Tìm x, biết: 5x(x – 1) = x – 1
Tìm x , biết
a) x (5x - 2y) + 2x ( x - 1) = 15
b) x^2 - 25x = 0
c) 5x (x - 1) = x - 1
b: \(\Leftrightarrow x\left(x-25\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=25\end{matrix}\right.\)
c: \(\Leftrightarrow\left(x-1\right)\left(5x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{5}\end{matrix}\right.\)
\(b,\Leftrightarrow x\left(x-25\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=25\end{matrix}\right.\\ c,\Leftrightarrow\left(x-1\right)\left(5x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{5}\end{matrix}\right.\)
tìm x biết :
2x^2-3(1-x).(x+1)=5x(x+1)
\(2x^2-3\left(1-x\right)\left(x+1\right)=5x\left(x+1\right)\\ \Leftrightarrow2x^2+3\left(x^2-1\right)=5x^2+5x\\ \Leftrightarrow2x^2+3x^2-3=5x^2+5x\\ \Leftrightarrow x=-\dfrac{3}{5}\\ \Rightarrow S=\left\{-\dfrac{3}{5}\right\}\)
Ta có: \(2x^2-3\left(1-x\right)\left(x+1\right)=5x\left(x+1\right)\)
\(\Leftrightarrow2x^2+3x^2-3-5x^2-5x=0\)
\(\Leftrightarrow-5x=3\)
hay \(x=-\dfrac{3}{5}\)
Câu 3: Tìm x, biết:
a) 5x( x-1)= x-1
b) 2(x+5)- x²- 5x= 0
a) 5x(x - 1) = x - 1
5x(x - 1) - (x - 1) = 0
(x - 1) (5x - 1) = 0
TH1: x - 1 = 0
x = 1
TH2: 5x - 1 = 0
5x = 1
x = 1/5
Vay x = 1 hoac x = 1/5.
b) 2(x + 5) - x2 - 5x = 0
2(x + 5) - x(x + 5) = 0
(x + 5) (2 - x) = 0
TH1: x + 5 = 0
x = -5
TH2: 2 - x = 0
x = 2
Vay x = -5 hoac x = 2
tìm x biết 25(x+3)^2+(1-5x)(1+5x)=0
\(25\left(x+3\right)^2+\left(1-5x\right)\left(1+5x\right)=0\)
\(25\left(x^2+6x+9\right)+1-25x^2=0\)
\(25x^2+150x+225+1-25x^2=0\)
\(150x=-226\)
\(x=-\frac{113}{75}\)
25 ( x + 3 )2 + ( 1 - 5x )( 1 + 5x ) = 0
25 ( x2 + 6x + 9 ) + 1 + 5x - 5x - 25x2 = 0
25x2 + 150x + 225 + 1 + 5x - 5x - 25x2 = 0
150x + 226 = 0
150x = -226
x = -226/150
Tìm hai số x,y biết:
5x.(x - 3)=(x - 2).(5x -1)-5
`5x(x-3)=(x-2)(5x-1)-5`
`\rightarrow 5x^2-15x= [x(5x-1)-2(5x-1)-5]`
`\rightarrow 5x^2-15x=(5x^2-x-10x+2-5)`
`\rightarrow 5x^2-15x=5x^2-11x-3`
`\rightarrow 5x^2-15x-5x^2+11x+3=0`
`\rightarrow -4x+3=0`
`\rightarrow 4x=3`
`\rightarrow x=`\(\dfrac{3}{4}\)
Vậy, `x=`\(\dfrac{3}{4}\)
Còn biến `y` thì mình k thấy bạn nhé!
Cho mk sửa lại từ dòng thứ 6 (tính cả đề)
`\rightarrow -4x+3=0`
`\rightarrow -4x=-3`
`\rightarrow x=-3/-4`
`\rightarrow x=3/4`
Vậy, `x=3/4`
Tìm x biết:
a) 5x(x-1)=x-1
b) 2(x+5)-x2-5x=0
a) \(5x\left(x-1\right)=x-1\)
\(\Rightarrow5x\left(x-1\right)-\left(x-1\right)=0\)
\(\Rightarrow\left(x-1\right).\left(5x-1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-1=0\\5x-1=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=1\\x=\frac{1}{5}\end{cases}}\)
Vậy \(x=1\) hoặc \(x=\frac{1}{5}\)
b) \(2\left(x+5\right)-x^2-5x\)
\(\Rightarrow2\left(x+5\right)-x\left(x+5\right)\)
\(\Rightarrow\left(x+5\right).\left(2-x\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x+5=0\\2-x=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=-5\\x=2\end{cases}}\)
Vậy \(x=-5\)hoặc \(x=2\)
Tìm x biết a) (x^2-4x+5)_(x^2-2x+1)=3 lớp 7
b)(4x^3-5X^2+3x-1)+(3-5x+5x^2-4x^3)=2
c)(3x-2)-(5x+4)=(x-3)-(X+5)
a, \(-4x+5+2x-1=3\Leftrightarrow-2x=-1\Leftrightarrow x=\dfrac{1}{2}\)
b, \(-2x+2=2\Leftrightarrow x=0\)
c, \(-2x-6=-8\Leftrightarrow x=1\)
tìm x biết
a) (5x-1)2-(5x-4)(5x+4)=7
b)5x2+4xy+4y2+4x+1=0
c)(x+2)3-x(x-1)(x+1)=6x2+21
tìm x ,biết
a, x^2 -4x =12
b, ( 5x+1)^2 - (5x -3)(5x+3)=30
c, (x-1) ( x^2 +x +1 ) + x(x +2) (2-x)=5
d, 5x(x-3)^2 -5 (x-1)^3 +15 (x+4) (x-4) =5