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1) |x + 2| = 4
\(\Leftrightarrow\orbr{\begin{cases}x+2=4\\x+2=-4\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=2\\x=-6\end{cases}}\)
2) 3 – |2x + 1| = (-5)
\(\Leftrightarrow\left|2x+1\right|=8\Leftrightarrow\orbr{\begin{cases}2x+1=8\\2x+1=-8\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{7}{2}\\x=\frac{-9}{2}\end{cases}}\)
3) 12 + |3 – x| = 9
\(\Leftrightarrow\left|3-x\right|=-3\)(vô lí)
=>\(x=\varnothing\)
1) I x+2 I=4
\(\Rightarrow\orbr{\begin{cases}x+2=4\\x+2=-4\end{cases}\Leftrightarrow\orbr{\begin{cases}x=2\\x=-6\end{cases}}}\)
2) \(3-|2x+1|=-5\)
\(\Leftrightarrow|2x+1|=8\)
\(\Leftrightarrow\orbr{\begin{cases}2x+1=8\\2x+1=-8\end{cases}\Leftrightarrow\orbr{\begin{cases}2x=7\\2x=-9\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{7}{2}\\x=\frac{-9}{2}\end{cases}}}\)
3) \(12+|3-x|=9\)
\(\Leftrightarrow|3-x|=-3\)(vô lí vì I 3-x I \(\ge\)0)
Bài 1:
a) \(\dfrac{9}{20}-\dfrac{8}{15}\times\dfrac{5}{12}\)
\(=\dfrac{9}{20}-\dfrac{2}{9}\)
\(=\dfrac{41}{180}\)
b) \(\dfrac{2}{3}\div\dfrac{4}{5}\div\dfrac{7}{12}\)
\(=\dfrac{2}{3}\times\dfrac{5}{4}\times\dfrac{12}{7}\)
\(=\dfrac{5}{6}\times\dfrac{12}{7}\)
\(=\dfrac{10}{7}\)
c) \(\dfrac{7}{9}\times\dfrac{1}{3}+\dfrac{7}{9}\times\dfrac{2}{3}\)
\(=\dfrac{7}{9}\times\left(\dfrac{1}{3}+\dfrac{2}{3}\right)\)
\(=\dfrac{7}{9}\times1\)
\(=\dfrac{7}{9}\)
Bài 2:
a) \(2\times\left(x-1\right)=4026\)
\(\left(x-1\right)=4026\div2\)
\(x-1=2013\)
\(x=2014\)
Vậy: \(x=2014\)
b) \(x\times3,7+6,3\times x=320\)
\(x\times\left(3,7+6,3\right)=320\)
\(x\times10=320\)
\(x=320\div10\)
\(x=32\)
Vậy: \(x=32\)
c) \(0,25\times3< 3< 1,02\)
\(\Leftrightarrow0,75< 3< 1,02\) ( S )
=> \(0,75< 1,02< 3\)
a. \(\dfrac{8}{7}-\dfrac{1}{7}:\left(\dfrac{x}{3}-2\right)=-1\)
\(-\dfrac{1}{7}:\left(\dfrac{x}{3}-2\right)=\dfrac{15}{7}\)
\(\dfrac{x}{3}-2=\dfrac{-1}{15}\)
\(\dfrac{x}{3}=\dfrac{29}{15}\)
\(x=5,8\)
b. \(\dfrac{5}{8}+\dfrac{1}{4}\left(2x-1\right)=\dfrac{5}{4}\)
\(\dfrac{1}{4}\left(2x-1\right)=\dfrac{5}{8}\)
\(2x-1=\dfrac{5}{2}\)
\(2x=\dfrac{7}{2}\)
\(x=\dfrac{7}{4}\)
Bài 1:
a: \(x=\dfrac{2}{3}:\dfrac{3}{5}=\dfrac{2}{3}\cdot\dfrac{5}{3}=\dfrac{10}{9}\)
b: \(x=\dfrac{17}{8}:\dfrac{7}{17}=\dfrac{17}{8}\cdot\dfrac{17}{7}=\dfrac{289}{56}\)
