A,(X-2)^11 = (x-2)^3 ; b, (x-5)^24 = (x-5)^9 ; c, (x-5)^25 = (x-5)^4
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\(4.3^{x-1}+2.3^{x+2}=4.3^6+2.3^9\)
\(3^{x-1}.\left(4+2.3^3\right)=3^6.\left(4+2.3^3\right)\)
\(\Leftrightarrow3^{x-1}=3^6\)
\(\Leftrightarrow x-1=6\)
\(\Leftrightarrow x=7\)
Vậy \(x=7\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(...\Rightarrow x.\left(2+5\right)=14\Rightarrow x.7=14\Rightarrow x=14:7=2\)
b) \(...\Rightarrow x.\left(9+1\right)=20\Rightarrow x.10=20\Rightarrow x=20:10=2\)
c) \(...\Rightarrow x.\left(\dfrac{2}{3}+\dfrac{1}{3}\right)=1999\Rightarrow x.\dfrac{3}{3}=1999\Rightarrow x=1999\)
d) \(...\Rightarrow11.x+22=5.x+40\Rightarrow11.x-5.x=40-22\Rightarrow6.x=18\Rightarrow x=18:6=3\)
e) \(...\Rightarrow11.x-66=4.x+11\Rightarrow11.x-4.x=11+66\Rightarrow7.x=77\Rightarrow x=77:7=11\)
f) \(...\Rightarrow\left(3.x-12\right):x=12-10\)
\(\Rightarrow3.x-12=2.x\)
\(\Rightarrow3.x-2.x=12\)
\(\Rightarrow x=12\)
g) \(...\Rightarrow\left(5.x+7\right):x=26-20\)
\(\Rightarrow5.x+7=6.x\)
\(\Rightarrow6.x-5.x=7\)
\(\Rightarrow x=7\)
h) \(...\Rightarrow x.\left(1999-1\right)=1999.\left(1997+1\right)\)
\(\Rightarrow x.1998=1999.1998\)
\(\Rightarrow x=1999.1998:1998\)
\(\Rightarrow x=1999\)
a, \(x\times\) 2 + \(x\times\) 5 = 14
\(x\) \(\times\) ( 2 + 5) = 14
\(x\) \(\times\) 7 = 14
\(x\) = 14: 7
\(x\) = 2
b, \(x\times9\) + \(x\)= 20
\(x\) \(\times\)( 9 + 1) = 20
\(x\) \(\times\) 10 = 20
\(x\) = 2
c, \(x\) : \(\dfrac{3}{2}\) + \(x\times\dfrac{1}{3}\) = 1999
\(x\times\) \(\dfrac{2}{3}\) + \(x\) \(\times\dfrac{1}{3}\) = 1999
\(x\times\) ( \(\dfrac{2}{3}\) + \(\dfrac{1}{3}\)) = 1999
\(x\) = 1999
d, 11\(\times\)(\(x+2\)) = 5 \(\times\) \(x\) + 40
11 \(\times\) \(x\) + 22 = 5 \(\times\) \(x\) + 40
11 \(\times\) \(x\) = 5 \(\times\) \(x\) + 40 - 22
11 \(\times\) \(x\) = 5 \(\times\) \(x\) + 18
11 \(\times\) \(x\) - 5 \(\times\) \(x\) = 18
\(x\) \(\times\) ( 11 - 5) = 18
\(x\) \(\times\) 6 = 18
\(x\) = 3
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Mình nghĩ câu A phải là 11/23.34 mới đúng.
