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a/ \(\dfrac{x^3-x^2y+xy^2-y^3}{x^2y+xy^2-x^3-y^3}\)
\(=\dfrac{x^2\left(x-y\right)+y^2\left(x-y\right)}{x^2\left(y-x\right)+y^2\left(x-y\right)}\)
\(=\dfrac{\left(x-y\right)\left(x^2+y^2\right)}{\left(x-y\right)\left(y-x\right)\left(y+x\right)}\)
\(=\dfrac{x^2+y^2}{\left(y-x\right)\left(y+x\right)}\)
Bài 1:
a) \(3x^2-2x(5+1,5x)+10=3x^2-(10x+3x^2)+10\)
\(=10-10x=10(1-x)\)
b) \(7x(4y-x)+4y(y-7x)-2(2y^2-3,5x)\)
\(=28xy-7x^2+(4y^2-28xy)-(4y^2-7x)\)
\(=-7x^2+7x=7x(1-x)\)
c)
\(\left\{2x-3(x-1)-5[x-4(3-2x)+10]\right\}.(-2x)\)
\(\left\{2x-(3x-3)-5[x-(12-8x)+10]\right\}(-2x)\)
\(=\left\{3-x-5[9x-2]\right\}(-2x)\)
\(=\left\{3-x-45x+10\right\}(-2x)=(13-46x)(-2x)=2x(46x-13)\)
Bài 2:
a) \(3(2x-1)-5(x-3)+6(3x-4)=24\)
\(\Leftrightarrow (6x-3)-(5x-15)+(18x-24)=24\)
\(\Leftrightarrow 19x-12=24\Rightarrow 19x=36\Rightarrow x=\frac{36}{19}\)
b)
\(\Leftrightarrow 2x^2+3(x^2-1)-5x(x+1)=0\)
\(\Leftrightarrow 2x^2+3x^2-3-5x^2-5x=0\)
\(\Leftrightarrow -5x-3=0\Rightarrow x=-\frac{3}{5}\)
\(2x^2+3(x^2-1)=5x(x+1)\)
Bài 1.
a) 2x - x2
= x(2 - x)
b) 16x2 - y2
= (4x + y)(4x - y)
c) xy + y2 - x - y
= (xy + y2) - (x + y)
= y(x + y) - (x + y)
= (y - 1)(x + y)
d) x2 - x - 12
= x2 + 3x - 4x - 12
= (x2 + 3x) - (4x + 12)
= x(x + 3) - 4(x + 3)
= (x - 4)(x + 3)
Bài 2.
(2x + 3y)(2x - 3y) - (2x - 1)2 + (3y - 1)2
= (2x + 3y)(2x - 3y) + [(3y - 1)2 - (2x - 1)2]
= (2x + 3y)(2x - 3y) + (3y - 1 + 2x - 1)(3y - 1 - 2x + 1)
= (2x + 3y)(2x - 3y) + (3y + 2x - 2)(3y - 2x)
= (2x + 3y)(2x - 3y) - (2x + 3y - 2)(2x - 3y)
= (2x - 3y)(2x + 3y - 2x - 3y + 2)
= 2.(2x + 3y)
Thay x = 1; y = -1 và biểu thức đại số, ta có:
2[2.1 + 3.(-1)]
= 2(2 - 3)
= 2.(-1) = -2
Bài 3
a) 9x2 - 3x = 0
3x(3x - 1) = 0
\(\Leftrightarrow\left\{{}\begin{matrix}3x=0\\3x-1=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0\\3x=1\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0\\x=\dfrac{1}{3}\end{matrix}\right.\)
b) x2 - 25 - (x + 5) = 0
(x2 - 25) - (x + 5) = 0
(x - 5)(x + 5) - (x + 5) = 0
(x - 5 - 1)(x + 5) = 0
(x - 6)(x + 5) = 0
\(\Leftrightarrow\left\{{}\begin{matrix}x-6=0\\x+5=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=6\\x=-5\end{matrix}\right.\)
