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a/
\(-24+\left(x+4\right)^4=10^3\)3
\(\Leftrightarrow-24+x^4+16x^3+96x^2+256x+256=10^3\)
<=>\(x^4+16x^3+96x^2+256x-768=0\)
Giải trên tập số phức ta được
\(x=-\sqrt{32}-4\)
\(x=\sqrt{32}-4\)
\(x=-\sqrt{32}i-4\)
\(x=\sqrt{32}1-4\)=> Phần a kog có giá trị nguyên nào của x thỏa mãn phương trình
b/
2(x+7)-3(6-x)=-24
<=> 2x+14-18+3x=-24
<=>5x=-20
<=>x=-4
Vậy x=-4
c/
\(3x-6x^2=0\)
\(\Leftrightarrow3x\left(1-2x\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}3x=0\\1-2x=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=\frac{1}{2}\end{cases}}\)(x = 1/2 kog thỏa mãn yêu cầu)
Vậy x=0
a/\(\left(x+4\right)^4=1000+24\)
\(\Rightarrow x^4+8x^2+4^4-1024=0\)
\(\Rightarrow x^4+8x^2-768\)
\(\Rightarrow x^4-24x+32x-768=0\)
\(\Rightarrow x.\left(x-24\right)+32.\left(x-24\right)\)
\(\Rightarrow\left(x+32\right).\left(x-24\right)\Rightarrow\orbr{\begin{cases}x=-32\\x=24\end{cases}}\)
b/2x+14-18+3x=-24
5x=-24-14+18
x=-20/5=-4
c/3x-6x\(^2\) =0
\(3x.\left(1-2x\right)=0\)
\(\Rightarrow\orbr{\begin{cases}3x=0\rightarrow x=0\\1-2x=0\rightarrow x=\frac{1}{2}\end{cases}}\)
KL bAN tu lam nhe
a) \(\dfrac{2}{3}x-\dfrac{1}{2}=\dfrac{1}{10}\)
\(\dfrac{2}{3}x=\dfrac{1}{10}+\dfrac{1}{2}=\dfrac{3}{5}\)
\(x=\dfrac{3}{5}:\dfrac{2}{3}=\dfrac{9}{10}\)
b) \(\dfrac{39}{7}:x=13\)
\(x=\dfrac{\dfrac{39}{7}}{13}=\dfrac{3}{7}\)
c) \(\left(\dfrac{14}{5}x-50\right):\dfrac{2}{3}=51\)
\(\dfrac{14}{5}x-50=51\cdot\dfrac{2}{3}=34\)
\(\dfrac{14}{5}x=34+50=84\)
\(x=\dfrac{84}{\dfrac{14}{5}}=30\)
d) \(\left(x+\dfrac{1}{2}\right)\left(\dfrac{2}{3}-2x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=0\\\dfrac{2}{3}-2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=\dfrac{1}{3}\end{matrix}\right.\)
e) \(\dfrac{2}{3}x-\dfrac{1}{2}x=\dfrac{5}{12}\)
\(\dfrac{1}{6}x=\dfrac{5}{12}\)
\(x=\dfrac{5}{12}:\dfrac{1}{6}=\dfrac{5}{2}\)
g) \(\left(x\cdot\dfrac{44}{7}+\dfrac{3}{7}\right)\dfrac{11}{5}-\dfrac{3}{7}=-2\)
\(\left(x\cdot\dfrac{44}{7}+\dfrac{3}{7}\right)\cdot\dfrac{11}{5}=-2+\dfrac{3}{7}=-\dfrac{11}{7}\)
\(x\cdot\dfrac{44}{7}+\dfrac{3}{7}=-\dfrac{11}{7}:\dfrac{11}{5}=-\dfrac{5}{7}\)
\(\dfrac{44}{7}x=-\dfrac{5}{7}-\dfrac{3}{7}=-\dfrac{8}{7}\)
\(x=-\dfrac{8}{7}:\dfrac{44}{7}=-\dfrac{2}{11}\)
h) \(\dfrac{13}{4}x+\left(-\dfrac{7}{6}\right)x-\dfrac{5}{3}=\dfrac{5}{12}\)
\(\dfrac{25}{12}x-\dfrac{5}{3}=\dfrac{5}{12}\)
\(\dfrac{25}{12}x=\dfrac{5}{12}+\dfrac{5}{3}=\dfrac{25}{12}\)
\(x=1\)
Mỏi tay woa bn làm nốt nha!!
A,-16+23+x=-16
<=>x=-16+16-23
<=>x=-23
B,10-2(4-3x)=-4
<=>10-8+6x=-4
<=>6x=-4-10+8
<=>6x=-6
<=>x=-1
C,-12+3(-x+7)=-18
<=>-12-3x+21=-18
<=>-3x=-18+12-21
<=>-3x=-27
<=>x=9
D,24:(3x-2)=-3
<=>24/(3x-2)=-3
<=>-3(3x-2)=24
<=>3x-2=-8
<=>3x=-6
<=>x=-2
E,|x+8|-7=8
<=>|x+8|=15
<=>x+8=+-15
Chia 2 TH:
TH1:x+8=15
<=>x=7
TH2:x+8=-15
<=>x=-23
F,-45:5(-3-2x)=3
<=>-9(-3-2x)=3
<=>27+18x=3
<=>18x=-24
<=>x=-4/3
G,|x-1|=0
<=>x-1=0
<=>x=1
H,-13|x|=-26
<=>|x|=2
<=>x=+-2
a) -16 + 23 + x = -16
x = -16 + 16 - 23
x = -23
Vậy x = -23
b) 10 - 2. (4 - 3x) = -4
10 - 2 . 4 + 2. 3x = -4
10 - 8 + 6x = -4
6x = -4 - 10 + 8
6x = -6
x = (-6) : 6
x = -1
Vậy x = -1
c) -12 + 3. (-x + 7) = -18
-12 + 3. (-x) + 3. 7 = -18
-12 + (-3x) + 21 = -18
-3x = -18 + 12 - 21
-3x = -27
x = (-27) : (-3)
x = -9
Vậy x = -9
d) 24 : (3x - 2) = -3
3x - 2 = 24 : (-3)
3x - 2 = -8
3x = -8 + 2
3x = -6
x = (-6) : 3
x = -2
Vậy x = -2
e) lx + 8l - 7 = 8
lx + 8l = 8 + 7
lx + 8l = 15
\(\Rightarrow\hept{\begin{cases}x+8=15\\x=18-5\\x=13\end{cases}hay\hept{\begin{cases}x+8=-15\\x=-15-8\\x=-23\end{cases}}}\)
Vậy \(x\in\left\{13;-23\right\}\)
f) (-45) : 5. (-3 - 2x) = 3
(-9). (-3 - 2x) = 3
(-9). (-3) + 9. 2x = 3
27 + 18x = 3
18x = 3 - 27
18x = -24
x = (-24) : 18
x =\(\frac{-4}{3}\)
Vậy x =\(\frac{-4}{3}\)
g) lx - 1l = 0
\(\Rightarrow x-1=0\)
x = 0 + 1
x = 1
Vậy x = 1
h) (-13). lxl = -26
lxl = (-26) : (-13)
lxl = 2
\(\Rightarrow x=2\)
Vậy x = 2
Chúc bạn học tốt!!!
