(x-3/2).(2x+1)>0
(-2x).(4/5-x)<0
Ai nhanh mình tích cho
1) (x + 1/2).(2/3 - 2x) = 0 2) 2/3x + 1/2x = 5/2 : 3 và 3/4 3) (2x - 3)(6 - 2x)= 0 4) -5(x + 1/5) - 1/2(x - 2/3) = 3/2x - 5/6
1: =>x+1/2=0 hoặc 2/3-2x=0
=>x=-1/2 hoặc x=1/3
2: =>7/6x=5/2:3,75=2/3
=>x=2/3:7/6=2/3*6/7=12/21=4/7
3: =>2x-3=0 hoặc 6-2x=0
=>x=3 hoặc x=3/2
4: =>-5x-1-1/2x+1/3=3/2x-5/6
=>-11/2x-3/2x=-5/6-1/3+1
=>-7x=-1/6
=>x=1/42
1) |2x - 1| = 5
2) |2x - 1| = |x + 5|
3) |3x + 1| = x - 2
4) |3 - 2x| = x + 2
5) |2x - 1| = 5 - x
6) |- 3x| = x - 2
7) |2 - 3x| = 2x + 1
8) |2x - 1| + |4x ^ 2 - 1| = 0
9) (2x + 5)/(x + 3) + 1 = 4/(x ^ 2 + 2x - 3) - (3x - 1)/(1 - x)
10) (x - 1)/(x + 3) - x/(x - 3) = (7x - 3)/(9 - x ^ 2)
11) 5 + 96/(x ^ 2 - 16) = (2x - 1)/(x + 4) + (3x - 1)/(x - 4)
12) (2x)/(2x - 1) + x/(2x + 1) = 1 + 4/((2x - 1)(2x + 1))
13) (x + 2)/(x - 2) - 1/x = 2/(x ^ 2 - 2x)
14) x/(2x - 6) + x/(2x + 2) = (2x + 4)/(x ^ 2 - 2x - 3)
a) (2x +1)(3 – x)(4 - 2x) = 0 b)2x(x – 3) + 5(x – 3) = 0
c) (x2 – 4) – (x – 2)(3 – 2x) = 0 d) x2 – 5x + 6 = 0
e) (2x + 5)2 = (x + 2)2 f) 2x3 + 6x2 = x2 + 3x
a: (2x+1)(3-x)(4-2x)=0
=>(2x+1)(x-3)(x-2)=0
hay \(x\in\left\{-\dfrac{1}{2};3;2\right\}\)
b: 2x(x-3)+5(x-3)=0
=>(x-3)(2x+5)=0
=>x=3 hoặc x=-5/2
c: =>(x-2)(x+2)+(x-2)(2x-3)=0
=>(x-2)(x+2+2x-3)=0
=>(x-2)(3x-1)=0
=>x=2 hoặc x=1/3
d: =>(x-2)(x-3)=0
=>x=2 hoặc x=3
e: =>(2x+5+x+2)(2x+5-x-2)=0
=>(3x+7)(x+3)=0
=>x=-7/3 hoặc x=-3
f: \(\Leftrightarrow2x^3+5x^2-3x=0\)
\(\Leftrightarrow x\left(2x^2+5x-3\right)=0\)
\(\Leftrightarrow x\left(x+3\right)\left(2x-1\right)=0\)
hay \(x\in\left\{0;-3;\dfrac{1}{2}\right\}\)
a) 2x-1/11+2x-2/12+2x-3/13=2x+5/5+2x+6/4+2x+7/3
b) x-1/2016+x-2/2015+x-3/2014+x-4/2013+x-5/2012 -5=0
c) x+2017/2+x+2015/3+x+2013/4+x+2011/5+8=0
1) (x+6)(3x-1)+x+6=0
2) (x+4)(5x+9)-x-4=0
3)(1-x)(5x+3)÷(3x-7)(x-1)
4)2x (2x-3)=(3-2x)(2-5x)
5)(2x-7)^2-6(2x-7)(x-3)=0
6)(x-2)(x+1)=x^2-4
7) x^2-5x+6=0
8)2x^3+6x^2=x^2+3x
9)(2x+5)^2=(x+2)^2
1) (x+6)(3x-1)+x+6=0
⇔(x+6)(3x-1)+(x+6)=0
⇔(x+6)(3x-1+1)=0
⇔3x(x+6)=0
2) (x+4)(5x+9)-x-4=0
⇔(x+4)(5x+9)-(x+4)=0
⇔(x+4)(5x+9-1)=0
⇔(x+4)(5x+8)=0
3)(1-x)(5x+3)÷(3x-7)(x-1)
=\(\frac{\left(1-x\right)\left(5x+3\right)}{\left(3x-7\right)\left(x-1\right)}=\frac{\left(1-x\right)\left(5x+3\right)}{\left(7-3x\right)\left(1-x\right)}=\frac{\left(5x+3\right)}{\left(7-3x\right)}\)
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
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 !
