giúp hộ mình
x(x+2)(x+4)(x+6)=9
tìm x nha mn
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BB
1
5 tháng 4 2022
a: \(\Leftrightarrow\left(2-x\right)\left(x-3\right)+\left(x-1\right)\left(x+3\right)=-4x\)
\(\Leftrightarrow2x-6-x^2+3x+x^2+3x-x-3=-4x\)
=>7x-9=-4x
=>11x=9
hay x=9/11
b: \(\Leftrightarrow\left(5-x\right)\left(x-4\right)+\left(x+2\right)\left(x+4\right)=-3x\)
\(\Leftrightarrow5x-20-x^2+4x+x^2+6x+8=-3x\)
=>15x-12=-3x
=>18x=12
hay x=2/3
NX
2
H9
HT.Phong (9A5)
CTVHS
31 tháng 10 2023
\(x+2\dfrac{1}{4}=4\dfrac{5}{6}\)
\(x=4\dfrac{5}{6}-2\dfrac{1}{4}\)
\(x=4+\dfrac{5}{6}-2-\dfrac{1}{4}\)
\(x=2+\dfrac{7}{12}\)
\(x=\dfrac{31}{12}\)
_____________
\(9\dfrac{3}{5}-x=3\dfrac{3}{8}\)
\(x=9\dfrac{3}{5}-3\dfrac{3}{8}\)
\(x=9+\dfrac{3}{5}-3-\dfrac{3}{8}\)
\(x=6+\dfrac{9}{40}\)
\(x=\dfrac{249}{40}\)
VN
Võ Ngọc Phương
VIP
31 tháng 10 2023
\(x+2\dfrac{1}{4}=4\dfrac{5}{6}\)
\(=>x+\dfrac{9}{4}=\dfrac{29}{6}\)
\(=>x=\dfrac{29}{6}-\dfrac{9}{4}=\dfrac{58}{12}-\dfrac{27}{12}\)
\(=>x=\dfrac{31}{12}\)
_______
\(9\dfrac{3}{5}-x=3\dfrac{3}{8}\)
\(=>\dfrac{48}{5}-x=\dfrac{27}{8}\)
\(=>x=\dfrac{48}{5}-\dfrac{27}{8}=\dfrac{384}{40}-\dfrac{135}{40}\)
\(=>x=\dfrac{249}{40}\)
M
4
28 tháng 1 2018
-7+x = 15
=> x = 15-(-7) = 22
2.x-10 = -26
=> 2.x = -26+10 = -16
=> x = -16 : 2 = -8
|x+7| = 4
=> x+7=-4 hoặc x+7=4
=> x=-11 hoặc x=-3
(x-9).(3x+6) = 0
=> x-9=0 hoặc 3x+6=0
=> x=9 hoặc x=-2
Tk mk nha
28 tháng 1 2018
1) Ta có: \(-7+x=15\)
\(\Rightarrow x=15-\left(-7\right)\)
\(\Rightarrow x=22\)
2) \(2x-10=-26\)
\(\Rightarrow2x=-26+10\)
\(\Rightarrow2x=-16\)
\(\Rightarrow x=-8\)
3) \(\left|x+7\right|=4\)
\(\Rightarrow x+7=\pm4\)
Nếu x + 7 = 4 thì x = -3
Nếu x + 7 = -4 thì x = -11
Vậy \(x=\left\{-3;-11\right\}\)
4) \(\left(x-9\right)\left(3x+6\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-9=0\\3x+6=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0+9\\3x=-6\end{cases}\Rightarrow}\orbr{\begin{cases}x=9\\x=-2\end{cases}}}\)
Vậy \(x=\left\{9;-2\right\}\)
16 tháng 2 2022
\(8x-48+4x-12-14=-x+4\)
\(\Leftrightarrow12x-75=-x+4\Leftrightarrow13x=79\Leftrightarrow x=\dfrac{79}{13}\)
\(-7\left(8-x\right)-6\left(x+9\right)=20-x\Leftrightarrow-56+7x-6x-54=20-x\)
\(\Leftrightarrow2x=130\Leftrightarrow x=65\)
\(9x-63-80+60x=-7x+15\Leftrightarrow76x=158\Leftrightarrow x=\dfrac{79}{38}\)
\(-96-16x-60+30x=-40x-16\Leftrightarrow54x=140\Leftrightarrow x=\dfrac{70}{27}\)
\(17x-102-14x-28=4x-24-2x+4\Leftrightarrow x=110\)
L
7 tháng 12 2017
