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Câu a:
(-15 + |\(x\)|) + (25 - |-\(x\)|)
= -15 + |\(x\)| + 25 - |\(x\)|
= (25- 15) + (|\(x\)| - |\(x\)|)
= 10 + 0
= 10
b; \(x-34\) - [(15 + \(x\)) -(23 - \(x\))]
\(x-34\) - 15 - \(x\) + 23 - \(x\)
(\(x-x-x\)) - (34 + 15 - 23)
= (0 - \(x\)) - (49 - 23)
= -\(x\) - 26
a ) Ta có : 4(x - 5) - 3(x + 7) = -19
<=> 4x - 20 - 3x - 21 = -19
=> x - 41 = -19
=> x = -19 + 41
=> x = 22
b) Ta có " 7(x - 3) - 5(3 - x) = 11x - 5
<=> 7x - 21 - 15 + 5x = 11x - 5
<=> 12x - 36 = 11x - 5
=> 12x - 11x = -5 + 36
=> x = 31
a) \(2\left(x-5\right)-3\left(x+7\right)=14\)
\(\Leftrightarrow2x-10-3x-21=14\)
\(\Leftrightarrow-x-31=14\)
\(\Leftrightarrow-x=45\Leftrightarrow x=-45\)
b) \(5\left(x-6\right)-2\left(x+3\right)=12\)
\(\Leftrightarrow5x-30-2x-6=12\)
\(\Leftrightarrow3x-36=12\)
\(\Leftrightarrow3x=48\Leftrightarrow x=16\)
c) \(3\left(x-4\right)-\left(8-x\right)=12\)
\(\Leftrightarrow3x-12-8+x=12\)
\(\Leftrightarrow4x-20=12\)
\(\Leftrightarrow4x=32\Leftrightarrow x=8\)
d) \(-7\left(3x-5\right)+2\left(7x-14\right)=28\)
\(\Leftrightarrow-21x+35+14x-28=28\)
\(\Leftrightarrow-7x+35=0\Leftrightarrow x=5\)
Câu a:
1 + {-2 - [-3 + (-4 + |x|)]} = 1 - 2 + [(-3 - 4)]
1 + {-2 -[-3 - 4 + |x|]} = 1 - 2 + [-7]
1 + {-2 - [- 7 + |x|]} = 1 - 2 - 7
1 + {- 2 + 7 - |x|} = - 1 - 7
1 + {5 - |x|} = - 8
5 - |x| = - 8 - 1
5 - |x| = - 9
|x| = 5 + 9
|x| = 14
x = - 14 hoặc x = 14
Vậy x ∈ {-14; 14}
Câu b:
34 + (9 - 21) = 3417 - (x + 3417)
34 + (-12) = 3417 - x - 3417
34 - 12 = 3417 - 3417 - x
22 = - x
x = 22 : (-1)
x = - 22
Vậy x = - 22
\(a)4-3x=5x-28\Leftrightarrow4+28=5x+3x\Leftrightarrow32=8x\Leftrightarrow4=x\)
\(b)\left(x-3\right)\left(x-16\right)=0\Leftrightarrow\hept{\begin{cases}x-3=0\\x-16=0\end{cases}\Leftrightarrow}\hept{\begin{cases}x=3\\x=16\end{cases}}\)
\(c)3\left(x-3\right)+2\left(5-x\right)=7\Leftrightarrow3x-9+10-2x=7\Leftrightarrow x+1=7\Leftrightarrow x=6\)
\(d)3\left(x+2\right)=2\left(8-x\right)\Leftrightarrow3x+6=16-2x\Leftrightarrow3x+2x=16-6\Leftrightarrow5x=10\Leftrightarrow x=2\)
Ta có :
TH1 : \(x< -4\) ; ta có :
\(Q=\left[-\left(x-3\right)\right]+\left[-\left(x+4\right)\right]+x-5\)
\(=3-x-x-4+x-5\)
\(=-6-x\)
TH2 : \(-4\le x< 3\) , ta có :
\(Q=\left[-\left(x-3\right)\right]+\left(4-x\right)+x-5\)
\(=3-x+4-x+x-5\)
\(=2-x\)
TH3 : \(x\ge3\) , ta có :
\(Q=\left(x-3\right)+\left(4+x\right)+x-5\)
\(=x-3+4+x+x-5\)
\(=3x-4\)
Q= \(\left|x-3\right|\) + \(\left|4+x\right|\) + x - 5
= x - 3 + 4 + x + x -5
= 3x - 4