tim x:
-3x-(-24)=-45
x/7=-6/21
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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
Bài 1: Tìm x, biết
a )24-(36+5)=x b)14-21=(13-x)-(15+8)
24-41=x (13-x)-23=-7
x=-17 13-x=(-7)+23
Vậy x=-17 13-x=16
x=13-16
x=-3 Vậy x=-3
Bài 2:Tìm x, biết
a)17-x=-25+(-16+9) b)3x-21=-19-(-2x)
17-x=-25+(-7) 3x-21=-19+2x
17-x=-32 3x-2x=-19+21
x=17-(-32) x=4
x=49 Vậy x=4
Vậy x=49
Bài 1:
a. 24 - (36+5) = x
=> 24 - 41 = x
=> -17 = x
=> x = -17
b. 14 - 21 = (13 - x) - (15 + 8)
=> -7 = 13 - x - 23
=> -7 - 13 + 23 = -x
=> 3 = -x
=> x = -3
Bài 2:
a. 17 - x = -25 + (-16 + 9)
=> 17 - x = -25 + (-7)
=> 17 - x = -32
=> 17 + 32 = x
=> x = 49
b. 3x - 21 = -19 - (-2x)
=> 3x - 21 = -19 + 2x
=> 3x - 2x = -19 + 21
=> x = 2
a) 219-7(x+1)=100
7(x+1)= 219-100
7(x+1) =119
x+1 = 119:7
x+1 = 17
=>x = 16.
b)(3x-6).3 = 24
3x-6 = 24:3
3x - 6 = 8
3x = 8+6
3x = 14
=> X không thỏa mãn vì x này là số thập phân
a)219-7(x+1)=100
212(x+1)=100
x+1=212:100
x+1=2,12
x=2,12-1
x=1,12
a,3 + 2x-1=24-[42-(22-1)]
3 + 2x-1=24-[42-(4-1)]
3 + 2x-1=24-(16-3)
3 + 2x-1=24-13
3 + 2x-1=11
3 + 2x=11+1
3 + 2x=12
2x=12-3
2x=9(đề sai)
b,(3x-24).75=2 .76
(3x-16).16807=2 .117649
(3x-16).16807=235298
3x-16=235298:16807
3x-16=14
3x=14+16
3x=30
x=30:3
x=10
Giải:
a) \(\left|48-3x\right|=0\)
\(\Leftrightarrow48-3x=0\)
\(\Leftrightarrow3x=48\)
\(\Leftrightarrow x=\dfrac{48}{3}=16\)
Vậy x = 16.
b) \(\left|-x-7\right|=24\)
\(\Leftrightarrow\left[{}\begin{matrix}-x-7=24\\-x-7=-24\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}-x=31\\-x=-17\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-31\\x=17\end{matrix}\right.\)
Vậy \(x=-31\) hoặc \(x=17\).
c) \(\left|4-x\right|=21\)
\(\Leftrightarrow\left[{}\begin{matrix}4-x=21\\4-x=-21\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-17\\x=25\end{matrix}\right.\)
Vậy \(x=-17\) hoặc \(x=25\).
d) \(\left|x+8\right|+12=0\)
\(\Leftrightarrow\left|x+8\right|=-12\)
\(\Leftrightarrow\left[{}\begin{matrix}x+8=-12\\x+8=-\left(-12\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-20\\x=4\end{matrix}\right.\)
Vậy \(x=-20\) hoặc \(x=4\).
e) \(6-\left|x\right|=2\)
\(\Leftrightarrow\left|x\right|=6-2=4\)
\(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-4\end{matrix}\right.\)
Vậy \(x=4\) hoặc \(x=-4\).
Chúc bạn học tốt!
1. |48 - 3x| = 0.
\(\Leftrightarrow\) 48 - 3x = 0.
\(\Leftrightarrow\) 3x = 48.
\(\Leftrightarrow\) x = \(\dfrac{48}{3}=16.\)
Vậy x = 16.
2. |-x - 7| = 24.
\(\Leftrightarrow\left[{}\begin{matrix}-x-7=24.\\-x-7=-24.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}-x=31.\\-x=-17.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-31.\\x=17.\end{matrix}\right.\)
Vậy \(\left[{}\begin{matrix}x=-31.\\x=17.\end{matrix}\right.\)
3. |4 - x| = 21.
\(\Leftrightarrow\left[{}\begin{matrix}4-x=21.\\4-x=-21.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-17.\\x=25.\end{matrix}\right.\)
Vậy \(\left[{}\begin{matrix}x=-17.\\x=25.\end{matrix}\right.\)
4. |x + 8| + 12 = 0.
|x + 8| = 0 - 12.
|x + 8| = -12.
\(\Leftrightarrow\left[{}\begin{matrix}x+8=-12.\\x+8=-\left(-12\right).\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x+8=-12.\\x+8=12.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-20.\\x=4.\end{matrix}\right.\)
Vậy \(\left[{}\begin{matrix}x=-20.\\x=4.\end{matrix}\right.\)
5. 6 - |x| = 2.
|x| = 6 - 2.
|x| = 4.
\(\Leftrightarrow\left[{}\begin{matrix}x=4.\\x=-4.\end{matrix}\right.\)
Vậy \(\left[{}\begin{matrix}x=4.\\x=-4.\end{matrix}\right.\)
Tìm x biết :
a) 3(5/3x-7)-2(1.5x+6)-(5-x)(x+4)=80+x^2
b) 4/5x^2(x/3-1/2)-(1/5x-2/3)(4x^2/3+1)=22/45x^2
`Answer:`
\(3\left(\frac{5}{3}x-7\right)-2\left(1.5x+6\right)-\left(5-x\right)\left(x+4\right)=80+x^2\)
\(\Leftrightarrow3\left(\frac{5x}{3}-7\right)-2\left(5x+6\right)-\left(5-x\right)\left(x+4\right)=80+x^2\)
\(\Leftrightarrow5x-21-10x-12-5x-20+x^2+4x=80+x^2\)
\(\Leftrightarrow5x-21-10x-12-5x-20+4x=80\)
\(\Leftrightarrow-6x-53=80\)
\(\Leftrightarrow-6x=133\)
\(\Leftrightarrow x=-\frac{133}{6}\)
\(\frac{4}{5}x^2\left(\frac{x}{3}-\frac{1}{2}\right)-\left(\frac{1}{5}x-\frac{2}{3}\right)\left(4\frac{x^2}{3}+1\right)=\frac{22}{45}x^2\)
\(\Leftrightarrow36x^2\left(\frac{x}{3}-\frac{1}{2}\right)-45\left(\frac{x}{5}-\frac{2}{3}\right)\left(\frac{4x^2}{3}+1\right)=22x^2\)
\(\Leftrightarrow12x^3-18x^2-12x^3-9x+40x^2+30=22x^2\)
\(\Leftrightarrow22x^2-9x+30=22x^2\)
\(\Leftrightarrow-9x+30=0\)
\(\Leftrightarrow-9x=-30\)
\(\Leftrightarrow x=\frac{10}{3}\)
Tìm x biết :
a) 3(5/3x-7)-2(1.5x+6)-(5-x)(x+4)=80+x^2
b) 4/5x^2(x/3-1/2)-(1/5x-2/3)(4x^2/3+1)=22/45x^2
-3x-(-24)=-45
- 3x + 24 = -45
- 3x =-45-24
- 3x =-69
x= 69:3
x= 23
Vậy...
\(\frac{x}{7}\)= \(-\frac{6}{21}\)
=> 21 x = -6.7
21x = -42
x= -42:21
x= -2
Vậy...