Tìm a thuộc Z biết : a) (x-3)(x=2)>0
b) (2x-4)(x+4)<0
c) ( x-2)^2 . (y-3) = -4
d) (x+2)^2 + 2(y-3)^2<4
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 1 Tìm X biết (x+4)²-81=0 Bài 2 cho biểu thức A=(x-3/x - x/x-3 + 9/x²-3x)2x-2/x A) tìm ĐKXĐ và rút gọn A B) tìm X thuộc Z để A thuộc Z Bài 3 A) x³-2x² B) y²-2y-x²+1 C) (x+1)²-25
\(\left(x+4\right)^2-81=0\Leftrightarrow\left(x+4\right)^2-9^2=0\)
\(\Leftrightarrow\left(x+4+9\right)\times\left(x+4-9\right)=0\)
\(\Leftrightarrow\left(x+13\right)\times\left(x-5\right)=0\)
\(\left[{}\begin{matrix}x+13=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-13\\x=5\end{matrix}\right.\)
tìm x biết
a x^2 (2x+15)+4(2x+15)=0
b 5x(x-2)-3(x-2)=0
c 2(x+3)-x^2-3x=0
a
\(x^2\left(2x+15\right)+4\left(2x+15\right)=0\\ \Leftrightarrow\left(2x+15\right)\left(x^2+4\right)=0\\ \Leftrightarrow2x+15=0\left(x^2+4>0\forall x\right)\\ \Leftrightarrow2x=-15\\ \Leftrightarrow x=-\dfrac{15}{2}\)
b
\(5x\left(x-2\right)-3\left(x-2\right)=0\\ \Leftrightarrow\left(x-2\right)\left(5x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-2=0\\5x-3=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0+2=2\\x=\dfrac{0+3}{5}=\dfrac{3}{5}\end{matrix}\right.\)
c
\(2\left(x+3\right)-x^2-3x=0\\ \Leftrightarrow2\left(x+3\right)-\left(x^2+3x\right)=0\\ \Leftrightarrow2\left(x+3\right)-x\left(x+3\right)=0\\ \Leftrightarrow\left(x+3\right)\left(2-x\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x+3=0\\2-x=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0-3=-3\\x=2-0=2\end{matrix}\right.\)
a: =>(2x+15)(x^2+4)=0
=>2x+15=0
=>2x=-15
=>x=-15/2
b; =>(x-2)(5x-3)=0
=>x=2 hoặc x=3/5
c: =>(x+3)(2-x)=0
=>x=2 hoặc x=-3
Tìm x , Biết
a) (x-4) x - (x-3)^2=0
b) 3x-6 = x^2-16
c) (2x-3)^2 - 49=0
d) 2x (x-5) - 7 (5-x)=0
a) \(\Leftrightarrow x^2-4x-x^2+6x-9=0\\ \Leftrightarrow2x=9\\ \Leftrightarrow x=4,5\)
b) \(\Leftrightarrow x^2-3x-10=0\\ \Leftrightarrow\left(x^2+2x\right)-\left(5x+10\right)=0\\ \Leftrightarrow x\left(x+2\right)-5\left(x+2\right)=0\\ \left(x-5\right)\left(x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=5\\x=-2\end{matrix}\right.\)
c) \(\Leftrightarrow\left(2x-3-7\right)\left(2x-3+7\right)=0\\ \Leftrightarrow\left(2x-10\right)\left(2x+4\right)=0\\ \Leftrightarrow\left(x-5\right)\left(x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=5\\x=-2\end{matrix}\right.\)
d) \(\Leftrightarrow\left(2x+7\right)\left(x-5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{7}{2}\\x=5\end{matrix}\right.\)
Tìm x thuộc Z biết :
A=x+4/x-2+2x-5/x-2 thuộc Z
Tìm x thuộc Z biết
a) 11+(15-2x)=0
b)-129-(35-x)=55
