Thực hiện phép tính\(\left(a^4+a^3b+a^2b^2+ab^3+b^4\right)\left(a-b\right)\)
thực hiện phép tính:
a,\(\left(x-y\right)^4:\left(x-2\right)^3\)
b,\(\dfrac{\left(3a^2b\right)^3\left(ab^3\right)^2}{\left(a^2b^2\right)^4}\)
Thực hiện các phép tính sau :
a. \(\dfrac{\sqrt{x}+1}{\sqrt{x}-2}+\dfrac{2\sqrt{x}}{\sqrt{x}+2}+\dfrac{2+5\sqrt{x}}{4-x}\)
b. \(\left(a^2b-3ab^2\right):\left(\dfrac{1}{2}ab\right)+\left(6b^3-5ab^2\right):b^2\)
Thực hiện các phép tính sau :
a. \(\dfrac{\sqrt{x}+1}{\sqrt{x}-2}+\dfrac{2\sqrt{x}}{\sqrt{x}+2}+\dfrac{2+5\sqrt{x}}{4-x}\)
b. \(\left(a^2b-3ab^2\right):\left(\dfrac{1}{2}ab\right)+\left(6b^3-5ab^2\right):b^2\)
\(a,=\dfrac{x+3\sqrt{x}+2+2x-4\sqrt{x}-2-5\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\\ =\dfrac{3x-6\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{3\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{3\sqrt{x}}{\sqrt{x}+2}\\ b,=2a-6b+6b-5a=-3a\)
Cho \(\left(a+b\right)\left(2b-a-4\right)=\left(a-2b\right)\left(5-a-b\right)\). Tính \(\frac{2a^2-3b^2}{ab+b^2}\)
1) Rút gọn :
\(B=\frac{\left(a+2b\right)^3-\left(a-2b\right)^3}{\left(2a+b\right)^3-\left(2a-b\right)^3}:\frac{3a^4+7a^2b^2+3b^4}{4a^4+7a^2b^2+3b^4}\)
thực hiện phép nhân
a) \(\left(X+1\right)\left(1+X-X^2+X^3-X^4\right)-\left(X-1\right)\left(1+X+X^2+X^3+X^4\right)\)
B) \(\left(2b^2-2-5b+6b^3\right)\left(3+3b^2-b\right)\)
c) \(\left(2ab+2a^2+b^2\right)\left(2ab^2+4a^3-4a^2b\right)\)
d) \(\left(2a^3-0,02a+0,4a^5\right)\left(0,5a^6-0,1a^2+0,03a^4\right)\)
Cho 2 số thực a , b . CMR \(2\left(a^4+b^4\right)\ge ab^3+a^3b+2a^2b^2\)
Áp dụng BĐT Bunhiacopxki, ta có:
\(2\left(a^4+b^4\right)\ge\left(a^2+b^2\right)^2\)\(\ge4a^2b^2\)(BĐT Cô-si)
Có: \(ab^3+a^3b=ab\left(a^2+b^2\right)\)
Áp dụng BĐT Cô-si, ta có:
\(ab\left(a^2+b^2\right)\ge2a^2b^2\)
\(\Rightarrow ab^3+a^3b+2a^2b^2\ge4a^2b^2\)
Vậy VT=VP.
Ta có đpcm.
thực hiện phép tính:
a,\(\left(x-y\right)^4:\left(x-2\right)^3\)
b,\(\dfrac{\left(3a^2b\right)^3\left(ab^3\right)^2}{\left(a^2b^2\right)^4}\)
a: Sửa đề: \(\left(x-2\right)^4:\left(x-2\right)^3\)
\(=\left(x-2\right)^{4-3}\)
=x-2
b: \(=\dfrac{27a^6b^3\cdot a^2b^6}{a^8b^8}=27b\)
Chứng minh \(^{\left(a^4-a^3b+a^2b^2-ab^3+b^4\right)\left(a+b\right)=a^5b^5}\)