(\(\dfrac{x}{2}\) - 2y)2
Những câu hỏi liên quan
phân tích đa thức \(\dfrac{1}{2}x^2-2y^2\) thành nhân tử
a. \(\dfrac{1}{2}x^2-2y^2=\dfrac{1}{2}\left(x^2-4y^2\right)=\dfrac{1}{2}\left(x-2y\right)\left(x+2y\right)\)
b. \(\dfrac{1}{2}x^2-2y^2=2\left(\dfrac{1}{4}x^2-y^2\right)=2\left(\dfrac{1}{2}x-y\right)\left(\dfrac{1}{2}x+y\right)\)
Cách phân tích nào đúng, a hay b ?
phân tích đa thức \(\dfrac{1}{2}x^2-2y^2\) thành nhân tử
a. \(\dfrac{1}{2}x^2-2y^2=\dfrac{1}{2}\left(x^2-4y^2\right)=\dfrac{1}{2}\left(x-2y\right)\left(x+2y\right)\)
b. \(\dfrac{1}{2}x^2-2y^2=2\left(\dfrac{1}{4}x^2-y^2\right)=2\left(\dfrac{1}{2}x-y\right)\left(\dfrac{1}{2}x+y\right)\)
Cách phân tích nào đúng, a hay b ?
cho 2 số thực `x,y` thỏa mãn `x>0,y>2,x`\(\ne\)`2y`. CMR: \(\left(\dfrac{x-y}{2y-x}-\dfrac{x^2+y^2+y-2}{x^2-xy-2y^2}\right)\left(2x^2+y+2\right):\dfrac{x^4+4x^2y^2+y^4-4}{x^2+y+xy+x}=\dfrac{x+1}{2y-x}\)
Đề bài sai, đề đúng thì phân thức đằng sau dấu chia phải là:
\(\dfrac{4x^4+4x^2y+y^2-4}{x^2+y+xy+x}\)
Đúng 0
Bình luận (0)
tính:
a, \(\dfrac{1}{1-x}+\dfrac{1}{1+x}+\dfrac{2}{1+x^2}+\dfrac{4}{1+x^2}+\dfrac{4}{1+x^4}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)
b, 2y - \(\dfrac{6xy+2y}{3x+2y}+\dfrac{2y-9x^2}{3x+2y}\)
Bài 1:Tìm đa thức M
a)\(\dfrac{^{x^3}+27}{x^2-3x+9}\)=\(\dfrac{x+3}{M}\)
b)\(\dfrac{M}{x+4}\)=\(\dfrac{x^2-8x+16}{16-x^2}\)
c)\(\dfrac{x-2y}{M}\)=\(\dfrac{3x^2-7xy+2y^2}{3x^2+5xy-2y^2}\)
a, \(\dfrac{x^3+27}{x^2-3x+9}=\dfrac{x+3}{M}\Leftrightarrow\dfrac{\left(x+3\right)\left(x^2-3x+9\right)}{x^2-3x+9}=\dfrac{x+3}{M}\)
\(\Rightarrow M=\dfrac{x+3}{x+3}=1\)
b, \(\dfrac{M}{x+4}=\dfrac{x^2-8x+16}{16-x^2}=\dfrac{\left(x-4\right)^2}{\left(4-x\right)\left(x+4\right)}=\dfrac{4-x}{x+4}\)
\(\Rightarrow M=\dfrac{\left(4-x\right)\left(x+4\right)}{x+4}=4-x\)
c, tương tự
Đúng 0
Bình luận (0)
Xem thêm câu trả lời
A=\(\dfrac{1}{5}x^2y^3+\dfrac{2}{3}x^2y^3-\dfrac{3}{4}x^2y^3+x^2y^3\)
B=\(\left(x^2y\right)^3.\left(\dfrac{1}{2}xy^2z\right)^2\)
Tính A+B,A-B
Helpp..
