B1: rut gon bieu thuc
a, (x+y)^2-4(x-y)^2
b, 2(x-y)(x+y)+(x+y)^2+(x-y)^2
B2: tim X
a, (2X-1)^2-4(X+2)^2=9
b, 3(X-1)^2-3X(X-5)=21
B3: Cho bieu thuc
M=(x+3)^3-(x-1)^3+12x(x-1)
a, Rut gon bieu thuc tren
b, Tinh gia tri M tai x=-2/3
c, Tim x de M=16
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
A) 2x2(1-3x)+6x3
=2x2*(1-3x)+2x2*3x
=2x2*(1-3x+3x)
=2x2
B) (x-y)2+(x+y)2+2(x-y)(x+y)
=2(x2-y2)+x2+2xy+y2+x2-2xy+y2
=2x2-2y2+x2+2xy+y2+x2-2xy+y2
=4x2
=x^3-xy-x^3-x^2y+x^2y--xy
=-2xy
thay x=1\2 va y bang 100 vao Bta duoc
B= -2.1\2.100=-100
a) A có nghĩa\(\Leftrightarrow x-y\ne0\Leftrightarrow x\ne y\)
b) \(A=\frac{x+y-2\sqrt{xy}}{x-y}=\frac{\left(\sqrt{x-\sqrt{y}}\right)^2}{\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}\right)}=\frac{\sqrt{x}-\sqrt{y}}{\sqrt{x}+\sqrt{y}}\)
a:
ĐKXĐ: x<>2
|2x-3|=1
=>\(\left[{}\begin{matrix}2x-3=1\\2x-3=-1\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=2\left(loại\right)\\x=1\left(nhận\right)\end{matrix}\right.\)
Thay x=1 vào A, ta được:
\(A=\dfrac{1+1^2}{2-1}=\dfrac{2}{1}=2\)
b: ĐKXĐ: \(x\notin\left\{-1;2\right\}\)
\(B=\dfrac{2x}{x+1}+\dfrac{3}{x-2}-\dfrac{2x^2+1}{x^2-x-2}\)
\(=\dfrac{2x}{x+1}+\dfrac{3}{x-2}-\dfrac{2x^2+1}{\left(x-2\right)\left(x+1\right)}\)
\(=\dfrac{2x\left(x-2\right)+3\left(x+1\right)-2x^2-1}{\left(x+1\right)\left(x-2\right)}\)
\(=\dfrac{2x^2-4x+3x+3-2x^2-1}{\left(x+1\right)\left(x-2\right)}\)
\(=\dfrac{-x+2}{\left(x+1\right)\left(x-2\right)}=-\dfrac{1}{x+1}\)
c: \(P=A\cdot B=\dfrac{-1}{x+1}\cdot\dfrac{x\left(x+1\right)}{2-x}=\dfrac{x}{x-2}\)
\(=\dfrac{x-2+2}{x-2}=1+\dfrac{2}{x-2}\)
Để P lớn nhất thì \(\dfrac{2}{x-2}\) max
=>x-2=1
=>x=3(nhận)
a, x.(x-y) +y.(x+y)
=x2-xy+xy+y2
=x2+y2
b, (x2-5).(2x+3)-2x.(x-3)
=2x3+3x2-10x-15-2x2+6x
=2x3-x2-4x-15
c, 8-5x.(x+2) +4 .( x-2) . (x+1) +2.( x+2)+ 2.(x-2)+10
=8-5x2-10x+4.(x2+x-2x-2)+2x+4+2x-4+10
=18-6x-5x2+4x2+4x-8x-8
=10-10x-x2
Hên xui thôi ( cái này không có chắc lắm )
\(\frac{x^3-xy^3+y^3z-yz^3+z^3x-x^3z}{x^2y-xy^2+y^2z-yz^2+z^2x-zx^2}\)
\(=xy-xy+xy-yz+zx-x^3\)\(z\)\(-\)\(zx^2\)
\(=xy-yz-zx-x^3\)\(z\)
phần trên sai rồi cho xin lỗi ( trình bày lại )
bạn ghi lại đề nha
= xy - xy + yz - yz + zx - x^3z - zx^2
= -zx - x^3z
\(a,\)\(2\left(x-y\right)\left(x+y\right)+\left(x-y\right)^2+\left(x+y\right)^2.\)
\(=\left[\left(x-y\right)+\left(x+y\right)\right]^2=\left(x-y+x+y\right)^2=x^2\)
\(b,\)\(\left(2x-3\right)\left(4x^2+6x+9\right)-\left(54+8x\right)\)
\(=8x^2-27-54-8x=8x^2-8x-81\)
\(c,\)\(\left(3x+y\right)\left(9x^2-3xy+y^2\right)-\left(3x-y\right)\left(9x^2+3xy+y^2\right)\)
\(=27x^3+y^3-\left(27x^3-y^3\right)=2y^3\)
\(d,\)\(\left(a+b+c\right)^2-\left(a-c\right)^2-2ab+2bc\)
\(=a^2+b^2+c^2+2ab+2bc+2ac-a^2+2ac-c^2-2ab+2bc\)
\(=b^2+4bc+4ac\)
\(P=2\left(x^2-y^2\right)-x^2+2xy-y^2+x^2+2xy+y^2-4y^2\)
\(=2\left(x^2-y^2\right)-4y^2+4xy\)
\(=2x^2-2y^2-4y^2+4xy\)
=2x^2+4xy-6y^2
1)a)=>x2+y2+2xy-4(x2-y2-2xy)
=>x2+y2+2xy-4.x2+4y2+8xy
=>-3.x2+5y2+10xy