bài 1 : tìm x
a) (x-3)2_(x+1) .(x-4) =0
b) x2 - 25 = 3x+15
c) x2 -10x +25 =2(x-5)
d)4x2-12x+9 =(1-x)2
Bài 2: cho x-y =2 , tính giá trị biểu thức
A= 2.(x3-y3)_ 3(x+y)2
bài 3 : phân tích đa tử thành nhân tử
a) x2 _y2 -2x -2y
b)x2 + 4x -y2 +4
c)x3-16x
giúp mình với các bạn ơi , mình đang rất cần, cảm ơn các bạn trước
Bài 1:
a) (x-3)\(^2\)-(x+1)(x-4)=0
<=>x\(^2\)-6x+9-x\(^2\)+4x-x+4=0
<=>-3x+13=0
<=>3x=13
<=> x=\(\dfrac{13}{3}\)
b)x\(^2\)-25=3x+15
<=>(x+5)(x-5)=3(x+5)
<=>(x+5)(x-5)-3(x+5)=0
<=>(x+5)[(x-5)-3]=0
<=>(x+5)(x-8)=0
<=> x+5=0 hoặc x-8=0
*x+5=0 *x-8=0
<=>x=-5 <=>x=8
c)x\(^2\)-10x+25=2(x-5)
<=>(x-5)\(^2\)=2(x-5)
<=>(x-5)\(^2\)-2(x-5)=0
<=>(x-5)[(x-5)-2]=0
<=>(x-5)(x-7)=0
<=>x-5=0 hoặc x-7=0
* x-5=0 *x-7=0
<=>x=5 <=>x=7
d)4x\(^2\)-12x+9=(1-x)\(^2\)
<=>4x\(^2\)-12x+9=1-2x+x\(^2\)
<=>4x\(^2\)-12x+9-1+2x-x\(^2\)=0
<=>3x\(^2\)-10x+9=0
Câu d đến đây mik chịu...
d)
\(4x^2-12x+9=\left(1-x\right)^2\)
\(\Leftrightarrow4x^2-12x+9-1+2x-x^2=0\)
\(\Leftrightarrow3x^2-10x+8=0\)
\(\Leftrightarrow3x^2-6x-4x+8=0\)
\(\Leftrightarrow3x\left(x-2\right)-4\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(3x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{4}{3}\end{matrix}\right.\)
Bài 3:
a)
\(x^2-y^2-2x-2y=x^2-2x+1-y^2-2y-1\)
\(=\left(x^2-2x+1\right)-\left(y^2+2y+1\right)\)
\(=\left(x-1\right)^2-\left(y+1\right)^2\)
\(=\left(x-1-y-1\right)\left(x-1+y+1\right)\)
\(=\left(x-y-2\right)\left(x+y\right)\)
b)
\(x^2+4x-y^2+4=x^2+4x+4-y^2\)
\(=\left(x+2\right)^2-y^2=\left(x+2-y\right)\left(x+2+y\right)\)
c)
\(x^3-16x=x\left(x^2-16\right)=x\left(x-4\right)\left(x+4\right)\)
Bài 2:
\(A=2\left(x^3-y^3\right)-3\left(x+y\right)^2\)
\(A=2\left(x-y\right)\left(x^2+xy+y^2\right)-3\left(x^2+2xy+y^2\right)\)
\(A=4x^2+4xy+4y^2-3x^2-6xy-3y^2\)
\(A=x^2-2xy+y^2\)
\(A=\left(x-y\right)^2=2^2=4\)