Tìm x:
x.(5-2x)+2x.(x-1)=15
B1: Tìm x:
1/ \(\dfrac{x+3}{15}\) = \(\dfrac{1}{3}\) - \(\dfrac{1}{15}\)
2/ (2x - 5) = (x - 3) = 0
3/ (3x - 4) - (2x - 5) = 3
4/ (2x + 1) x (\(\dfrac{1}{2}\)x - 1) = 0
1) PT \(\Leftrightarrow\dfrac{x+3}{15}=\dfrac{4}{15}\) \(\Rightarrow x+3=4\) \(\Rightarrow x=1\)
Vậy ...
2) Mạnh dạn đoán đề là \(\left(2x-5\right)\left(x-3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}2x-5=0\\x-3=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=3\end{matrix}\right.\)
Vậy ...
3) PT \(\Rightarrow3x-4-2x+5=3\)
\(\Rightarrow x=2\)
Vậy ...
4) PT \(\Rightarrow\left[{}\begin{matrix}2x+1=0\\\dfrac{1}{2}x-1=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=2\end{matrix}\right.\)
Vậy ...
3) Ta có: \(\left(3x-4\right)-\left(2x-5\right)=3\)
\(\Leftrightarrow3x-4-2x+5=3\)
\(\Leftrightarrow x+1=3\)
hay x=2
Tìm x, biết: x(5 – 2x) + 2x(x - 1) = 15
x(5 – 2x) + 2x(x – 1) = 15
(x.5 – x.2x) + (2x.x – 2x.1) = 15
5x – 2x2 + 2x2 – 2x = 15
(2x2 – 2x2) + (5x – 2x) = 15
3x = 15
x = 5.
Vậy x = 5.
Tìm x, biết:
a) 16x2-(4x-5)2=15 b) (2x+1)(1-2x)+(1-2x)2=18
c) (x-5)2-x(x-4)=9 d) (x-5)2+(x-4)(1-x)=0
a) <=> (4x - 4x + 5)(4x + 4x - 5) = 15 <=> 40x = 15 <=> x = 3/8
a) <=> (4x - 4x + 5)(4x + 4x - 5) = 15 <=> 5(8x-5) = 15
<=> 40x = 40 <=> x = 1
Cái này mới chuẩn
b) (2x+1)(1-2x)+(1-2x)2=18 <=> 1 - 4x2 + 4x2 - 4x + 1 = 18
<=> -4x = 16 <=> x = -4
Tìm x : (2x−5)(x 2)−2x(x−1)=15
PT: x(5-2x)+2x(x-1)=15
<=> 5x - 2x² + 2x² - 2x = 15
<=> 3x = 15
<=> x = 5.
\(\left(2x-5\right)\left(x+2\right)-2x\left(x-1\right)=15\)
\(\Leftrightarrow\left(2x^2+4x-5x-10\right)-2\left(x^2-x\right)-15=0\)
\(\Leftrightarrow2x^2-x-10-2x^2+2x-15=0\)
\(\Leftrightarrow x=25\)
(2x-5)(x2)-2x(x-1)=15
\(2x^2-10-2x^2+2x=15\)
\(-10+2x=15\)
\(-8x=15\)
\(x=-\frac{15}{8}\)
Tìm x, biết: x( 5 - 2x ) + 2x( x - 1 ) = 15
x5-x22+2x2-2x=15
x(5-2)+x2(2-2)=15
=>x3+x20=15
=>x3=15
=>x=15:3
=>x=5
vậy x=5
hơi dài dòng 1 chút nhưng chắc ko sao
\(\Leftrightarrow5x-2x^2+2x^2-2x=15\Leftrightarrow3x=15\Leftrightarrow x=5\)
Ta có:
x( 5 - 2x ) + 2x( x - 1 ) = 15
=> 5x-2x2+2x2-2x=15
=> 3x=15
=> x=5
bạn chỉ cần phân tích ra là đuợc :)
1,Tìm x
5^x-5^x+2=650
2x-3/11+2x-4/12+2x-5/13=2x-6/14+2x-7/15+2x-8/6
5^x + 5^ ( x + 2 ) = 650
5x + 5x . 52 = 650
5x .( 1 + 25 ) = 650
5x . 26 = 650
5x = 650 : 26
5x = 25
5x = 52
=> x = 2
Vậy x = 2
The Reflection Of Light~ Ủa đề là 5^x-5^x+2= 650 sao ông lại đổi thành 5^x-5^(x+2) zậy
Tìm x
x^10=1^x
x^10= x
(2x - 15)^5 = (2x-15)^3
Tìm x biết : x .(5-2x)+2x.(x-1)=15
x(5 - 2x) + 2x(x - 1) = 15
=> 5x - 2x\(^2\)+ 2x\(^2\)- 2x = 15
=> 5x - 2x = 15
=> 3x = 15
=> x = 5
K cho mình nhé! Cảm ơn
\(x\left(5-2x\right)+2x\left(x-1\right)=15\)
\(\Rightarrow5x-2x^2+2x^2-2x=15\)
\(\Rightarrow\left(5x-2x\right)+\left(-2x^2+2x^2\right)=15\)
\(\Rightarrow3x=15\)
\(\Rightarrow x=15:3\)
\(\Rightarrow x=5\)
Vậy x = 5
_Chúc bạn học tốt_
1. Tìm số tự nhiên x biết 1) 3^x . 3=243 2) 7.2^x=56 3) x^3=8 4)x^20=x 5) 2^x-15=17 6) (2x+1)^3=9.81 7) x^6:x^3=115 8) ( 2x-15)^5=(2x-15)^3 9) 3^x+2-5.3^x=36 10) 7.4^x-1+4^x+1=23 Giúp tớ bài này với
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
Tìm x,biết :
a) 2x^2-7x+5=0
b) x(2x-5) - 4x+10=0
c) (x-5)(x+5) - x(x-2)=15
