Tìm x biết:4|x-5|+2|3x-4|-12=0
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a) \(5x\left(x+4\right)-x\left(5x+1\right)=0\)
\(\Leftrightarrow x\left[5\left(x+4\right)-5x-1\right]=0\)
\(\Leftrightarrow x\left(5x+20-5x-1\right)=0\Leftrightarrow x=0\)
b) \(3x\left(5-x\right)+4\left(x-5\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left(4-3x\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=5\\x=\frac{4}{3}\end{cases}}\)
c) \(x\left(x-3\right)+4x-12=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+4\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=3\\x=-4\end{cases}}\)
d) \(x^2-36=0\)
\(\Leftrightarrow\left(x+6\right)\left(x-6\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=6\\x=-6\end{cases}}\)
e) \(x^2+3x+1=2\)
\(\Leftrightarrow x^2+3x+1-2=0\)
\(\Leftrightarrow x^2+3x-1=0\)
\(\Leftrightarrow x^2+3x+\frac{3}{2}-\frac{5}{2}=0\)
\(\Leftrightarrow\left(x+\frac{3}{2}\right)^2-\frac{5}{2}=0\)
\(\Leftrightarrow\left(x+\frac{3}{2}+\frac{\sqrt{5}}{\sqrt{2}}\right)\left(x+\frac{3}{2}-\frac{\sqrt{5}}{\sqrt{2}}\right)=0\)
Còn lại ........... Tự lm nất nha
\(\left(x+2\right)-2=0\)
\(\Rightarrow x+2-2=0\)
\(\Rightarrow x=0\)
\(\left(x+3\right)+1=7\)
\(\Rightarrow x+3+1=7\)
\(\Rightarrow x+4=7\)
\(\Rightarrow x=3\)
\(\left(3x-4\right)+4=12\)
\(\Rightarrow3x-4+4=12\)
\(\Rightarrow3x=12\)
\(\Rightarrow x=4\)
\(\left(5x+4\right)-1=13\)
\(\Rightarrow5x+4-1=13\)
\(\Rightarrow5x+3=13\)
\(\Rightarrow5x=10\)
\(\Rightarrow x=2\)
\(\left(4x-8\right)-3=5\)
\(\Rightarrow4x-8-3=5\)
\(\Rightarrow4x-11=5\)
\(\Rightarrow4x=16\)
\(\Rightarrow x=4\)
\(8-\left(2x+4\right)=2\)
\(\Rightarrow8-2x-4=2\)
\(\Rightarrow4-2x=2\)
\(\Rightarrow2x=2\)
\(\Rightarrow x=1\)
\(7+\left(5x+2\right)=14\)
\(\Rightarrow7+5x+2=14\)
\(\Rightarrow9+5x=14\)
\(\Rightarrow5x=5\)
\(\Rightarrow x=1\)
\(5-\left(3x-11\right)=1\)
\(\Rightarrow5-3x+11=1\)
\(\Rightarrow16-3x=1\)
\(\Rightarrow3x=15\)
\(\Rightarrow x=5\)
b: \(3x^2-2x-1=0\)
=>\(3x^2-3x+x-1=0\)
=>\(\left(x-1\right)\left(3x+1\right)=0\)
=>\(\left[{}\begin{matrix}x-1=0\\3x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{1}{3}\end{matrix}\right.\)
a: Bạn ghi lại đề đi bạn
bài 1:
a) 1/3x=-5/3
x=-5
b) x+12/5=37/15
x=1/15
c) x-1/7=-36/7
x=-5
d) 3x-1/2=0
3x=1/2
x=1/6
e) 2/5+x=1/4
x=-3/20
a: \(x^3-4x^2-x+4=0\)
=>\(\left(x^3-4x^2\right)-\left(x-4\right)=0\)
=>\(x^2\left(x-4\right)-\left(x-4\right)=0\)
=>\(\left(x-4\right)\left(x^2-1\right)=0\)
=>\(\left[{}\begin{matrix}x-4=0\\x^2-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x^2=1\end{matrix}\right.\Leftrightarrow x\in\left\{2;1;-1\right\}\)
b: Sửa đề: \(x^3+3x^2+3x+1=0\)
=>\(x^3+3\cdot x^2\cdot1+3\cdot x\cdot1^2+1^3=0\)
=>\(\left(x+1\right)^3=0\)
=>x+1=0
=>x=-1
c: \(x^3+3x^2-4x-12=0\)
=>\(\left(x^3+3x^2\right)-\left(4x+12\right)=0\)
=>\(x^2\cdot\left(x+3\right)-4\left(x+3\right)=0\)
=>\(\left(x+3\right)\left(x^2-4\right)=0\)
=>\(\left(x+3\right)\left(x-2\right)\left(x+2\right)=0\)
=>\(\left[{}\begin{matrix}x+3=0\\x-2=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=2\\x=-2\end{matrix}\right.\)
d: \(\left(x-2\right)^2-4x+8=0\)
=>\(\left(x-2\right)^2-\left(4x-8\right)=0\)
=>\(\left(x-2\right)^2-4\left(x-2\right)=0\)
=>\(\left(x-2\right)\left(x-2-4\right)=0\)
=>(x-2)(x-6)=0
=>\(\left[{}\begin{matrix}x-2=0\\x-6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=6\end{matrix}\right.\)
a.Ta có: 2|x| + 3|1-x| - |x-3| =0
2x + 3 - 3x - x +3 =0
-2x + 6 =0
-2x =-6
x =3
b.Ta có:4|3x-1| + |x| - 2|x-5| +7|x-3| =12
12x-4 + x -2(x-5) + 7x - 3 =12
12x-4+ x - 2x +10 + 7x - 3=12
18x + 3 =12
18x =12-3
18x =9
x =1/2
a. 2 |x |+ 3 | 1- x| - 5 | x - 3= 0
2 x + 3 - 3 x - x + 3 = 0
-2x + 6 = 0
- 2 x = - 6
x = 3
a)(x+2).(x+3)-(x-2).(x+5)=10
( x^2 +3x+2x+6)-(x^2 +5x-2x-10)=10
x^2 +3x+2x+6-x^2 -5x+2x+10-10=0
2x+6=0
2x=-6
x=-3