1, tìm x biết:
a, |2,7-x|=|-0,3|
b,12/5 -|x+1,5|=0
c,2.|2x-3|=1/2
d, 7,5-3.|5-2x|=-4,5
e, |3x-4|+|3y+15|=0
ê, |x-y|+ |y+9/25|=0
tìm x biết:
a, |2,7-x|=|-0,3|
b, 12/5 -|x+1,5|=0
c, 2. |2x-3|=1/3
d, 7,5-3.|5-2x|=-4/5
e, |3x-4|+|3y+15|=0
ê, |x-y|+ |y+9/25|=0
a: =>x-2,7=0,3 hoặc x-2,7=-0,3
=>x=3 hoặc x=2,4
b: =>|x+1,5|=2,4
=>x+1,5=2,4 hoặc x+1,5=-2,4
=>x=-3,9 hoặc x=0,9
c: =>|2x-3|=1/6
=>2x-3=1/6 hoặc 2x-3=-1/6
=>2x=19/6 hoặc 2x=17/6
=>x=17/12 hoặc x=19/12
d: =>3|2x-5|=7,5+0,8=8,3
=>|2x-5|=83/30
=>2x-5=83/30 hoặc 2x-5=-83/30
=>2x=233/30 hoặc 2x=67/30
=>x=233/60 hoặc x=67/60
e: =>x-y=0 và y+9/25=0
=>x=y=-9/25
Tìm x, biết:
a) 7x2 - 28 = 0
b) \(\dfrac{2}{3}\)x(x2 - 4) = 0
c) 2x(3x - 5) - (5 - 3x) = 0
d) (2x - 1)2 - 25 = 0
a) Ta có: \(7x^2-28=0\)
\(\Leftrightarrow7\left(x^2-4\right)=0\)
\(\Leftrightarrow7\left(x-2\right)\left(x+2\right)=0\)
mà 7>0
nên (x-2)(x+2)=0
hay \(\left[{}\begin{matrix}x-2=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)
Vậy: \(x\in\left\{2;-2\right\}\)
b) Ta có: \(\dfrac{2}{3}x\left(x^2-4\right)=0\)
\(\Leftrightarrow\dfrac{2}{3}x\left(x-2\right)\left(x+2\right)=0\)
mà \(\dfrac{2}{3}>0\)
nên x(x-2)(x+2)=0
hay \(\left[{}\begin{matrix}x=0\\x-2=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x=-2\end{matrix}\right.\)
Vậy: \(x\in\left\{0;-2;2\right\}\)
c) Ta có: \(2x\left(3x-5\right)-\left(5-3x\right)=0\)
\(\Leftrightarrow2x\left(3x-5\right)+\left(3x-5\right)=0\)
\(\Leftrightarrow\left(3x-5\right)\left(2x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-5=0\\2x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=5\\2x=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{3}\\x=-\dfrac{1}{2}\end{matrix}\right.\)
Vậy: \(x\in\left\{\dfrac{5}{3};-\dfrac{1}{2}\right\}\)
d) Ta có: \(\left(2x-1\right)^2-25=0\)
\(\Leftrightarrow\left(2x-1-5\right)\left(2x-1+5\right)=0\)
\(\Leftrightarrow\left(2x-6\right)\left(2x+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-6=0\\2x+4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=6\\2x=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)
Vậy: \(x\in\left\{3;-2\right\}\)
a,7x2 - 28 = 0
=> 7x2 = 28 => x2 = 4 => x = 2
b,2/3x(x2 - 4) = 0
=>2/3x(x - 2)(x + 2) = 0
=> x ∈ {0 ; 2 ; -2}