\(A=x^2y^3\left(\dfrac{1}{5}+\dfrac{2}{3}-\dfrac{3}{4}+1\right)=\dfrac{67}{60}x^2y^3\)
\(B=x^6y^3\cdot\dfrac{1}{4}x^2y^4z^2=\dfrac{1}{4}x^8y^7z^2\)
\(A+B=\dfrac{67}{60}x^2y^3+\dfrac{1}{4}x^8y^7z^2\)
\(A-B=\dfrac{67}{60}x^2y^3-\dfrac{1}{4}x^8y^7z^2\)
Đúng 3
Bình luận (0)
B=x6y3⋅14x2y4z2=14x8y7z2B=x6y3⋅14x2y4z2=14x8y7z2
A−B=6760x2y3−14x8y7z2
Đúng 1
Bình luận (0)
\(A+B=\dfrac{67}{60}x^2y^3+\left(x^6y^3\right)\left(\dfrac{1}{4}x^2y^4z^2\right)\)
\(=\dfrac{67}{60}x^2y^3+\dfrac{1}{4}x^8y^7z^2\)
\(A-B=\dfrac{67}{60}x^2y^3-\dfrac{1}{4}x^8y^7z^2\)
Đúng 2
Bình luận (0)
cm đẳng thức\(a.\dfrac{x}{x+y}+\dfrac{4}{x^2+3xy+2y^2}+\dfrac{-3x}{x+2y}=\dfrac{-2x^2-xy+4}{\left(x+y\right)\left(x+2y\right)}\) với x ≠ -y; x ≠ -2y
b. \(\dfrac{x+y}{x-y}=\dfrac{x^2+2xy+y^2}{x^2-y^2}\)
\(a,VT=\dfrac{x^2+2xy+4-3x^2-3xy}{\left(x+y\right)\left(x+2y\right)}=\dfrac{-2x^2-xy+4}{\left(x+y\right)\left(x-2y\right)}=VP\\ b,VP=\dfrac{\left(x+y\right)^2}{\left(x-y\right)\left(x+y\right)}=\dfrac{x+y}{x-y}=VT\)
Đúng 3
Bình luận (0)
Tính giá trị của biểu thức
A=
\(\dfrac{1}{5}x^2y^3+\dfrac{2}{3}x^2y^3-\dfrac{3}{4}x^2y^3+x^2y^3\)
B=\(\left(x^2y\right)^3.\left(\dfrac{1}{2}xy^2z\right)^2\)
\(A=\dfrac{1}{5}x^2y^3+\dfrac{2}{3}x^2y^3-\dfrac{3}{4}x^2y^3+x^2y^3=\left(\dfrac{1}{5}+\dfrac{2}{3}-\dfrac{3}{4}+1\right)x^2y^3=\dfrac{67}{60}x^2y^3\\ B=\left(x^2y\right)^3\left(\dfrac{1}{2}xy^2z\right)^2=x^6y^3.\dfrac{1}{4}x^2y^4z^2=\dfrac{1}{4}x^8y^7z^2\)
Đúng 4
Bình luận (0)
Tính
a) \(\dfrac{1}{1-x}+\dfrac{1}{1+x}+\dfrac{2}{1+x^2}+\dfrac{4}{1+x^4}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)
b) \(2y+\dfrac{6xy+2y}{3x+2y}+\dfrac{2y-9x^2}{3x+2y}\)
cho mik sửa lại câu
b) \(2y-\dfrac{6xy+2y}{3x+2y}+\dfrac{2y-9x^2}{3x+2y}\)
Đúng 0
Bình luận (0)
b) \(2y-\dfrac{6xy+2y}{3x+2y}+\dfrac{2y-9x^2}{3x+2y}\)
\(=\dfrac{2y\left(3x+2y\right)}{3x+2y}-\dfrac{6xy+2y}{3x+2y}+\dfrac{2y-9x^2}{3x+2y}\)
\(=\dfrac{2y\left(3x+2y\right)-\left(6xy+2y\right)+\left(2y-9x^2\right)}{3x+2y}\)
\(=\dfrac{6xy+4y^2-6xy-2y+2y-9x^2}{3x+2y}\)
\(=\dfrac{4y^2-9x^2}{3x+2y}\)
\(=\dfrac{-\left(9x^2-4y^2\right)}{3x+2y}\)
\(=\dfrac{-\left[\left(3x\right)^2-\left(2y\right)^2\right]}{3x+2y}\)
\(=\dfrac{-\left(3x-2y\right)\left(3x+2y\right)}{3x+2y}\)
\(=-\left(3x-2y\right)\)
\(=-3x+2y\)
Đúng 0
Bình luận (0)
a)\(\dfrac{1}{1-x}+\dfrac{1}{1+x}+\dfrac{2}{1+x^2}+\dfrac{4}{1+x^4}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)
