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a: \(=\dfrac{4x^2+4x+1-4x^2+4x-1}{\left(2x+1\right)\left(2x-1\right)}\cdot\dfrac{5\left(2x-1\right)}{4x}\)
\(=\dfrac{8x\cdot5}{4x\left(2x+1\right)}=\dfrac{10}{2x+1}\)
b: \(=\left(\dfrac{1}{x^2+1}+\dfrac{x-2}{x+1}\right):\dfrac{1+x^2-2x}{x}\)
\(=\dfrac{x+1+x^3+x-2x^2-2}{\left(x+1\right)\left(x^2+1\right)}\cdot\dfrac{x}{\left(x-1\right)^2}\)
\(=\dfrac{x^3-2x^2+2x-1}{\left(x+1\right)\left(x^2+1\right)}\cdot\dfrac{x}{\left(x-1\right)^2}\)
\(=\dfrac{\left(x-1\right)\left(x^2-x+1\right)}{\left(x+1\right)\left(x^2+1\right)}\cdot\dfrac{x}{\left(x-1\right)^2}\)
\(=\dfrac{x\left(x^2-x+1\right)}{\left(x^2-1\right)\left(x^2+1\right)}\)
c: \(=\dfrac{1}{x-1}-\dfrac{x^3-x}{x^2+1}\cdot\left(\dfrac{1}{\left(x-1\right)^2}-\dfrac{1}{\left(x-1\right)\left(x+1\right)}\right)\)
\(=\dfrac{1}{x-1}-\dfrac{x\left(x-1\right)\left(x+1\right)}{x^2+1}\cdot\dfrac{x+1-x+1}{\left(x-1\right)^2\cdot\left(x+1\right)}\)
\(=\dfrac{1}{x-1}-\dfrac{x}{x^2+1}\cdot\dfrac{2}{\left(x-1\right)}\)
\(=\dfrac{x^2+1-2x}{\left(x-1\right)\left(x^2+1\right)}=\dfrac{x-1}{x^2+1}\)
Câu 2:
a: \(=\dfrac{\left(x^2-1\right)\left(x^2+1\right)-2x\left(x^2-1\right)}{x^2-1}=x^2-2x+1=\left(x-1\right)^2\)
b: \(=\dfrac{x^3-3x^2+2x^2-6x-x+3}{x-3}=x^2+2x-1\)
1) 2x2-8xy-5x+20y
=2x(x-4y)-5(x-4y)
=(2x-5)(x-4y)
2) x3-x2y-xy+y2
=x2(x-y)-y(x-y)
=(x2-y)(x-y)
3) x2-2xy-4z2+y2
=(x-y)2-(2z)2
=(x-y-2z)(x-y+2z)
4) a3+a2b-a2c-abc
=a2(a+b)-ac(a+b)
=(a2-ac)(a+b)
=a(a-c)(a+b)
5) x3+y3+3x2y+3xy2-x-y
=(x+y)(x2-xy+y2)+3xy(x+y)-(x+y)
=(x+y)(x2-xy+y2+3xy-1)
=(x+y)[(x+y)2-1)]
=(x+y)(x+y+1)(x+y-1)
6) x3+x2y-x2z-xyz
=x2(x+y)-xz(x+y)
=(x2-xz)(x+y)
=x(x-z)(x+y)
7) =[x(y+z)2-2xyz]+[y(z+x)2-2xyz]+z(x+y)2
=x(y2+z2)+y(z2+x2)+z(x+y)2
=xy(x+y)+z2(x+y)+z(x+y)2
=(x+y)(xy+z2+zx+zy)
=(x+y)(x+z)(y+z)
8) x3(z-y)+y3(x-z)+z3(y-x)
Tách x-z= -[z-y+y-x]
Bài 1:
1.1
a) Ta có: \(A=\left(x+y\right)\left(x-y\right)+x\left(2x-1\right)+y\left(y+1\right)\)
\(=x^2-y^2+2x^2-x+y^2+y\)
\(=3x^2-x+y\)
b) Thay x=1 và y=2018 vào biểu thức \(A=3x^2-x+y\), ta được:
\(A=3\cdot1^2-1+2018\)
\(=2+2018=2020\)
Vậy: Khi x=1 và y=2018 thì A=2020
1.2
a) Ta có: \(2x^2\left(x^2-3x+1\right)\)
\(=2x^2\cdot x^2-2x^2\cdot3x+2x^2\cdot1\)
\(=2x^4-6x^3+2x^2\)
b) Ta có: \(\left(2x-1\right)\left(6x^2+3x-3\right)\)
\(=2x\cdot6x^2+2x\cdot3x-2x\cdot3-6x^2-3x+3\)
\(=12x^3+6x^2-6x-6x^2-3x+3\)
\(=12x^3-9x+3\)
1.3
a) Ta có: \(x^3-2x^2+x\)
\(=x\left(x^2-2x+1\right)\)
\(=x\left(x-1\right)^2\)
b) Ta có: \(x^2-xy-8x+8y\)
\(=x\left(x-y\right)-8\left(x-y\right)\)
\(=\left(x-y\right)\left(x-8\right)\)
1.1
a) A= (x+y).(x-y) + x(2x-1) + y(y+1)
= x2- x.y + x.y - y2 + 2x2 - x +y2 + y = 3x2 - x + y
b) Ta có A= 3x2 - x + y; thay x=1,y=2018 vào biểu thức:
A= 3.12 - 1+ 2018 = 2020
1.3
a)x3 - 2x2 + x = x.( x2 - 2x + 1) = x.(x-1)2
b) x2 - xy - 8x + 8y = x.(x - y) - 8.(x - y)= (x - y).(x-8).