c,2x(3x - 5) - (5 - 3x) = 0
= 2x(3x - 5) + (3x - 5)
= (3x - 5)(2x + 1) = 0
=> x ∈ { 5/3 ; -1/2}
d, (2x - 1)2 - 25 = 0
=> (2x - 4)(2x - 6) = 0
=> x ∈ {2 ;3}
a,7x2 - 28 = 0
=> 7x2 = 28 => x2 = 4 => x = 2
b,2/3x(x2 - 4) = 0
=>2/3x(x - 2)(x + 2) = 0
=> x ∈ {0 ; 2 ; -2}
c,2x(3x - 5) - (5 - 3x) = 0
= 2x(3x - 5) + (3x - 5)
= (3x - 5)(2x + 1) = 0
=> x ∈ { 5/3 ; -1/2}
d, (2x - 1)2 - 25 = 0
=> (2x - 4)(2x - 6) = 0
=> x ∈ {2 ;3}
Bài 1: Phân tích đa thức sau :
a)2x(xy+y^2-3)
b)(x-y)(2x+y)
c)(x-2y)^2
d)(2x-y)(y+2x)
bài 2: Phân tích các đơn thức thành nhân tử
a)3x^2-3xy
b)x^2-4y^2
c)3x-3y+xy-y^2
d)x^2-1+2y-y^2
Bài 3: Tìm x biết:
a)3x^2-6x=0
b)Tìm x,y thuộc z biết: x^2+4y^2-2xy=4
Bài 2:
a: \(3x^2-3xy=3x\left(x-y\right)\)
b: \(x^2-4y^2=\left(x-2y\right)\left(x+2y\right)\)
c: \(3x-3y+xy-y^2=\left(x-y\right)\left(3+y\right)\)
d: \(x^2-y^2+2y-1=\left(x-y+1\right)\left(x+y-1\right)\)
bài 3: Tìm x, biết:
a)(2x-1)2-25=0 b)8x3-50x=0
c)2(x+3)-x2-3x=0 d)x3+27+(x+3)(x-9)=0
mình cần gấppppppppp
bài 3: Tìm x, biết:
a)(2x-1)2-25=0 b)8x3-50x=0
c)2(x+3)-x2-3x=0 d)x3+27+(x+3)(x-9)=0
mình cần gấpppppppppp
a: Ta có: \(\left(2x-1\right)^2-25=0\)
\(\Leftrightarrow\left(2x-6\right)\left(2x+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)
Tìm tập hợp giá trị của x biết:
a,(x-1)(x-2)>0
b,2x-3<0
c,(2x-4)(9-3x)>0
d,2x/3-3/4<0
Bài 1 : Thực hiện phép tính :
a. ( 3x+5)^2
b.(2x-3y)^2
c.(4x^2-5y)^2
d.(2xy+3y)^2
e.(3x+5)^2-9(x-2)^2
f.(x+y)^2+(x-y)^2
g.(2a-b)^2-(2a+b)^2
Bài 2 : tìm x
a.(3x-4)(3x+4)-(3x+1)^2=0
b.(2x-5)^2-(2x+1)(2x-1)=10
c.(3x-1)^2+2(3x-1)(3-x)+(3-x)^2=25
Tìm x biết:
a, 16x² – 9(x + 1)²= 0
b, x2 (x – 1) – 4x2 + 8x – 4 = 0
c, x(2x – 3) – 2(3 – 2x) = 0
d, (x – 3)(x² + 3x + 9) – x(x + 2)(x – 2) = 1
e, 4x² + 4x – 6 = 2
f, 2x² + 7x + 3 = 0
e: ta có: \(4x^2+4x-6=2\)
\(\Leftrightarrow4x^2+4x-8=0\)
\(\Leftrightarrow\left(x+2\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=1\end{matrix}\right.\)
f: Ta có: \(2x^2+7x+3=0\)
\(\Leftrightarrow\left(x+3\right)\left(2x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-\dfrac{1}{2}\end{matrix}\right.\)
Tìm x,y biết:
a) 3.(x-1/2)-5(x+3/5)=-x+1/5
b) 3.(3x-1/2)^3 +1/9=0
c) 60%x+2/3x=1/3.6/1/3
d)x/2=-3y/4 và x-2y=3
e) 2x/5=3y/7 và 2x-y=5