Tìm X ∈ N biết:
a,2x . 2x + 1=128
b,3x+1 . 3x=243
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a: =>1/3x-2/5x=5
=>-1/15x=5
=>x=-75
b: =>4x=4
=>x=1
c: =>6*3^x-5*3^x=243
=>3^x=243
=>x=5
\(a,\Rightarrow5x+3x^2-3x^2-x+2=6\\ \Rightarrow4x=4\Rightarrow x=1\\ b,\Rightarrow\left(2x+\dfrac{1}{2}-1+2x\right)\left(2x+\dfrac{1}{2}+1-2x\right)=2\\ \Rightarrow\dfrac{3}{2}\left(4x-\dfrac{1}{2}\right)=2\\ \Rightarrow6x-\dfrac{3}{4}=2\\ \Rightarrow6x=\dfrac{11}{4}\\ \Rightarrow x=\dfrac{11}{24}\\ c,\Rightarrow\left(x+3\right)\left(x-2\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=2\\x=-3\end{matrix}\right.\)
a) Ta có: \(6x\left(x-5\right)+3x\left(7-2x\right)=18\)
\(\Leftrightarrow6x^2-30x+21x-6x^2=18\)
\(\Leftrightarrow-9x=18\)
hay x=-2
Vậy: S={-2}
b) Ta có: \(2x\left(3x+1\right)+\left(4-2x\right)\cdot3x=7\)
\(\Leftrightarrow6x^2+2x+12x-6x^2=7\)
\(\Leftrightarrow14x=7\)
hay \(x=\dfrac{1}{2}\)
Vậy: \(S=\left\{\dfrac{1}{2}\right\}\)
c) Ta có: \(0.5x\left(0.4-4x\right)+\left(2x+5\right)\cdot x=-6.5\)
\(\Leftrightarrow0.2x-2x^2+2x^2+5x=-6.5\)
\(\Leftrightarrow5.2x=-6.5\)
hay \(x=-\dfrac{5}{4}\)
Vậy: \(S=\left\{-\dfrac{5}{4}\right\}\)
d) Ta có: \(\left(x+3\right)\left(x+2\right)-\left(x-2\right)\left(x+5\right)=6\)
\(\Leftrightarrow x^2+5x+6-\left(x^2+3x-10\right)=6\)
\(\Leftrightarrow x^2+5x+6-x^2-3x+10=6\)
\(\Leftrightarrow2x+16=6\)
\(\Leftrightarrow2x=-10\)
hay x=-5
Vậy: S={-5}
e) Ta có: \(3\left(2x-1\right)\left(3x-1\right)-\left(2x-3\right)\left(9x-1\right)=0\)
\(\Leftrightarrow3\left(6x^2-5x+1\right)-\left(18x^2-29x+3\right)=0\)
\(\Leftrightarrow18x^2-15x+3-18x^2+29x-3=0\)
\(\Leftrightarrow14x=0\)
hay x=0
Vậy: S={0}
Bài 1:
\(a,=6x^2+6x\\ b,=15x^3-10x^2+5x\\ c,=6x^3+12x^2\\ d,=15x^4+20x^3-5x^2\\ e,=2x^2+3x-2x-3=2x^2+x-3\\ f,=3x^2-5x+6x-10=3x^2+x-10\)
Bài 2:
\(a,\Leftrightarrow3x^2+3x-3x^2=6\\ \Leftrightarrow3x=6\Leftrightarrow x=2\\ b,\Leftrightarrow6x^2+3x-6x^2+9x-2x-3=10\\ \Leftrightarrow10x=13\Leftrightarrow x=\dfrac{13}{10}\)
a: Ta có: \(x\left(2x-3\right)-\left(2x-1\right)\left(x+5\right)=17\)
\(\Leftrightarrow2x^2-3x-2x^2-10x+x+5=17\)
\(\Leftrightarrow-12x=12\)
hay x=-1
a) \(2^x\cdot4=128\)
\(\Rightarrow2^x\cdot2^2=2^7\)
\(\Rightarrow2^{x+2}=2^7\)
\(\Rightarrow x+2=7\)
\(\Rightarrow x=5\)
b) \(\left(2x+1\right)^3=125\)
\(\Rightarrow\left(2x+1\right)^3=5^3\)
\(\Rightarrow2x+1=5\)
\(\Rightarrow2x=4\)
\(\Rightarrow x=4:2\)
\(\Rightarrow x=2\)
c) \(2x-2^6=6\)
\(\Rightarrow2x-64=6\)