\(=\dfrac{1+x}{\left(1-x\right)\left(1+x\right)}+\dfrac{1-x}{\left(1-x\right)\left(1+x\right)}+\dfrac{2}{1+x^2}+\dfrac{4}{1+x^4}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)
\(=\dfrac{\left(1+x\right)+\left(1-x\right)}{\left(1-x\right)\left(1+x\right)}+\dfrac{2}{1+x^2}+\dfrac{4}{1+x^4}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)
\(=\dfrac{2}{1-x^2}+\dfrac{2}{1+x^2}+\dfrac{4}{1+x^4}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)
\(=\dfrac{2\left(1+x^2\right)}{\left(1-x^2\right)\left(1+x^2\right)}+\dfrac{2\left(1-x^2\right)}{\left(1-x^2\right)\left(1+x^2\right)}+\dfrac{4}{1+x^4}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)
\(=\dfrac{2\left(1+x^2\right)+2\left(1-x^2\right)}{\left(1-x^2\right)\left(1+x^2\right)}+\dfrac{4}{1+x^4}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)
\(=\dfrac{2+2x^2+2-2x^2}{1-x^4}+\dfrac{4}{1+x^4}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)
\(=\dfrac{4}{1-x^4}+\dfrac{4}{1+x^4}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)
\(=\dfrac{4\left(1+x^4\right)}{\left(1+x^4\right)\left(1-x^4\right)}+\dfrac{4\left(1-x^4\right)}{\left(1+x^4\right)\left(1-x^4\right)}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)
\(=\dfrac{4\left(1+x^4\right)+4\left(1-x^4\right)}{\left(1+x^4\right)\left(1-x^4\right)}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)
\(=\dfrac{4+4x^4+4-4x^4}{1-x^8}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)
\(=\dfrac{8}{1-x^8}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)
\(=\dfrac{8\left(1+x^8\right)}{\left(1-x^8\right)\left(1+x^8\right)}+\dfrac{8\left(1-x^8\right)}{\left(1-x^8\right)\left(1+x^8\right)}+\dfrac{16}{1+x^{16}}\)
\(=\dfrac{8\left(1+x^8\right)+8\left(1-x^8\right)}{\left(1-x^8\right)\left(1+x^8\right)}+\dfrac{16}{1+x^{16}}\)
\(=\dfrac{8+8x^8+8-8x^8}{1-x^{16}}+\dfrac{16}{1+x^{16}}\)
\(=\dfrac{16}{1-x^{16}}+\dfrac{16}{1+x^{16}}\)
\(=\dfrac{16\left(1+x^{16}\right)}{\left(1-x^{16}\right)\left(1+x^{16}\right)}+\dfrac{16\left(1-x^{16}\right)}{\left(1-x^{16}\right)\left(1+x^{16}\right)}\)
\(=\dfrac{16\left(1+x^{16}\right)+16\left(1-x^{16}\right)}{\left(1-x^{16}\right)\left(1+x^{16}\right)}\)
\(=\dfrac{16+16x^{16}+16-16x^{16}}{1-x^{32}}\)
\(=\dfrac{32}{1-x^{32}}\)
Đúng 0
Bình luận (0)
\(\dfrac{x}{x-2y}+\dfrac{x}{x+2y}+\dfrac{4xy}{4y^2-x^2}\)
