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\(\frac{2^{15}\cdot9^4}{6^3.8^3}=\frac{2^{15}.3^8}{2^3.3^3.2^9}=\frac{2^{15}.3^8}{2^{12}.3^3}=2^3.3^5=8.243=1944\)
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\(\left|x+\frac{4}{15}\right|-\left|-3,75\right|=-\left|-2,15\right|\)
=> \(\left|x+\frac{4}{15}\right|-3,75=-2,15\)
=> \(\left|x+\frac{4}{15}\right|=-2,15+3,75\)
=> \(\left|x+\frac{4}{15}\right|=1,6=\frac{8}{5}\)
=> \(\orbr{\begin{cases}x+\frac{4}{15}=\frac{8}{5}\\x+\frac{4}{15}=\frac{-8}{5}\end{cases}}\)=> \(\orbr{\begin{cases}x=\frac{4}{3}\\x=\frac{-28}{15}\end{cases}}\)
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Zế trái luôn ở tổng của các số giá trị tuyệt đối nên tổng ko âm
Nên phá đc dấu giá trị
=> x+19+x+5x+x+2020=5x
=>x+x+5x+x-5x=-19-2020
=>3x=-2039
=> x=-2039/3
TH1: x > 0 => l x + 19 l + l x + 5 l + l x + 2020 l = 5x TH2: x < 0 => l x + 19 l + l x + 5 l + l x + 2020 l = 5x
x + 19 + x + 5 + x + 2020 = 5x (-x) + 19 + (-x) + 5 + (-x) + 2020 = 5x
(x + x + x) + (19 + 5 + 2020) = 5x [(-x) + (-x) + (-x)] + (19 + 5 + 2020) = 5x
3x + 2044 = 5x 3(-x) + 2044 = 5x
3x + 2044 = 3x + 2x 3(-x) + 2044 = 3(-x) + 2(-x)
=> 2x = 2044 => 2(-x) = 2044
x = 2044 : 2 -x = 2044 : 2
x = 1022 -x = 1022
x = -1022
Vậy x = 1022 hoặc x = -1022 (mình cũng ko chắc lắm đâu)
Bài 1:
a) \(\frac{-5}{21}+\frac{-2}{21}+\frac{8}{24} \)
\(=\frac{-7}{21}+\frac{1}{3}\)
\(=\frac{-1}{3}+\frac{1}{3}\)
\(=0\)
b) \(\frac{-3}{5}.\frac{2}{7}+\frac{-3}{5}.\frac{5}{7}+2\frac{3}{5}\)
\(=\frac{-3}{5}.\left(\frac{2}{7}+\frac{5}{7}\right)+\frac{13}{5}\)
\(=\frac{-3}{5}.1+\frac{13}{5}\)
\(=\frac{-3}{5}+\frac{13}{5}\)
\(=\frac{10}{5}\)
\(=2\)
c) \(75\%-1\frac{1}{2}+0,5\div\frac{5}{12}-\left(-\frac{1}{2}\right)^2\)
\(=\frac{3}{4}-\frac{3}{2}+\frac{1}{2}.\frac{12}{5}-\frac{1}{4}\)
\(=\frac{3}{4}-\frac{3}{2}+\frac{6}{5}-\frac{1}{4}\)
\(=\frac{3}{4}+\frac{-3}{2}+\frac{6}{5}+\frac{-1}{4}\)
\(=\left(\frac{3}{4}+\frac{-1}{4}\right)+\frac{-3}{2}+\frac{6}{5}\)
\(=\frac{1}{2}+\frac{-3}{2}+\frac{6}{5}\)
\(=-1+\frac{6}{5}\)
\(=\frac{-5}{5}+\frac{6}{5}\)
\(=\frac{1}{5}\)
Bài 2:
a) \(1\frac{1}{3}+\frac{2}{3}x=5\frac{1}{3}\)
\(\frac{4}{3}+\frac{2}{3}x=\frac{16}{3}\)
\(\frac{2}{3}x=\frac{16}{3}-\frac{4}{3}\)
\(\frac{2}{3}x=4\)
\(x=4\div\frac{2}{3}\)
\(x=4.\frac{3}{2}\)
\(x=6\)
Vậy \(x=6\)
b) \(\left(x+\frac{1}{3}\right)\left(\frac{3}{4}-2x\right)=0\)
TH1: \(x+\frac{1}{3}=0\)
\(x=0-\frac{1}{3}\)
\(x=\frac{-1}{3}\)
TH2 : \(\frac{3}{4}-2x=0\)
\(2x=\frac{3}{4}-0\)
\(2x=\frac{3}{4}\)
\(x=\frac{3}{4}\div2\)
\(x=\frac{3}{4}.\frac{1}{2}\)
\(x=\frac{3}{8}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{-1}{3}\\x=\frac{3}{8}\end{cases}}\)
c) \(\frac{-4}{5}x+1\frac{1}{3}x=-1\frac{1}{3}\)
\(\frac{-4}{5}x+\frac{4}{3}x=\frac{-4}{3}\)
\(\left(\frac{-4}{5}+\frac{4}{3}\right).x=\frac{-4}{3}\)
\(\frac{8}{15}.x=\frac{-4}{3}\)
\(x=\frac{-4}{3}\div\frac{8}{15}\)
\(x=\frac{-4}{3}.\frac{15}{8}\)
\(x=\frac{-5}{2}\)
Vậy \(x=\frac{-5}{2}\)