Do  A = 98 99 + 1 98 89 + 1 > 1 nên  A = 98 99 + 1 98 89 + 1 > 98 99 + 1 + 97 98 89 + 1 + 97 = 98 98 98 + 1 98 98 88 + 1 = 98 98 + 1 98 88 + 1 = B

Vậy A > B

a) Do A = 98 99 + 1 98 89 + 1 > 1  nên

A = 98 99 + 1 98 89 + 1 > 98 99 + 1 + 97 98 89 + 1 + 97 = 98 ( 98 98 + 1 ) 98 ( 98 88 + 1 ) = 98 98 + 1 98 88 + 1 = B

Vậy A > B

b) Do C = 100 2008 + 1 100 2018 + 1  < 1 nên

C= 100 2008 + 1 100 2018 + 1 > 100 2008 + 1 + 99 100 2018 + 1 + 99 = 100 ( 100 2007 + 1 ) 100 ( 100 2017 + 1 ) = 100 2007 + 1 100 2017 + 1 = D

Vậy C > D.

Sửa đề: \(C=\dfrac{17^{99}+1}{17^{99}-1}\)

\(C=\dfrac{17^{99}-1+2}{17^{99}-1}=1+\dfrac{2}{17^{99}-1}\)

\(D=\dfrac{17^{98}-1+2}{17^{98}-1}=1+\dfrac{2}{17^{98}-1}\)

17^99>17^98

=>17^99-1>17^98-1

=>C<D

a) \(1010+1111+1212+.....+9898+9999\)\(=\frac{\left(1010+9999\right)\cdot\left(\frac{9999-1010}{1111-1010}+1\right)}{2}\)\(=\frac{11009\cdot\left(\frac{8989}{101}+1\right)}{2}\)\(=\frac{11009\cdot\left(89+1\right)}{2}\)\(=\frac{11009\cdot90}{2}\)\(=\frac{990810}{2}\)\(=495405\)

a) Khoảng cách của dãy số là:

1111-1010=101;1212-1111=101;...

Số số hạng của dãy số là:

(9999-1010):101+1=90(số)

Tổng:

(1010+9999)*90:2=495405

Đ/s:495405

Nhớ k mk nha!

a: Thay x=3 vào P, ta được:

\(P=\sqrt3-\frac{1}{\sqrt3}=\frac{3-1}{\sqrt3}=\frac{2}{\sqrt3}=\frac{2\sqrt3}{3}\)

b: \(Q=\frac{\sqrt{x}-1}{\sqrt{x}}+\frac{1-\sqrt{x}}{x+\sqrt{x}}\)

\(=\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)+1-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}\)

\(=\frac{x-1+1-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}=\frac{x-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}\)

\(=\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}=\frac{\sqrt{x}-1}{\sqrt{x}+1}\)

c: \(Q-1=\frac{\sqrt{x}-1}{\sqrt{x}+1}-1\)

\(=\frac{\sqrt{x}-1-\sqrt{x}-1}{\sqrt{x}+1}=\frac{-2}{\sqrt{x}+1}<0\)

=>Q<1

\(a,ĐK:x>0;x\ne4\\ E=\dfrac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}\cdot\dfrac{\sqrt{x}-2}{2}=\dfrac{\sqrt{x}-2}{2\sqrt{x}}\\ b,x=19-8\sqrt{3}=\left(4-\sqrt{3}\right)^2\\ \Leftrightarrow E=\dfrac{4-\sqrt{3}-2}{2\left(4-\sqrt{3}\right)}=\dfrac{\left(2-\sqrt{3}\right)\left(4+\sqrt{3}\right)}{26}=\dfrac{5-2\sqrt{3}}{26}\\ c,E=-1\Leftrightarrow\sqrt{x}-2=-2\sqrt{x}\\ \Leftrightarrow3\sqrt{x}=2\Leftrightarrow\sqrt{x}=\dfrac{2}{3}\Leftrightarrow x=\dfrac{4}{9}\left(tm\right)\\ d,E=\dfrac{1}{\sqrt{x}}\Leftrightarrow\dfrac{\sqrt{x}-2}{2}=1\Leftrightarrow\sqrt{x}=4\Leftrightarrow x=16\left(tm\right)\)

\(e,E>0\Leftrightarrow\sqrt{x}-2>0\left(2\sqrt{x}>0\right)\Leftrightarrow x>4\\ f,E=\dfrac{\sqrt{x}-2}{2\sqrt{x}}=\dfrac{1}{2}-\dfrac{1}{\sqrt{x}}< \dfrac{1}{2}\left(-\dfrac{1}{\sqrt{x}}< 0\right)\\ g,\dfrac{1}{E}=\dfrac{2\sqrt{x}}{\sqrt{x}-2}=\dfrac{2\left(\sqrt{x}-2\right)+4}{\sqrt{x}-2}\in Z\\ \Leftrightarrow\sqrt{x}-2\inƯ\left(4\right)=\left\{-1;0;1;2;4\right\}\left(\sqrt{x}-2>-2\right)\\ \Leftrightarrow\sqrt{x}\in\left\{1;2;3;4;6\right\}\\ \Leftrightarrow x\in\left\{1;9;16;36\right\}\left(x\ne4\right)\\ h,x>4\Leftrightarrow\sqrt{x}-2>0\\ \Leftrightarrow E=\dfrac{\sqrt{x}-2}{2\sqrt{x}}>0\Leftrightarrow E\ge\sqrt{E}\)

a: \(B=\frac{\sqrt{x}}{x+\sqrt{x}}\)

\(=\frac{\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}=\frac{1}{\sqrt{x}+1}\)

\(A=\frac{1}{\sqrt{x}}+\frac{\sqrt{x}}{\sqrt{x}+1}=\frac{x+\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}\)

P=A:B

\(=\frac{x+\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}:\frac{1}{\sqrt{x}+1}=\frac{x+\sqrt{x}+1}{\sqrt{x}}\)

Thay x=4 vào P, ta được:

\(P=\frac{4+2+1}{2}=\frac72\)

b: A<=3B

=>\(\frac{x+\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}\le\frac{3}{\sqrt{x}+1}=\frac{3\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}\)

=>\(x+\sqrt{x}+1\le3\sqrt{x}\)

=>\(x-2\sqrt{x}+1\le0\)

=>\(\left(\sqrt{x}-1\right)^2\le0\)

=>\(\sqrt{x}-1=0\)

=>x=1(nhận)c

c: \(B-1=\frac{1}{\sqrt{x}+1}-1=\frac{1-\sqrt{x}-1}{\sqrt{x}+1}=\frac{-\sqrt{x}}{\sqrt{x}+1}<0\forall x\) thỏa mãn ĐKXĐ
=>B<1∀x thỏa mãn ĐKXĐ

=9900

9910