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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_
\(\frac{11}{4}:\frac{3}{2}:\left|4x-\frac{1}{3}\right|=\frac{7}{2}\)
\(\Leftrightarrow\frac{3}{2}:\left|4x-\frac{1}{3}\right|=\frac{11}{4}:\frac{7}{2}\)
\(\Leftrightarrow\frac{3}{2}:\left|4x-\frac{1}{3}\right|=\frac{11}{14}\)
\(\Leftrightarrow\left|4x-\frac{1}{3}\right|=\frac{3}{2}:\frac{11}{14}\)
\(\Leftrightarrow\left|4x-\frac{1}{3}\right|=\frac{21}{11}\)
\(\Leftrightarrow\orbr{\begin{cases}4x-\frac{1}{3}=\frac{21}{11}\\4x-\frac{1}{3}=-\frac{21}{11}\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{37}{66}\\x=-\frac{13}{33}\end{cases}}\)
Bài dưới tương tự
\(\left(x-2\right)^2=\left(x-4\right)^2\)
\(\left(x-2\right)^2=0\)
\(x-2=0\)
\(x=2\)
\(770\div\left[\left(20x+10\right)\div x\right]=35\)
\(\frac{20x+10}{x}=22\Rightarrow20x+10=22x\Rightarrow2x=10\Rightarrow x=5\)
\(2\frac{2}{3}:\left\{\left[\left(3,72-0,02.x\right)\frac{10}{37}\right]:\frac{5}{6}+2,8\right\}-\frac{7}{15}=0,2\)
\(2\frac{2}{3}:\left\{\left[\left(3,75-0,02.x\right)\frac{10}{37}\right]:\frac{5}{6}+2,8\right\}=\frac{2}{3}\)
\(\left\{\left[\left(3,72-0,02.x\right)\frac{10}{37}\right]:\frac{5}{6}+2,8\right\}=4\)
\(\left[\left(3,72-0,02.x\right)\frac{10}{37}\right]:\frac{5}{6}=\frac{6}{5}\)
\(\left[\left(3,72-0,02.x\right)\frac{10}{37}\right]=1\)
\(\left(3,72-0,02.x\right)=\frac{37}{10}\)
\(0,02.x=0,02\)
\(x=1\)
\(2\frac{2}{3}:\left\{\left[\left(3,72-0,02.x\right)\frac{10}{37}\right]:\frac{5}{6}+2,8\right\}-\frac{7}{15}=0,2\)
\(\Rightarrow\frac{8}{3}:\left\{\left[\left(\frac{93}{25}-\frac{1}{50}.x\right)\frac{10}{37}\right]:\frac{5}{6}+\frac{14}{5}\right\}-\frac{7}{15}=\frac{1}{5}\)
\(\Rightarrow\left\{\left[\frac{93}{25}-\frac{1}{50}.x\right]:\frac{5}{6}+\frac{14}{5}\right\}-\frac{7}{15}=\frac{8}{3}:\frac{1}{5}=\frac{40}{3}\)
\(\Rightarrow\left[\frac{93}{25}-\frac{1}{50}.x\right]:\frac{5}{6}+\frac{14}{5}=\frac{40}{3}+\frac{7}{15}=\frac{69}{5}\)
\(\Rightarrow\left[\frac{93}{25}-\frac{1}{50}.x\right]:\frac{5}{6}=\frac{69}{5}-\frac{14}{5}=11\)
\(\Rightarrow\frac{93}{25}-\frac{1}{50}.x=11.\frac{5}{6}=\frac{55}{6}\)
\(\Rightarrow\frac{1}{50}.x=\frac{93}{25}-\frac{55}{6}=\frac{-817}{150}\)
\(\Rightarrow x=\frac{-817}{150}:\frac{1}{50}=\frac{-817}{3}\)
Ủng hộ tớ nha m.n?
B=\(\left(1-\dfrac{1}{1+2}\right)\). \(\left(1-\dfrac{1}{1+2+3}\right)\).....\(\left(1-\dfrac{1}{1+2+...+100}\right)\)
B=\(\left(1-\dfrac{1}{3}\right)\cdot\left(1-\dfrac{1}{6}\right)\cdot...\cdot\left(1-\dfrac{1}{\left(1+100\right)\cdot100:2}\right)\)
B=\(\dfrac{2}{3}\cdot\dfrac{5}{6}\cdot...\cdot\dfrac{101\cdot100:2-1}{101\cdot100:2}\)
B=\(\dfrac{4}{6}\cdot\dfrac{10}{12}\cdot...\cdot\dfrac{\left(101.100:2-1\right).2}{101.100}\)
B=\(\dfrac{1.4}{2.3}.\dfrac{2.5}{3.4}\cdot...\cdot\dfrac{99.102}{100.101}\)
B=\(\dfrac{1.2.3.4.....99}{3.4.5....100}.\dfrac{4.5.6.....102}{3.4.5.....101}\)
B=\(\dfrac{2}{100}\).\(\dfrac{102}{3}\)
B=\(\dfrac{17}{25}\)
Ta có S = ( 1/2 - 1) : ( 1/3 - 1) : (1/4 - 1) :... : ( 1/50 - 1)
S = -1/2 : ( -2/3) : ( -3/4) : ... : ( -49/ 50)
S= -1/2 x (-3/2) x ( -4/3) x ... x (-50/49)
S= -1/2 x 1/3 x 50
S= -25/3
a) \(\left(-7\right)-\left[\left(-19\right)+\left(-21\right)\right].\left(-3\right)-\left[\left(+32\right)+\left(-7\right)\right]\)
\(=\left(-7\right)-\left(-40\right).\left(-3\right)-25\)
\(=\left(-7\right)-120-25\)
\(=-152\)
b) \(\left(-2\right)^3.3-\left(1^{10}+8\right):\left(-3\right)^2\)
\(=\left(-8\right).3-\left(1+8\right):9\)
\(=\left(-24\right)-9:9\)
\(=\left(-24\right)-1\)
\(=-25\)
Bài giải
a, \(\left(-7\right)-\left[\left(-19\right)+\left(-21\right)\right]\cdot\left(-3\right)-\left[\left(+32\right)+\left(-7\right)\right]\)
\(=\left(-7\right)-\left(-40\right)\cdot\left(-3\right)-25\)
\(=-7-120-25\)
\(=-127-25\)
\(=-152\)
b, \(\left(-2\right)^3\cdot3-\left(1^{10}+8\right)\text{ : }\left(-3\right)^2\)
\(=-8\cdot3-\left(1+8\right)\text{ : }9\)
\(=-24-9\text{ : }9\)
\(=-24-1\)
\(=-25\)
bn đổi ngược hai vế cho nhau là ra 1 bài toán bình thường thôi
Bài nỳ tuy rất dài nhưng cũng dễ
Chí cần cậu cuyển VT sang VP rồi tìm x bình thường
Chúc cậu học tốt
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:
\(\left(x.100\right)+5050=205550\\ \\ x.100=205550-5050\\ \\x.100=20500\\ \\x=20500:100\\ \\\Rightarrow x=2005\)