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a) 2x2 - 28x + 98
= 2.(x2 - 2.x.7 + 72)
= 2.(x - 7)2
b) 5x2 - 20y2
= 5.[x2 - (2y)2 ]
= 5.(x - 2y).(x + 2y)
c) 4x2 - 324
= 4.(x2 - 92)
= 4.(x - 9).(x + 9)
d) 3a2 + 6ab + 3b2
= 3.(a2 + 2ab + b2)
= 3.(a + b)2
e) (a - 4)2 - 4y2
= (a - 4)2 - (2y)2
= (a - 2y - 4).(a + 2y - 4)
f) 2x3 + 16
= 2.(x3 + 23)
= 2.(x + 2).(x2 + 2x + 4)
a ) \(x^3+3x^2+6x+4\)
\(=x^3+3x^2+3x+1+3x+3\)
\(=\left(x+1\right)^3+3\left(x+1\right)\)
\(=\left(x+1\right)\left[\left(x+1\right)^2+3\right]\)
\(=\left(x+1\right)\left(x^2+2x+4\right)\)
b ) \(3a^2c^2+bd+3abc+acd\)
\(=\left(3a^2c^2+3abc\right)+\left(bd+acd\right)\)
\(=3ac\left(ac+b\right)+d\left(ac+b\right)\)
\(=\left(3ac+d\right)\left(ac+b\right)\)
c ) \(3a^2-6ab+3b^2-12c^2\)
\(=3\left(a^2-2ab+b^2\right)-3\left(2c\right)^2\)
\(=3\left[a^2-2ab+b^2-\left(2c\right)^2\right]\)
\(=3\left[\left(a-b\right)^2-\left(2c\right)^2\right]\)
\(=3\left(a-b-2c\right)\left(a-b+2c\right)\)
d ) \(x^2+y^2-x^2y^2+xy-x-y\)
\(=-x^2y^2+x^2+y^2-y+xy-x\)
\(=-x^2\left(y^2-1\right)+y\left(y-1\right)+x\left(y-1\right)\)
\(=-x^2\left(y+1\right)\left(y-1\right)+\left(x+y\right)\left(y-1\right)\)
\(=\left(y-1\right)\left[-x^2\left(y+1\right)+x+y\right]\)
\(=\left(y-1\right)\left[-x^2y-x^2+x+y\right]\)
\(=\left(y-1\right)\left[x\left(1-x\right)+y\left(1-x^2\right)\right]\)
\(=\left(y-1\right)\left[x\left(1-x\right)+y\left(1+x\right)\left(1-x\right)\right]\)
\(=\left(y-1\right)\left[x+y\left(1+x\right)\right]\left(1-x\right)\)
e ) \(a^6-b^6=\left(a^3\right)^2-\left(b^3\right)^2=\left(a^3-b^3\right)\left(a^3+b^3\right)\) \(=\left(a-b\right)\left(a^2+ab+b^2\right)\left(a+b\right)\left(a^2-ab+b^2\right)\)
a, \(x^3+3x^2+6x+4\)
\(=\left(x^3+x^2\right)+\left(2x^2+2x\right)+\left(4x+4\right)\)
\(=x^2\left(x+1\right)+2x\left(x+1\right)+4\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2+2x+4\right)\)
a) x3 - 4x2 + 4x
= x(x2 - 4x + 4)
= x(x - 2)2
b) 2xy - x2 - y2 + 16
= 16 -x2 + 2xy - y2
= 16 - (x2 - 2xy + y2)
= 42 - (x - y)2
= [4 - (x - y)].(4 + x - y)
= (4 - x + y)(4 + x - y)
c) x2 - y2 - 2yz - z2
= x2 - (y2 + 2yz + z2)
= x2 - (y + z)2
= [x -(y + z)].(x + y +z)
=(x - y - z)(x + y + z)
d) 3a2 - 6ab + 3b2 - 12c2
= 3(a2 - 2ab + b2 - 4c2)
= 3[(a2 - 2ab + b2) - (2c)2]
= 3[(a - b)2 - (2c)2]
= 3(a - b - c)(a - b + c)
con D bạn chép sai đề bài rồi, phải là +3b2 chứ. tích cho mik nha, ko thì lần sau mik ko giúp đâu ihihihi.....!!!!!!!!!