c: \(x=-\dfrac{3}{4}:\dfrac{7}{12}=\dfrac{-3}{4}\cdot\dfrac{12}{7}=\dfrac{-63}{28}=-\dfrac{9}{4}\)
d: \(\Leftrightarrow x\cdot\dfrac{1}{6}=\dfrac{3}{8}-\dfrac{1}{4}=\dfrac{1}{4}\)
hay \(x=\dfrac{1}{4}:\dfrac{1}{6}=\dfrac{3}{2}\)
e: \(\Leftrightarrow\dfrac{1}{2}:x=-4-\dfrac{1}{3}=-\dfrac{17}{3}\)
hay \(x=-\dfrac{1}{2}:\dfrac{17}{3}=\dfrac{-3}{34}\)
3:
\(=\dfrac{1}{7}\cdot\dfrac{3}{5}\cdot\dfrac{5}{6}\cdot\dfrac{5}{8}=\dfrac{1}{7}\cdot\dfrac{1}{2}\cdot\dfrac{5}{8}=\dfrac{5}{112}\)
4:
=>2/3:x=-2-1/3=-7/3
=>x=-2/3:7/3=-2/7
5:
AC=CB=12/2=6cm
IB=6/2=3cm
Bài 1 : Bài giải
\(-2\left(-3-4x\right)-3\left(3x-7\right)=31\)
\(6+8x-9x-21=31\)
\(-x-15=31\)
\(-x=31+15\)
\(-x=46\)
\(x=-46\)
b, \(10\left(x-7\right)=8\left(x-4\right)+x\)
\(10x-70=8x-32+x\)
\(10x-70=9x-32\)
\(10x-9x=-32+70\)
\(x=38\)
c, \(2\left|x-1\right|=3\cdot\left(5-1\right)\)
\(2\left|x-1\right|=15-3\)
\(2\left|x-1\right|=12\)
\(\left|x-1\right|=12\text{ : }2\)
\(\left|x-1\right|=6\)
\(\Rightarrow\orbr{\begin{cases}x-1=-6\\x-1=6\end{cases}}\Rightarrow\orbr{\begin{cases}x=-5\\x=7\end{cases}}\)
\(\Rightarrow\text{ }x\in\left\{-5\text{ ; }7\right\}\)
d, \(2\left(x+3\right)-2\left(x-3\right)=x\)
\(2\left(x+3-x-3\right)=x\)
\(2\cdot0=x\)
\(x=0\)
e, \(\left(x+3\right)\left(x-4\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x+3=0\\x-4=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=-3\\x=4\end{cases}}\)
\(\Rightarrow\text{ }x\in\left\{-3\text{ ; }4\right\}\)
a, \(-2\left(-3-4x\right)-3\left(3x-7\right)=31\)
\(6+8x-9x-21=31\)
\(-x-15=31\)
\(-x=31+15\)
\(-x=46\)
\(x=-46\)
b, \(10\left(x-7\right)=8\left(x-4\right)+x\)
\(10x-70=8x-32+x\)
\(10x-70=9x-32\)
\(10x-9x=-32+70\)
\(x=38\)
c, \(2\left|x-1\right|=3\cdot\left(5-1\right)\)
\(2\left|x-1\right|=15-3\)
\(2\left|x-1\right|=12\)
\(\left|x-1\right|=12\text{ : }2\)
\(\left|x-1\right|=6\)
\(\Rightarrow\orbr{\begin{cases}x-1=-6\\x-1=6\end{cases}}\Rightarrow\orbr{\begin{cases}x=-5\\x=7\end{cases}}\)
\(\Rightarrow\text{ }x\in\left\{-5\text{ ; }7\right\}\)
d, \(2\left(x+3\right)-2\left(x-3\right)=x\)
\(2\left(x+3-x-3\right)=x\)
\(2\cdot0=x\)
\(x=0\)
e, \(\left(x+3\right)\left(x-4\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x+3=0\\x-4=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=-3\\x=4\end{cases}}\)
\(\Rightarrow\text{ }x\in\left\{-3\text{ ; }4\right\}\)
