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14-(7-x+3)=5-[4-(5-3)] 14-(7-x+3)=5-[4-2] 14-(7-x+3)=5-2 14-(7-x+3)=3 7-x+3=14-3 7-x+3=11 7-x=11-3 7-x=8 x=7-8 x=-1 vậy x=-1
14-(7-x+3)=5-[4-(5-3] 14-(7-x+3)=5-[4-2] 14-(7-x+3)=5-2 14-(7-x+3)=3 7-x +3=14-3 7-x+3=11 7-x=11-3 7-x=8 x=7-8 x=-1
\(71+\left(26-3x\right)\div5=75\)
\(\left(26-3x\right)\div5=4\)
\(26-3x=20\)
\(3x=6\Leftrightarrow x=2\)
\(20-\left[7\left(x-3\right)+4\right]=2\)
\(7x-21+4=18\)
\(7x-17=18\)
\(7x=25\Leftrightarrow x=\frac{25}{7}\)
a, 3.(2\(x\) + 4) + 198 = (-3)2.10
3.(2\(x\) + 4) + 198 = 90
3.(2\(x\) + 4) = 90 - 198
3.(2\(x\) + 4) = - 108
2\(x\) + 4 = -108 : 3
2\(x\) + 4 = -36
2\(x\) = - 36 - 4
2\(x\) = - 40
\(x\) = -40 : 2
\(x\) = - 20
b, 2.(\(x\) + 7) - 6 = 18
2.(\(x\) + 7) = 18 + 6
2.(\(x\) + 7) =24
\(x\) + 7 = 24 : 2
\(x\) + 7 = 12
\(x\) = 12 - 7
\(x\) = 5
Bài 1:
Ta có: \(4-2\left(x+1\right)=2\)
\(\Leftrightarrow2\left(x+1\right)=2\)
\(\Leftrightarrow x+1=1\)
hay x=0
Bài 2:
Ta có: \(\left|2x-3\right|-1=2\)
\(\Leftrightarrow\left|2x-3\right|=3\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-3=3\\2x-3=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=6\\2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=0\end{matrix}\right.\)
b1
A,=\(\frac{35}{54}\) B,=4
b2
A,x=\(\frac{7}{15}\) B=\(\frac{-9}{7}\)
CHÚC BẠN HỌC GIỎI
\(2\left(x-3\right)-5\left(x-4\right)=-7+14.\left(-3\right)\)
\(2x-6-5x+20=-7+\left(-42\right)\)
\(\left(2x-5x\right)+\left(-6+20\right)=-49\)
\(-3x+14=-49\)
\(-3x=-63\)
\(x=21\)
thank you