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a)
\(\begin{array}{l}x:{\left( {\frac{{ - 1}}{2}} \right)^3} = - \frac{1}{2}\\x = - \frac{1}{2}.{\left( {\frac{{ - 1}}{2}} \right)^3}\\x = {\left( {\frac{{ - 1}}{2}} \right)^4}\\x = \frac{1}{{16}}\end{array}\)
Vậy \(x = \frac{1}{{16}}\).
b)
\(\begin{array}{l}x.{\left( {\frac{3}{5}} \right)^7} = {\left( {\frac{3}{5}} \right)^9}\\x = {\left( {\frac{3}{5}} \right)^9}:{\left( {\frac{3}{5}} \right)^7}\\x = {\left( {\frac{3}{5}} \right)^2}\\x = \frac{9}{{25}}\end{array}\)
Vậy \(x = \frac{9}{{25}}\).
c)
\(\begin{array}{l}{\left( {\frac{{ - 2}}{3}} \right)^{11}}:x = {\left( {\frac{{ - 2}}{3}} \right)^9}\\x = {\left( {\frac{{ - 2}}{3}} \right)^{11}}:{\left( {\frac{{ - 2}}{3}} \right)^9}\\x = {\left( {\frac{{ - 2}}{3}} \right)^2}\\x = \frac{4}{9}.\end{array}\)
Vậy \(x = \frac{4}{9}\).
d)
\(\begin{array}{l}x.{\left( {0,25} \right)^6} = {\left( {\frac{1}{4}} \right)^8}\\x.{\left( {\frac{1}{4}} \right)^6} = {\left( {\frac{1}{4}} \right)^8}\\x = {\left( {\frac{1}{4}} \right)^8}:{\left( {\frac{1}{4}} \right)^6}\\x = {\left( {\frac{1}{4}} \right)^2}\\x = \frac{1}{{16}}\end{array}\)
Vậy \(x = \frac{1}{{16}}\).
Ta có: \(A=\left[6.\left(-\frac{1}{3}\right)^2-3.\left(-\frac{1}{3}\right)+1\right]:\left(\frac{-1}{3}-1\right)\)
\(=\left(6.\frac{1}{9}-\left(-1\right)+1\right):\left(\frac{-4}{3}\right)\)
\(=\left(\frac{2}{3}+2\right).\left(\frac{-3}{4}\right)\)
\(=\frac{8}{3}.\left(-\frac{3}{4}\right)\)
\(=-2\)
\(B=\left(729-1^3\right)\left(729-3^3\right)...\left(729-125^3\right)\)
\(\Rightarrow B=\left(729-1^3\right)\left(729-3^3\right)...\left(729-9^3\right)...\left(729-125^3\right)\)
\(\Rightarrow B=\left(729-1^3\right)\left(729-3^3\right)...0...\left(729-125^3\right)\)
\(\Rightarrow B=0\)
Vì -2 < 0 nên A < B
Vậy A < B
Câu a) số lớn lắm
b) \(3^{-3}\cdot3^5\cdot3^x=3^8\)
=> \(\frac{1}{27}\cdot3^5\cdot3^x=3^8\)
=> \(\frac{1}{27}\cdot3^x=3^3\)
=> \(3^x=3^3:\frac{1}{27}=3^3:\left(\frac{1}{3}\right)^3=3^3:\frac{1^3}{3^3}=3^3\cdot3^3=3^6\)
=> x = 6
b) \(\left(7x+2\right)^{-1}=3^{-2}\)
=> \(\frac{1}{7x+2}=\frac{1}{9}\)
=> 7x + 2 = 9
=> 7x = 7
=> x = 1
Bài 2:
a) \(3^4\cdot\frac{1}{729}\cdot81^3\cdot\frac{1}{9^2}\)
\(=3^4\cdot\left(\frac{1}{3}\right)^6\cdot\left(3^4\right)^3\cdot\left(\frac{1}{3}\right)^4\)
\(=3^4\cdot\left(\frac{1}{3}\right)^6\cdot3^{12}\cdot\left(\frac{1}{3}\right)^4=3^{16}\cdot\left(\frac{1}{3}\right)^{10}=\frac{3^{16}}{3^{10}}=3^6\)
b) \(\left(8\cdot2^5\right):\left(2^4\cdot\frac{1}{32}\right)=\left(2^3\cdot2^5\right):\left(2^4\cdot\left(\frac{1}{2}\right)^5\right)\)
\(=2^8:\left(2^4\cdot\frac{1^5}{2^5}\right)=2^8:\left(\frac{2^4}{2^5}\right)=2^8:2^{-1}=512\)
c) \(12^8\cdot9^{12}=\left(2^2\cdot3\right)^8\cdot\left(3^2\right)^{12}=2^{16}\cdot3^8\cdot3^{24}=2^{16}\cdot3^{32}\)
d) Tương tự
a)(x-2)^4=4^4
=(x-2)=4 sr x=4+2
sr x=6
xin lỗi nha mình chỉ biết làm bài a thôi mong cậu thông cảm và kb với mình nhé
Mình biết bài nào làm bài đó thôi nhé
a) (x-2)4 = 256
=> x-2 = 4
x = 4+2
x = 6
b) \(\frac{x^4}{256}=81\)
\(\Rightarrow\frac{x^4}{4^4}=3^4\)
Từ đây có thể làm theo 2 cách khác nhau
C1 : \(\frac{x^4}{4^4}=81\)
\(\Rightarrow\left(\frac{x}{4}\right)^4=81\)
\(\Rightarrow\frac{x}{4}=3\)
x = 3.4
x = 12
C2 : \(\frac{x^4}{4^4}=3^4\)
=> x4 = 34.44
x4 = (3.4)4
x4 = 124
<=> x = 4
c) \(125.\left(x+\frac{4}{5}\right)^3=729\)
\(\left(x+\frac{4}{5}\right)^3=729:125\)
\(\left(x+\frac{4}{5}\right)^3=5,832\)
\(\Rightarrow x+\frac{4}{5}=1,8=\frac{9}{5}\)
\(x=\frac{9}{5}-\frac{4}{5}\)
\(x=\frac{5}{5}=1\)
\(\Leftrightarrow\dfrac{1}{2}x^2-3x-\dfrac{9}{2}-\dfrac{4}{3}\left(x^2+4x+4\right)-\dfrac{5}{4}\left(x^2-1\right)=\dfrac{3}{2}x\left(x-2\right)-x-4\)
\(\Leftrightarrow\dfrac{1}{2}x^2-3x-\dfrac{9}{2}-\dfrac{4}{3}x^2-\dfrac{16}{3}x-\dfrac{16}{3}-\dfrac{5}{4}x^2+\dfrac{5}{4}=\dfrac{3}{2}x^2-3x-x-4\)
\(\Leftrightarrow x^2\cdot\dfrac{-25}{12}-\dfrac{25}{3}x-\dfrac{103}{12}-\dfrac{3}{2}x^2+4x+4=0\)
\(\Leftrightarrow\dfrac{-43x^2}{12x}-\dfrac{13x}{3}-\dfrac{55}{12}=0\)
\(\Leftrightarrow43x^2+52x+55=0\)
\(\text{Δ}=52^2-4\cdot43\cdot55=-6756< 0\)
Do đó: Phương trình vô nghiệm
\(\left(\frac{3^x}{3}\right)^2=\frac{1}{729}\Rightarrow3^{x-1}=\frac{1}{27}\Rightarrow x-1=-3\Rightarrow x=-2\)
\(x=-2\)