x - 3x = 16
Tìm x:
3x(x - 16) - (x - 16) = 0
\(\left\{{}\begin{matrix}x-16=0\\3x-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=16\\x=\dfrac{1}{3}\end{matrix}\right.\)
\(\Leftrightarrow\left(x-16\right)\left(3x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-16=0\\3x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=16\\x=\dfrac{1}{3}\end{matrix}\right.\)
=>x-16=0 hoặc 3x-1=0
x = 16 x = ⅓
Bài 1:Tìm đa thức M
a)\(\dfrac{^{x^3}+27}{x^2-3x+9}\)=\(\dfrac{x+3}{M}\)
b)\(\dfrac{M}{x+4}\)=\(\dfrac{x^2-8x+16}{16-x^2}\)
c)\(\dfrac{x-2y}{M}\)=\(\dfrac{3x^2-7xy+2y^2}{3x^2+5xy-2y^2}\)
a, \(\dfrac{x^3+27}{x^2-3x+9}=\dfrac{x+3}{M}\Leftrightarrow\dfrac{\left(x+3\right)\left(x^2-3x+9\right)}{x^2-3x+9}=\dfrac{x+3}{M}\)
\(\Rightarrow M=\dfrac{x+3}{x+3}=1\)
b, \(\dfrac{M}{x+4}=\dfrac{x^2-8x+16}{16-x^2}=\dfrac{\left(x-4\right)^2}{\left(4-x\right)\left(x+4\right)}=\dfrac{4-x}{x+4}\)
\(\Rightarrow M=\dfrac{\left(4-x\right)\left(x+4\right)}{x+4}=4-x\)
c, tương tự
Rút gọn biểu thức: P=12−12(x:16−14)−2|3x−2|P=12−12(x:16−14)−2|3x−2|
a, x lớn hơn hoặc bằng 2323
b, x <23
Bạn ghi rõ đề ra đi bạn. Khó hiểu quá!
Rút gọn biểu thức: 2323
b, x <
(3x - 2)(x^2 + 16) = (2x - 7)(x^2 + 16) giúp với ạ
\(\left(3x-2\right)\left(x^2+16\right)=\left(2x-7\right)\left(x^2+16\right)\)
\(\Leftrightarrow\left(3x-2\right)\left(x^2+16\right)-\left(2x-7\right)\left(x^2+16\right)=0\)
\(\Leftrightarrow\left(x^2+16\right)\left(3x-2-2x+7\right)=0\)
\(\Leftrightarrow\left(x^2+16\right)\left(x+5\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x^2+16=0\\x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\pm\sqrt{-16}\\x=-5\end{matrix}\right.\)
\(\left(3\text{x}-2\right)\left(x^2+16\right)=\left(2\text{x}-7\right)\left(x^2+16\right)\)
\(\Leftrightarrow\left(3\text{x}-2\right)=\left(2\text{x}-7\right)\)
\(\Leftrightarrow3\text{x}-2x=-7+2\)
\(\Leftrightarrow x=-5\)
3x+2/4=16/3x+2 tìm x
\(\dfrac{3x+2}{4}=\dfrac{16}{3x+2}\)
`=> (3x+2)^2 =4.16`
`=> (3x+2)^2 = 64`
`=> (3x+2)^2 = +- 8^2`
\(\Rightarrow\left[{}\begin{matrix}3x+2=8\\3x+2=-8\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}3x=6\\3x=-10\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=2\\x=-\dfrac{10}{3}\end{matrix}\right.\)
Bài 1: tìm x
a, (3x-5)2 - (x-1)2 = 0
b, 16(2-3x) + x2(3x-2) =0
Bài 2:
a: \(x^2-4x+3=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=3\end{matrix}\right.\)
a, (3x-5)^2 - (x-1)^2 = 0
(3x-5-x+1)(3x-5+x-1) =0
(2x-4)(4x-6)=0
Do đó: 2x-4=0 hoặc 4x-6=0
Th1: 2x-4=0 => 2x=4
=> x=2
Th2: 4x-6=0 => 4x=6
=> x = 4/6 =2/3
Vậy x = 2 ; 2/3
1. √x^2-8x+16 +|x+2|=0
2. √x^2-x-6 = √3x+5
3.√x^2-x =√3x+5
1: =>|x-4|+|x+2|=0
=>x-4=0 và x+2=0
=>\(x\in\varnothing\)
2: =>x^2-x-6=3x+5
=>x^2-4x-11=0
=>x^2-4x+4-15=0
=>(x-2)^2-15=0
=>x=căn 15+2 hoặc x=-căn 15+2
3: =>x^2-x=3x+5
=>x^2-4x-5=0
=>(x-5)(x+1)=0
=>x=-1 hoặc x=5
Giải phương trình :
1) √x2+x+2 + 1/x= 13-7x/2
2) x2 + 3x = √1-x + 1/4
3) ( x+3)√48-x2-8x= 28-x/ x+3
4) √-x2-2x +48= 28-x/x+3
5) 3x2 + 2(x-1)√2x2-3x +1= 5x + 2
6) 4x2 +(8x - 4)√x -1 = 3x+2√2x2 +5x-3
7) x3/ √16-x2 + x2 -16 = 0
Tìm x biết:
1,
a,3x(x+1) - 2x(x+2) = -x-1
b,2x(x-2020) - x+2020 = 0
c,(x-4)2 - 36 = 0
d,x2 + 8x - 16 = 0
e,x(x+6) - 7x - 42 = 0
f,25x2 - 16 = 0
2,
a,3x3 - 12x = 0
b,x2 + 3x - 10 = 0
Bài 1:
a) \(\Rightarrow3x^2+3x-2x^2-4x+x+1=0\)
\(\Rightarrow x^2=-1\left(VLý\right)\Rightarrow S=\varnothing\)
b) \(\Rightarrow\left(x-2020\right)\left(2x-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=2020\\x=\dfrac{1}{2}\end{matrix}\right.\)
c) \(\Rightarrow\left(x-10\right)\left(x+2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=10\\x=-2\end{matrix}\right.\)
d) \(\Rightarrow\left(x+4\right)^2=0\Rightarrow x=-4\)
e) \(\Rightarrow\left(x+6\right)\left(x-7\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=-6\\x=7\end{matrix}\right.\)
f) \(\Rightarrow\left(5x-4\right)\left(5x+4\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{4}{5}\\x=-\dfrac{4}{5}\end{matrix}\right.\)
Bài 2:
a) \(\Rightarrow3x\left(x^2-4\right)=0\Rightarrow3x\left(x-2\right)\left(x+2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=2\\x=-2\end{matrix}\right.\)
b) \(\Rightarrow x\left(x-2\right)+5\left(x-2\right)=0\Rightarrow\left(x-2\right)\left(x+5\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=2\\x=-5\end{matrix}\right.\)
a) (2x-3)²=9
b) (x²-16)²-16(x-4)²=0
c) (2x/5-3/4)²-3x/5-1/4=0
d) 1/16(x-2)²=9/25(5/3x-5)²