c) x2 + 4x + 3 = 0
x2 + x + 3x + 3 = 0
(x2 + x) + (3x + 3) = 0
x(x + 1) + 3(x + 1) = 0
(x + 3)(x + 1) = 0
\(\Leftrightarrow\left\{{}\begin{matrix}x+3=0\\x+1=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-3\\x=-1\end{matrix}\right.\)
d) (3x - 1)(2x - 7) - (x + 1)(6x - 5) = 16
6x2 - 21x - 2x + 7 - 6x2 + 5x - 6x + 5 - 16 = 0
-24x - 4 = 0
\(\Rightarrow\)-24x = 4
\(\Rightarrow\) x = \(\dfrac{-1}{6}\)
Bài 1:Phân tích đa thức thành nhân tử
a,2x−x2
=x(2-x)
b,
16x2−y2
=(4x-y)(4x+y)
c,xy+y2−x−y
=(xy+y2)-(x+y)
=y(x+y)-(x+y)
=(x+y)(y-1)
d,
x2−x−12
=x2-4x+3x-12
=(x2-4x)+(3x-12)
=x(x-4)+3(x-4)
=(x-4)(x+3)
\(1.\)
\(a.\)
\(\dfrac{8}{\left(x^2+3\right)\left(x^2-1\right)}+\dfrac{2}{x^2+3}+\dfrac{1}{x+1}\)
\(=\dfrac{8}{\left(x^2+3\right)\left(x^2-1\right)}+\dfrac{2\left(x^2-1\right)}{\left(x^2+3\right)\left(x^2-1\right)}+\dfrac{1\left(x-1\right)\left(x^2+3\right)}{\left(x^2-1\right)\left(x^2+3\right)}\)
\(=\dfrac{8}{\left(x^2+3\right)\left(x^2-1\right)}+\dfrac{2x^2-2}{\left(x^2+3\right)\left(x^2-1\right)}+\dfrac{x^3-x^2+3x-3}{\left(x^2-1\right)\left(x^2+3\right)}\)
\(=\dfrac{8+2x^2-2+x^3-x^2+3x-3}{\left(x^2+3\right)\left(x^2-1\right)}\)
\(=\dfrac{x^3+x^2+3x+3}{\left(x^2+3\right)\left(x^2-1\right)}\)
\(=\dfrac{x^2\left(x+1\right)+3\left(x+1\right)}{\left(x^2+3\right)\left(x^2-1\right)}\)
\(=\dfrac{\left(x^2+3\right)\left(x+1\right)}{\left(x^2+3\right)\left(x^2-1\right)}\)
\(=x-1\)
\(b.\)
\(\dfrac{x+y}{2\left(x-y\right)}-\dfrac{x-y}{2\left(x+y\right)}+\dfrac{2y^2}{x^2-y^2}\)
\(=\dfrac{x+y}{2\left(x-y\right)}-\dfrac{x-y}{2\left(x+y\right)}+\dfrac{2y^2}{\left(x-y\right)\left(x+y\right)}\)
\(=\dfrac{\left(x+y\right)^2}{2\left(x^2-y^2\right)}-\dfrac{\left(x-y\right)^2}{2\left(x^2-y^2\right)}+\dfrac{4y^2}{2\left(x^2-y^2\right)}\)
\(=\dfrac{x^2+2xy+y^2}{2\left(x^2-y^2\right)}-\dfrac{x^2-2xy+y^2}{2\left(x^2-y^2\right)}+\dfrac{4y^2}{2\left(x^2-y^2\right)}\)
\(=\dfrac{x^2+2xy+y^2-x^2+2xy-y^2+4y^2}{2\left(x^2-y^2\right)}\)
\(=\dfrac{4xy+4y^2}{2\left(x^2-y^2\right)}\)
\(=\dfrac{4y\left(x+y\right)}{2\left(x^2-y^2\right)}\)
\(=\dfrac{2y}{\left(x-y\right)}\)