a) X x 2 + X x 3 = 4
X x (2 + 3) = 4
X x 5 = 4
x = 4 : 5
x = 0,8
b) X x 9 - X x 5 = 23
X x (9 - 5) = 23
X x 4 = 23
X = 23 : 4
X = 5,75
7 tháng 12 2017
a, 2x+3x=4 \(\Rightarrow\)5x=4 =>x=\(\frac{4}{5}\)
b, 9x-5x=23 => 4x=23 => x=\(\frac{23}{4}\)
24 tháng 7 2017
a) 32.x+2=1342176728
32.x=134217728-2
32.x=134217726
x=134217726:32
x=4194303,938
12 tháng 6 2023
2:
=>x^3-1-2x^3-4x^6+4x^6+4x=6
=>-x^3+4x-7=0
=>x=-2,59
4: =>8x-24x^2+2-6x+24x^2-60x-4x+10=-50
=>-62x+12=-50
=>x=1
GN
GV Nguyễn Trần Thành Đạt
Giáo viên
19 tháng 7 2023
\(l,\\ 2^x=1=2^0\\ Vậy:x=0\\ m,\\ 3^x=81=3^4\\ Vậy:x=4\\ n,\\ 3^x=37=3^3\\ Vậy:x=3\\ o,\\ 9^x=3^4=\left(3^2\right)^2=9^2\\ Vậy:x=2\)
H9
HT.Phong (9A5)
CTVHS
13 tháng 7 2023
\(6\cdot x+9=15\)
\(\Rightarrow6\cdot x=15-9\)
\(\Rightarrow6\cdot x=6\)
\(\Rightarrow x=\dfrac{6}{6}=1\)
_______________
\(43+2x=49\)
\(\Rightarrow2x=49-43\)
\(\Rightarrow2x=6\)
\(\Rightarrow x=\dfrac{6}{2}=3\)
____________
\(x:2+5=11\)
\(\Rightarrow x:2=11-5\)
\(\Rightarrow x:2=6\)
\(\Rightarrow x=6\cdot2=12\)
______________
\(77-11x=0\)
\(\Rightarrow11x=77\)
\(\Rightarrow x=\dfrac{77}{11}\)
\(\Rightarrow x=7\)
_______________
\(12-4:x=8\)
\(\Rightarrow4:x=12-8\)
\(\Rightarrow4:x=4\)
\(\Rightarrow x=\dfrac{4}{4}=1\)
_____________
\(x:3+8=11\)
\(\Rightarrow x:3=11-8\)
\(\Rightarrow x:3=3\)
\(\Rightarrow x=3\cdot3=9\)
13 tháng 7 2023
a: 6x+9=15
=>6x=6
=>x=1
b: 2x+43=49
=>2x=6
=>x=3
c: x:2+5=11
=>x:2=6
=>x=12
d: 77-11x=0
=>7-x=0
=>x=7
e: 12-4:x=8
=>4:x=4
=>x=1
f: x:3+8=11
=>x:3=3
=>x=9
\(x\left(x+2\right)\left(x+4\right)\left(x+6\right)=9\)
\(\Leftrightarrow x\left(x+2\right)\left(x+4\right)\left(x+6\right)-9=0\)
Đặt \(A=x\left(x+2\right)\left(x+4\right)\left(x+6\right)-9\)
\(A=x\left(x+6\right)\left(x+2\right)\left(x+4\right)-9\)
\(=\left(x^2+6x\right)\left(x^2+6x+8\right)-9\)(1)
Đặt \(a=x^2+6x\)
\(\Rightarrow\left(1\right)=a\left(a+8\right)-9=a^2+8a-9\)
\(=\left(a+4\right)^2-25=\left(a+4-5\right)\left(a+4+5\right)\)
\(=\left(a-1\right)\left(a+9\right)=\left(x^2+6x-1\right)\left(x^2+6x+9\right)\)
\(=\left(x^2+6x-1\right)\left(x+3\right)^2\)
\(pt\Leftrightarrow\left(x^2+6x-1\right)\left(x+3\right)^2=0\)
\(TH1:x^2+6x-1=0\)
\(\Leftrightarrow\left(x+3\right)^2-10=0\)
\(\Leftrightarrow\left(x+3\right)^2=10\)
\(\Leftrightarrow\orbr{\begin{cases}x+3=\sqrt{10}\\x+3=-\sqrt{10}\end{cases}}\Leftrightarrow x=\pm\sqrt{10}+3\)
\(TH2:\left(x+3\right)^2=0\Leftrightarrow x+3=0\Leftrightarrow x=-3\)
Vậy \(x\in\left\{\pm\sqrt{10}+3;-3\right\}\)