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bài 1:
a) 1/3x=-5/3
x=-5
b) x+12/5=37/15
x=1/15
c) x-1/7=-36/7
x=-5
d) 3x-1/2=0
3x=1/2
x=1/6
e) 2/5+x=1/4
x=-3/20
1b)\(\frac{7}{19}x\frac{8}{11}+\frac{3}{11}:\frac{19}{7}-\frac{2}{-19}=\frac{7}{19}x\frac{8}{11}+\frac{3}{11}x\frac{7}{19}+\frac{2}{19}=\left(\frac{8}{11}+\frac{3}{11}\right)\frac{7}{19}+\frac{2}{19}=\frac{7}{19}+\frac{2}{19}=\frac{9}{19}\)
c)\(4\left(\frac{4}{9}+\frac{7}{11}-\frac{4}{9}\right)=4\frac{7}{11}\)
từ rồi làm tiếp
\(\frac{10}{3}.x+\frac{67}{4}=\frac{53}{4}\)
\(\frac{10}{3}.x=\frac{53}{4}-\frac{67}{4}\)
\(\frac{10}{3}.x=-\frac{7}{2}\)
\(x=-\frac{7}{2}:\frac{10}{3}\)
\(x=-\frac{21}{20}\)
Tìm số nguyên x, biết:
a, (-4).|x+2|= -8
|x+2| = -8 : (-4)
|x+2| = 2
\(\Rightarrow\left[{}\begin{matrix}x+2=2\\x+2=-2\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=0\\x=-4\end{matrix}\right.\)
Vậy x = 0 ; x = -4
b, 4.|x+1|= 8.(-2)-8.(-5)
4.|x+1|= 8. [ (-2) - (-5) ]
4.|x+1|= 8. 3
4.|x+1|= 24
|x+1| = 6
\(\Rightarrow\left[{}\begin{matrix}x+1=6\\x+1=-6\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=6\\x=-7\end{matrix}\right.\)
Vậy x = 6; x = -7
c, (2.x-4).(x+6)=0
\(\Rightarrow\) 2.x-4 = x + 6
\(\Rightarrow\) 2x : x = 6 + 4
\(\Rightarrow\) x = 10
d,4.(x-5)-3(x+7)= -19
\(\Rightarrow\) 4x - 20 - 3x - 21 = -19
\(\Rightarrow\) ( 4x - 3x ) - ( 20 - 21 ) = -19
\(\Rightarrow\) x - (-1) = -19
\(\Rightarrow\) x + 1 = -19
\(\Rightarrow\) x = -19 - 1 = -20
e, 3.(x-4)-(8-x)=12
\(\Rightarrow\) 3x - 12 - 8 + x = 12
\(\Rightarrow\) 4x - ( 12 + 8 ) = 12
\(\Rightarrow\) 4x - 20 = 12
\(\Rightarrow\) 4x = 32
\(\Rightarrow\) x = 8
f, 7.(x-3)-5.(3-x)=11.x-5
\(\Rightarrow\) 7x - 21 - 5x - 15 = 11x - 5
\(\Rightarrow\) 12x - 36 = 11x - 5
\(\Rightarrow\) 12x - 11x = -5 + 36
\(\Rightarrow\) x = 31
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Tập xác định của phương trình
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Biến đổi vế trái của phương trình
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3
Phương trình thu được sau khi biến đổi
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4
Lời giải thu được
\(2\left(x-5\right)+3\left(2-3x\right)=5x+7\)
\(\Leftrightarrow2x-10+6-9x=5x+7\)
\(\Leftrightarrow\left(2x-9x\right)+\left(6-10\right)=5x+7\)
\(\Leftrightarrow-7x-4=5x+7\)
