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\(\frac{3}{x-5}=-\frac{4}{x+2}\)
=> 3 ( x + 2 ) = 4 ( x - 5 )
=> 3x + 6 = 4x - 20
=> 3x - 4x = - 6 - 20
=> - 1x = - 26
=> x = 26
1) 3/x-5=-4/x+2
3x+6=-4x+20
7x . =14
x . =2
2) x+3/-4=-9/x+3
(x+3)^2=36
Ta có hai trường hợp:
*x+3=6=)x=3
*x+3=-6=)x=-9
<=> 2x^2 +x-4x-2-5x-15=2x^2-6x+4+8x-2-2x
2x^2-8x-17-2x^2-2=0
-8x-19=0
x=-19/8
`2x-15=-25`
`2x=-10`
`x=-5`
___________
`3/5<x/10<4/5`
`3/5=(3xx10)/(5xx10)=30/50`
`x/10=(5x)/(10xx5)=(5x)/50`
`4/5=(4xx10)/(5xx10)=40/50`
`=>30/50<(5x)/50<40/50`
`=>30<5x<40`
`=>x=7`
`#3107`
b)
`2.3^x = 162`
`\Rightarrow 3^x = 162 \div 2`
`\Rightarrow 3^x = 81`
`\Rightarrow 3^x = 3^4`
`\Rightarrow x = 4`
Vậy, `x = 4`
c)
`(2x - 15)^5 = (2 - 15)^3`
\(\Rightarrow \)`(2x - 15)^5 - (2x - 15)^3 = 0`
\(\Rightarrow \)`(2x - 15)^3 . [ (2x - 15)^2 - 1] = 0`
\(\Rightarrow\left[{}\begin{matrix}\left(2x-15\right)^3=0\\\left(2x-15\right)^2-1=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}2x-15=0\\\left(2x-15\right)^2=1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}2x=15\\\left(2x-15\right)^2=\left(\pm1\right)^2\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{15}{2}\\2x-15=1\\2x-15=-1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{15}{2}\\2x=16\\2x=-14\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{15}{2}\\x=8\\x=-7\end{matrix}\right.\)
Vậy, `x \in`\(\left\{-7;8;\dfrac{15}{2}\right\}.\)
`d)`
\(3^{x+2}-5.3^x=?\) Bạn ghi tiếp đề nhé!
`e)`
\(7\cdot4^{x-1}+4^{x-1}=23?\)
\(4^{x-1}\cdot\left(7+1\right)=23\\ \Rightarrow4^{x-1}\cdot8=23\\ \Rightarrow4^{x-1}=\dfrac{23}{8}\)
Bạn xem lại đề!
`f)`
\(2\cdot2^{2x}+4^3\cdot4^x=1056\)
\(\Rightarrow2\cdot2^{2x}+\left(2^2\right)^3\cdot\left(2^2\right)^x=1056\\ \Rightarrow2\cdot2^{2x}+2^6\cdot2^{2x}=1056\\ \Rightarrow2^{2x}\cdot\left(2+2^6\right)=1056\\ \Rightarrow2^{2x}\cdot66=1056\\ \Rightarrow2^{2x}=1056\div66\\ \Rightarrow2^{2x}=16\\ \Rightarrow2^{2x}=2^4\\ \Rightarrow2x=4\\ \Rightarrow x=2\)
Vậy, `x = 2`
_____
\(10 -{[(x \div 3+17) \div 10+3.2^4] \div 10}=5\)
\(\Rightarrow\left[\left(x\div3+17\right)\div10+48\right]\div10=10-5\)
\(\Rightarrow\left[\left(x\div3+17\right)\div10+48\right]\div10=5\)
\(\Rightarrow\left(x\div3+17\right)\div10+48=50\)
\(\Rightarrow\left(x\div3+17\right)\div10=2\)
\(\Rightarrow x\div3+17=20\)
\(\Rightarrow x\div3=3\\ \Rightarrow x=9\)
Vậy, `x = 9.`
a) 2.(x+1) - 3 = 11
2.(x+1) = 14
x + 1 = 7
x = 6
b) 60 - 5.(x-3) = 50
5.(x-3) = 10
x-3 = 2
x = 5
c) (10+2x).43 = 45
10 + 2x = 16
2x = 6
x =3
\(a,2\cdot\left(x+1\right)-3\left(11\right)=0.\)
\(\left(2\cdot\left(x+1\right)-3\right)-11=0.\)
\(2x-12=2\cdot\left(x-6\right)\)
\(2\cdot\left(x-6\right)=0.\)
\(x-6=0.\)
\(x=6.\)
\(b,\left(60-5\cdot\left(x-3\right)\right)-50=0\)
\(25-5x=-5\cdot\left(x-5\right)\)
\(-5x\left(x-5\right)=0.\)
\(x-5=0\)
\(x=5\)
Câu c khó quá bn tự làm nhé.
học tốt
a)
\(\frac{x-3}{10}=\frac{4}{x-3}\)
=> ( x - 3 )2 = 4 . 10.
( x - 3 )2 = 40
Mà x - 3 thuộc Z ( vì x thuộc Z ) nên ( x - 3 )2 là số chính phương.
Do 40 không là số chính phương.
=> Ko tìm được x thuộc Z thỏa mãn đề bài.
b)
\(\frac{x+5}{9}=\frac{4}{x+5}\)
=> ( x + 5 )2 = 4 . 9
( x + 5 )2 = 36
=> x + 5 = 6 hoặc x + 5 = -6.
+) x + 5 = 6
x = 1.
+) x + 5 = -6
x = -11.
Vậy x = 1; x = -11.
1) \(\Rightarrow\left[{}\begin{matrix}x=-3\\x=5\end{matrix}\right.\)
2) \(\Rightarrow5\left(x-2\right).3\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)
3) \(\Rightarrow2\left(x-4\right)\left(x-7\right)=0\Rightarrow\left[{}\begin{matrix}x=4\\x=7\end{matrix}\right.\)
\(2x-3-3x+3.5=4.3-4.x-18\)
<=> \(2x-3-3x+15=12-4x-18\)
<=> \(2x-3x+4x=12-18-15+3\)
<=>\(3x=-18\)
<=> x=-18/3=-6
\(2x-3-3\left(x-5\right)=4\left(3-x\right)-18\)
\(\Leftrightarrow2x-3-3x+15=12-4x-18\)
\(\Leftrightarrow\left(2x-3x\right)+\left(15-3\right)=\left(12-18\right)-4x\)
\(\Leftrightarrow-x+12=-6-4x\)
\(\Leftrightarrow12=-6-3x\)
\(\Leftrightarrow3x=-6-12\)
\(\Leftrightarrow3x=-18\)
\(\Leftrightarrow x=\frac{-18}{3}\)
\(\Leftrightarrow x=-6\)
\(\text{Vậy }x=-6\)