bài1:
a)|x+3|=15
b)|x-7|+13=25
c)|x-3|-16=-4
d)26-|x+9|=-13
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a) 32.x+2=1342176728
32.x=134217728-2
32.x=134217726
x=134217726:32
x=4194303,938
a) \(\dfrac{2}{5}+\dfrac{11}{15}=\dfrac{6}{15}+\dfrac{11}{15}=\dfrac{17}{15}\)
b) \(\dfrac{7}{8}-\dfrac{7}{9}=\dfrac{63}{72}-\dfrac{56}{72}=\dfrac{7}{72}\)
c) \(\dfrac{11}{13}\cdot\dfrac{26}{31}=\dfrac{26}{13}\cdot\dfrac{11}{31}=\dfrac{22}{31}\)
d) \(\dfrac{1}{2}:\dfrac{1}{3}\cdot\dfrac{2}{5}=\dfrac{1}{2}\cdot3\cdot\dfrac{2}{5}=\dfrac{3}{5}\)
a)
x-3=15
x = 15+3=18
b)
x+7-13=13
x+7 =13-13=0
x = 0-7=-7
c)
x+3-16=-4
x+3 =-4+16=12
x = 12-3=9
d)
26-x-9=-13
-x-9= -13-26=-39
-x =-39+9=-30
x=30
a: =>x=-7/6+5/8=-13/24
b: =>x=-14/25-3/4=-131/100
c: \(x=\dfrac{-33}{26}:\dfrac{-9}{13}=\dfrac{33}{26}\cdot\dfrac{13}{9}=\dfrac{11}{3}\cdot\dfrac{1}{2}=\dfrac{11}{6}\)
d: \(x=\dfrac{4}{9}:\dfrac{5}{3}=\dfrac{4}{9}\cdot\dfrac{3}{5}=\dfrac{12}{45}=\dfrac{4}{15}\)
a, \(\left|x+3\right|=15\)
TH1 : \(x+3=15\Leftrightarrow x=12\)
TH2 : \(x+3=-15\Leftrightarrow x=-18\)
b, \(\left|x-7\right|+13=15\Leftrightarrow\left|x-7\right|=2\)
TH1 : \(x-7=2\Leftrightarrow x=9\)
TH2 : \(x-7=-2\Leftrightarrow x=5\)
c, \(\left|x-3\right|-16=-4\Leftrightarrow\left|x-3\right|=12\)
TH1 : \(x-3=12\Leftrightarrow x=15\)
TH2 : \(x-3=-12\Leftrightarrow x=-9\)
d, \(26-\left|x+9\right|=-13\Leftrightarrow\left|x+9\right|=39\)
TH1 : \(x+9=39\Leftrightarrow x=30\)
TH2 : \(x+9=-39\Leftrightarrow x=-48\)
a)|x-7|+13=25
|x-7| = 25-13
|x-7| = 12
|x| = 12+7
=> x = 19
b)|x-3|-16=-4
|x-3| = -4+16
|x-3| = 12
|x| = 12+3
=> x = 15
c)26-|x+9|=-13
|x+9| = -13-26
|x+9| = -39
|x| = -39 -9
=> x = 48
Bài 1:
\(\dfrac{2}{3}\) + \(\dfrac{3}{4}\) - \(\dfrac{2}{5}\)
=\(\dfrac{40}{60}\) + \(\dfrac{45}{60}\) - \(\dfrac{24}{60}\)
= \(\dfrac{61}{60}\)
b; \(\dfrac{12}{13}\) x \(\dfrac{3}{4}\) - \(\dfrac{7}{13}\)
= \(\dfrac{9}{13}\) - \(\dfrac{7}{13}\)
= \(\dfrac{2}{13}\)
c; \(\dfrac{15}{17}\) : \(\dfrac{19}{34}\) - \(\dfrac{17}{19}\)
= \(\dfrac{15}{17}\) x \(\dfrac{34}{19}\) - \(\dfrac{17}{19}\)
= \(\dfrac{30}{19}\) - \(\dfrac{17}{19}\)
= \(\dfrac{13}{19}\)
Bài 2:
a; \(\dfrac{x}{5}\) x \(\dfrac{3}{7}\) = \(\dfrac{9}{35}\)
\(\dfrac{x}{5}\) = \(\dfrac{9}{35}\) : \(\dfrac{3}{7}\)
\(\dfrac{x}{5}\) = \(\dfrac{3}{5}\)
\(x\) = \(\dfrac{3}{5}\) x 5
\(x\) = 3
\(\left(1+x\right)+\left(5+x-4\right)+\left(9+x-8\right)+...=501501\)
\(\left(1+x\right)+\left(x+1\right)+\left(x+1\right)+...=501501\) có x :2 dấu ngoặc
\(\left(1+x\right).x:2=501501\)
\(\left(1+x\right).x=1003002=1002.1001\Rightarrow x=1001\)
\(\left|x+3\right|=15\)
\(\Rightarrow\orbr{\begin{cases}x+3=15\\x+3=-15\end{cases}\Rightarrow\orbr{\begin{cases}x=15-3\\x=-15-3\end{cases}\Rightarrow}}\orbr{\begin{cases}x=12\\x=-18\end{cases}}\)
Vậy .........
a, bạn Forevër™ ( 30 - 03 - 2007 ) giải rồi nhé
b, Ta có :\(|x-7|+13=25\)
<=> : \(|x-7|=25-13\)
<=> : \(|x-7|=12\)
=> : \(x-7=12\) :
< => : \(x=19\)
: \(x=-5\)
Vậy\(x=\left\{-5;19\right\}\)
C,Ta có : \(|x-3|-16=-4\)
\(< =>|x-3|=-4+16\)
\(< =>|x-3|=12\)
\(< =>x-3=12\)hoặc \(x-3=-12\)
\(< =>x=15\)hoặc \(x=-9\)
Vậy\(x=\left\{15;-9\right\}\)