Tìm x: a)(2x+1)(1-2x)+(1-2x)^2=18
b) 2(x+1)^2-(x-3)(x+3)-(x-4)^2=0
c) (x-5)^2-x(x-4)=9
d) (x-5)^2+(x-4)(1-x)=0
a) (2x + 1)(1 - 2x) + (1 - 2x)2 = 18
= ( 1 - 2x) \(\left[\left(2x+1+1-2x\right)\right]\) = 18
= 2(1 - 2x) - 18 = 0
= 2 - 4x - 18 = 0
= -16 - 4x = 0
= -4x = 16
= x = \(\dfrac{16}{-4}=-4\)
b) 2(x + 1)2 -(x - 3)(x + 3) - (x - 4)2 = 0
= 2 (x2 + 2x + 1) - (x2 - 9) - (x2 - 8x + 16) = 0
= 2x2 + 4x + 2 - x2 + 9 - x2 + 8x - 16 = 0
= 12x - 5 = 0
= 12x = 5
= x = \(\dfrac{5}{12}\)
c) (x - 5)2 - x(x - 4) = 9
= x2 - 10x + 25 - x2 + 4x - 9 = 0
= -6x + 16 = 0
= -6x = -16
= x = \(\dfrac{-16}{-6}=\dfrac{8}{3}\)
d) (x - 5)2 + (x - 4)(1 - x)
= x2 - 10x + 25 + 5x - x2 - 4 = 0
= -5x + 21 = 0
= -5x = -21
= x = \(\dfrac{-21}{-5}=\dfrac{21}{5}\)
Chúc bạn học tốt
Tìm x
1) (2x-1)(x+3)(2-x)=0
2)x^3 + x^2 + x + 1 = 0
3) 2x(x-3)+5(x-3) =0
4)x(2x-7)-(4x-14)=0
5) 2x^3 + 3x^2 + 2x + 3 = 0
1) (2x-1)(x+3)(2-x)=0
=>2x-1 =0 hoặc x+3=0 hoặc 2-x=0
=>x=1/2 hoặc x=-3 hoặc x=2
2)x^3 + x^2 + x + 1 = 0
=>.x^2(x+1)+(x+1)=0
=>(x^2+1)(x+1)=0
=>x^2+1=0 hoặc x+1=0
=> x =-1
3) 2x(x-3)+5(x-3) =0
=>(2x+5)(x-3)=0
=>2x+5=0 hoặc x-3=0
=>x=-5/2 hoặc x=3
4)x(2x-7)-(4x-14)=0
=> (x-2)(2x-7)=0
=> x-2 =0 hoặc 2x-7=0
=>x=2 hoặc x=7/2
5)2x^3+3x^2+2x+3=0
=>x^2(2x+3)+2x+3=0
=>(x^2+1)(2x+3)=0
=>x^2+1=0 hoặc 2x+3=0
=> x =-3/2
Câu 1. Giải các phườn trình sau:
a, 3x+6=0
b, 2x-10=0
c, 3x-7=11
d, 3x-9=0
e, 3x(2-x) =15(x-2)
f, (x+5)(x+4)=0
g, x(x+4)=0
h, (2x -4)(x-2)=0
i, (x+1/5)(2x-3)=0
k, x²-4x=0
m, 4x²-1=0
n, x²-6x+9=0
l, (3x-5)²-(x+4)²=0
o, 7x(x+2)-5(x+2)=0
p, 3x(2x-5)-4x+10=0
q, (2-2x)-x²+1=0
r, x(1-3x)=5(1-3x)
s, 2x-3/4+x+1/6=3
t, x-3/4-2x+1/3=x/6
u, x+1/13+x+2/12=x+3/11+x+4/10
v, 2x+1/15+2x+2/14=2x+3/13+2x+4/12
Giúp e nha mn. E cảm ơn trc ạ!
