1. Tính \(\frac{12^4\cdot\left(10^2\right)}{3^4\cdot4^5\cdot5^2}\)
2. Tìm x a,\(\left(2^3:4\right)\cdot2^{x+1=64}\)
b,,\(\left|3-2x\right|-3=\left(-3\right)\)
c,\(4-3x=\frac{-5}{3}\)
d, \(\frac{3}{4}+\frac{1}{4}:x=\frac{2}{5}\)
3.Tìm số hữu tỉ x biêt,
a,\(4\frac{1}{3}:x=-6:0,3\)
b, \(\left|3-\frac{3}{4}\right|-9=\frac{1}{2}\)
Giải giúp mình với nha thanks.
Bài 1
\(\frac{12^4\cdot10^2}{3^4\cdot4^5\cdot5^2}\)
\(=\frac{3^4\cdot2^8\cdot2^2\cdot5^2}{3^4\cdot2^{10}\cdot5^2}=1\)
Bài 2
\(\left(2^3:4\right)\cdot2^{x+1}=64\)
\(2\cdot2^{x+1}=64\)
\(2^{x+2}=2^6\)
\(x+2=6\)
\(x=4\)
\(b,\left|3-2x\right|-3=\left(-3\right)\)
\(\left|3-2x\right|=0\)
\(3-2x=0\)
\(2x=3\)
\(x=\frac{3}{2}\)
\(c,4-3x=-\frac{5}{3}\)
\(-3x=-4-\frac{5}{3}\)
\(3x=\frac{17}{3}\)
\(x=\frac{17}{9}\)
\(d,\frac{3}{4}+\frac{1}{4}:x=\frac{2}{5}\)
\(\frac{1}{4}:x=-\frac{7}{20}\)
\(x=-\frac{5}{7}\)
Bài 3
\(a,4\frac{1}{3}:x=-6:0,3\)
\(\frac{13}{3}:x=-20\)
\(x=-\frac{13}{60}\)
\(b,\left|3-\frac{3}{4}\right|-9=\frac{1}{2}\)
không có x đề sai
1.\(\frac{12^4\cdot10^2}{3^4\cdot4^5\cdot5^2}=\frac{2^8\cdot3^4\cdot2^2\cdot5^2}{3^4\cdot2^{10}\cdot5^2}=\frac{2^{10}\cdot3^4\cdot5^2}{3^4\cdot2^{10}\cdot5^2}=1\)
2. a,\(\left(2^3:4\right)\cdot2^{x+1}=64\)
\(\left(8:4\right)\cdot2^{x+1}=64\)
\(2\cdot2^{x+1}=64\)
\(2^{x+2}=2^6\)
\(\Rightarrow x+2=6\)
\(x=4\)
Vậy....
b, \(\left|3-2x\right|-3=\left(-3\right)\)
\(\left|3-2x\right|=\left(-3\right)+3\)
\(\left|3-2x\right|=0\)
\(3-2x=0\)
\(2x=3\)
\(x=\frac{3}{2}\)
c, \(4-3x=-\frac{5}{3}\)
\(3x=4-\left(-\frac{5}{3}\right)\)
\(3x=\frac{17}{3}\)
\(x=\frac{17}{9}\)
Vậy...
d, \(\frac{3}{4}+\frac{1}{4}:x=\frac{2}{5}\)
\(\frac{1}{4}:x=\frac{2}{5}-\frac{3}{4}\)
\(\frac{1}{4}:x=-\frac{7}{20}\)
\(x=\frac{1}{4}:\left(-\frac{7}{20}\right)\)
\(x=-\frac{5}{7}\)
Vậy...
3. a, \(4\frac{1}{3}:x=-6:0,3\)
\(\frac{13}{3}:x=-20\)
\(x=\frac{13}{3}:\left(-20\right)\)
\(x=-\frac{13}{60}\)
Vậy...
b, x đâu bn 'v' ?