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a) Ta có: \(\left(2^2\right)^3\cdot4^5\)
\(=2^6\cdot2^{10}\)
\(=2^{16}=65536\)
b) Ta có: \(\left[\left(-4\right)^2\right]^2\cdot6\)
\(=16^2\cdot6\)
\(=256\cdot6=1536\)
c) Ta có: \(\frac{16}{25}\cdot\left(\frac{4}{5}\right)^3\)
\(=\left(\frac{4}{5}\right)^2\cdot\left(\frac{4}{5}\right)^3\)
\(=\left(\frac{4}{5}\right)^5\)
\(=\frac{1024}{3125}\)
d) Ta có: \(\left(\frac{121}{64}\right)^2\cdot\left(-\frac{64}{11}\right)^2\)
\(=\frac{121^2}{64^2}\cdot\frac{64^2}{11^2}\)
\(=11^2=121\)
e) Ta có: \(\left[\left(-3\right)^3\right]^3\cdot271:125\)
\(=\left(-27\right)^3\cdot\frac{271}{125}\)
\(=\frac{-5334093}{125}\)
Bạn ơi cả 3x-5 chia cho 12 hay chỉ có 5 chia cho 12 thui z ạ
\(\left(3x-\frac{5}{12}\right)^2-\frac{121}{64}=0\)
\(\left(3x-\frac{5}{12}\right)^2=\frac{121}{64}\)
\(3x-\frac{5}{12}=\sqrt{\frac{121}{64}}\)
\(3x-\frac{5}{12}=\frac{11}{8}\)
\(3x=\frac{11}{8}+\frac{5}{12}\)
\(3x=\frac{33}{24}+\frac{10}{24}\)
\(3x=\frac{43}{24}\)
\(x=\frac{43}{24}:3\)
\(x=\frac{43}{24}\cdot\frac{1}{3}\)
\(x=\frac{43}{72}\)
a)
\(\left(x+5\right)^3=-64=\left(-4\right)^3\)
=> x + 5 = -4
=> x = -9
b)
\(\left(2x-3\right)^2=9=\left(\pm3\right)^2\)
+) 2x - 3 = 3
2x = 6
x = 3
+) 2x - 3 = -3
2x = 0
x = 0
c)
\(M=\frac{8^{10}+4^{10}}{8^4+4^{11}}\)
\(M=\frac{2^{30}+2^{20}}{2^{12}+2^{22}}\)
\(M=\frac{2^{20}\left(2^{10}+1\right)}{2^{12}\left(1+2^{10}\right)}\)
\(M=\frac{2^{12}\cdot2^8}{2^{12}}\)
\(M=2^8=256\)
a) (\(x-\dfrac{3}{5}\))\(^2\)=64
=> \(x-\dfrac{3}{5}\)=\(\sqrt{64}\)
=> \(x-\dfrac{3}{5}\) =8 hoặc \(x-\dfrac{3}{5}\)=-8
Trường hợp 1:
\(x-\dfrac{3}{5}\) = 8=>\(x\) =\(8+\dfrac{3}{5}\)=>\(x\)=\(\dfrac{43}{5}\)
Trường hợp 2:
\(x-\dfrac{3}{5}\)=\(-8\)=>\(x\)=\(-8+\dfrac{3}{5}\)=>\(x=-\dfrac{37}{5}\)
b: y/2=z/3 nên y/6=z/9
=>x/4=y/6=z/9=(x+y+z)/(4+6+9)=121/19
=>x=484/19; y=726/19; z=1089/19
Đáp án B