Bài 1: Tìm x
a, 23 .2x=128
b, 2x+2x+3=144
c, x-287:285=125
d,3x+4-3x=720
e, 812x:27x=95
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
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) \(\Rightarrow2^x=32\Rightarrow2^x=2^5\Rightarrow x=5\)
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: x^3-7x-6
=x^3-x-6x-6
=x(x-1)(x+1)-6(x+1)
=(x+1)(x^2-x-6)
=(x-3)(x+2)(x+1)
b: =2x^3+x^2-2x^2-x+6x+3
=x^2(2x+1)-x(2x+1)+3(2x+1)
=(2x+1)(x^2-x+3)
c: =2x^3-3x^2-2x^2+3x+2x-3
=x^2(2x-3)-x(2x-3)+(2x-3)
=(2x-3)(x^2-x+1)
d: =2x^3+x^2+2x^2+x+2x+1
=(2x+1)(x^2+x+1)
e: =3x^3+x^2-3x^2-x+6x+2
=(3x+1)(x^2-x+2)
f: =27x^3-9x^2-18x^2+6x+12x-4
=(3x-1)(9x^2-6x+4)
a) \(x^3-7x-6\)
\(=x^3-x-6x-6\)
\(=\left(x^3-x\right)-\left(6x+6\right)\)
\(=x\left(x^2-1\right)-6\left(x+1\right)\)
\(=x\left(x+1\right)\left(x-1\right)-6\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2-x-6\right)\)
b) \(2x^3-x^2+5x+3\)
\(=2x^3+x^2-2x^2-x+6x+3\)
\(=\left(2x^3+x^2\right)-\left(2x^2+x\right)+\left(6x+3\right)\)
\(=x^2\left(2x+1\right)-x\left(2x+1\right)+3\left(2x+1\right)\)
\(=\left(x^2-x+3\right)\left(2x+1\right)\)
c) \(2x^3-5x^2+5x+1\)
\(=2x^3-3x^2-2x^2+3x+2x-3\)
\(=\left(2x^3-3x^2\right)-\left(2x^2-3x\right)+\left(2x-3\right)\)
\(=x^2\left(2x-3\right)-x\left(2x-3\right)+\left(2x-3\right)\)
\(=\left(x^2-x+1\right)\left(2x-3\right)\)
d) \(2x^3+3x^2+3x+1\)
\(=2x^3+x^2+2x^2+x+2x+1\)
\(=\left(2x^3+x^2\right)+\left(2x^2+x\right)+\left(2x+1\right)\)
\(=x^2\left(2x+1\right)+x\left(2x+1\right)+\left(2x+1\right)\)
\(=\left(2x+1\right)\left(x^2+x+1\right)\)
e) \(3x^3-2x^2+5x+2\)
\(=3x^3+x^2-3x^2-x+6x+2\)
\(=\left(3x^3+x^2\right)-\left(3x^2+x\right)+\left(6x+2\right)\)
\(=x^2\left(3x+1\right)-x\left(3x+1\right)+2\left(3x+1\right)\)
\(=\left(3x-1\right)\left(x^2-x+2\right)\)
f) \(27x^3-27x^2+18x-4\)
\(=27x^3-9x^2-18x^2+6x+12x-4\)
\(=\left(27x^3-9x^2\right)-\left(18x^2-6x\right)+\left(12x-4\right)\)
\(=9x^2\left(3x-1\right)-6x\left(3x-1\right)+4\left(3x-1\right)\)
\(=\left(3x-1\right)\left(9x^2-6x+4\right)\)
2: Tìm x
a) Ta có: x+25=40
nên x=40-25=15
Vậy: x=15
b) Ta có: 198-(x+4)=120
\(\Leftrightarrow x+4=198-120=78\)
hay x=78-4=74
Vậy: x=74
c) Ta có: \(\left(2x-7\right)\cdot3=125\)
\(\Leftrightarrow2x-7=\dfrac{125}{3}\)
\(\Leftrightarrow2x=\dfrac{125}{3}+7=\dfrac{125}{3}+\dfrac{21}{3}=\dfrac{146}{3}\)
\(\Leftrightarrow x=\dfrac{146}{3}:2=\dfrac{146}{6}=\dfrac{73}{3}\)
Vậy: \(x=\dfrac{73}{3}\)
d) Ta có: \(x+16⋮x+1\)
\(\Leftrightarrow x+1+15⋮x+1\)
mà \(x+1⋮x+1\)
nên \(15⋮x+1\)
\(\Leftrightarrow x+1\inƯ\left(15\right)\)
\(\Leftrightarrow x+1\in\left\{1;-1;3;-3;5;-5;15;-15\right\}\)