c)27-2.5^x-2=5^8:5^6
cíu mik zới , mai mik phải nộp bài r
1: =>15-2x=-11
=>2x=26
hay x=13
bài 4 :Tìm x ∈ Z, biết:
a) -6 < x < 0
b) -3 ≤ x < 7
c) -5 < x < 5
d) -3 ≤ x < 3
\(a,x\in\left\{-5;-4;-3;-2;-1\right\}\\ b,x\in\left\{-3;-2;-1;...;5;6\right\}\\ c,x\in\left\{-4;-3;...;3;4\right\}\\ d,x\in\left\{-3;-2;-1;0;1;2\right\}\)
Tìm x thuộc Z , biết :
a) 3 ( x - 12 ) - 2 ( 2x - 4 ) = 3 - 2 ( x - 11 )
b) 5 ( x - 7 ) - 4 ( x - 8 ) = 5 - 2 ( x - 4 ) + 2x
\(a,3\left(x-12\right)-2\left(2x-4\right)=3-2\left(x-11\right)\)
\(3x-36-4x+8=3-2x+22\)
\(3x-4x+2x=3+22+36-8\)
\(x=53\)
\(a,3\left(x-12\right)-2\left(2x-4\right)=3-2\left(x-11\right)\)
\(3x-36-4x+8=3-2x+22\)
\(3x-4x+2x=3+22+36-8\)
\(x=53\)
\(b,5\left(x-7\right)-4\left(x-8\right)=5-2\left(x-4\right)+2x\)
\(5x-35-4x+32=5-2x+8+2x\)
\(5x-4x+2x-2x=5+8+35-32\)
\(x=16\)
a, x =53 b,x= 16 k tui nha hinh nen ban de thuong qua
Tìm x biết:
a) 7x.(2x - 3) - (4x2 - 9) = 0
b) (2x - 7).(x - 2).(x2 - 4) = 0
c) (9x2 - 25) - (6x - 10) = 0
a) \(7x\left(2x-3\right)-\left(4x^2-9\right)=0\Rightarrow7x\left(2x-3\right)-\left(2x-3\right)\left(2x+3\right)=0\Rightarrow\left(2x-3\right)\left(7x-2x+3\right)=0\Rightarrow\left[{}\begin{matrix}2x-3=0\\5x+3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{3}{5}\end{matrix}\right.\)
b) \(\left(2x-7\right).\left(x-2\right)\left(x^2-4\right)=0\Rightarrow\left(2x-7\right)\left(x-2\right)^2\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}2x-7=0\\\left(x-2\right)^2=0\\x+2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}\\x=2\\x=-2\end{matrix}\right.\)
c)\(\left(9x^2-25\right)-\left(6x-10\right)=0\Rightarrow\left(3x-5\right)\left(3x+5\right)-2\left(3x-5\right)=0\Rightarrow\left(3x-5\right)\left(3x+5-2\right)=0\Rightarrow\left[{}\begin{matrix}3x-5=0\\3x+3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{5}{3}\\x=1\end{matrix}\right.\)
a: Ta có: \(7x\left(2x-3\right)-\left(4x^2-9\right)=0\)
\(\Leftrightarrow7x\left(2x-3\right)-\left(2x-3\right)\left(2x+3\right)=0\)
\(\Leftrightarrow\left(2x-3\right)\left(5x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=\dfrac{3}{5}\end{matrix}\right.\)
b: Ta có: \(\left(2x-7\right)\left(x-2\right)\left(x^2-4\right)=0\)
\(\Leftrightarrow\left(2x-7\right)\left(x-2\right)^2\cdot\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}\\x=2\\x=-2\end{matrix}\right.\)
c: Ta có: \(\left(9x^2-25\right)-\left(6x-10\right)=0\)
\(\Leftrightarrow\left(3x-5\right)\left(3x+5-2\right)=0\)
\(\Leftrightarrow\left(3x-5\right)\left(3x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{3}\\x=-1\end{matrix}\right.\)