Tìm x biết (x^2-1)(x-5^2)(x^2-8)(x^2-15)< 0 và x thuộc Z
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1: x=3/4-1/2=3/4-2/4=1/4
2: x-1/5=2/11
=>x=2/11+1/5=21/55
3: x-5/6=16/42-8/56
=>x-5/6=8/21-4/28=5/21
=>x=5/21+5/6=15/14
4: x/5=5/6-19/30
=>x/5=25/30-19/30=6/30=1/5
=>x=1
5: =>|x|=1/3+1/4=7/12
=>x=7/12 hoặc x=-7/12
6: x=-1/2+3/4
=>x=3/4-1/2=1/4
11: x-(-6/12)=9/48
=>x+1/2=3/16
=>x=3/16-1/2=-5/16
1)x= 1/4
2)x= 2/11+ 1/5
x= 21/55
3)x - 5/6 = 5/21
x = 5/21+5/6
x = 15/14
4)x/5 = 5/6 + -19/30
x:5 = 1/5
x = 1/5.5
x = 1
5) |x| - 1/4 = 6/18
|x| = 6/18 - 1/4
|x| =7/12
⇒x= 7/12 hoặc -7/12
6)x = -1/2 +3/4
x= 1/4
7) x/15 = 3/5 + -2/3
x:15 = -1/15
x = -1/15. 15
x = -1
8)11/8 + 13/6 = 85/x
85/24 = 85/x
⇒ x = 24
9) x - 7/8 = 13/12
x = 13/12 + 7/8
x = 47/24
10)x - -6/15 = 4/27
x = 4/27 + (-6/15)
x = -34/135
11) -(-6/12)+x = 9/48
x= 9/48 - 6/12
x = -5/16
12) x - 4/6 = 5/25 + -7/15
x -4/6 = -4/15
x = -4/15 + 4/6
x = 2/5
Bài 1:
a) \(=\dfrac{8}{15}\left(\dfrac{7}{13}+\dfrac{6}{13}\right)=\dfrac{8}{15}.1=\dfrac{8}{15}\)
b) \(=\dfrac{3.3-7-2.4}{12}=-\dfrac{6}{12}=-\dfrac{1}{2}\)
Bài 2:
\(\dfrac{x}{2,7}=-\dfrac{2}{3,6}\Rightarrow x=\dfrac{\left(-2\right).2,7}{3,6}\Rightarrow x=-\dfrac{3}{2}\)
Bài 3:
\(\dfrac{x}{2}=\dfrac{y}{5}=\dfrac{x+y}{2+5}=-\dfrac{21}{7}=-3\)
\(\Rightarrow\left\{{}\begin{matrix}x=\left(-3\right).2=-6\\y=\left(-3\right).5=-10\end{matrix}\right.\)
a: 2x(x+1)-135=-200
=>2(x^2+x)=-65
=>2x^2+2x+65=0
=>x^2+x+32,5=0
=>x^2+x+0,25+32,25=0
=>(x+0,5)^2+32,25=0(vô lý)
b: 4x-5(x-1)+15=13
=>4x-5x+5=-2
=>5-x=-2
=>x=5+2=7
c: 2/3x-1/4=3/5-7/8
=>2/3x=3/5-7/8+1/4=24/40-35/40+10/40=-1/40
=>x=-1/40:2/3=-1/40*3/2=-3/80
d: 1/2(2x-3)+105/2=-137/2
=>1/2(2x-3)=-137/2-105/2=-242/2=-121
=>2x-3=-242
=>2x=-239
=>x=-239/2
1: =>3^x=81
=>x=4
2: =>2^x=8
=>x=3
3: =>x^3=2^3
=>x=2
4: =>x^20-x=0
=>x(x^19-1)=0
=>x=0 hoặc x=1
5: =>2^x=32
=>x=5
6: =>(2x+1)^3=9^3
=>2x+1=9
=>2x=8
=>x=4
7: =>x^3=115
=>\(x=\sqrt[3]{115}\)
8: =>(2x-15)^5-(2x-15)^3=0
=>(2x-15)^3*[(2x-15)^2-1]=0
=>2x-15=0 hoặc (2x-15)^2-1=0
=>2x-15=0 hoặc 2x-15=1 hoặc 2x-15=-1
=>x=15/2 hoặc x=8 hoặc x=7
1. Tìm số tự nhiên x biết:
1) \(3^x.3=243\)
\(3^x=243:3\)
\(3^x=81\)
\(3^x=3^4\)
\(\Rightarrow x=4\)
_____
2) \(7.2^x=56\)
\(2^x=56:7\)
\(2^x=8\)
\(2^x=2^3\)
\(\Rightarrow x=3\)
_____
3) \(x^3=8\)
\(x^3=2^3\)
\(\Rightarrow x=3\)
_____
4) \(x^{20}=x\)
\(x^{20}-x=0\)
\(x\left(x^{19}-1\right)=0\)
\(\Rightarrow x=0\) hoặc \(x=1\)
5) \(2^x-15=17\)
\(2^x=17+15\)
\(2^x=32\)
\(2^x=2^5\)
\(\Rightarrow x=5\)
_____
6) \(\left(2x+1\right)^3=9.81\)
\(\left(2x+1\right)^3=729=9^3\)
\(\rightarrow2x+1=9\)
\(2x=9-1\)
\(2x=8\)
\(x=8:2\)
\(\Rightarrow x=4\)
_____
7) \(x^6:x^3=125\)
\(x^3=125\)
\(x^3=5^3\)
\(\Rightarrow x=5\)
_____
8) \(\left(2x-15\right)^5=\left(2x-15\right)^3\)
\(\rightarrow\left(2x-15\right)^5-\left(2x-15\right)^3=0\)
\(\left(2x-15\right)^3.\left[\left(2x-15\right)^2-1\right]=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(2x-15\right)^3=0\\\left(2x-15\right)^2-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{15}{2}\\x=7\\x=8\end{matrix}\right.\)