hay \(x\in\left\{0;-2;2;-4;4;-6;14;-16\right\}\)
Vậy: \(x\in\left\{0;-2;2;-4;4;-6;14;-16\right\}\)
\(a,x+25=40\\ \Rightarrow x=40-25\\ \Rightarrow x=15\\ b,198-\left(x+4\right)=120\\ \Rightarrow-\left(x+4\right)=120-198\\ \Rightarrow-\left(x+4\right)=-78\\ \Rightarrow x+4=78\\ \Rightarrow x=78-4\\ \Rightarrow x=74\\ c,\left(2x-7\right).3=125\\ \Rightarrow2x-7=\dfrac{125}{3}\\ \Rightarrow2x=\dfrac{125}{3}+7\\ \Rightarrow2x=\dfrac{146}{3}\\ \Rightarrow x=\dfrac{146}{3}:2\Rightarrow x=\dfrac{73}{3}\\ d,\left(x+16\right)⋮\left(x+1\right)\\ \Rightarrow\left[\left(x+1\right)+15\right]⋮\left(x+1\right)\\ mà:\left(x+1\right)⋮\left(x+1\right)\\ \Rightarrow15⋮\left(x+1\right)\\ \Rightarrow\left(x+1\right)\inƯ\left(15\right)\\ \Rightarrow\left(x+1\right)\in\left\{-15;-1;1;15\right\}\\ \Rightarrow x\in\left\{-16;-2;0;14\right\}\)
Tự kết luận nhé bạn
a) \(\left(2x-1\right)^2-25=0\)
⇔ \(\left(2x-1\right)^2-5^2=0\)
⇔ \(\left(2x-1-5\right)\left(2x-1+5\right)=0\)
⇒ \(2x-1-5=0\) hoặc \(2x-1+5=0\)
⇔ \(x=3\) hoặc \(x=-2\)
Bài 1: Tìm x
a) (2x-1) ² - 25 = 0
<=> (2x-1)2 = 25
<=> 2x-1 = 5 hay 2x-1 =-5
<=> 2x= 6 hay 2x=-4
<=> x=3 hay x= -2
Vậy S={3; -2}
b) 3x (x-1) + x - 1 = 0
<=> (x-1)(3x+1)=0
<=> x-1=0 hay 3x+1=0
<=> x=1 hay 3x=-1
<=> x=1 hay x=\(\dfrac{-1}{3}\)
Vậy S={1;\(\dfrac{-1}{3}\)}
c) 2(x+3) - x ² - 3x = 0
<=> 2(x+3)- x(x+3)=0
<=> (x+3)(2-x)=0
<=> x+3=0 hay 2-x=0
<=> x=-3 hay x=2
Vậy S={-3;2}
d) x(x - 2) + 3x - 6 = 0
<=> x(x-2)+3(x-2)=0
<=> (x-2)(x+3)=0
<=> x-2=0 hay x+3=0
<=> x=2 hay x=-3
Vậy S={2;-3}
e) 4x ² - 4x +1 = 0
<=> (2x-1)2=0
<=> 2x-1=0
<=> 2x=1
<=> x=\(\dfrac{1}{2}\)
Vậy S={\(\dfrac{1}{2}\)}
f) x +5x2 = 0
<=> x(1+5x)=0
<=>x=0 hay 1+5x=0
<=> x=0 hay 5x=-1
<=> x=0 hay x= \(\dfrac{-1}{5}\)
Vậy S={0;\(\dfrac{-1}{5}\)}
g) x ²+ 2x -3 = 0
<=> x2-x+3x-3=0
<=> x(x-1)+3(x-1)=0
<=> (x-1)(x+3)=0
<=> x-1=0 hay x+3=0
<=> x=1 hay x=-3
Vậy S={1;-3}
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) 3x(4x-3)-2x(5-6x)=0
\(\Leftrightarrow12x^2-9x-10x+12x^2=0\)
\(\Leftrightarrow24x^2-19x=0\)
\(\Leftrightarrow x\left(24x-19\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\24x-19=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\24x=19\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{19}{24}\end{matrix}\right.\)
Vậy x=0 hoặc x=\(\dfrac{19}{24}\)
a)23 .2x=128
=>23+x=128
=>23+x=27
=>3+x=7
=>x=4
b)2x+2x+3=144
=>2x(1+23)=144
=>2x*9=144
=>2x=16
=>2x=24
=>x=4
c)x-287:285=125
=>x-287-85=125
=>x-4=125
=>x=129
d)3x+4-3x=720
=>3x(34-1)=720
=>3x*80=720
=>3x=9
=>3x=32
=>x=2
e)812x:27x=95
=>(34)2x:(33)x=95
=>38x:33x=95
=>38x-3x=(32)5
=>35x=310
=>5x=10
=>x=2
a) 23 .2x=128
=>23+x = 27
=>3+x=7
=>x=4