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Ta có : x3 - 7x + 6
= x3 - x - 6x + 6
= x(x2 - 1) - 6(x - 1)
= x(x + 1)(x - 1) - 6(x - 1)
= (x - 1) [x(x + 1) - 6]
= (x - 1) (x2 + x - 6) .
CÁC Ý SAU TƯƠNG TỰ
Giải như sau.
(1)+(2)⇔x2−2x+1+√x2−2x+5=y2+√y2+4⇔(x2−2x+5)+√x2−2x+5=y2+4+√y2+4⇔√y2+4=√x2−2x+5⇒x=3y(1)+(2)⇔x2−2x+1+x2−2x+5=y2+y2+4⇔(x2−2x+5)+x2−2x+5=y2+4+y2+4⇔y2+4=x2−2x+5⇒x=3y
⇔√y2+4=√x2−2x+5⇔y2+4=x2−2x+5, chỗ này do hàm số f(x)=t2+tf(x)=t2+t đồng biến ∀t≥0∀t≥0
Công việc còn lại là của bạn !
\(\left(x+6\right)\left(2x+1\right)=0\)
<=> \(\orbr{\begin{cases}x+6=0\\2x+1=0\end{cases}}\)
<=> \(\orbr{\begin{cases}x=-6\\x=-\frac{1}{2}\end{cases}}\)
Vậy....
hk tốt
^^
a) \(x^2-2x-4y^2-4y=\left(x^2-2x+1\right)-\left(4y^2+4y+1\right)\)
\(=\left(x-1\right)^2-\left(2y+1\right)^2=\left(x-1-2y-1\right)\left(x-1+2y+1\right)\)
\(=\left(x-2y-3\right)\left(x+2y\right)\)
b) \(x^2-4x^2y^2+y^2+2xy=\left(x^2+2xy+y^2\right)-4x^2y^2\)
\(=\left(x+y\right)^2-4x^2y^2=\left(x+y-2xy\right)\left(x+y+2xy\right)\)
c) \(x^6-x^4+2x^3+2x^2=\left(x^6+2x^3+1\right)-\left(x^4-2x^2+1\right)\)
\(=\left(x^3+1\right)^2-\left(x^2-1\right)^2=\left(x^3+1-x^2+1\right)\left(x^3+1+x^2-1\right)=x^2\left(x^3-x^2+2\right)\left(x+1\right)\)
d) \(x^3+3x^2+3x+1-8y^3=\left(x+1\right)^3-8y^3=\left(x+1-2y\right)\left(x^2+2x+1+2xy+2y+4y^2\right)\)
1
x3-7x+6
=x3+0x2-7x +6
= x3-x2+x2-x-6x+6
=(x3-x2)+(x2-x)-(6x-6)
=x2(x-1)+x(x-1)-6(x-1)
=(x-1)(x2+x-6)
=(x-1)(x2+3x-2x-6)
=(x-1)[x(x+3)-2(x+3)]
=(x-1)(x-2)(x+3)
7) (x+2)(x+3)(x+4)(x+5)-24
=(x+2)(x+5) (x+3)(x+4)-24
=[x(x+5)+2(x+5)][x(x+4)+3(x+4)]-24
=[x2+5x+2x+10][x2+4x+3x+12]-24
=[x2+7x+10][x2+7x+12]-24
đặt a=x2+7x+10
=>x2+7x+12=a+2
=a(a+2)-24
=a2+2a-24
=a2+6a-4a-24
=(a2+6a)-(4a+24)
=a(a+6)-4(a+6)
=(a+6)(a-4)
thay a= x2+7x+10 vào ta được
(x2+7x+10+6)(x2+7x+10-4)
=(x2+7x+16)(x2+7x+6)
A) \(\left(x-3\right)^2-\left(x+2\right)^2\)
\(=\left(x-3-x-2\right)\left(x-3+x+2\right)\)
\(=-5.\left(2x-1\right)\)
B) \(\left(4x^2+2xy+y^2\right)\left(2x-y\right)-\left(2x+y\right)\left(4x^2-2xy+y^2\right)\)
\(=\left(2x\right)^3-y^3-\left[\left(2x\right)^3+y^3\right]\)
\(=8x^3-y^3-8x^3-y^3\)
\(=-2y^3\)
C) \(x^2+6x+8\)
\(=x^2+6x+9-1\)
\(=\left(x+3\right)^2-1\)
\(=\left(x+3-1\right)\left(x+3+1\right)\)
\(=\left(x+2\right)\left(x+4\right)\)
bài 3 A) \(x^2-16=0\)
\(\left(x-4\right)\left(x+4\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-4=0\\x+4=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=4\\x=-4\end{cases}}\)
vậy \(\orbr{\begin{cases}x=4\\x=-4\end{cases}}\)
B) \(x^4-2x^3+10x^2-20x=0\)
\(x^3\left(x-2\right)+10x\left(x-2\right)=0\)