Tương tự các câu còn lại
Bài 2:
a: \(=6x^2+30x+x+5-\left(6x^2-3x-10x+5\right)\)
\(=6x^2+31x+5-6x^2+13x-5=18x⋮6\)
b: \(=x^3+2x^2+3x^2+6x-x-2-x^3+2\)
\(=5x^2+5x=5x\left(x+1\right)⋮2\)
Bài 1:
\(a, \dfrac{1}{2}x(2-x)=x-\dfrac{1}{2}x^2\)
\(b, \dfrac{x-5}{5-x}\)\(=-\dfrac{x-5}{x-5}\)\(=-1\)
Bài 2:
\(a, x+y-x^2+y^2=(x+y)-(x^2-y^2)=(x+y)-(x-y)(x+y)\)
\(=(x+y)(1-x+y)\)
\(b, x(x-3)+3x-1=0 \)
\(⇔x^2-3x+3x-1=0 \)
\(⇔x^2-1=0 \)
\(⇔(x-1)(x+1)=0 \)
\(⇔\left[\begin{array}{} x-1=0\\ x+1=0 \end{array}\right.\)
\(⇔\left[\begin{array}{} x=1\\ x=-1 \end{array}\right.\)
Bài 3:
\(a,A=\dfrac{x(x+2)-x(x-2)+8}{x^2-4}:\dfrac{4}{x-2}\)
\(A=\dfrac{4x+8}{(x-2)(x+2)}.\dfrac{x-2}{4}\)
\(A=\dfrac{4(x+2)}{(x-2)(x+2)}.\dfrac{x-2}{4}\)
\(A=1\)
\(b, B=(1-\dfrac{a+b}{a-b})(1-\dfrac{2b}{a+b})\)
\(B=\dfrac{-2b}{a-b}.\dfrac{a-b}{a+b}\)
\(B=\dfrac{-2b}{a+b}\)
Bài 4:
\(C=(2+1)(2^2+1)(2^4+1)(2^8+1)(2^{16}+1)(2^{32}+1)\)
\(C=(2^2-1)(2^2+1)(2^4+1)(2^8+1)(2^{16}+1)(2^{32}+1)\)
\(C=(2^4-1)(2^4+1)(2^8+1)(2^{16}+1)(2^{32}+1)\)
\(C=(2^8-1)(2^8+1)(2^{16}+1)(2^{32}+1)\)
\(C=(2^{16}-1)(2^{16}+1)(2^{32}+1)\)
\(C=(2^{32}-1)(2^{32}+1)=2^{64}-1\)
1,
a,\(2x\left(3x^2-5x+3\right)\)
\(=6x^3-10x^2+6x\)
b,\(-2x\left(x^2+5x-3\right)\)
\(=-2x^3-10x^2+6x\)
c,\(-\dfrac{1}{2}x\left(2x^3-4x+3\right)\)
\(=-x^4+2x^2-\dfrac{3}{2}x\)
Bài 2:
a) \(\left(2x-1\right)\left(x^2-5-4\right)\)
\(=\left(2x-1\right)\left(x^2-9\right)\)
\(=2x^3-18x-x^2+9\)
b) \(-\left(5x-4\right)\left(2x+3\right)\)
\(=-\left(10x^2+15x-8x-12\right)\)
\(=-10x^2-7x+12\)
c) \(\left(2x-y\right)\left(4x^2-2xy+y^2\right)\)
\(=8x^3-y^3\)
1)Nhân vào ta sẽ đc VT=\(x^4-y^4+x^2y^2-x^2y^2+xy^3-x^3y-xy^3+x^3y=x^4-y^4\)
2) \(x\left(x+2\right)\left(x^2+2x+2\right)+1=\left(x^2+2x\right)\left(x^2+2x+2\right)\)
Đặt y=\(x^2+2x\).Ta sẽ đc : \(y\left(y+2\right)+1=y^2+2y+1=\left(y+1\right)^2=\left(x^2+2x+1\right)^2=\left(x+1\right)^4\)
3/Theo đề ta có: \(a^2+b^2+c^2=ab+bc+ca\)
\(\Leftrightarrow2\left(a^2+b^2+c^2\right)-2\left(ab+bc+ca\right)=0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}a-b=0\\b-c=0\\c-a=0\end{matrix}\right.\)\(\Leftrightarrow a=b=c\)
Vậy Ta có \(a^4+b^4+c^4=3a^4=3\Rightarrow a=b=c=1\)