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Bài 2:
a,\(|x-2|-3=-7\)
\(\Rightarrow|x-2|=-4\)
\(\Rightarrow x\in\varnothing\)
b,\(|x-5|+3=8\)
\(\Rightarrow|x-5|=5\)
\(\Rightarrow\left[{}\begin{matrix}x-5=5\\x-5=-5\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=10\\x=0\end{matrix}\right.\)
c,\(|x+3|-2+2x=5\)
\(\Rightarrow|x+3|+2x=7\)
\(\Rightarrow\left[{}\begin{matrix}x+3=2x=7,x+3\ge0\\-\left(x+3\right)+2x=7,x+3< 0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{4}{3}\\x=10,x< -3\end{matrix}\right.,x\ge-3\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{4}{3}\\x\in\varnothing\end{matrix}\right.\)
d,\(|x-3,1|+|y-1,2|=0\)
Thay y =0
\(\Rightarrow|x-3,1|+|0-1,2|=0\)
\(\Rightarrow|x-3,1|+|-1,2|=0\)
\(\Rightarrow|x-3,1|+1,2=0\)
\(\Rightarrow|x-3,1|=-1,2\)
\(\Rightarrow x\in\varnothing\)
Bài 3:
a,(-2,7)+9-3,5)-1,2(-2)
=>-2,7-3,5+2,4
=>-3,8
b, (-2,3)(-0,4)-(-3,1)(-2,1)-13,2
=>0,92+3,1(-2,1)-13,2
=>0,92-6,51-13,2
=-18,79
\(1,2:\frac{3}{5}+\left(1\frac{1}{15}-\frac{8}{15}\right)x2\frac{2}{4}\)
\(=\frac{6}{5}x\frac{5}{3}+\left(\frac{16}{15}-\frac{8}{15}\right)x\frac{10}{4}\)
\(=2+\frac{8}{15}x\frac{10}{4}=2+\frac{2}{3}\)
\(=2\frac{2}{3}\)
\(1,2:\frac{3}{5}+\left(1\frac{1}{15}-\frac{8}{15}\right)\cdot2\frac{2}{4}\)
\(=\frac{6}{5}\cdot\frac{5}{3}+\left(\frac{16}{15}-\frac{8}{15}\right)\cdot\frac{10}{4}\)
\(=\frac{30}{15}+\frac{8}{15}\cdot\frac{10}{4}\)
\(=\frac{30}{15}+\frac{80}{60}=\frac{200}{60}\)
a) Ta có: \(-3\dfrac{1}{4}\cdot x-75\%+\dfrac{3x}{2}=-1.2:\dfrac{-9}{10}-1\dfrac{1}{4}\)
\(\Leftrightarrow\dfrac{-13x}{4}-\dfrac{3}{4}+\dfrac{3x}{2}=\dfrac{-6}{5}\cdot\dfrac{10}{-9}-\dfrac{5}{4}\)
\(\Leftrightarrow\dfrac{-13x-3+6x}{4}=\dfrac{4}{3}-\dfrac{5}{4}\)
\(\Leftrightarrow\dfrac{-7x-3}{4}=\dfrac{1}{12}\)
\(\Leftrightarrow-7x-3=\dfrac{1}{3}\)
\(\Leftrightarrow-7x=\dfrac{10}{3}\)
hay \(x=-\dfrac{10}{21}\)
b) Ta có: \(\dfrac{5}{3}+\dfrac{5}{15}+\dfrac{5}{35}+...+\dfrac{5}{x\left(x+2\right)}=2\dfrac{8}{17}\)
\(\Leftrightarrow\dfrac{5}{2}\left(\dfrac{2}{3}+\dfrac{2}{15}+\dfrac{2}{35}+...+\dfrac{2}{x\left(x+2\right)}\right)=2\dfrac{8}{17}\)
\(\Leftrightarrow\dfrac{5}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{x}-\dfrac{1}{x+2}\right)=2+\dfrac{8}{17}\)
\(\Leftrightarrow\left(1-\dfrac{1}{x+2}\right)=\dfrac{42}{17}:\dfrac{5}{2}\)
\(\Leftrightarrow\dfrac{x+1}{x+2}=\dfrac{42}{17}\cdot\dfrac{2}{5}=\dfrac{84}{85}\)
\(\Leftrightarrow85x+85=84x+168\)
\(\Leftrightarrow x=83\)
a) \(\left(\frac{-1}{2}\right)+\left|\frac{-7}{6}\right|+\left(\frac{-7}{8}\right)=\frac{-1}{2}+\frac{7}{6}+\left(\frac{-7}{8}\right)=\frac{-5}{24}\)
b) \(\frac{3}{2}.\left|2+x\right|=\frac{5}{4}\)
\(\Rightarrow\left|2+x\right|=\frac{5}{4}:\frac{3}{2}=\frac{5}{6}\)
\(\Rightarrow2+x=\pm\frac{5}{6}\)
\(\Rightarrow\orbr{\begin{cases}2+x=\frac{5}{6}\Rightarrow x=\frac{5}{6}-2=\frac{-7}{6}\\2+x=\frac{-5}{6}\Rightarrow x=\frac{-5}{6}-2=\frac{-17}{6}\end{cases}}\)
Vậy \(x=\left\{\frac{-7}{6};\frac{-17}{6}\right\}\)
c) Ta có: \(\frac{3,7}{x}=\frac{-5}{1,2}\Rightarrow3,7.1,2:\left(-5\right)=x\)
\(\Rightarrow x=-0,888\)
d) Tự làm
A=(\(\frac{-1}{2}\))+|\(\frac{-7}{6}\)|+(\(\frac{-7}{8}\))
=(\(\frac{-1}{2}\))+\(\frac{-7}{6}\)+(\(\frac{-7}{8}\))
=\(\frac{2}{3}\)+(\(\frac{-7}{8}\))
=\(\frac{-5}{24}\)
Vậy x=\(\frac{-5}{24}\)
B=\(\frac{3}{2}\)x|2+x|=\(\frac{5}{4}\)
=|2+x|=\(\frac{5}{4}\):\(\frac{3}{2}\)
=\(\orbr{\begin{cases}2+x=\frac{5}{6}\\2+x=\frac{-5}{6}\end{cases}}\)
=\(\orbr{\begin{cases}x=\frac{5}{6}-2\\x=\frac{-5}{6}-2\end{cases}}\)
=\(\orbr{\begin{cases}x=-1\frac{1}{6}\\x=-2\frac{5}{6}\end{cases}}\)
Vậy x\(\in\){\(-1\frac{1}{6}\);\(-2\frac{5}{6}\)}
2A = 1 + \(\dfrac{1}{2}\)+\(\dfrac{1}{2^2}\)+\(\dfrac{1}{2^3}\)+...+\(\dfrac{1}{2^{99}}\)
2A - A= 1- \(\dfrac{1}{2^{100}}\)
A= 1
x = 2,075