\(\Leftrightarrow-7x-5x=4+7\)
\(\Leftrightarrow-12x=11\)
\(\Leftrightarrow x=\frac{-11}{12}\)
\(3x-5\left(x-2\right)+7=4x-12\)
\(\Leftrightarrow3x-5x-10+7=4x-12\)
\(\Leftrightarrow\left(3x-5x\right)-\left(10-7\right)=4x-12\)
\(\Leftrightarrow-2x-3=4x-12\)
\(\Leftrightarrow-2x-4x=3-12\)
\(\Leftrightarrow-6x=-15\)
\(\Leftrightarrow x=\frac{-15}{-6}=\frac{5}{2}\)
a ) \(-5\times\left(-x+7\right)-3\times\left(-x-5\right)=-4\times\left(12-x\right)+48\)
\(\Leftrightarrow5x-35+3x+15=-48+4x+48\)
\(\Leftrightarrow5x-3x+4x=35-15-48+48\)
\(\Leftrightarrow2x=20\)
\(\Leftrightarrow x=10\)
b ) \(-2\times\left(15-3x\right)-4\times\left(-7x+8\right)=-5-9\times\left(-2x+1\right)\)
\(\Leftrightarrow-30+6x-28x-32=-5+18x-9\)
\(\Leftrightarrow6x-28x-18x=30+32-5-9\)
\(\Leftrightarrow-40x=48\)
\(\Leftrightarrow x=-1.2\)
a) \(-\dfrac{3}{7}-x=-\dfrac{1}{2}\\ x=-\dfrac{3}{7}-\left(-\dfrac{1}{2}\right)\\ x=\dfrac{-3}{7}+\dfrac{1}{2}=\dfrac{-6}{14}+\dfrac{7}{14}=\dfrac{1}{14}\)
b) \(x-\dfrac{4}{5}=\dfrac{1}{2}-\dfrac{1}{3}\\ x-\dfrac{4}{5}=\dfrac{3}{6}-\dfrac{2}{6}=\dfrac{1}{6}\\ x=\dfrac{1}{6}+\dfrac{4}{5}=\dfrac{5}{30}+\dfrac{24}{30}\\ x=\dfrac{29}{30}\)
c) \(-x-\dfrac{3}{4}=-\dfrac{8}{11}\\ -x=-\dfrac{8}{11}+\dfrac{3}{4}\\ -x=-\dfrac{32}{44}+\dfrac{33}{44}=\dfrac{1}{44}\\ x=-\dfrac{1}{44}\)
d) \(\dfrac{11}{12}-\left(\dfrac{2}{5}+x\right)=\dfrac{2}{3}\\ \dfrac{11}{12}-\dfrac{2}{5}-x=\dfrac{2}{3}\\ x=\dfrac{11}{12}-\dfrac{2}{5}-\dfrac{2}{3}\\ x=\dfrac{55}{60}-\dfrac{24}{60}-\dfrac{40}{60}\\ x=-\dfrac{9}{60}\)
a) \(\dfrac{-3}{7}-x=\dfrac{-1}{2}\)
\(\Rightarrow x=\dfrac{1}{2}-\dfrac{3}{7}\)
\(\Rightarrow x=\dfrac{1}{14}\)
Vậy \(x=\dfrac{1}{14}\)
b) \(x-\dfrac{4}{5}=\dfrac{1}{2}-\dfrac{1}{3}\)
\(\Rightarrow x-\dfrac{4}{5}=\dfrac{1}{6}\)
\(\Rightarrow x=\dfrac{1}{6}+\dfrac{4}{5}\)
\(\Rightarrow x=\dfrac{29}{30}\)
Vậy \(x=\dfrac{29}{30}\)
c) \(-x-\dfrac{3}{4}=\dfrac{-8}{11}\)
\(\Rightarrow-x=\dfrac{3}{4}-\dfrac{8}{11}\)
\(\Rightarrow-x=\dfrac{1}{44}\)
\(\Rightarrow x-\dfrac{1}{44}\)
Vậy \(x=-\dfrac{1}{44}\)
d) \(\dfrac{11}{12}-\left(\dfrac{2}{5}+x\right)=\dfrac{2}{3}\)
\(\Rightarrow\dfrac{2}{5}+x=\dfrac{11}{12}-\dfrac{2}{3}\)
\(\Rightarrow\dfrac{2}{5}+x=\dfrac{1}{4}\)
\(\Rightarrow x=\dfrac{1}{4}-\dfrac{2}{5}\)
\(\Rightarrow x=\dfrac{-3}{20}\)
Vậy \(x=\dfrac{-3}{20}\)