15-(49-4x)=3x-27
[70-(x-37)-(-61)=14-2x
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17x + 3. ( -16x – 37) = 2x + 43 - 4x
<=>17x-48x-111=-2x+43
<=>-29x=154
<=> \(x=-\frac{154}{29}\)
-3. (2x + 5) -16 < -4. (3 – 2x)
\(\Leftrightarrow-6x-31< -12+8x.\)
\(\Leftrightarrow-14x< 19\Rightarrow x< -\frac{19}{14}\)
Câu 1:
a: =>-2x-x+17=34+x-25
=>-3x+17=x+9
=>-4x=-8
hay x=2
b: =>17x+16x+27=2x+43
=>33x+27=2x+43
=>31x=16
hay x=16/31
c: =>-2x-3x+51=34+2x-50
=>-5x+51=2x-16
=>-7x=-67
hay x=67/7
e: 3x-32>-5x+1
=>8x>33
hay x>33/8
a) 3x - ( 15 + 8 ) = 2x- [ -10:(-5) ]
<=> 3x - 23 = 2x - 2
<=> 3x - 2x = 23 - 2
<=> x = 21
b) -3 |x| = 15-12
<=> -3 |x| = 3
<=> |x| = 3: (-3)
<=> |x| = -1 Vô lý
Không có giá trị nào của x thỏa mãn
c) -7 |x| = -49 + 70
<=> -7 |x| = 21
<=> |x| = 21 : (-7)
<=> |x| = - 3
Không có giá trị nào của x thỏa mãn
d) -4x - ( 20.-18) = -(3x-40)
<=> -4x -2 = -3x +40
<=> 4x - 3x = -2 -40
<=> x = - 42
Nhanh nhanh dùm mình nha mình đang gấp
\(a,\Leftrightarrow\left(3x-7\right)\left(3x+7\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{3}\\x=-\dfrac{7}{3}\end{matrix}\right.\\ b,\Leftrightarrow\left(x+2\right)\left(x-1-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\\ c,\Leftrightarrow4x^2-7x-2-4x^2+4x+3=7\\ \Leftrightarrow-3x=6\Leftrightarrow x=-2\\ d,\Leftrightarrow3x^2+2x+x^2+2x+1-4x^2+25=0\\ \Leftrightarrow4x=-26\Leftrightarrow x=-\dfrac{13}{2}\\ e,\Leftrightarrow x^3+27-x^3+x-27=0\\ \Leftrightarrow x=0\\ f,\Leftrightarrow\left(4x-3\right)\left(4x-3+3x\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=\dfrac{3}{7}\end{matrix}\right.\)
a) 9x2-49=0
(3x)2-72=0
<=> (3x-7)(3x+7)=0
th1: 3x-7=0
<=>3x=7
<=>x=\(\dfrac{7}{3}\)
th2: 3x+7=0
<=>3x=-7
<=>x=\(-\dfrac{7}{3}\)
1) \(2x\left(x-3\right)+5x-15=0\)
\(2x\left(x-3\right)+5\left(x-3\right)=0\)
\(\left(x-3\right)\left(2x+5\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-3=0\\2x+5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{-5}{2}\end{matrix}\right.\)
2) \(x\left(2x-7\right)-4x+14=0\)
\(x\left(2x-7\right)-2\left(2x-7\right)=0\)
\(\left(2x-7\right)\left(x-2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}2x-7=0\\x-2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}\\x=2\end{matrix}\right.\)
3) \(x^2-12x+36=0\)
\(\left(x-6\right)^2=0\)
\(x-6=0\)
\(x=6\)
4) \(\left(x+3\right)\left(x^2-3x+9\right)-x\left(x-1\right)\left(x+1\right)-27=0\)
\(\left(x^3+3^3\right)-x\left(x^2-1\right)-27=0\)
\(x^3+27-x^3+x-27=0\)
\(x=0\)
\(a)5-\left(x-6\right)=4\left(3-2x\right)\)
\(\Leftrightarrow5-x+6=12-8x\)
\(\Leftrightarrow-x+8x=12-5-6\)
\(\Leftrightarrow7x=1\Leftrightarrow x=\frac{1}{7}\)
a) 5-(x-6)=4(3-2x)
<=>5-x-6=12-8x
<=>-x+8x=2-5-6
<=>7x=1
<=>x=1/7
a) 15-(49-4x)=3x-27
<=> 15-49+4x=3x-27
<=> -34+4x-3x=-27
<=> x=-27+34=7
Vậy x=7
b) 70-(x-37)-(-61)=14-2x
<=> 70-x+34+61=14-2x
<=> 120-x+2x=14
<=> x=120-14=106
Vậy x=106