Theo đề mới: A)
11/12 = 11/1.12
Vậy A = 11/1.12 + 11/12.23 + ... + 11/89.100
= 1 - 1/12 + 1/12 - 1/23 + .... + 1/89 - 1/100
= 99/1000
Vậy x là: 2/3 - 99/100 = -97/300
B) 2/11.13 + ... + 2/19.21 = 1/11 - 1/13 + .... + 1/19 - 1/21 = 10/231
=> 10/231 - x = 4/3 - 221/223 = 229/669
=> x = 10/231 - 229/669
=> x = 6690/154539 - 52899/154539
=> x = -46209/154539 = -15403/51513
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a,Cách 1: (6/11 + 5/11 ) x 3/7 Cách 2 :(6/11 + 5/11) x 3/7
= 1 x 3/7 =6/11 x 3/7 + 5/11 x 3/7
= 3/7 = 18/77 + 15/77
= 3/7
b, Cách 1:3/5 x 7/9 - 3/5 x 2/9 Cách 2 :3/5 x 7/9 - 3/5 x 2/9
= 7/15 - 2/15 = 3/5 x (7/9 - 2/9 )
= 1/3 = 3/5 x 5/9
= 1/3
c, Cách 1:(6/7 - 4/7) : 2/5 Cách 2: ( 6/7- 4/7 ) : 2/5
= 2/7 : 2/5 = 6/7 : 2/5 - 4/7 : 2/5
= 5/7 = 15/7 - 10/7
= 5/7
d,Cách 1:8/15 : 2/11 + 7/15 : 2/11 Cách 2:8/15 : 2/11 +7/15 : 2/11
= 88/30 + 77/30 =(8/15+7/15) :2/11
= 11/2 = 1 : 2/11
= 11/2
Bài 2 cậu tự làm nhé !
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\dfrac{a}{b}\) x( \(\dfrac{16}{11}\) -\(\dfrac{2}{11}\)-\(\dfrac{3}{11}\)+9) = \(\dfrac{5}{3}\)
\(\dfrac{a}{b}\) x (1 + 9) =\(\dfrac{5}{3}\)
\(\dfrac{a}{b}\) x 10 = \(\dfrac{5}{3}\)
\(\dfrac{a}{b}\) = \(\dfrac{5}{3}\) : 10
a = \(\dfrac{1}{6}\)
lần sau bạn hãy nhớ giữ gìn sự trong sáng của tiếng việt nhé, học sinh của mình mà hỏi mình trống không thế này mình cho nghỉ vô thời hạn
![](https://rs.olm.vn/images/avt/0.png?1311)
a) 2/5 x 1/4 + 3/4 x 2/5
= 2/5 x (1/4 + 3/4)
= 2/5 x 1
= 2/5
b) 6/11 : 2/3 + 5/11 : 2/3
= (6/11 + 5/11) : 2/3
= 1 : 2/3
= 3/2
a: =>(x-2)^3*[(x-2)^8-1]=0
=>(x-2)(x-3)(x-1)=0
=>\(x\in\left\{2;3;1\right\}\)
b: (x-5)^24=(x-5)^9
=>\(\left(x-5\right)^9\cdot\left[\left(x-5\right)^{15}-1\right]=0\)
=>x-5=0 hoặc x-5=1
=>x=6 hoặc x=5
c: =>(x-5)^4*[(x-5)^21-1]=0
=>x-5=0 hoặc x-5=1
=>x=5 hoặc x=6
a) \(\left(x-2\right)^{11}=\left(x-2\right)^3\)
\(\Rightarrow\left(x-2\right)^{11}-\left(x-2\right)^3=0\)
\(\Rightarrow\left(x-2\right)^3\left[\left(x-2\right)^8-1\right]=0\)
\(\Rightarrow\left[{}\begin{matrix}\left(x-2\right)^3=0\\\left(x-2\right)^8-1=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x-2=0\\\left(x-2\right)^8=1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=2\\x-2=1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=2\\x=3\end{matrix}\right.\)
b) \(\left(x-5\right)^{24}=\left(x-5\right)^9\)
\(\Rightarrow\left(x-5\right)^{24}-\left(x-5\right)^9=0\)
\(\Rightarrow\left(x-5\right)^9\left[\left(x-5\right)^{15}-1\right]=0\)
\(\Rightarrow\left[{}\begin{matrix}\left(x-5\right)^9=0\\\left(x-5\right)^{15}-1=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x-5=0\\\left(x-5\right)^{15}=1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=5\\x-5=1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=5\\x=6\end{matrix}\right.\)
c) \(\left(x-5\right)^{25}=\left(x-5\right)^4\)
\(\Rightarrow\left(x-5\right)^{25}-\left(x-5\right)^4\)
\(\Rightarrow\left(x-5\right)^4\left[\left(x-5\right)^{21}-1\right]=0\)
\(\Rightarrow\left[{}\begin{matrix}\left(x-5\right)^4=0\\\left(x-5\right)^{21}-1=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x-5=0\\\left(x-5\right)^{21}=1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=5\\x-5=1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=5\\x=6\end{matrix}\right.\)