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(1 + x) + (2 + x) + (3 + x) + ... + (10 + x) = 75
=> (1 + 2 + 3 + ... + 10) + (x + x + x + ... + x) = 75
=> 55 + 10x = 75
=> 10x = 20
=> x = 2
vậy_
x : [(1800 + 600) : 30] = 560 : (315 - 35)
=> x : [2400 : 30] = 560 : 280
=> x : 80 = 2
=> x = 160
vậy_
a/ \(3+2^{x-1}=24-\left[4^2-\left(2^2-1\right)\right]\\3+2^{x+1}=24-\left[16-\left(4-1\right)\right]\)
\(3+2^{x+1}=24-\left(16-3\right)\\ 3+2^{x-1}=24-13\\ 3+2^{x-1}=11\\ 2^{x+1}=11-3\\ 2^{x-1}=8\)
\(2^{x-1}=2^3\\ \Rightarrow x-1=3\\x=3+1\\ x=4\)
\(\left(x+1\right)+\left(x+2\right)+\left(x+3\right)+...+\left(x+100\right)=205550\)
\(\left(x.100\right)+\left(1+2+3+....+100\right)=205550\)
Ta tính tổng \(1+2+3+...+100\\ \) trước
Số các số hạng: \(\left[\left(100-1\right):1+1\right]=100\)
Tổng :\(\left[\left(100+1\right).100:2\right]=5050\)
Thay số vào ta có được:
\(\left(x.100\right)+5050=205550\\ \\ x.100=205550-5050\\ \\x.100=20500\\ \\x=20500:100\\ \\\Rightarrow x=2005\)
\(30-\left[4\left(x-2\right)+15\right]=3\) \(\left(8x-120:4\right).3^3=3^6\)
\(4\left(x-2\right)+15=27\) \(\left(8x-120:4\right)=27\)
\(4\left(x-2\right)=12\) \(8x-30=27\)
\(x-2=3\) \(8x=57\)
\(x=5\) \(x=\frac{57}{8}\)
\(2\frac{2}{3}:\left\{\left[\left(3,72-0,02.x\right)\frac{10}{37}\right]:\frac{5}{6}+2,8\right\}-\frac{7}{15}=0,2\)
\(2\frac{2}{3}:\left\{\left[\left(3,75-0,02.x\right)\frac{10}{37}\right]:\frac{5}{6}+2,8\right\}=\frac{2}{3}\)
\(\left\{\left[\left(3,72-0,02.x\right)\frac{10}{37}\right]:\frac{5}{6}+2,8\right\}=4\)
\(\left[\left(3,72-0,02.x\right)\frac{10}{37}\right]:\frac{5}{6}=\frac{6}{5}\)
\(\left[\left(3,72-0,02.x\right)\frac{10}{37}\right]=1\)
\(\left(3,72-0,02.x\right)=\frac{37}{10}\)
\(0,02.x=0,02\)
\(x=1\)
\(2\frac{2}{3}:\left\{\left[\left(3,72-0,02.x\right)\frac{10}{37}\right]:\frac{5}{6}+2,8\right\}-\frac{7}{15}=0,2\)
\(\Rightarrow\frac{8}{3}:\left\{\left[\left(\frac{93}{25}-\frac{1}{50}.x\right)\frac{10}{37}\right]:\frac{5}{6}+\frac{14}{5}\right\}-\frac{7}{15}=\frac{1}{5}\)
\(\Rightarrow\left\{\left[\frac{93}{25}-\frac{1}{50}.x\right]:\frac{5}{6}+\frac{14}{5}\right\}-\frac{7}{15}=\frac{8}{3}:\frac{1}{5}=\frac{40}{3}\)
\(\Rightarrow\left[\frac{93}{25}-\frac{1}{50}.x\right]:\frac{5}{6}+\frac{14}{5}=\frac{40}{3}+\frac{7}{15}=\frac{69}{5}\)
\(\Rightarrow\left[\frac{93}{25}-\frac{1}{50}.x\right]:\frac{5}{6}=\frac{69}{5}-\frac{14}{5}=11\)
\(\Rightarrow\frac{93}{25}-\frac{1}{50}.x=11.\frac{5}{6}=\frac{55}{6}\)
\(\Rightarrow\frac{1}{50}.x=\frac{93}{25}-\frac{55}{6}=\frac{-817}{150}\)
\(\Rightarrow x=\frac{-817}{150}:\frac{1}{50}=\frac{-817}{3}\)
Ủng hộ tớ nha m.n?
\(\frac{11}{4}:\frac{3}{2}:\left|4x-\frac{1}{3}\right|=\frac{7}{2}\)
\(\Leftrightarrow\frac{3}{2}:\left|4x-\frac{1}{3}\right|=\frac{11}{4}:\frac{7}{2}\)
\(\Leftrightarrow\frac{3}{2}:\left|4x-\frac{1}{3}\right|=\frac{11}{14}\)
\(\Leftrightarrow\left|4x-\frac{1}{3}\right|=\frac{3}{2}:\frac{11}{14}\)
\(\Leftrightarrow\left|4x-\frac{1}{3}\right|=\frac{21}{11}\)
\(\Leftrightarrow\orbr{\begin{cases}4x-\frac{1}{3}=\frac{21}{11}\\4x-\frac{1}{3}=-\frac{21}{11}\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{37}{66}\\x=-\frac{13}{33}\end{cases}}\)
Bài dưới tương tự
\(\left(x+\frac{1}{4}-\frac{1}{3}\right)\div\left(2+\frac{1}{6}-\frac{1}{4}\right)=\frac{7}{46}\)
\(\Leftrightarrow\left(x+\frac{1}{4}-\frac{1}{6}\right)\div\frac{23}{12}=\frac{7}{46}\)
\(\Leftrightarrow x+\frac{1}{4}-\frac{1}{6}=\frac{7}{46}\times\frac{23}{12}\)
\(\Leftrightarrow x+\frac{1}{4}-\frac{1}{6}=\frac{7}{24}\)
\(\Leftrightarrow x+\frac{1}{4}=\frac{7}{24}+\frac{1}{6}\)
\(\Leftrightarrow x+\frac{1}{4}=\frac{11}{24}\)
\(\Leftrightarrow x=\frac{11}{24}-\frac{1}{4}\)
\(\Leftrightarrow x=\frac{5}{24}\)
Học tốt !
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Chúc cậu học tốt
\(\left(x-2\right)^2=\left(x-4\right)^2\)
\(\left(x-2\right)^2=0\)
\(x-2=0\)
\(x=2\)
\(770\div\left[\left(20x+10\right)\div x\right]=35\)
\(\frac{20x+10}{x}=22\Rightarrow20x+10=22x\Rightarrow2x=10\Rightarrow x=5\)