Tìm x
(x-3)2-4=0
x2-2x=24
(2x-1)2+(x+3)2-5(x+7)(x-7)=0
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
1) \(\left(x-3\right)^2-4=0\)
\(\Leftrightarrow\left(x-3-2\right)\left(x-3+2\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=1\end{matrix}\right.\)
2) \(x^2-2x=24\)
\(\Leftrightarrow x^2-2x-24=0\)
\(\Leftrightarrow x^2+4x-6x-24=0\)
\(\Leftrightarrow x\left(x+4\right)-6\left(x+4\right)=0\)
\(\Leftrightarrow\left(x+4\right)\left(x-6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-4\end{matrix}\right.\)
a/ (x-3)2 - 4 = 0
=> (x-3-2)(x-3+2)=0
=> (x-5)(x-1)=0
=> x = 5 hoặc x=1
a:x=3/5-7/8=24/40-35/40=-11/40
b: =>1/3:(2x-1)=-1/6
=>2x-1=-2
=>2x=-1
=>x=-1/2
c: =>x-3/4=17/2+7/4=34/4+7/4=41/4
=>x=11
d: =>3x+2=0 hoặc 2/5x+7=0
=>x=-2/3 hoặc x=-7:2/5=-35/2
tìm x biết:
(3x-1) [- 1/2x+5]=0
1/4+1/3:(2x-1)=-5
[2x+3/5]2 - 9/25=0
-5(x+1/5)-1/2(x-2/3)=3/2x - 5 /6
[x+1/2]x [2/3-2x]=0
17/2-|2x-3/4|=-7/4
2/3x-1/2x =5/12
(x+1/5)2+17/25=26/25
[x.44/7+3/7].11/5-3/7=-2
3[3x-1/2]+1/9=0
Toán lớp 6Tìm x
Trả lời Câu hỏi tương tự
Chưa có ai trả lời câu hỏi này,bạn hãy là người đâu tiên giúp nguyenvanhoang giải bài toán này !
1: =>(x+3)(x-5)=0
=>x=5 hoặc x=-3
2: =>(x-1)(5x-1)=0
=>x=1/5 hoặc x=1
5: =>(x-4)*x=0
=>x=0 hoặc x=4
10: =>(x+5)(x-3)=0
=>x=3 hoặc x=-5
9: =>(x-2)(x-4)=0
=>x=2 hoặc x=4
7: =>(x-6)(2x-1)=0
=>x=1/2 hoặc x=6
8: =>(2x-1)(3x-12)=0
=>x=4 hoặc x=1/2
\(a,x^2-2x=24\)
\(x^2-2x-24=0\)
\(x^2-2x+1-25=0\)
\(\left(x-1\right)^2=5^2=\left(-5\right)^2\)
\(x-1=5\) hoặc \(x-1=-5\)
\(\Rightarrow\hept{\begin{cases}x=6\\x=-4\end{cases}}\)
\(b,\left(2x-1\right)^2+\left(x+3\right)^2-5\left(x+7\right)\left(x-7\right)=0\)
\(4x^2-4x+1+x^2+6x+9-5\left(x^2-49\right)=0\)
\(4x^2-4x+1+x^2+6x+9-5x^2+245=0\)
\(2x+255=0\)
\(2x=-255\)
\(x=-\frac{255}{2}\)
a/ \(x^2-2x=24\)
<=> \(x^2-2x+1-1=24\)
<=> \(\left(x-1\right)^2=25\)
<=> \(\orbr{\begin{cases}x-1=25\\x-1=-25\end{cases}}\)<=> \(\orbr{\begin{cases}x=26\\x=-24\end{cases}}\)
b/ \(\left(2x-1\right)^2+\left(x+3\right)^2-5\left(x+7\right)\left(x-7\right)=0\)
<=> \(4x^2-4x+1+x^2+6x+9-5\left(x^2-49\right)=0\)
<=> \(4x^2-4x+1+x^2+6x+9-5x^2+245=0\)
<=> \(2x+255=0\)
<=> \(2x=-255\)
<=> \(x=-\frac{255}{2}\)
x^2 -2x = 24
=> x^2 - 2x - 24=0
=>x^2 -8x+6x - 24 = 0
=> ( x^2- 8x)+( 6x-24) = 0
=> x(x-8) + 6(x-8) = 0
=> (x+6)(x-8)=0
=>\(\orbr{\begin{cases}x=-6\\x=8\end{cases}}\)
a) \(\left(x-3\right)^2-4=0\)
\(\left(x-7\right)\left(x+1\right)=0\)
\(\orbr{\begin{cases}x=7\\x=-1\end{cases}}\)
b) \(x^2-2x=24\)
\(x^2-2x-24=0\)
\(\left(x-6\right)\left(x+4\right)=0\)
\(\orbr{\begin{cases}x=6\\x=-4\end{cases}}\)
c) \(\left(2x-1\right)^2+\left(x+3\right)^2-5\left(x+7\right)\left(x-7\right)=0\)
\(4x^2+4x+1+x^2+6x+9-5\left(x^2-49\right)=0\)
\(5x^2+10x+10-5x^2+245=0\)
\(10x+255=0\)
\(x=-25.5\)
A) \(\left(x-3\right)^2-4=0\)
\(\left(x-3\right)^2=4\Rightarrow\left(x-3\right)^2=\left(-2\right)^2;2^2\)
th1\(\left(x-3\right)^2=2^2\)
\(\Rightarrow x-3=2\)
\(\Rightarrow x=2+3\)
\(\Rightarrow x=5\)
th2: \(\left(x-3\right)^2=\left(-2\right)^2\)
\(\Rightarrow x-3=-2\)
\(\Rightarrow x=-2+3\)
\(\Rightarrow x=1\)
\(\Leftrightarrow x\in\left\{1;5\right\}\)