( 3x-1).(2x-3).(x+5)= 0
Những câu hỏi liên quan
1)4x-20=0 ; 2) 5x+15=0 ; 3) 3x-5=7x+2 ; 4) 4x-(x-1)=2(1+x) ; 5) x2 -2x=0 ; 6) 2(3x-5)-3(x-2)=3(x+4) ; 7) (x+3)(2x-7)=0
8) 5x(x-3)+2x-6=0 ; 9) (3x-1)(2x-1)-(3x-1)(x+2)=0
10)|2x-1|+1=8 ; 11) |x-2|=3x+1 ; 12) |2x|=21-x
Giải các phương trình nha mọi người ^_^
1) |2x - 1| 52) |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)
Đọc tiếp
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)
. Tìm x, biết:
a) 6x.(x – 5) + 3x.(7 – 2x) = 18 b) 2x.(3x + 1) + (4 – 2x).3x = 7 c) 0,5x.(0,4 – 4x) + (2x + 5).x = -6,5 | d) (x + 3)(x + 2) – (x - 2)(x + 5) = 6 e) 3(2x - 1)(3x - 1) – (2x - 3)(9x - 1) = 0 |
a) Ta có: \(6x\left(x-5\right)+3x\left(7-2x\right)=18\)
\(\Leftrightarrow6x^2-30x+21x-6x^2=18\)
\(\Leftrightarrow-9x=18\)
hay x=-2
Vậy: S={-2}
b) Ta có: \(2x\left(3x+1\right)+\left(4-2x\right)\cdot3x=7\)
\(\Leftrightarrow6x^2+2x+12x-6x^2=7\)
\(\Leftrightarrow14x=7\)
hay \(x=\dfrac{1}{2}\)
Vậy: \(S=\left\{\dfrac{1}{2}\right\}\)
c) Ta có: \(0.5x\left(0.4-4x\right)+\left(2x+5\right)\cdot x=-6.5\)
\(\Leftrightarrow0.2x-2x^2+2x^2+5x=-6.5\)
\(\Leftrightarrow5.2x=-6.5\)
hay \(x=-\dfrac{5}{4}\)
Vậy: \(S=\left\{-\dfrac{5}{4}\right\}\)
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d) Ta có: \(\left(x+3\right)\left(x+2\right)-\left(x-2\right)\left(x+5\right)=6\)
\(\Leftrightarrow x^2+5x+6-\left(x^2+3x-10\right)=6\)
\(\Leftrightarrow x^2+5x+6-x^2-3x+10=6\)
\(\Leftrightarrow2x+16=6\)
\(\Leftrightarrow2x=-10\)
hay x=-5
Vậy: S={-5}
e) Ta có: \(3\left(2x-1\right)\left(3x-1\right)-\left(2x-3\right)\left(9x-1\right)=0\)
\(\Leftrightarrow3\left(6x^2-5x+1\right)-\left(18x^2-29x+3\right)=0\)
\(\Leftrightarrow18x^2-15x+3-18x^2+29x-3=0\)
\(\Leftrightarrow14x=0\)
hay x=0
Vậy: S={0}
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d) (3x – 5)(7 – 5x) – (5x + 2)(2 – 3x) = 4 g) 3(2x - 1)(3x - 1) - (2x - 3)(9x - 1) =0 j) (2x – 1)(3x + 1) – (4 – 3x)(3 – 2x) = 3 k) (2x + 1)(x + 3) – (x – 5)(7 + 2x) = 8 m) 2(3x – 1)(2x + 5) – 6(2x – 1)(x + 2) = - 6
g: Ta có: \(3\left(2x-1\right)\left(3x-1\right)-\left(2x-3\right)\left(9x-1\right)=0\)
\(\Leftrightarrow3\left(6x^2-5x+1\right)-\left(18x^2-29x+3\right)=0\)
\(\Leftrightarrow18x^2-15x+3-18x^2+29x-3=0\)
\(\Leftrightarrow14x=0\)
hay x=0
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Câu 1. Giải các phườn trình sau:
a, 3x+60
b, 2x-100
c, 3x-711
d, 3x-90
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, x²-4x0
m, 4x²-10
n, x²-6x+90
l, (3x-5)²-(x+4)²0
o, 7x(x+2)-5(x+2)0
p, 3x(2x-5)-4x+100
q, (2-2x)-x²+10
r, x(1-3x)5(1-3x)
s, 2x-3/4+x+1/63
t, x-3/4-2x+1/3x/6
u, x+1/13+x+2/12x+3/11+x+4/10
v, 2x+1/15+2x+2/142x+3/13+2x+4/12
Giúp e nha mn. E cảm ơn trc ạ!
