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B1: Phân tích thành nhân tử:
a) \(6x^2+9x=3x\left(2x+3\right)\)
b) \(4x^2+8x=4x\left(x+2\right)\)
c) \(5x^2+10x=5x\left(x+2\right)\)
d) \(2x^2-8x=2x\left(x-4\right)\)
e) \(5x-15y=5\left(x-3y\right)\)
f) \(x\left(x^2-1\right)+3\left(x^2-1\right)=\left(x^2-1\right)\left(x+3\right)\)
\(=\left(x-1\right)\left(x+1\right)\left(x+3\right)\)
g) \(x^2-2x+1-4y^2=\left(x-1\right)^2-4y^2\)
\(=\left(x-1-2y\right)\left(x-1+2y\right)\)
h) \(x^2-100=\left(x-10\right)\left(x+10\right)\)
i) \(9x^2-18x+9=\left(3x-3\right)^2\)
k) \(x^3-8=\left(x-2\right)\left(x^2+2x+4\right)\)
l) \(x^2+6xy^2+9y^4=\left(x+3y\right)^2\)
m) \(4xy-4x^2-y^2=-\left(4x^2-4xy+y^2\right)\)
\(=-\left(2x-y\right)^2\)
n) \(\left(x-15\right)^2-16=\left(x-15-16\right)\left(x-15+16\right)\)
\(=\left(x-31\right)\left(x+1\right)\)
o) \(25-\left(3-x\right)^2=\left(5-3+x\right)\left(5+3+x\right)\)
\(=\left(2+x\right)\left(8+x\right)\)
p) \(\left(7x-4\right)^2-\left(2x+1\right)^2\)
\(=\left(7x-4-2x-1\right)\left(7x-4+2x+1\right)\)
\(=\left(5x-5\right)\left(9x-3\right)\)
Bài 1 :
a ) \(6x^2+9x=3x\left(x+3\right)\)
b ) \(4x^2+8x=4x\left(x+2\right)\)
c ) \(5x^2+10x=5x\left(x+2\right)\)
d ) \(2x^2-8x=2x\left(x-4\right)\)
e ) \(5x-15y=5\left(x-3y\right)\)
f ) \(x\left(x^2-1\right)+3\left(x^2-1\right)=\left(x^2-1\right)\left(x+3\right)\)
g ) \(x^2-2x+1-4y^2=\left(x-1\right)^2-\left(2y\right)^2=\left(x-1-2y\right)\left(x-1+2y\right)\)
h ) \(x^2-100=x^2-10^2=\left(x-10\right)\left(x+10\right)\)
i ) \(9x^2-18x+9=\left(3x-3\right)^2\)
k ) \(x^3-8=\left(x-2\right)\left(x^2+2x+2^2\right)\)
l ) \(x^2+6xy^2+9y^4=\left(x+3y^2\right)^2\)
m ) \(4xy-4x^2-y^2=-\left(2x-y\right)^2\)
n ) \(\left(x-15\right)^2=x^2-30x+15^2\)
o ) \(25-\left(3-x\right)^2=\left(5-3+x\right)\left(5+3-x\right)=\left(2+x\right)\left(8-x\right)\)
p ) \(\left(7x-4\right)^2-\left(2x+1\right)^2=\left(7x-4-2x-1\right)\left(7x-4+2x+1\right)=\left(5x-5\right)\left(9x-3\right)\)
Bài 2 :
a ) \(3x^3-6x^2+3x^2y-6xy=3x\left(x^2-2x+xy-2y\right)\)
b ) \(x^2-2x+xy-2y=x\left(x-2\right)+y\left(x-2\right)=\left(x-2\right)\left(x+y\right)\)
c ) \(2x+x^2-2y-2xy=......................\)
d ) \(x^2-2xy+y^2-9=\left(x-y\right)^2-3^2=\left(x-y-3\right)\left(x-y+3\right)\)
e ) \(x^2+y^2-2xy-4=\left(x-y\right)^2-2^2=\left(x-y-2\right)\left(x-y+2\right)\)
f )\(2xy-x^2-y^2+9=-\left(x-y\right)^2+9=3^2-\left(x-y\right)^2=\left(3-x+y\right)\left(3+x-y\right)\)
Câu 1:
a: \(C=a^2+b^2=\left(a+b\right)^2-2ab=23^2-2\cdot132=265\)
b: \(D=x^3+y^3+3xy\)
\(=\left(x+y\right)^3-3xy\left(x+y\right)+3xy\)
\(=1-3xy+3xy=1\)
Bài 3:
\(a,x^2-81=0\)
\(\Rightarrow x^2-9^2=0\)
\(\Rightarrow\left(x-9\right)\left(x+9\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-9=0\\x+9=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=9\\x=-9\end{matrix}\right.\)
\(b,x^2-x-6=0\)
\(\Rightarrow x^2-3x+2x-6=0\)
\(\Rightarrow\left(x^2+2x\right)-\left(3x+6\right)=0\)
\(\Rightarrow x\left(x+2\right)-3\left(x+2\right)=0\)
\(\Rightarrow\left(x+3\right)\left(x+2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x+3=0\\x+2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-3\\x=-2\end{matrix}\right.\)
bài 1: rút gọn biểu thức:
B = (x−2y)2- (x+2y)2+ (4y + 1) ( 1 - 4y)
= x2 - 4xy+ 4y2 - x2 +4xy+4y2+4y- 16y2 +1-4y
=2x2- 8y2+1
E = (2x−3)2 - (3x+1)2 - 5 (x-2) (x+2)
=4x2- 12x+ 9- 9x2+ 6x+ 1- 5x2+20
= - 10x2- 6x+ 30
1. \(x^3-x^2+x-1=(x^3-x^2)+(x-1)\)
\(=x^2(x-1)+(x-1)=(x^2+1)(x-1)\)
2. \(6x^2y-2xy^2+3x-y=2xy(3x-y)+(3x-y)\)
\(=(3x-y)(2xy+1)\)
3. \(4x^2+1\) thì còn cái gì để phân tích hả bạn? Hay ý bạn là \(4x^4+1\)?
