Giúp em vs !!!
Rút gọn | x-2016 |+| x-2017| với 2016 <= x<=2017
<= lớn hơn hoặc bằng
Rút gọn 2016/2017 nhân 2017/2018 nhân 2018/2019 nhân 2019/2020
\(\frac{2016}{2017}\times\frac{2017}{2018}\times\frac{2018}{2019}\times\frac{2019}{2020}\)=
\(0,998109801980198\)
Đổi ra ta sẽ có !
\(\frac{504}{505}\)
Vậy là : ...................
So sánh : B = 2016/2017 + 2017/2018 + 2018/2016 với 3
Giúp mk nhé ! Mik tick cho
\(B=\frac{2016}{2017}+\frac{2017}{2018}+\frac{2018}{2016}\)
\(B=1-\frac{1}{2017}+1-\frac{1}{2018}+1+\frac{2}{2016}\)
\(B=\left(1+1+1\right)-\left(\frac{1}{2017}+\frac{1}{2018}-\frac{2}{2016}\right)\)
\(B=3-\left(...\right)< 3\)
P/s :
\(\left(...\right)la`\left(\frac{1}{2017}+\frac{1}{2018}-\frac{2}{2016}\right)\)
quên ^^
Tìm x, biết:
2017+ 2016+ 2015+ ...+ ( x+1 )+ x= 2017 (x\(\in\) Z)
giúp mk vs cc bn ơi
rút gọn biểu thức A=(x+y)(x^2+y^2)(x^4+y^4)...(x^64+y^64) với x-y=1.
các bn lm giúp mk nhá.
\(A=1.\left(x+y\right)\left(x^2+y^2\right)...\left(x^{64}+y^{64}\right)\)
\(=\left(x-y\right)\left(x+y\right)\left(x^2+y^2\right)...\left(x^{64}+y^{64}\right)\)
\(=\left(x^2-y^2\right)\left(x^2+y^2\right)\left(x^4+y^4\right)...\left(x^{64}+y^{64}\right)\)
\(=\left(x^4-y^4\right)...\left(x^{64}+y^{64}\right)\)
\(=...=\left(x^{64}-y^{64}\right)\left(x^{64}+y^{64}\right)=x^{128}-y^{128}\)
Rút gọn \(\frac{x^{\text{4}}-x^3-x+1}{x^4+x^3+3x^2+2x+2}\)
Ai giúp mình với mình cảm ơn
\(\frac{x^4-x^3-x+1}{x^4+x^3+3x^2+2x+2}\)
\(=\frac{x^3\left(x-1\right)-\left(x-1\right)}{x^4+x^3+x^2+2x^2+2x+2}\)
\(=\frac{\left(x-1\right)\left(x^3-1\right)}{x^2\left(x^2+x+1\right)+2\left(x^2+x+1\right)}\)
\(=\frac{\left(x-1\right)\left(x-1\right)\left(x^2+x+1\right)}{\left(x^2+x+1\right)\left(x^2+2\right)}\)
\(=\frac{\left(x-1\right)^2}{\left(x^2+2\right)}\)
\(A=\frac{x^4-x^3-x+1}{x^4+x^3+3x^2+2x+2}\)
A, Rút gọn
B, Chứng minh A ko âm
giúp với ạ
Câu a đơn giản
b)
\(A=\frac{x^4-x^3-x+1}{x^4+x^3+3x^2+2x+2}=\frac{\left(x^4-x^3\right)-\left(x-1\right)}{\left(x^4+x^3+\frac{x^2}{4}\right)+\left(\frac{11}{4}x^2+2x+\frac{4}{11}\right)+1-\frac{4}{11}}\)
\(=\frac{\left(x-1\right)\left(x^3-1\right)}{\left(x^2+\frac{x}{2}\right)^2+\left(\frac{\sqrt{11}}{2}+\frac{2}{\sqrt{11}}\right)^2+\frac{7}{11}}\)
\(=\frac{\left(x-1\right)^2\left(x^2+x+1\right)}{\left(x^2+\frac{x}{2}\right)^2+\left(\frac{\sqrt{11}}{2}+\frac{2}{\sqrt{11}}\right)^2+\frac{7}{11}}\)
\(=\frac{\left(x-1\right)^2\left[\left(x^2+x+0,25\right)+0,75\right]}{\left(x^2+\frac{x}{2}\right)^2+\left(\frac{\sqrt{11}}{2}+\frac{2}{\sqrt{11}}\right)^2+\frac{7}{11}}\)
\(=\frac{\left(x-1\right)^2\left[\left(x+0,5\right)^2+0,75\right]}{\left(x^2+\frac{x}{2}\right)^2+\left(\frac{\sqrt{11}}{2}+\frac{2}{\sqrt{11}}\right)^2+\frac{7}{11}}\)
Vì \(\left(x-1\right)^2\left[\left(x+0,5\right)^2+0,75\right]>0\)và \(\left(x^2+\frac{x}{2}\right)^2+\left(\frac{\sqrt{11}}{2}+\frac{2}{\sqrt{11}}\right)^2+\frac{7}{11}>0\)
nên \(A>0\)hay A ko âm
Nhớ k nha !
So sánh 2016^10 + 2016^9 với 2017^10
Làm ơn giúp mình nha,càng nhanh càng tốt.Ai đúng mình sẽ tick!!
Ta có:
\(2016^{10}+2016^9=2016^9.2016+2016^9=2016^9(2016+1)=2017.2016^9\)
\(2017^{10}=2017.2017^9\)
Xét thấy: \(2016<2017\Rightarrow 2016^9<2017^9\Rightarrow 2017.2016^9<2017.2017^9\)
\(\Rightarrow 2016^{10}+2016^9<2017^{10}\)
Rút gọn các phân thức sau:
1) ( x + 2 )^2 - 16/ x^2 - 4x + 4
2) 6xy - 2x + 9y - 3/ y^2 - 3y^3 + 3y - 1
Giúp mình với ạ!
rút gọn
a, (x-3)3-(x-3)*(xx2+3x+9)+6×(x+1)2
Giúp mik với mik tick chooooo nha ^^^^