(x-2)3-x(x-1)(x+1)+x(7x-6)=0
Tìm x
x ^ 2- ( 2 + √2 + √3 ) x + 1 + √2 + √3 + √6 = 0
Tìm x
giúp e với ạ
đánh đề bằng latex cho rõ đi bạn, không biết nào dấu nào biến:v
\(\dfrac{1}{3}x\)+\(\dfrac{2}{3}\)(x-1)=0
tìm x
\(\dfrac{1}{3}x+\dfrac{2}{3}\left(x-1\right)=0\\ \dfrac{1}{3}x+\dfrac{2}{3}x-\dfrac{2}{3}=0\\ x=\dfrac{2}{3}\)
`1/3x + 2/3(x-1) =0`
` 1/3x + 2/3x -2/3 = 0`
` ( 1/3 + 2/3) x -2/3 = 0`
` 3/3x -2/3 = 0`
` 1x-2/3 = 0`
`1/x = 0 + 2/3`
` 1x = 2/3`
` x = 2/3`
(x - 1 ) . ( x +2) . (-x - 3) = 0
Tìm các số nguyên x , thỏa mãn :
( x - 7 ) ( x + 3 ) < 0
a: (x-1)(x+2)(-x-3)=0
=>(x-1)(x+2)(x+3)=0
=>\(\left[{}\begin{matrix}x-1=0\\x+2=0\\x+3=0\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=1\\x=-2\\x=-3\end{matrix}\right.\)
b: (x-7)(x+3)<0
TH1: \(\left\{{}\begin{matrix}x-7>0\\x+3< 0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x>7\\x< -3\end{matrix}\right.\)
=>\(x\in\varnothing\)
TH2: \(\left\{{}\begin{matrix}x-7< 0\\x+3>0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x< 7\\x>-3\end{matrix}\right.\)
=>-3<x<7
mà x nguyên
nên \(x\in\left\{-2;-1;0;1;2;3;4;5;6\right\}\)
Tìm x biết: (x + 2)^2 - (x + 2)(x - 3) = 0
Tìm x biết :
a,(x+2)^2-(x+2)(x-3)=0
b,2x^3-4x^2+2x=0
c,(x-1)^2-(2x+1)^2=0
\(a,\Leftrightarrow\left(x+2\right)\left(x+2-x+3\right)=0\\ \Leftrightarrow5\left(x+2\right)=0\Leftrightarrow x=-2\\ b,\Leftrightarrow2x\left(x-1\right)^2=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\\ c,\Leftrightarrow\left(x-1-2x-1\right)\left(x-1+2x+1\right)=0\\ \Leftrightarrow3x\left(-x-2\right)=0\Leftrightarrow-3x\left(x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\)
7x(x-20)+10(x-20)=0
tìm x
giúp với mọi người ơi
\(\Leftrightarrow\left[{}\begin{matrix}x=20\\x=-\dfrac{10}{7}\end{matrix}\right.\)
\(\left(x-20\right).\left(7x+10\right)=0\)
\(=>\left[{}\begin{matrix}x-20=0\\7x+10=0\end{matrix}\right.\)
\(=>\left[{}\begin{matrix}x=20\\x=-\dfrac{10}{7}\end{matrix}\right.\)
7x(x-20)+10(x-20)=0
(x-20)x(7x+10)=0
(x-20)=0 hoặc (7x+10)=0
x=20 hoặc 7x=-10
x=20 hoặc x=-10/7
x2-4x+7 = 0 ⇔ x2 -4x + 4 + 3 = 0
⇔ (x-2)2+3=0 ⇔ (x-2)2=-3 (vô lí)
Vậy pt vô nghiệm
*Chứng minh phương trình \(x^2-4x+7=0\) vô nghiệm
Ta có: \(x^2-4x+7=0\)
\(\Leftrightarrow x^2-4x+4+3=0\)
\(\Leftrightarrow\left(x-2\right)^2+3=0\)
mà \(\left(x-2\right)^2+3\ge3>0\forall x\)
nên \(x\in\varnothing\)(đpcm)
x(x+1)-(x-2)(x+1)=0
tìm x
giúp mình với ạ
x(x+1)-(x-2)(x+1)=0
\(\left(x+1\right)\left(x-x+2\right)=0\\ \left(x+1\right)\cdot2=0\\ =>x+1=0\\ x=0-1\\ x=-1\)
=>(x+1)(x-x+2)=0
=>x+1=0
=>x=-1
a) 6-7x+7=-8x+12
b) \(\dfrac{\text{4x-1}}{8}\)\(\)=\(\dfrac{6-x}{2}\)-\(\dfrac{1}{4}\)
c) 2x(x-3)+7x-21+0
d) \(\dfrac{x}{x-3}\)+\(\dfrac{x-2}{x+3}\)\(\dfrac{2\left(x^2+6\right)}{x^2-9}\)
Tìm x :
1/ |x-5|-7(x+4)=5-7x
2/3|x+4|-2(x+1)=7-2x
3/2|x-6|+7x-2=|x-6|+7x
4/|x+8|+(x-2)=5x-10:2
5/|x+2|-6(x-4)=20-6x
1: =>|x-5|=5-7x+7x+28=33
=>x-5=33 hoặc x-5=-33
=>x=38 hoặc x=-28
3: 2|x-6|+7x-2=|x-6|+7x
=>|x-6|=2
=>x-6=2 hoặc x-6=-2
=>x=8 hoặc x=4