\(3a^2+3b^2=10ab\)
\(\Rightarrow3a^2-10ab+3b^2=0\)
\(\Rightarrow3a^2-ab-9ab+3b^2=0\)
\(\Rightarrow\left(3a^2-ab\right)-\left(9ab-3b^2\right)=0\)
\(\Rightarrow a\left(3a-b\right)-3b\left(3a-b\right)=0\)
\(\Rightarrow\left(3a-b\right)\left(a-3b\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}3a-b=0\\a-3b=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}b=-3a\\b=\dfrac{a}{3}\end{matrix}\right.\)
Với \(b=-3a,\)có :
\(P=\dfrac{-3a-a}{-3a+a}=\dfrac{-4a}{-2a}=2\)
Với \(b=\dfrac{a}{3},\)có :
\(P=\dfrac{\dfrac{a}{3}-a}{\dfrac{a}{3}+a}=\dfrac{\dfrac{a}{3}-\dfrac{3a}{3}}{\dfrac{a}{3}+\dfrac{3a}{3}}=\dfrac{-\dfrac{2a}{3}}{\dfrac{4a}{3}}=-\dfrac{2a}{3}.\dfrac{3}{4a}=-\dfrac{1}{2}\)
( Nếu sai thì cho mk xin lỗi nha bn , tại mk ko chắc lắm )
Lời giải:
Đặt $ab=x,bc=y, ca=z$. Điều kiện đề bài tương đương với: Cho $x,y,z\neq 0$ thỏa mãn:
\(x^3+y^3+z^3=3xyz\)
\(\Leftrightarrow (x+y)^3-3xy(x+y)+z^3=3xyz\)
\(\Leftrightarrow (x+y)^3+z^3-3xy(x+y+z)=0\)
\(\Leftrightarrow (x+y+z)[(x+y)^2-z(x+y)+z^2]-3xy(x+y+z)=0\)
\(\Leftrightarrow (x+y+z)(x^2+y^2+z^2-xy-yz-xz)=0\)
\(\Rightarrow \left[\begin{matrix} x+y+z=0(1)\\ x^2+y^2+z^2-xy-yz-xz=0(2)\end{matrix}\right.\)
Với (1):\(\Leftrightarrow ab+bc+ac=0\)
\(A=(1+\frac{a}{b})(1+\frac{b}{c})(1+\frac{c}{a})=\frac{(a+b)(b+c)(c+a)}{abc}=\frac{(ab+bc+ac)(a+b+c)-abc}{abc}=\frac{0-abc}{abc}=-1\)
Với (2) \(\Leftrightarrow \frac{(x-y)^2+(y-z)^2+(z-x)^2}{2}=0\)
\(\Leftrightarrow (x-y)^2+(y-z)^2+(z-x)^2=0\)
Ta thấy $(x-y)^2; (y-z)^2; (z-x)^2\geq 0, \forall x,y,z$ nên để tổng của chúng bằng $0$ thì:
\((x-y)^2=(y-z)^2=(z-x)^2=0\Rightarrow x=y=z\)
\(\Leftrightarrow ab=bc=ac\Leftrightarrow a=b=c\) (do $a,b,c\neq 0$)
\(\Rightarrow A=(1+1)(1+1)(1+1)=8\)
Vậy...........
