dfrac{2^4\cdot2^2}{2}\cdot\dfrac{9}{3^3\cdot3^4}=\dfrac{2^5}{3^5}=\left(\dfrac{2}{3}\right)^5\)
b: \(=2^8\cdot\dfrac{3^4}{2^4}=3^4\cdot2^4=6^4\)
c: \(=\dfrac{\dfrac{1}{2^3}\cdot\dfrac{1}{2^2}\cdot8}{\left(-8\right)^2\cdot2^4}\cdot2^6=\dfrac{1}{2^2}\cdot2^6:2^{10}=\dfrac{2^4}{2^{10}}=\dfrac{1}{2^6}=\left(\dfrac{1}{8}\right)^2\)
a: \(=\dfrac{3^3\cdot2^6}{3^{-4}\cdot2^6}=3^7\)
b: \(=\left(\dfrac{3}{7}\right)^5\cdot\left(\dfrac{3}{7}\right)\cdot\dfrac{5^6}{3^6}:\left(\dfrac{625}{343}\right)^2\)
\(=\dfrac{3^6}{7^6}\cdot\dfrac{5^6}{3^6}:\dfrac{5^8}{7^6}\)
\(=\dfrac{1}{5^2}\)
c: \(=5^{4+3}\cdot\left(\dfrac{5}{2}\right)^{-5}\cdot\dfrac{1}{25}\)
\(=5^5\cdot\left(\dfrac{2}{5}\right)^5=2^5\)
làm bài 3 BĐT
theo bảng xét dấu
còn bài 1,2 ở trên là 1.1 và 1.2 đều trg bài 1.2
bài 1.2 (tức bài 2 ở trên )làm a,b,c,d
\còn bài 2( tức bài 2 ở trên) làm hết
a: \(=\dfrac{3^3\cdot2^6}{3^{-4}\cdot2^6}=3^7\)
b: \(=\left(\dfrac{3}{7}\cdot\dfrac{5}{3}\right)^6\cdot\dfrac{5}{3}\cdot\dfrac{3}{7}:\left(\dfrac{7^3}{5^4}\right)^{-2}\)
\(=\left(\dfrac{5}{7}\right)^6\cdot\dfrac{5}{7}\cdot\left(\dfrac{5}{7}\right)^6\cdot5^2\)
\(=\left(\dfrac{5}{7}\right)^{13}\cdot5^2\)
c: \(=5^4\cdot2.5^{-5}\cdot125\cdot0.04\)
\(=5^4\cdot5\cdot\left(\dfrac{5}{2}\right)^{-5}\)
\(=5^5\cdot\left(\dfrac{2}{5}\right)^5=2^5\)
1.Tính
a.\(\dfrac{7}{23}\left[(-\dfrac{8}{6})-\dfrac{45}{18}\right]=\dfrac{7}{23}.-\dfrac{12}{6}=-\dfrac{7}{6}\)
b.\(\dfrac{1}{5}\div\dfrac{1}{10}-\dfrac{1}{3}(\dfrac{6}{5}-\dfrac{9}{4})=2-(-\dfrac{7}{20})=\dfrac{47}{20}\)
c.\(\dfrac{3}{5}.(-\dfrac{8}{3})-\dfrac{3}{5}\div(-6)=-\dfrac{3}{2}\)
d.\(\dfrac{1}{2}.(\dfrac{4}{3}+\dfrac{2}{5})-\dfrac{3}{4}.(\dfrac{8}{9}+\dfrac{16}{3})=-\dfrac{19}{5}\)
e.\(\dfrac{6}{7}\div(\dfrac{3}{26}-\dfrac{3}{13})+\dfrac{6}{7}.(\dfrac{1}{10}-\dfrac{8}{5})=-\dfrac{61}{7}\)
Bài 2
a.\(1^2_5x+\dfrac{3}{7}=\dfrac{4}{5}\)
\(x=\dfrac{13}{49}\)
b.\(\left|x-1,5\right|=2\)
Xảy ra 2 trường hợp
TH1
\(x-1,5=2\)
\(x=3,5\)
TH2
\(x-1,5=-2\)
\(x=-0,5\)
Vậy \(x=3,5\) hoặc \(x=-0,5\) .