\(4x^4+1=(2x^2)^2+1=(2x^2)^2+1+4x^2-4x^2\)
\(=(2x^2+1)^2-(2x)^2=(2x^2+1-2x)(2x^2+1+2x)\)
4. \(x^2-9x+8=(x^2-x)-(8x-8)\)
\(=x(x-1)-8(x-1)=(x-1)(x-8)\)
5. \(x^3-2x^2y+3xy^2=x(x^2-2xy+3y^2)\)
6. \(x^2-6x+y-y^2\) (sai đề)
7. \(x^2-xy-2x+2y=(x^2-xy)-(2x-2y)\)
\(=x(x-y)-2(x-y)=(x-y)(x-2)\)
\(a,x^2+y^2-x-y=8\)
\(\Rightarrow x^2-x+\frac{1}{4}+y^2-y+\frac{1}{4}-8,5=0\)
\(\Rightarrow\left(x-\frac{1}{2}\right)^2+\left(y-\frac{1}{2}\right)^2-8,5=0\)
Ta có : \(\left(x-\frac{1}{2}\right)^2+\left(y-\frac{1}{2}\right)^2-8,5\ge-8,5\forall x;y\)
Để VP=0 và là các số nguyên
=>\(\left(x-\frac{1}{2}\right)^2+\left(y-\frac{1}{2}\right)^2=8,5\)
a/ x^2 + y^2 - x - y = 8
<=> 4x^2 + 4y^2 - 4x - 4y = 32
<=> (2x - 1)^2 + (2y - 1)^2 = 34
<=> (2x - 1)^2 = 9 và (2y - 1)^2 = 25
Hoặc (2x - 1)^2 = 25 và (2y - 1)^2 = 9
Bài 3 :
a ) \(x\left(x-1\right)+x-1=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)
Vậy...........
b ) \(3\left(x-3\right)-4x+12=0\)
\(\Leftrightarrow3\left(x-3\right)-4\left(x-3\right)=0\)
\(\Leftrightarrow\) \(\left(x-3\right)=0\Rightarrow x=3\)
Vậy............
Các câu sau tương tự
a) x2-y2-2x+2y
=(x+y)(x-y)-2(x-y)
=(x-y)(x+y-2)
b) 2x + 2y - x2 -xy
=2(x+y) - x(x+y)
=(x+y)(2-x)
c) 3a2 - 6ab + 3b2 - 12c2
= 3(a2+b2) -3(2ab+4c2)
= 3(a2+b2-2ab-4c2)
d) x2 - 25 + y2 + 2xy
= x2 + 2xy + y2 -25
= (x+y)2 - 52
= (x+y+5)(x+y-5)
e) x2y - x3 - 9y + 9x
= (9x - x3)+(x2y -9y)
= x(9-x2)+y(x2 - 9)
= x(9-x2)-y(9-x2)
= (9-x2)(x-y)
f) x2-2x-4y2-4y
= x2-4y2-2(x+2y)
=(x+2y)(x-4y)-2(x+2y)
=(x+2y)(x-4y-2)
câu g trùng với câu e
h) x2(x-1)+16(1-x)
= x2(x-1)-16(x-1)
= (x2-16)(x-1)
= (x+4)(x-4)(x-1)
\(1.A=\left(x-2y\right)^3-\left(x+2y\right)^3+12x^2y-xy\\ =x^3-3x^2.2y+3x\cdot\left(2y\right)^2-\left(2y\right)^3-\left[x^3+3x^2\cdot2y+3x\cdot\left(2y\right)^2+\left(2y\right)^3\right]+12x^2y-xy\)
\(=x^3-6x^2y+12xy^2-8y^3-x^3-6x^2y-12xy^2-8y^3+12x^2y-xy\\ =xy-16y^3\)
2 . Tìm x :
\(a)\left(2x-3\right)^2-\left(x+2\right)^2=0\\ \left[\left(2x-3\right)-\left(x+2\right)\right]\cdot\left[\left(2x-3\right)+\left(x+2\right)\right]=0\\ \left(2x-3-x-2\right)\cdot\left(2x-3+x+2\right)=0\\ \left(x-5\right)\left(3x-1\right)=0\\ 4x-x-15x+5=0\\ -18x=-5\Rightarrow x=\dfrac{5}{18}\)
\(b)\left(x-3\right)^3=x\cdot\left(x^2-9x+2\right)\\ x^3-9x^2+27x-27=x^3-9x^2+2x\\ \Leftrightarrow27x-27=2x\\ \Leftrightarrow27x-2x=27\\ \Leftrightarrow25x=27\\ \Leftrightarrow x=\dfrac{27}{25}\)
\(c)36x^2-49=0\\ \Leftrightarrow36x^2=49\\ \Leftrightarrow x^2=\dfrac{49}{36}\\ \Leftrightarrow x^2=\left(\pm\dfrac{7}{6}\right)^2\\ \Rightarrow x=\pm\dfrac{7}{6}\)