_____
9) \(3^{x+2}-5.3^x=36\)
\(3^x.\left(3^2-5\right)=36\)
\(3^x.\left(9-5\right)=36\)
\(3^x.4=36\)
\(3^x=36:4\)
\(3^x=9\)
\(3^x=3^2\)
\(\Rightarrow x=2\)
_____
10) \(7.4^{x-1}+4^{x+1}=23\)
\(\rightarrow7.4^{x-1}+4^{x-1}.4^2=23\)
\(4^{x-1}.\left(7+4^2\right)=23\)
\(4^{x-1}.\left(7+16\right)=23\)
\(4^{x-1}.23=23\)
\(4^{x-1}=23:23\)
\(4^{x-1}=1\)
\(4^{x-1}=4^1\)
\(\rightarrow x-1=0\)
\(x=0+1\)
\(\Rightarrow x=1\)
Chúc bạn học tốt
Bài 1:
a) \(\dfrac{9}{20}-\dfrac{8}{15}\times\dfrac{5}{12}\)
\(=\dfrac{9}{20}-\dfrac{2}{9}\)
\(=\dfrac{41}{180}\)
b) \(\dfrac{2}{3}\div\dfrac{4}{5}\div\dfrac{7}{12}\)
\(=\dfrac{2}{3}\times\dfrac{5}{4}\times\dfrac{12}{7}\)
\(=\dfrac{5}{6}\times\dfrac{12}{7}\)
\(=\dfrac{10}{7}\)
c) \(\dfrac{7}{9}\times\dfrac{1}{3}+\dfrac{7}{9}\times\dfrac{2}{3}\)
\(=\dfrac{7}{9}\times\left(\dfrac{1}{3}+\dfrac{2}{3}\right)\)
\(=\dfrac{7}{9}\times1\)
\(=\dfrac{7}{9}\)
Bài 2:
a) \(2\times\left(x-1\right)=4026\)
\(\left(x-1\right)=4026\div2\)
\(x-1=2013\)
\(x=2014\)
Vậy: \(x=2014\)
b) \(x\times3,7+6,3\times x=320\)
\(x\times\left(3,7+6,3\right)=320\)
\(x\times10=320\)
\(x=320\div10\)
\(x=32\)
Vậy: \(x=32\)
c) \(0,25\times3< 3< 1,02\)
\(\Leftrightarrow0,75< 3< 1,02\) ( S )
=> \(0,75< 1,02< 3\)
1) |x + 2| = 4
\(\Leftrightarrow\orbr{\begin{cases}x+2=4\\x+2=-4\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=2\\x=-6\end{cases}}\)
2) 3 – |2x + 1| = (-5)
\(\Leftrightarrow\left|2x+1\right|=8\Leftrightarrow\orbr{\begin{cases}2x+1=8\\2x+1=-8\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{7}{2}\\x=\frac{-9}{2}\end{cases}}\)
3) 12 + |3 – x| = 9
\(\Leftrightarrow\left|3-x\right|=-3\)(vô lí)
=>\(x=\varnothing\)
1) I x+2 I=4
\(\Rightarrow\orbr{\begin{cases}x+2=4\\x+2=-4\end{cases}\Leftrightarrow\orbr{\begin{cases}x=2\\x=-6\end{cases}}}\)
2) \(3-|2x+1|=-5\)
\(\Leftrightarrow|2x+1|=8\)
\(\Leftrightarrow\orbr{\begin{cases}2x+1=8\\2x+1=-8\end{cases}\Leftrightarrow\orbr{\begin{cases}2x=7\\2x=-9\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{7}{2}\\x=\frac{-9}{2}\end{cases}}}\)
3) \(12+|3-x|=9\)
\(\Leftrightarrow|3-x|=-3\)(vô lí vì I 3-x I \(\ge\)0)
Bài 1:
|\(x\)| = - (-12) + (-15) + 8
|\(x\)| = 12 - 15 + 8
|\(x\)| = - 3 + 8
|\(x\)| = 5
\(\left[{}\begin{matrix}x=5\\x=-5\end{matrix}\right.\)
Vậy \(x\) \(\in\) {- 5; 5}
Bài 2: |\(x\)| = - 8 + (5 - 9) - 5 + 8
|\(x\)| = - 8 - 4 - 5 + 8
|\(x\)| = - 12 - 5 + 8
|\(x\)| = - 17 + 8
|\(x\)| = -9 (vô lý)
vậy không có giá trị nào của \(x\) thỏa mãn đề bài
Kết luận: \(x\) \(\in\) \(\varnothing\)
nhiều quá :((
\(a,2\left(x-5\right)-3\left(x+7\right)=14\)
\(2x-10-3x-21=14\)
\(-x-31=14\)
\(-x=45\)
\(x=45\)
\(b,5\left(x-6\right)-2\left(x+3\right)=12\)
\(5x-30-2x-6=12\)
\(3x-36==12\)
\(3x=48\)
\(x=16\)
\(c,3\left(x-4\right)-\left(8-x\right)=12\)
\(3x-12-8+x=0\)
\(4x-20=0\)
\(4x=20\)
\(x=5\)
Cố nốt nha bn !
cảm ơn, bn nha:)))
mà hình như bạn TOP 3 trả lời câu hỏi pải ko nhỉ???
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