\(\Rightarrow2x=70\)
\(\Rightarrow x=70:2\)
\(\Rightarrow x=35\)
d) \(64\cdot4^x=45\)
\(\Rightarrow4^3\cdot4^x=45\)
\(\Rightarrow4^{x+3}=45\)
Xem lại đề
e) \(27\cdot3^x=243\)
\(\Rightarrow3^3\cdot3^x=3^5\)
\(\Rightarrow3^{x+3}=3^5\)
\(\Rightarrow x+3=5\)
\(\Rightarrow x=2\)
g) \(49\cdot7^x=2401\)
\(\Rightarrow7^2\cdot7^x=7^4\)
\(\Rightarrow7^{x+2}=7^4\)
\(\Rightarrow x+2=4\)
\(\Rightarrow x=2\)
h) \(3^x=81\)
\(\Rightarrow3^x=3^4\)
\(\Rightarrow x=4\)
k) \(3^4\cdot3^x=3^7\)
\(\Rightarrow3^{x+4}=3^7\)
\(\Rightarrow x+4=7\)
\(\Rightarrow x=3\)
n) \(3^x+25=26\cdot2^2+2\cdot3^0\)
\(\Rightarrow3^x+25=104+2\)
\(\Rightarrow3^x+25=106\)
\(\Rightarrow3^x=81\)
\(\Rightarrow3^x=3^4\)
\(x=4\)
`@` `\text {Ans}`
`\downarrow`
`a)`
`2^x*4 = 128`
`=> 2^x = 128 \div 4`
`=> 2^x = 2^7 \div 2^2`
`=> 2^x = 2^5`
`=> x = 5`
Vậy, `x = 5.`
`b)`
\(\left(2x+1\right)^3=125\)
`=> (2x + 1)^3 = 5^3`
`=> 2x + 1 = 5`
`=> 2x = 5-1`
`=> 2x = 4`
`=> x = 4 \div 2`
`=> x = 2`
Vậy, `x = 2`
`c)`
\(2x-2^6=6\)
`=> 2x = 6+2^6`
`=> 2x = 70`
`=> x = 70 \div 2`
`=> x = 35`
Vậy, `x = 35`
`d)`
\(64\cdot4^x=45\) Bạn xem lại đề
`e)`
`27*3^x = 243`
`=> 3^3 * 3^x = 3^5`
`=> 3^(3 + x) = 3^5`
`=> 3 + x = 5`
`=> x = 5 - 3`
`=> x = 2`
Vậy, `x = 2`
`g)`
`49* 7^x = 2401`
`=> 7^2 * 7^x = 7^4`
`=> 7^(2 + x) = 7^4`
`=> 2 + x = 4`
`=> x = 4 - 2`
`=> x = 2`
Vậy, `x = 2`
`h)`
`3^x = 81`
`=> 3^x = 3^4`
`=> x = 4`
Vậy, `x = 4`
`k)`
`3^4 * 3^x = 3^7`
`=> 3^(4 + x) = 3^7`
`=> 4 + x = 7`
`=> x = 7 - 4`
`=> x = 3`
Vậy, `x = 3`
`n)`
`3^x + 25 = 26*2^2 + 2*3^0`
`=> 3^x + 25 = 104 + 2`
`=> 3^x + 25 = 106`
`=> 3^x = 106 - 25`
`=> 3^x = 81`
`=> 3^x = 3^4`
`=> x = 4`
Vậy, `x = 4.`
\(#48Cd\)
a) \(2x\left(x-5\right)-x\left(3+2x\right)=26\)
\(\Rightarrow2x^2-10x-3x-2x^2=26\)
\(\Rightarrow-13x=26\Rightarrow x=-2\)
b) \(3x\left(1-2x\right)+2\left(3x+7\right)=29\)
\(\Rightarrow3x-6x^2+6x+14=29\)
\(\Rightarrow-6x^2+9x-15=0\)
\(\Rightarrow-6\left(x^2-\dfrac{3}{2}x+\dfrac{9}{16}\right)-\dfrac{93}{8}=0\)
\(\Rightarrow-6\left(x-\dfrac{3}{4}\right)^2-\dfrac{93}{8}=0\)(vô lý)
Vậy \(S=\varnothing\)
\(a,3\left(2x-3\right)+2\left(2-x\right)=-3\\ \Leftrightarrow6x-9+4-2x=-3\\ \Leftrightarrow4x=2\\ \Leftrightarrow x=\dfrac{1}{2}\\ b,x\left(5-2x\right)+2x\left(x-1\right)=13\\ \Leftrightarrow5x-2x^2+2x^2-2x=13\\ \Leftrightarrow3x=13\\ \Leftrightarrow x=\dfrac{13}{3}\\ c,5x\left(x-1\right)-\left(x+2\right)\left(5x-7\right)=6\\ \Leftrightarrow5x^2-5x-5x^2-3x+14=6\\ \Leftrightarrow-8x=-8\\ \Leftrightarrow x=1\\ d,3x\left(2x+3\right)-\left(2x+5\right)\left(3x-2\right)=8\\ \Leftrightarrow6x^2+9x-6x^2-11x+10=8\\ \Leftrightarrow-2x=-2\\ \Leftrightarrow x=1\)
