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e: =>-40+3+33+40-x=47
=>36-x=47
=>x=-11
f: =>x(x-3)(11-x)(11+x)=0
hay \(x\in\left\{0;3;11;-11\right\}\)
g: =>-62-38-x+2x=-100
=>x-100=-100
hay x=0
i: =>x-12-2x-31=6
=>-x-43=6
=>x+43=-6
hay x=-49
h: =>(x+1)=0
=>x=-1
f: =>x(x-3)(x+11)(x-11)=0
hay \(x\in\left\{0;3;-11;11\right\}\)
l) (x + 9) . (x2 – 25) = 0
<=> (x + 9) . (x – 5) . (x + 5) = 0
<=> \(\left[{}\begin{matrix}\text{x + 9 = 0}\\x-5=0\\x+5=0\end{matrix}\right.\left[{}\begin{matrix}x=-9\\x=5\\x=-5\end{matrix}\right.\)
Vậy S = \(\left\{-9,5,-5\right\}\)
e) |x - 4 |< 7
<=> \(\left[{}\begin{matrix}x-4=7\\x-4=-7\end{matrix}\right.< =>\left[{}\begin{matrix}x=11\\x=-3\end{matrix}\right.\)
Vậy S = \(\left\{11;-3\right\}\)
I,(x+9).(x^2-25)=0
tương đương:x+9=0
x^2-25=0
tương đương : x=-9
x=5
e,\(\left|x-4\right|\)=7
tương đương x-4=4
x-4=-4
tương đương :x=0
x=-8
Bài 1:
a) Ta có: \(-5x+32=\left(-2\right)^3\)
\(\Leftrightarrow-5x+32=-8\)
\(\Leftrightarrow-5x=-40\)
hay x=8
Vậy: x=8
b) Ta có: \(-2x+36=6\)
\(\Leftrightarrow-2x=6-36=-30\)
hay x=15
Vậy: x=15
e) Ta có: \(\left(x+9\right)\left(x^2-25\right)=0\)
\(\Leftrightarrow\left(x+9\right)\left(x-5\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+9=0\\x-5=0\\x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-9\\x=5\\x=-5\end{matrix}\right.\)
Vậy: \(x\in\left\{-9;5;-5\right\}\)
b,-2x+36=6
tương đương -2x=-30
tương đương x=15
a, -5x+32=(-2)^3
tương đương -5x+32=8
tương đương -5x=-24
tương đương x=24/5
\(1\)) \(5-\left(10-x\right)=7\)
\(10-x=5-7\)
\(10-x=-2\)
\(x=10-\left(-2\right)\)
\(x=12\)
\(2\)) \(-32-\left(x-5\right)=0\)
\(x-5=-32-0\)
\(x-5=-32\)
\(x=-32+5\)
\(x=-27\)
1) Ta có: \(\left(-86-x\right)-\left(3+2x\right)=-4-15\)
\(\Leftrightarrow-86-x-3-2x+4+15=0\)
\(\Leftrightarrow-3x-70=0\)
\(\Leftrightarrow-3x=70\)
hay \(x=-\dfrac{70}{3}\)
Vậy: \(x=-\dfrac{70}{3}\)
2) Ta có: \(18+\left(-x\right)-\left(40-28\right)=-32-\left(-18\right)\)
\(\Leftrightarrow18-x-40+28+32-18=0\)
\(\Leftrightarrow-x+20=0\)
\(\Leftrightarrow-x=-20\)
hay x=20
Vậy: x=20
3) Ta có: \(-27-\left(-31+x\right)-25=-5-17\)
\(\Leftrightarrow-27+31-x-25+5+17=0\)
\(\Leftrightarrow-x+1=0\)
\(\Leftrightarrow-x=-1\)
hay x=1
Vậy: x=1
4) Ta có: \(-9-14-x+42-38=-5+13\)
\(\Leftrightarrow-x-19=8\)
\(\Leftrightarrow-x=27\)
hay x=-27
Vậy: x=-27
Bài 1: Tìm x
a) 2x - 15 = -27
b) 2 (x + 1) – 3 = 7
c) 14 – (40 – x) = -27
d) 96 – 2(4 – 5x) = -12
e) – 40 – (– 3 – 33) + (40 – x) = – (– 47)
f) x(3x – 9). (121 – x2) = 0
g) – 62 – (38 + x) + 2x = – 100
h) (x + 1)2.(x2 + 1) = 0
i) (x – 12) – (2x + 31) = 6
k) 17/ (x + 3)3 : 3 – 1 = – 10
Bài 1: a) 2x - 15 = 27 => 2x = 27 + 15 = 42 => x = 42 : 2 = 21. b) 2(x + 1) - 3 = 7 => 2(x + 1) = 7 + 3 = 10 => x + 1 = 10 : 2 = 5 => x = 5 - 1 = 4. c) 14 - (40 - x) = -27 => 40 - x = 14 - (-27) = 41 => x = 40 - 41 = -1. d) 96 - 2(4 - 5x) = -12 => 2(4 - 5x) = 96 - (-12) = 108 => 4 - 5x = 108 : 2 = 54 => 5x = 4 - 54 = -50 => x = (-50) : 5 = -10.