\(\left(x^3+10x\right)\left(x-2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x^3+10x=0\\x-2=0\end{cases}}\Rightarrow\orbr{\begin{cases}x\left(x^2+10\right)=0\\x=2\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=2\end{cases}}\)
vậy \(\orbr{\begin{cases}x=0\\x=2\end{cases}}\)
a) \(4x\left(a-b\right)+6xy\left(b-a\right)\)
\(=4x\left(a-b\right)-6xy\left(a-b\right)\)
\(=\left(4x-6xy\right)\left(a-b\right)\)
\(=2x\left(2-3y\right)\left(a-b\right)\)
9) \(\left(a+b\right)^3-\left(a-b\right)^3\)
\(=\left(a+b-a+b\right)\left[\left(a+b\right)^2+\left(a+b\right)\left(a-b\right)+\left(a-b\right)^2\right]\)
\(=b^2\left[a^2+2ab+b^2+a\left(a-b\right)+b\left(a-b\right)+a^2-2ab+b^2\right]\)
\(=b^2\left(a^2+2ab+b^2+a^2-ab+ab-b^2+a^2-2ab+b^2\right)\)
\(=b^2\left(3a^2+b^2\right)\)
10) \(\left(6x-1\right)^2-\left(3x+2\right)^2\)
\(=\left(6x-1-3x-2\right)\left(6x-1+3x+2\right)\)
\(=\left(3x-3\right)\left(9x+1\right)\)
11) \(x^2-4x^2y^2+y^2+2xy\)
\(=\left(x^2+2xy+y^2\right)-4x^2y^2\)
\(=\left(x+y\right)^2-\left(2xy\right)^2\)
\(=\left(x+y-2xy\right)\left(x+y+2xy\right)\)
12) \(\left(x^2-25\right)^2-\left(x-5\right)^2\)
\(=\left(x^2-25-x+5\right)\left(x^2-25+x-5\right)\)
\(=\left(x^2-x-20\right)\left(x^2-30+x\right)\)
13) \(x^6-x^4+2x^3+2x^2\)
\(=x^6-x^4+2x^3+2x^2-1+1\)
\(=\left(x^6+2x^3+1\right)-\left(x^4-2x^2+1\right)\)
\(=\left[\left(x^3\right)^2+2x^3.1+1^2\right]-\left[\left(x^2\right)^2-2x^2.1+1^2\right]\)
\(=\left(x^3+1\right)^2-\left(x^2-1\right)^2\)
\(=\left(x^3+1-x^2+1\right)\left(x^3+1+x^2-1\right)\)
\(=\left(x^3-x^2+2\right)\left(x^3+x^2\right)\)
1) \(\left(x+y\right)^2-25\)
\(=\left(x+y\right)^2-5^2\)
\(=\left(x+y-5\right)\left(x+y+5\right)\)
2) \(100-\left(3x-y\right)^2\)
\(=10^2-\left(3x-y\right)^2\)
\(=\left(10-3x+y\right)\left(10+3x-y\right)\)
3) \(64x^2-\left(8a+b\right)^2\)
\(=\left(8x\right)^2-\left(8a+b\right)^2\)
\(=\left(8x-8a-b\right)\left(8x+8a+b\right)\)
4) \(4a^2b^4-c^4d^2\)
\(=\left(2ab^2\right)^2-\left(c^2d\right)^2\)
\(=\left(2ab^2-c^2d\right)\left(2ab^2+c^2d\right)\)
5) Đề đúng ko vậy ạ?
6) \(16x^3+54y^3\)
\(=2\left(8x^3+27y^3\right)\)
\(=2\left[\left(2x\right)^3+\left(3y\right)^3\right]\)
\(=2\left(2x+3y\right)\left[\left(2x\right)^2-2x.3y+\left(3y\right)^2\right]\)
\(=2\left(2x+3y\right)\left(4x^2-6xy+9y^2\right)\)
7) \(8x^3-y^3\)
\(=\left(2x\right)^3-y^3\)
\(=\left(2x-y\right)\left[\left(2x\right)^2+2xy+y^2\right]\)
\(=\left(2x-y\right)\left(4x^2+2xy+y^2\right)\)
8) \(\left(a+b\right)^2-\left(2ab-b\right)^2\)
\(=\left(a+b-2ab+b\right)\left(a+b+2ab-b\right)\)
\(=\left(a+2b-2ab\right)\left(a+2ab\right)\)