\(e,2\left(5x-8\right)-3\left(4x-5\right)=4\left(3x-4\right)+11\\ \Leftrightarrow10x-16-12x+15=12x-16+11\\ \Leftrightarrow-14x=-4\\ \Leftrightarrow x=\dfrac{2}{7}\\ f,2x\left(6x-2x^2\right)+3x^2\left(x-4\right)=8\\ \Leftrightarrow12x^2-4x^3+3x^3-12x^2=8\\ \Leftrightarrow-x^3-8=0\\ \Leftrightarrow-\left(x^3+8\right)=0\\ \Leftrightarrow-\left(x+2\right)\left(x^2-2x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-2\\x\in\varnothing\left(x^2-2x+4=\left(x-1\right)^2+3>0\right)\end{matrix}\right.\)
Bài 4:
a: Ta có: \(3\left(2x-3\right)-2\left(x-2\right)=-3\)
\(\Leftrightarrow6x-9-2x+4=-3\)
\(\Leftrightarrow4x=2\)
hay \(x=\dfrac{1}{2}\)
b: Ta có: \(x\left(5-2x\right)+2x\left(x-1\right)=13\)
\(\Leftrightarrow5x-2x^2+2x^2-2x=13\)
\(\Leftrightarrow3x=13\)
hay \(x=\dfrac{13}{3}\)
c: Ta có: \(5x\left(x-1\right)-\left(x+2\right)\left(5x-7\right)=6\)
\(\Leftrightarrow5x^2-5x-5x^2+7x-10x+14=6\)
\(\Leftrightarrow-8x=-8\)
hay x=1
a) \(\Rightarrow2^x=32\Rightarrow2^x=2^5\Rightarrow x=5\)
b) \(\Rightarrow\left(2x+1\right)^3=5^3\)
\(\Rightarrow2x+1=5\Rightarrow x=2\)
c) \(\Rightarrow2^x=32\Rightarrow x=5\)
d) \(\Rightarrow4^3.4^x=4^5\Rightarrow4^x=4^2\Rightarrow x=2\)
e) \(\Rightarrow3^3.3^x=3^5\Rightarrow3^x=3^2\Rightarrow x=2\)
f) \(\Rightarrow7^2.7^x=7^4\Rightarrow7^x=7^2\Rightarrow x=2\)
a. 2x . 4 = 128
<=> 2x + 2 = 27
<=> x + 2 = 7
<=> x = 5
b. (2x + 1)3 = 125
<=> (2x + 1)3 - 53 = 0
<=> (2x + 1 - 5)\(\left[\left(2x+1\right)^2+\left(2x+1\right).5+25\right]=0\)
<=> (2x - 4)(4x2 + 4x + 1 + 10x + 5 + 25) = 0
<=> (2x - 4)(4x2 + 14x + 31) = 0
<=> \(\left[{}\begin{matrix}2x-4=0\\4x^2+14x+31=0\end{matrix}\right.\)
<=> \(\left[{}\begin{matrix}x=2\\VôNghiệm\end{matrix}\right.\)
c. 2x - 26 = 6
<=> 2x = 32
<=> x = 5
d. 64 . 4x = 45
<=> 43 . 4x = 45
<=> 43 + x = 45
<=> 3 + x = 5
<=> x = 2
e. 27 . 3x = 243
<=> 33 . 3x = 35
<=> 33 + x = 35
<=> 3 + x = 5
<=> x = 2
g. 49 . 7x = 2401 (Bn xem lại đề câu này)
<=> 72 . 7x = 74
<=> 72 + x = 74
<=> 2 + x = 4
<=> x = 2
a)2x.2x+1=128
\(4x^2+1=128\)
\(\Rightarrow4x^2=127\)
\(\Rightarrow x^2=127\div4\)
\(\Rightarrow x^2=3.175\)
mà \(x\in N\) hay\(x^2\in N\)
\(\Rightarrow\)Không có x
b) 3x+1.3x=243
\(\Rightarrow\)3x+3x=243
\(\Rightarrow\)6x=243
\(\Rightarrow\)x=40,5
Mà \(x\in N\)
\(\Rightarrow\)Không có x