Lời giải:
Đặt $ab=x,bc=y, ca=z$. Điều kiện đề bài tương đương với: Cho $x,y,z\neq 0$ thỏa mãn:
\(x^3+y^3+z^3=3xyz\)
\(\Leftrightarrow (x+y)^3-3xy(x+y)+z^3=3xyz\)
\(\Leftrightarrow (x+y)^3+z^3-3xy(x+y+z)=0\)
\(\Leftrightarrow (x+y+z)[(x+y)^2-z(x+y)+z^2]-3xy(x+y+z)=0\)
\(\Leftrightarrow (x+y+z)(x^2+y^2+z^2-xy-yz-xz)=0\)
\(\Rightarrow \left[\begin{matrix} x+y+z=0(1)\\ x^2+y^2+z^2-xy-yz-xz=0(2)\end{matrix}\right.\)
Với (1):\(\Leftrightarrow ab+bc+ac=0\)
\(A=(1+\frac{a}{b})(1+\frac{b}{c})(1+\frac{c}{a})=\frac{(a+b)(b+c)(c+a)}{abc}=\frac{(ab+bc+ac)(a+b+c)-abc}{abc}=\frac{0-abc}{abc}=-1\)
Với (2) \(\Leftrightarrow \frac{(x-y)^2+(y-z)^2+(z-x)^2}{2}=0\)
\(\Leftrightarrow (x-y)^2+(y-z)^2+(z-x)^2=0\)
Ta thấy $(x-y)^2; (y-z)^2; (z-x)^2\geq 0, \forall x,y,z$ nên để tổng của chúng bằng $0$ thì:
\((x-y)^2=(y-z)^2=(z-x)^2=0\Rightarrow x=y=z\)
\(\Leftrightarrow ab=bc=ac\Leftrightarrow a=b=c\) (do $a,b,c\neq 0$)
\(\Rightarrow A=(1+1)(1+1)(1+1)=8\)
Vậy...........
a) \(x^4-4x^3+8x^2-16x+16\)
\(=x^4-2x^3-2x^3+4x^2+4x^2-8x-8x+16\)
\(=x^3\left(x-2\right)-2x^2\left(x-2\right)+4x\left(x-2\right)-8\left(x-2\right)\)
\(=\left(x-2\right)\left(x^3-2x^2+4x-8\right)\)
\(=\left(x-2\right)\left[x^2\left(x-2\right)+4\left(x-2\right)\right]\)
\(=\left(x-2\right)\left(x-2\right)\left(x^2+4\right)\)
\(=\left(x-2\right)^2\left(x^2+4\right)\)
b) \(3a^2-6ab+3b^2-12c^2\)
\(=3\left(a^2-2ab+b^2-4c^2\right)\)
\(=3\left[\left(a-b\right)^2-\left(2c\right)^2\right]\)
\(=3\left(a-b-2c\right)\left(a-b+2c\right)\)
c/ \(a^2+2ab+b^2-ac-bc\)
\(=\left(a+b\right)^2-c\left(a+b\right)\)
\(=\left(a+b\right)\left(a+b-c\right)\)
d/ \(ac-bc-a^2+2ab-b^2\)
\(=c\left(a-b\right)-\left(a^2-2ab+b^2\right)\)
\(=c\left(a-b\right)-\left(a-b\right)^2\)
\(=\left(a-b\right)\left(c-a+b\right)\)
e/ \(\left(x-y+5\right)^2-2\left(x-y+5\right)+1\)
\(=\left(x-y+5\right)\left(x-y+5-2\right)+1\)
\(=\left(x-y+5\right)\left(x-y+3\right)+1\)
f/ \(2x^2+7x+5\)
\(=2x^2+2x+5x+5\)
\(=2x\left(x+1\right)+5\left(x+1\right)\)
\(=\left(x+1\right)\left(2x+5\right)\)
b: \(=abx^2+ab+a^2x+b^2x\)
\(=bx\left(ax+b\right)+a\left(ax+b\right)\)
\(=\left(ax+b\right)\left(bx+a\right)\)
c: \(=\left(x-1\right)^3-8y^3\)
\(=\left(x-1-2y\right)\left(x^2-2x+1+2xy-2y+4y^2\right)\)
d: \(=\left(x^2-3\right)\left(x^2+3\right)+3x\left(x^2-3\right)\)
\(=\left(x^2-3\right)\left(x^2+3x+2\right)\)
e: \(=\left(x+3\right)^2-y^2=\left(x+3+y\right)\left(x+3-y\right)\)