Đọc tiếp
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
Xem thêm câu trả lời
Tìm x:
a, 3x (4x -3) - 2x (5-6x) = 0
b, 5 (2x-3) + 4x (x-2) + 2x (3-2x) = 0
c, 3x (2-x) + 2x (x-1) = 5x (x+3)
d, 3x (x+1) - 5x (3-x) + 6(x2 + 2x + 3) = 0
a) 3x(4x - 3) - 2x(5 - 6x) = 0
=> 6x2 - 9x - 10x + 12x2 = 0
=> 18x2 - 19x = 0
=> x(18x - 19) = 0
=> \(\orbr{\begin{cases}x=0\\18x-19=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=0\\x=\frac{19}{18}\end{cases}}\)
b) 5(2x - 3) + 4x(x - 2) + 2x(3 - 2x) = 0
=> 10x - 15 + 4x2 - 8x + 6x - 4x2 = 0
=> 8x - 15 = 0
=> 8x = 15
=> x = 15 : 8 = 15/8
c) 3x(2 - x) + 2x(x - 1) = 5x(x + 3)
=> 6x - 3x2 + 2x2 - 2x = 5x2 + 15x
=> 4x - x2 - 5x2 - 15x = 0
=> -6x2 - 11x = 0
=> -x(6x - 11) = 0
=> \(\orbr{\begin{cases}-x=0\\6x-11=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=0\\x=\frac{11}{6}\end{cases}}\)
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a) \(3x\left(4x-3\right)-2x\left(5-6x\right)=0\)
\(\Leftrightarrow12x^2-9x-10x+12x^2=0\)
\(\Leftrightarrow-19x=0\Leftrightarrow x=0\)
b) \(5\left(2x-3\right)+4x\left(x-2\right)+2x\left(3-2x\right)=0\)
\(\Leftrightarrow10x-15+4x^2-8x+6x-4x^2=0\)
\(\Leftrightarrow8x-15=0\Leftrightarrow x=\frac{15}{8}\)
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d) \(3x\left(x+1\right)-5x\left(3-x\right)+6\left(x^2+2x+3\right)=0\)
\(\Leftrightarrow3x^2+3x-15x+5x^2+6x^2+12x+18=0\)
\(\Leftrightarrow14x^2+18=0\)
Mà \(14x^2+18>0\)nên pt vô nghiệm
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Xem thêm câu trả lời
Tìm x, biết:
a) 3x(4x-3) - 2x(5-6x) = 0
b) 5(2x-3) + 4x(x-2) + 2x(3-2x) = 0
c) 3x(2-x) + 2x(x-1) = 5x(x+3)
d) 3x(x+1) - 5x(3-x) + 6(x2 + 2x + 3) = 0
a) Ta có: 3x(4x-3)-2x(5-6x)=0
\(\Leftrightarrow12x^2-9x-10x+12x^2=0\)
\(\Leftrightarrow24x^2-19x=0\)
\(\Leftrightarrow x\left(24x-19\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\24x-19=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\24x=19\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\frac{19}{24}\end{matrix}\right.\)
Vậy: \(x\in\left\{0;\frac{19}{24}\right\}\)
b) Ta có: \(5\left(2x-3\right)+4x\left(x-2\right)+2x\left(3-2x\right)=0\)
\(\Leftrightarrow10x-15+4x^2-8x+6x-4x^2=0\)
\(\Leftrightarrow8x-15=0\)
\(\Leftrightarrow8x=15\)
hay \(x=\frac{15}{8}\)
Vậy: \(x=\frac{15}{8}\)
c) Ta có: \(3x\left(2-x\right)+2x\left(x-1\right)=5x\left(x+3\right)\)
\(\Leftrightarrow6x-3x^2+2x^2-2x=5x^2+15x\)
\(\Leftrightarrow-x^2+4x-5x^2-15x=0\)
\(\Leftrightarrow-6x^2-11x=0\)
\(\Leftrightarrow6x^2+11x=0\)
\(\Leftrightarrow x\left(6x+11\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\6x+11=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\6x=-11\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\frac{-11}{6}\end{matrix}\right.\)