Ngại làm quá trời ơi,lần sau bn tách ra nhá làm vậy mỏi tay quá.
câu 1 \(A=\dfrac{3^2}{5^2}.5^2-\dfrac{9^3}{4^3}:\dfrac{3^3}{4^3}+\dfrac{1}{2}\)
\(A=\dfrac{3^2}{5^2}.5^2-\dfrac{\left(3^2\right)^3}{4^3}.\dfrac{4^3}{3^3}+\dfrac{1}{2}\)
\(A=\dfrac{3^2}{5^2}.5^2-\dfrac{3^6}{4^3}.\dfrac{4^3}{3^3}+\dfrac{1}{2}=3^2-3^3+\dfrac{1}{2}=-18+\dfrac{1}{2}=-\dfrac{35}{2}\)
\(B=\left[\dfrac{4}{11}+\dfrac{7}{22}.2\right]^{2010}-\left(\dfrac{1}{2^2}.\dfrac{4^4}{8^2}\right)^{2009}\)
\(B=\left[\dfrac{4}{11}+\dfrac{7}{11}\right]^{2010}-\left(\dfrac{1}{2^2}.\dfrac{\left(2^2\right)^4}{\left(2^3\right)^2}\right)^{2009}\)
\(B=1^{2010}-\left(\dfrac{1}{2^2}.\dfrac{2^8}{2^6}\right)^{2009}\)
\(B=1^{2010}-\left(\dfrac{2^8}{2^8}\right)^{2009}\)
\(B=1^{2010}-1^{2009}=1-1=0\)
câu 2
a) \(2x-\dfrac{5}{4}=\dfrac{20}{15}\)
\(\Leftrightarrow2x=\dfrac{4}{3}+\dfrac{5}{4}\)
\(\Leftrightarrow2x=\dfrac{31}{12}\)
\(\Leftrightarrow x=\dfrac{31}{24}\)
b) \(\left(x+\dfrac{1}{3}\right)^3=\left(-\dfrac{1}{2}\right)^3\)
\(\Leftrightarrow x+\dfrac{1}{3}=-\dfrac{1}{2}\)
\(\Leftrightarrow x=-\dfrac{1}{2}-\dfrac{1}{3}\)
\(\Leftrightarrow x=-\dfrac{5}{6}\)
a: \(=\left|\dfrac{3}{2}-\dfrac{7}{3}\right|^2+\dfrac{1}{4}=\dfrac{17}{18}\)
b: \(=\left|1-2-\dfrac{1}{3}\right|+\dfrac{5}{6}=1+\dfrac{1}{3}+\dfrac{5}{6}=\dfrac{13}{6}\)
c: \(=\left|\dfrac{3}{2}-\dfrac{7}{4}\right|-\dfrac{7}{4}=-\dfrac{3}{2}\)
d: =x-5+8-x=3
a) \(x+\dfrac{3}{10}=\dfrac{-2}{5}\)
\(x=\dfrac{-2}{5}-\dfrac{3}{10}\)
\(x=\dfrac{-7}{10}\)
b) \(x+\dfrac{5}{6}=\dfrac{2}{5}-\left(-\dfrac{2}{3}\right)\)
\(x+\dfrac{5}{6}=\dfrac{2}{5}+\dfrac{2}{3}\)
\(x+\dfrac{5}{6}=\dfrac{16}{15}\)
\(x=\dfrac{16}{15}-\dfrac{5}{6}\)
\(x=\dfrac{7}{30}\)
c) \(1\dfrac{2}{5}x+\dfrac{3}{7}=-\dfrac{4}{5}\)
\(\dfrac{7}{5}x+\dfrac{3}{7}=-\dfrac{4}{5}\)
\(\dfrac{7}{5}x=-\dfrac{4}{5}-\dfrac{3}{7}\)
\(\dfrac{7}{5}x=\dfrac{-43}{35}\)
\(\Rightarrow x=\dfrac{-43}{49}\)
d) \(\left[x+\dfrac{3}{4}\right]-\dfrac{1}{3}=0\)
\(\left[x+\dfrac{3}{4}\right]=0+\dfrac{1}{3}\)
\(\left[x+\dfrac{3}{4}\right]=\dfrac{1}{3}\)
\(x=\dfrac{1}{3}-\dfrac{3}{4}\)
\(x=\dfrac{-5}{12}\)
e) \(\left[x+\dfrac{4}{5}\right]-\left(-3,75\right)=-\left(-2,15\right)\)
\(\left[x+\dfrac{4}{5}\right]+3,75=2,15\)
\(x+\dfrac{4}{5}=2,15-3,75\)
\(x+\dfrac{4}{5}=-\dfrac{8}{5}\)
\(x=\dfrac{-8}{5}-\dfrac{4}{5}\)
\(x=\dfrac{-12}{5}\)
f) \(\left(x-2\right)^2=1\)
\(\Rightarrow x=1\)
Sức chịu đựng có giới hạn -.-
- Mình tiếp tục cho Nguyễn Phương Trâm nhé.
