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a) \(x^4-2x^2+1=\left(x^2-1\right)^2=\left(x-1\right)^2\left(x+1\right)^2\)
b) \(x^2-y^2-5x+5y=\left(x-y\right)\left(x+y\right)-5\left(x-y\right)=\left(x-y\right)\left(x+y-5\right)\)
c) \(2x^3-x^2-8x+4\)
\(=x^2\left(2x-1\right)-4\left(2x-1\right)\)
\(=\left(x-1\right)\left(x+1\right)\left(2x-1\right)\)
d) \(x\left(x-y\right)^2+y\left(x-y\right)^2-xy+x^2\)
\(=\left(x+y\right)\left(x-y\right)^2+x\left(x-y\right)\)
\(=\left(x-y\right)\left(x^2-y^2+x\right)\)
e) \(2x^2-5x+2\)
\(=\left(2x^2-x\right)-\left(4x-2\right)\)
\(=x\left(2x-1\right)-2\left(2x-1\right)\)
\(=\left(x-2\right)\left(2x-1\right)\)
1.a) \(\Leftrightarrow\) 3x+10-2x =0
\(\Leftrightarrow\text{ 3x-2x=-10}\)
\(\Leftrightarrow x=-10\)
b) coi lại có thiếu ngoặc ko nhé
cứ nhân vào dấu ngoặc rồi làm như thường
6) c) x3 - x2 + x = 1
<=> x3 - x2 + x - 1 = 0
<=> (x3 - x2) + (x - 1) = 0
<=> x2 (x - 1) + (x - 1) = 0
<=> (x - 1) (x2 + 1) = 0
=> x - 1 = 0 hoặc x2 + 1 = 0
* x - 1 = 0 => x = 1
* x2 + 1 = 0 => x2 = -1 => x = -1
Vậy x = 1 hoặc x = -1
Bài 5:
a) Đặt \(A=\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(\Rightarrow8A=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(\Rightarrow8A=\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(\Rightarrow8A=\left(3^8-1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(\Rightarrow8A=\left(3^{16}-1\right)\left(3^{16}+1\right)\)
\(\Rightarrow8A=3^{32}-1\)
\(\Rightarrow A=\frac{3^{32}-1}{8}\)
b) (7x+6)2 + (5-6x)2 - (10-12x)(7x+6)
=(7x+6)2 + (5-6x)2 - 2(5-6x)(7x+6)
\(=\left(7x+6-5+6x\right)^2\)
\(=\left(13x+1\right)^2\)
1 ) Thực hiện phép tính :
a ) \(-\frac{1}{3}xz\left(-9xy+15yz\right)+3x^2\left(2yz^2-yz\right)\)
\(=3x^2yz-5xyz^2+6x^2yz^2-3x^2yz\)
\(=-5xyz^2+6x^2yz^2\)
b ) \(\left(x-2\right)\left(x^2-5x+1\right)-x\left(x^2+11\right)\)
\(=x^3-5x^2-x-2x^2+10x-2-x^3-11x\)
\(=-7x^2-2x-2-x^3\)
c ) \(\left(x^3+5x^2-2x+1\right)\left(x-7\right)\)
\(=x^4+5x^3-2x^2+x-7x^3-35x^2+14x-7\)
\(=x^4-2x^3-37x^2+15x-7\)
d ) \(\left(2x^2-3xy+y^2\right)\left(x+y\right)\)
\(=2x^3-3x^2y+xy^2+2x^2y-3xy^2+y^3\)
\(=2x^3-x^2y-2xy^2+y^3\)
e ) \(\left[\left(x^2-2xy+2y^2\right)\left(x+2y\right)-\left(x^2-4y^2\right)\left(x-y\right)\right]2xy\)
( để xem lại )
2 Tìm x
a ) \(6x\left(5x+3\right)+3x\left(1-10x\right)=7\)
\(\Leftrightarrow30x^2+18x+3x-30x^2=7\)
\(\Leftrightarrow21x=7\)
\(\Leftrightarrow x=3\)
b ) Sai đề
c ) \(\left(x+1\right)\left(x+2\right)\left(x+5\right)-x^2\left(x+8\right)=27\)
( Để xem lại )
mình chép đúng theo đề cô cho mà sao lại sai được ,hay cô cho sai đề
a: \(\dfrac{2x^4-x^3-x^2+7x-4}{x^2+x-1}\)
\(=\dfrac{2x^4+2x^3-2x^2-3x^3-3x^2+3x+4x^2+4x-4}{x^2+x-1}\)
=2x^2-3x+4
b: \(=\dfrac{y}{x\left(2x-y\right)}+\dfrac{4x}{y\left(y-2x\right)}\)
\(=\dfrac{y^2-4x^2}{xy\left(2x-y\right)}=\dfrac{-\left(2x-y\right)\left(2x+y\right)}{xy\left(2x-y\right)}=\dfrac{-2x-y}{xy}\)
c: \(=\dfrac{6\left(x+8\right)}{7\left(x-1\right)}\cdot\dfrac{\left(x-1\right)^2}{\left(x-8\right)\left(x+8\right)}=\dfrac{6\left(x-1\right)}{7\left(x-8\right)}\)