e, 3x(2-x) =15(x-2)
\(\Leftrightarrow3x\left(2-x\right)-15\left(x-2\right)=0\)
\(\Leftrightarrow-3x\left(x-2\right)-15\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(-3x-15\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-2=0\\-3x-15=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\x=5\end{matrix}\right.\)
Vậy..
f, (x+5)(x+4)=0
\(\Leftrightarrow\left\{{}\begin{matrix}x+5=0\\x+4=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-5\\x=-4\end{matrix}\right.\)
Vậy..
g, x(x+4)=0
\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\x+4=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\x=-4\end{matrix}\right.\)
,h, (2x -4)(x-2)=0
\(\Leftrightarrow2\left(x-2\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(2-1\right)=0\)
\(\Leftrightarrow x-2=0\Leftrightarrow x=2\)
i, (x+1/5)(2x-3)=0
\(\Leftrightarrow\left\{{}\begin{matrix}x+\frac{1}{5}=0\\2x-3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\frac{-1}{5}\\x=\frac{3}{2}\end{matrix}\right.\)
k, x²-4x=0
\(\Leftrightarrow x\left(x-2\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)
m, 4x²-1=0
\(\Leftrightarrow\left(2x\right)^2-1^2=0\)
\(\Leftrightarrow\left(2x-1\right)\left(2x+1\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x-1=0\\2x+1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x=1\\2x=-1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\frac{1}{2}\\x=\frac{-1}{2}\end{matrix}\right.\)
n, x²-6x+9=0
\(\Leftrightarrow x^2-2.x.3+3^2=0\)
\(\Leftrightarrow\left(x-3\right)^2=0\Leftrightarrow x-3=0\)
<=> x=3
l, (3x-5)²-(x+4)²=0
\(\Leftrightarrow\left(3x-5-x-4\right)\left(3x-5+x+4\right)=0\)
\(\Leftrightarrow\left(2x-9\right)\left(4x-1\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x-9=0\\4x-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x=9\\4x=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\frac{9}{2}\\x=\frac{1}{4}\end{matrix}\right.\)
Vậy ..
o, 7x(x+2)-5(x+2)=0
\(\Leftrightarrow\left(x+2\right)\left(7x-5\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+2=0\\7x-5=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-2\\7x=5\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=-2\\x=\frac{5}{7}\end{matrix}\right.\)
Vậy....
p, 3x(2x-5)-4x+10=0
\(\Leftrightarrow3x\left(2x-5\right)-\left(4x-10\right)=0\)
\(\Leftrightarrow3x\left(2x-5\right)-2\left(2x-5\right)=0\)
\(\Leftrightarrow\left(2x-5\right)\left(3x-2\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x-5=0\\3x-2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x=5\\3x=2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\frac{5}{2}\\x=\frac{2}{3}\end{matrix}\right.\)
Vậy...
q, (2-2x)-x²+1=0
\(\Leftrightarrow2\left(1-x\right)-\left(x^2-1^2\right)=0\)
\(\Leftrightarrow2\left(1-x\right)-\left(x-1\right)\left(x+1\right)=0\)
\(\Leftrightarrow2\left(1-x\right)+\left(1-x\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left(1-x\right)\left(2+x+1\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}1-x=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-1\\x=-3\end{matrix}\right.\)
Vậy ....