g, \(\left(2x-1\right)^3=-27\)
\(\Rightarrow\left(2x-1\right)^3=\left(-3\right)^3\)
\(\Rightarrow2x-1=-3\)
\(\Rightarrow2x=-2\)
=> \(x=-1\)
- Vậy x = -1
h,\(\dfrac{x-1}{-15}=-\dfrac{60}{x-1}\)
\(\Rightarrow\left(x-1\right)^2=-60.\left(-15\right)\)
\(\Rightarrow\left(x-1\right)^2=900 \)
\(\Rightarrow\left(x-1\right)^2=30^2\Rightarrow x-1=30\)
=> x = 31
i,\(x:\left(\dfrac{-1}{2}\right)^3=\dfrac{-1}{2}\)
=> \(x:\left(-\dfrac{1}{8}\right)=-\dfrac{1}{2}\)
\(\Rightarrow x=\dfrac{1}{16}\)
- Vậy x=\(\dfrac{1}{16}\)
j, \(\left(\dfrac{3}{4}\right)^5.x=\left(\dfrac{3}{4}\right)^7\)
\(\Rightarrow \left(\dfrac{3}{4}\right).x=\left(\dfrac{3}{4}\right)^2\)
\(\Rightarrow x=\left(\dfrac{3}{4}\right)^2:\dfrac{3}{4}\)
\(\Rightarrow x=\dfrac{3}{4}\)
- Vạy x = \(\dfrac{3}{4}\)
k, \(8^x:2^x=4\Rightarrow\left(8:2\right)^x=4\)
=>\(4^x=4\)
=> x = 1
- Vậy x = 1
1.
a) x : \(\left(\dfrac{3}{4}\right)^3\) =\(\left(\dfrac{3}{4}\right)^3\)
x = \(\left(\dfrac{3}{4}\right)^3.\left(\dfrac{3}{4}\right)^3\)
x = \(\dfrac{3}{4}^{3+3}\)
x = \(\dfrac{3}{4}^6\)
x = \(\dfrac{729}{4096}\)
b) \(\left(\dfrac{2}{5}\right)^5.x=\left(\dfrac{2}{5}\right)^8\)
x = \(\left(\dfrac{2}{5}\right)^8:\left(\dfrac{2}{5}\right)^5\)
x = \(\dfrac{2}{5}^{8-5}\)
x = \(\dfrac{2}{5}^3\)
x = \(\dfrac{8}{5}\)
2.
(0,36)\(^8\) \([\left(0,6\right)^3]^8\) = (0,6)\(^{3.8}\) = ( 0,6)\(^{24}\)
( 0,216)\(^4\) = \([\left(0,6\right)^3]^4\) = (0.6)\(^{3.4}\) = ( 0,6)\(^{12}\)
\(x:\left(\dfrac{3}{4}\right)^3=\left(\dfrac{3}{4}\right)^2\)
\(x=\left(\dfrac{3}{4}\right)^2.\left(\dfrac{3}{4}\right)^3\) <=> \(x=\left(\dfrac{3}{4}\right)^{2+3}\)
=> \(x=\left(\dfrac{3}{4}\right)^5\)
b, \(\left(\dfrac{2}{5}\right)^5.x=\left(\dfrac{2}{5}\right)^8\)
\(x=\left(\dfrac{2}{5}\right)^8:\left(\dfrac{2}{5}\right)^5\Leftrightarrow x=\left(\dfrac{2}{5}\right)^{8-5}\)
=>\(x=\left(\dfrac{2}{5}\right)^3\)
bài 2 : Với bài này ta cần áp dụng quy tắc: \(\left(x^m\right)^n=x^{m.n}\)
\(0,36^8=\left[\left(0,6\right)^2\right]^8=\left(0,6\right)^{16}\)
\(0,216^4=\left[\left(0,6\right)^3\right]^4=\left(0,6\right)^{12}\)