d) x^2(2x-3) - 12+8x=0
e) x(x - 1)+5x - 5=0
f) (2x-3)^2 - 4x(x - 1)=5
g) x(5 - 2x)+2x(x - 1)=13
h)2(x+5)(2x - 5)+(x - 1)(5 - 2x)=0
\(2x^2-7x+5=0\)
\(2x^2-2x-5x+5=0\)
\(2x\left(x-1\right)-5\left(x-1\right)=0\)
\(\left(x-1\right)\left(2x-5\right)=0\)
\(\left[\begin{array}{nghiempt}x-1=0\\2x-5=0\end{array}\right.\)
\(\left[\begin{array}{nghiempt}x=1\\2x=5\end{array}\right.\)
\(\left[\begin{array}{nghiempt}x=1\\x=\frac{5}{2}\end{array}\right.\)
\(x\left(2x-5\right)-4x+10=0\)
\(x\left(2x-5\right)-2\left(2x-5\right)=0\)
\(\left(2x-5\right)\left(x-2\right)=0\)
\(\left[\begin{array}{nghiempt}x-2=0\\2x-5=0\end{array}\right.\)
\(\left[\begin{array}{nghiempt}x=2\\2x=5\end{array}\right.\)
\(\left[\begin{array}{nghiempt}x=2\\x=\frac{5}{2}\end{array}\right.\)
\(\left(x-5\right)\left(x+5\right)-x\left(x-2\right)=15\)
\(x^2-25-x^2+2x=15\)
\(2x=15+25\)
\(2x=40\)
\(x=\frac{40}{2}\)
\(x=20\)
\(x^2\left(2x-3\right)-12+8x=0\)
\(x^2\left(2x-3\right)+4\left(2x-3\right)=0\)
\(\left(2x-3\right)\left(x^2+4\right)=0\)
\(2x-3=0\) (vì \(x^2\ge0\Rightarrow x^2+4\ge4>0\))
\(2x=3\)
\(x=\frac{3}{2}\)
\(x\left(x-1\right)+5x-5=0\)
\(x\left(x-1\right)+5\left(x-1\right)=0\)
\(\left(x-1\right)\left(x+5\right)=0\)
\(\left[\begin{array}{nghiempt}x-1=0\\x+5=0\end{array}\right.\)
\(\left[\begin{array}{nghiempt}x=1\\x=-5\end{array}\right.\)
\(\left(2x-3\right)^2-4x\left(x-1\right)=5\)
\(4x^2-12x+9-4x^2+4x=5\)
\(-8x=5-9\)
\(-8x=-4\)
\(x=\frac{4}{8}\)
\(x=\frac{1}{2}\)
\(x\left(5-2x\right)+2x\left(x-1\right)=13\)
\(5x-2x^2+2x^2-2x=13\)
\(3x=13\)
\(x=\frac{13}{3}\)
\(2\left(x+5\right)\left(2x-5\right)+\left(x-1\right)\left(5-2x\right)=0\)
\(\left(2x+10\right)\left(2x-5\right)-\left(x-1\right)\left(2x-5\right)=0\)
\(\left(2x-5\right)\left(2x+10-x+1\right)=0\)
\(\left(2x-5\right)\left(x+11\right)=0\)
\(\left[\begin{array}{nghiempt}2x-5=0\\x+11=0\end{array}\right.\)
\(\left[\begin{array}{nghiempt}2x=5\\x=-11\end{array}\right.\)
\(\left[\begin{array}{nghiempt}x=\frac{5}{2}\\x=-11\end{array}\right.\)
\(a,2x^2-7x+5=0\Leftrightarrow2x^2-2x-5x+5=0\Leftrightarrow2x\left(x-1\right)-5\left(x-1\right)=0\Leftrightarrow\left(x-1\right)\left(2x-5\right)=0\Rightarrow\left[{}\begin{matrix}x-1=0\\2x-5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\2x=5\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=2,5\end{matrix}\right.\)\(b,x\left(2x-5\right)-4x+10=0\Rightarrow x\left(2x-5\right)-2\left(2x-5\right)=0\Leftrightarrow\left(x-2\right)\left(2x-5\right)=0\Rightarrow\left[{}\begin{matrix}x-2=0\\2x-5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2\\2x=5\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2\\x=2,5\end{matrix}\right.\)\(c,\left(x-5\right)\left(x+5\right)-x\left(x-2\right)=15\Leftrightarrow x^2-25-x^2+2x-15=0\Leftrightarrow2x-40=0\Rightarrow2x=40\Rightarrow x=20\)\(d,x^2\left(2x-3\right)-12+8x=0\Rightarrow2x^3-3x^2-12+8x=0\Leftrightarrow2x^3+8x-3x^2-12=0\Leftrightarrow2x\left(x^2+4\right)-2\left(x^2+4\right)=0\Leftrightarrow\left(2x-2\right)\left(x^2+4\right)=0\Rightarrow\left[{}\begin{matrix}2x-2=0\\x^2+4=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x=2\\x^2=-4\end{matrix}\right.\Rightarrow x=1\)