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câu 1.
P= 2(x+y)(x-y)+(x-y)^2+(x+y)^2-4y^2
P= (x+y+x-y)^2-(2y)^2
P=(2x-2y)(2x+2y)
P=4(x^2-y^2)
câu 2.
a, x^3-2x^2-4xy^2+x= x(x^2-2x+1)-4xy^2
=x(x-1)^2-4xy^2
=x(x-1-2y)(x-1+2y)
b, (x+1)(x+2)(x+3)(x+4)-24= (x^2+5x+4)(x^2+5x+6)-24
Đặt x^2+5x+4= a
Lúc đó: (x+1)(x+2)(x+3)(x+4)-24= a(a+2)-24
= a^2+2a-24
=a^2+2a+1-25
= (a+1)^2-5^2
= (a+1-5)(a+1+5)
= (a-4)(a+6)
mà ta đặt x^2+5x+4=a => (x+1)(x+2)(x+3)(x+4)-24= (x^2+5x+4-4)(x^2+5x+4+6)
= (x^2+5x)(x^2+5x+10)
câu3. (x+2)^2= 4-x^2
=> (x+2)^2-4+x^2=0
=>. (x+2)^2-(2-x)(2+x)=0
=> (x+2)(x+2-2+x)=0
=> (x+2)2x=0
=> x+2=0 hoặc 2x=0
=> x=-2 hoặc x=0
1)P=2(x^2-y^2)+x^2-2xy+y^2+x^2+2xy+y^2-4y^2=2x^2-2y^2+2x^2+2y^2-4y^2=4x^2-4y^2 . 3) <=> x^2+4x+4-4+x^2=0
<=> 2x^2+4x=0 <=>2x(x+2)=0 <=>2x=0 hay x+2=0 <=>x=0 hay x=-2
\(1,\Leftrightarrow x=10\\ 2,=x^2-4x+4-x^2+4x=4\\ 3,=\left(x+y\right)^2-49=\left(x+y+7\right)\left(x+y-7\right)\)
Câu 1:
a) 2x(3x+2) - 3x(2x+3) = 6x^2+4x - 6x^2-9x = -5x
b) \(\left(x+2\right)^3+\left(x-3\right)^2-x^2\left(x+5\right)\)
\(=x^3+6x^2+12x+8+x^2-6x+9-x^3-5x^2\)
\(=2x^2+6x+17\)
c) \(\left(3x^3-4x^2+6x\right)\div\left(3x\right)=x^2-\dfrac{4}{3}x+2\)
câu 1:
a,x2+2x-4z2+1
=x2+2x.1+12-(2z)2
=(x+1)2-(2z)2
=(x+1-2z)(x+1+2z)
1.
\(A=\dfrac{2x-9}{\left(x-2\right)\left(x-3\right)}-\dfrac{\left(x-3\right)\left(x+3\right)}{\left(x-2\right)\left(x-3\right)}+\dfrac{\left(2x+4\right)\left(x-2\right)}{\left(x-2\right)\left(x-3\right)}\)
\(=\dfrac{2x-9-\left(x^2-9\right)+\left(2x^2-8\right)}{\left(x-2\right)\left(x-3\right)}\)
\(=\dfrac{x^2+2x-8}{\left(x-2\right)\left(x-3\right)}=\dfrac{\left(x-2\right)\left(x+4\right)}{\left(x-2\right)\left(x-3\right)}\)
\(=\dfrac{x+4}{x-3}\)
b.
\(A=2\Rightarrow\dfrac{x+4}{x-3}=2\Rightarrow x+4=2\left(x-3\right)\)
\(\Rightarrow x=10\) (thỏa mãn)
2.
\(x^4+2x^2y+y^2-9=\left(x^2+y\right)^2-3^2=\left(x^2+y-3\right)\left(x^2+y+3\right)\)
(x+y+z)^3 - x^3 - y^3 - z^3
\(=x^3+y^3+z^3+3xy\left(x+y\right)+3yz\left(y+z\right)+3xz\left(x+z\right)-x^3-y^3-z^3\)
\(=3x^2y+3xy^2+3y^2z+3yz^2+3x^2z+3xz^2\)
\(=3\left(x^2y+xy^2+y^2z+yz^2+x^2z+xz^2\right)\)
3.(2^2 +1 ).(2^4 +1).(2^8 +1).(2^16 +1)
\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^{16}-1\right)\left(2^{16}+1\right)=2^{32}-1\)
Mấy câu này bạn áp dụng hằng đẳng thức số 3 là ra ngay nha!
Câu 1:
x2 - y2 + 4y - 4x = (x - y)(x+y) + 4 (y-x) = (x - y) ( x +y - 4)
Câu 2:
x2 - 4 + (x-2)2 = (x - 2)(x+2) + (x - 2) (x - 2) = (x - 2) (x+2 + x - 2) = 2x(x-2)
Câu 3: (x-y)2 - 9 = (x-y-3) (x-y+3)
Câu 4: (3x-1)2 - 16 = (3x-1-4) (3x-1+4) = (3x -5) (3x+3) = 3 (3x-5)(x+1)
Câu 5: (x2+1)2 - 25 = (x2 +1 - 5) (x2+1+5) = (x2 -4)( x2+6) = (x - 2) (x+2) (x2+6)
Vậy....
Câu 3 :
( x + 2 ) 2 = 4 - x 2
\(\Leftrightarrow\) ( x + 2 ) 2 = ( 2 - x ) ( 2 + x )
\(\Leftrightarrow\) ( x + 2 ) 2 - ( 2 - x ) ( 2 + x ) = 0
\(\Leftrightarrow\) ( x + 2 ) ( x + 2 - 2 + x ) = 0
\(\Leftrightarrow\) 2x . ( x + 2 ) = 0
\(\Leftrightarrow\left[{}\begin{matrix}2x=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\)
Vậy phương trình có nghiệm x = 0 hoặc x = -2 .
phynit bài này đúng không ạ