Xin lỗi nha, tớ không biết làm bài 1.2.
Chúc bạn học tốt!!
\(\Leftrightarrow x^2-2.3.x+9+1=\left(x-3\right)^2+1\Rightarrow\hept{\begin{cases}\left(x-3\right)^2\ge0\\1>0\end{cases}}\Rightarrow\left(x-3\right)^2+1>0\)
\(\Leftrightarrow x^2-2.\frac{3}{2}.x+\frac{9}{4}+\frac{7}{4}=\left(x-\frac{3}{2}\right)^2+\frac{7}{4}\Leftrightarrow\hept{\begin{cases}\left(x-\frac{3}{2}\right)^2\ge0\\\frac{7}{4}>0\end{cases}}\Rightarrow\left(x-\frac{3}{2}\right)^2+\frac{7}{4}>0\)
\(\Leftrightarrow2.\left(x^2+xy+y^2+1\right)=x^2+2xy+y^2+x^2+y^2+2=\left(x+y\right)^2+x^2+y^2+2\)
ta có \(\left(x+y\right)^2\ge0,x^2\ge0,y^2\ge0,2>0\Rightarrow\left(x+y\right)^2+x^2+y^2+2>0\)
\(\Leftrightarrow x^2-2xy+y^2+x^2-2.1x+1+y^2+2.2.y+4+3\)\(=\left(x-y\right)^2+\left(x-1\right)^2+\left(y+2\right)^2+3\)
Ta có \(=\left(x-y\right)^2\ge0,\left(x-1\right)^2\ge0,\left(y+2\right)^2\ge0,3>0\)\(\Rightarrow=\left(x-y\right)^2+\left(x-1\right)^2+\left(y+2\right)^2+3>0\)
T i c k cho mình 1 cái nha mới bị trừ 50 đ
a, Ta có: 4x2-2x+1 = (x2 -2x+1)+ 3x2=(x-1)2 +3x2>0 (thay x=1 và x=0 thì biểu thức vãn lớn hơn 0)
b, x4-3x2+9=x4- 6x2 +32 +3x2=(x2-3)2 +3x2 >0
c, x2+y2-2x-2y+2xy+2=(x+y)2 -1 -2(x+y-1) +1 =(x+y -1)(x+y+1) - 2(x+y-1)+1=(x+y-1)(x+y+1-2) + 1=(x+y-1)2 +1 >0
d, 2(x2+3xy+3y2)=2x2+6xy+6y2=(x2+2xy+y2) +(x2+4xy+4y2)+y2=(x+y)2+(x+2y)2+y2>0
e, 2x2+y2+2x(y-1)+2= (x2+2xy+y2) +(x2-2x+1)+1=(x+y)2+(x-1)+1>0
nhớ bấm đúng cho mình nhé!
a) \(\left(x+2y\right)^2=x^2+4xy+4y^2\)
b) \(\left(3x-\frac{1}{8}y\right)^2=9x^2-\frac{3}{4}xy+\frac{1}{64}y^2\)
c) \(\left(-6x-\frac{2}{5}\right)^2=36x^2+\frac{24}{5}x+\frac{4}{25}\)
d) \(\left(xy^2+1\right)\left(xy^2-1\right)=x^2y^4-1\)
e) \(\left(x-y\right)^2\left(x+y\right)^2=\left(x^2-y^2\right)^2=x^4-2x^2y^2+y^4\)
f) \(\left(\frac{1}{2}x-\frac{1}{3}y-1\right)^2=\frac{1}{4}x^2+\frac{1}{9}y^2+1-\frac{1}{3}xy-x+\frac{2}{3}y\)
1.
\(a,\left(-xy\right)\left(-2x^2y+3xy-7x\right)\)
\(=2x^3y^2-3x^2y^2+7x^2y\)
\(b,\left(\dfrac{1}{6}x^2y^2\right)\left(-0,3x^2y-0,4xy+1\right)\)
\(=-\dfrac{1}{20}x^4y^3-\dfrac{1}{15}x^3y^3+\dfrac{1}{6}x^2y^2\)
\(c,\left(x+y\right)\left(x^2+2xy+y^2\right)\)
\(=\left(x+y\right)^3\)
\(=x^3+3x^2y+3xy^2+y^3\)
\(d,\left(x-y\right)\left(x^2-2xy+y^2\right)\)
\(=\left(x-y\right)^3\)
\(=x^3-3x^2y+3xy^2-y^3\)
2.
\(a,\left(x-y\right)\left(x^2+xy+y^2\right)\)
\(=x^3-y^3\)
\(b,\left(x+y\right)\left(x^2-xy+y^2\right)\)
\(=x^3+y^3\)
\(c,\left(4x-1\right)\left(6y+1\right)-3x\left(8y+\dfrac{4}{3}\right)\)
\(=24xy+4x-6y-1-24xy-4x\)
\(=\left(24xy-24xy\right)+\left(4x-4x\right)-6y-1\)
\(=-6y-1\)
#Toru