r, x(1-3x)=5(1-3x)
\(\Leftrightarrow x\left(1-3x\right)-5\left(1-3x\right)=0\)
\(\Leftrightarrow\left(1-3x\right)\left(x-5\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}1-3x=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-3x=-1\\x=5\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\frac{1}{3}\\x=5\end{matrix}\right.\)
s, 2x-3/4+x+1/6=3
\(\Leftrightarrow x-\frac{7}{12}=3\Leftrightarrow x=3+\frac{7}{12}=\frac{43}{12}\)
r, x(1-3x)=5(1-3x)
➜x(1-3x)-5(1-3x)=0
➜(x-5)(1-3x)=0
➜\(\left[{}\begin{matrix}x-5=0\\1-3x=0\end{matrix}\right.\text{➜}\left[{}\begin{matrix}x=5\\x=\frac{1}{3}\end{matrix}\right.\)
Mk lười lắm mai nha!!!~~~~~~~~~~~~
Làm dần:
a, 3x+6=0
➜3x=-6
➜x=2
b, 2x-10=0
➜2x=10
➜x=5
c, 3x-7=11
➜3x=11+7
➜3x=18
➜x=6
d, 3x-9=0
➜3x=9
➜x=3
Bài 1:Giải phương trình sau:
1, x-8=3-2 (x+4)
2, 2 (x+3)-3 (x-1)=2
3, 4(x-5)-(3x-1)=x-19
4, 7-(x-2)=5 (2x-3)
5, 32-4(0,5y-5)=3y +2
6, 3 (x-1)-x=2x-3
Bài 2: Giải phương trình sau:
1, 2-x/3=3-2x/5
2, 3-4x/4=x+2/5
3, 2x-1/3+x=x+4/2
4, 1+2x-5/6=3-x/4
5, x-3/5+1-2x/3=-6
6, 3x-5/5+x/4=1/20
Bài 3: Giải các phương trình sau:
1, x^2-7x=0
2, 2x(x+3)+5(x+3)=0
3, 3x(x-1)+6 (x-1)=0
4, 3x(2x-8)-(2x-8)^2=0
Bài 1:
1. \(x-8=3-2\left(x+4\right)\)
\(x-8=3-2x-8\)
\(3x=3\Rightarrow x=1\)
2. \(2\left(x+3\right)-3\left(x-1\right)=2\)
\(2x+6-3x+3=2\)
\(-x+9=2\Rightarrow x=7\)
3. \(4\left(x-5\right)-\left(3x-1\right)=x-19\)
\(4x-20-3x+1=x-19\)
\(0x=0\Rightarrow x=0\)
4. \(7-\left(x-2\right)=5\left(2x-3\right)\)
\(7-x+2=10x-15\)
\(-11x=-24\Rightarrow x=\frac{24}{11}\)
5. \(32-4\left(0,5y-5\right)=3y+2\)
\(32-2y+20=3y+2\)
\(-5y=-50\Rightarrow y=10\)
6. \(3\left(x-1\right)-x=2x-3\)
\(3x-3-x=2x-3\)
\(0x=0\Rightarrow x=0\)
Bài 2:
1. \(\frac{2-x}{3}=\frac{3-2x}{5}\)
\(\frac{\left(2-x\right)5}{15}-\frac{\left(3-2x\right)3}{15}=0\)
\(\frac{10-5x-9+6x}{15}=0\)
\(x+1=0\Rightarrow x=-1\)
2. \(\frac{3-4x}{4}=\frac{x+2}{5}\)
\(\frac{5\left(3-4x\right)}{20}-\frac{4\left(x+2\right)}{20}=0\)
\(\frac{15-20x-4x-8}{20}=0\)
\(7-24x=0\)
\(24x=7\Rightarrow x=\frac{7}{24}\)