Vậy: \(x\in\left\{0;\frac{-11}{6}\right\}\)
d) Ta có: \(3x\left(x+1\right)-5x\left(3-x\right)+6\left(x^2+2x+3\right)=0\)
\(\Leftrightarrow3x^2+3x-15x+5x^2+6x^2+12x+18=0\)
\(\Leftrightarrow14x^2+18=0\)
\(\Leftrightarrow14x^2=-18\)
mà \(14x^2\ge0\forall x\)
nên \(x\in\varnothing\)
Vậy: \(x\in\varnothing\)
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Giải phương trình :
a)(2x-5)^3-(3x-4)^x+(x+1)^3=0
b)(x-1)^3+(2x-3)^3+(3x-5)^3 - 3(x-1)(2x-3)(3x-5) = 0
c)(x^2+3x-4)^3 + (3x^2+7x+4)^3 = (4x^2+10x)^3
Tìm x
a) 1/3x + 2/5( x - 1 ) = 0
b) (2x - 3 )(6 - 2x ) =0
c) 2|1/2x - 1/3 | - 3/2 = 1/4
d) 3/4 - 2 . | 2x - 2/3 | = 2
e) ( 3x - 1)(-1/2x + 5 ) = 0
g)-5(x + 1/5 ) - 1/2(x - 2/3 ) = 2/3x - 5/6
h) 3(x - 1/2 ) - 5(x + 3/5 ) = -3 + 1/5
i) 60%x + 2/3x = 1/3. 6và1/3 ( 6và1/3 là hỗn số )
tìm x thuộc z
1)(-3x+2)-(5-3x)=-3
2) 3+x-(3x-1)=6-2x
3) (x-5).(3x+4)=0
4) 7x.(2x-1)=0
5) (3x-1).2x=0
giúp mik với mai mik đi học rùi :((
\(\left(-3x+2\right)-\left(5-3x\right)=-3\)
\(\Rightarrow-3x+2-5+3x=-3\)
\(\Rightarrow-3x+3x=-3+5-2\)
\(\Rightarrow0x=0\Rightarrow x\in Z\)
\(3+x-\left(3x-1\right)=6-2x\)
\(\Rightarrow3+x-3x+1=6-2x\)
\(\Rightarrow x-3x+2x=6-1-3\)
\(\Rightarrow0x=2\left(loại\right)\)
\(\left(x-5\right)\left(3x+4\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-5=0\\3x+4=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=5\\x=-\frac{4}{3}\end{cases}}}\)
\(7x\left(2x-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}7x=0\\2x-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=\frac{1}{2}\end{cases}}}\)
\(\left(3x-1\right)2x=0\)
\(\Leftrightarrow\orbr{\begin{cases}3x-1=0\\2x=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{3}\\x=0\end{cases}}}\)
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khó hiểu quá
bạn ghi bằng số luôn đừng ghi phần
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\(\left(-3x+2\right)-\left(5-3x\right)=-3\)
\(\Rightarrow-3x+2-5+3x=-3\)
\(\Rightarrow-3=-3\)
\(\forall x\in Z\)
\(3+x-\left(3x-1\right)=6-2x\)
\(\Rightarrow2=6\left(vl\right)\)
\(\left(x-5\right)\left(3x+4\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-5=0\\3x+4=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=5\\x=\frac{4}{3}\end{cases}}\)
\(7x\left(2x-1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}7x=0\\2x-1=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x=\frac{1}{2}\end{cases}}\)
\(\left(3x-1\right)2x=0\)
\(\Rightarrow\orbr{\begin{cases}3x-1=0\\2x=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{1}{3}\\x=0\end{cases}}\)
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