Phân tích mẫu thức thành nhân tử ta có : 

1/(x+1)(x+2)+1/(x+2)(x+3)+...+1/(x+7)(x+8)=1/14

1/(x+1)-1/(x+2)+1/(x+2)-1/(x+3)+...+1/(x+7)-1/(x+8)=1/14

1/(x+1)-1/(x+8)=1/14

7/(x+1)(x+8)=1/14

Nhân chéo ta có x^2+9x+8=98

x^2+9x-90=0

(x+15)(x-6)=0

Suy ra x=-15 hoặc x=6

Bài này dễ mà!

Ta có : \(C=\frac{\frac{1}{3}-\frac{1}{7}-\frac{1}{13}}{\frac{2}{3}-\frac{2}{7}-\frac{2}{13}}\cdot\frac{\frac{1}{3}-0,25+0,2}{1\frac{1}{6}-0,875+0,7}+\frac{6}{7}\) 

\(\Rightarrow C=\frac{\frac{1}{3}-\frac{1}{7}-\frac{1}{13}}{2\left(\frac{1}{3}-\frac{1}{7}-\frac{1}{13}\right)}\cdot\frac{\frac{1}{3}-\frac{1}{4}+\frac{1}{5}}{\frac{7}{6}-\frac{7}{8}+\frac{7}{10}}+\frac{6}{7}\) 

\(\Rightarrow C=\frac{1}{2}\cdot\frac{2\left(\frac{1}{6}-\frac{1}{8}+\frac{1}{10}\right)}{7\left(\frac{1}{6}-\frac{1}{8}+\frac{1}{10}\right)}+\frac{6}{7}\)  

\(\Rightarrow C=\frac{1}{2}.\frac{2}{7}+\frac{6}{7}=\frac{1}{7}+\frac{6}{7}=1\) 

Nhớ t.i.c.k đúng nha!

khó v

bit lm bài này k giup tui

Như Ý lớp mấy mà thông minh thế

Bài đó dễ lắm mà!!!

hệ pt <=> 4x + 8y = 8 ; 5x-8y = 3

<=> 4x+8y+5x-8y = 11

<=> 9x = 11

<=> x=11/9

<=> y = 7/18

Vậy ............

Tk mk nha

qua de dang nhe

S=1/(1+2)+1/(1+2+3)+1/(1+2+3+4)+...+1/(1+2+3+4+...+10)

S=1/(2*3/2)+1/(3*4/2)+1/(4*5/2)+...+1/(10*11/2)

S=2(1/(2*3)+1/(3*4)+1/(4*5)+1/(5*6)+...+1/(10*11)

S=2(1/2-1/3+1/3-1/4+1/4-1/5+...+1/10-1/11)

S=2(1/2-1/11)

S=2*9/22

S=9/11

nho k cho minh voi nha

\(S=\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+...+\frac{1}{1+2+3+...+10}\)

\(S=\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{55}\)

\(S=\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{110}\)

\(S=2\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{110}\right)\)

\(S=2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{10.11}\right)\)

\(S=2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{10}-\frac{1}{11}\right)\)

\(S=2\left(\frac{1}{2}-\frac{1}{11}\right)\)

\(S=2.\frac{9}{22}\)

\(S=\frac{9}{11}\)

cho bài cm hình đi

vd như Cho hình bình hành ABCD. trên các cạnh AB, BC, CD, DA theo thứ tự lấy các điểm M, N, P, Q sao cho AM = BN = CP = DQ. Chứng minh tứ giác MNPQ là hình bình hành

Chả biết đề có đúng không nữa nhưng mà nếu thử x = 0 ; y = -1 thì VT = 1,5 > 1 :)

Đặt \(\left(\frac{1}{x};\frac{1}{y}\right)=\left(a;b\right)\).Từ gt\(\Rightarrow\frac{\left(x+1\right)\left(y+1\right)}{xy}=4\Rightarrow\left(1+\frac{1}{x}\right)\left(1+\frac{1}{y}\right)=4\)

\(\Rightarrow\left(a+1\right)\left(b+1\right)=4\Rightarrow ab+a+b=3\)

Mà \(ab+a+b=\left(\sqrt{a}-\sqrt{b}\right)^2+ab+2\sqrt{ab}\ge ab+2\sqrt{ab}\)

\(\Rightarrow3\ge ab+2\sqrt{ab}\Rightarrow4\ge\left(\sqrt{ab}+1\right)^2\)\(\Rightarrow\sqrt{ab}+1\le2\Rightarrow ab\le1\)

Ta có:\(\frac{1}{\sqrt{3x^2+1}}=\frac{\frac{1}{x}}{\sqrt{3+\frac{1}{x^2}}}=\frac{a}{\sqrt{ab+a+b+a^2}}=\frac{a}{\sqrt{\left(a+1\right)\left(a+b\right)}}\)\(\le\frac{1}{2}\left(\frac{a}{a+1}+\frac{a}{a+b}\right)\)

Tương tự:\(\frac{1}{\sqrt{3y^2+1}}\le\frac{1}{2}\left(\frac{b}{b+1}+\frac{b}{a+b}\right)\)

Cộng theo vế BĐT ta có:\(\frac{1}{\sqrt{3x^2+1}}+\frac{1}{\sqrt{3y^2+1}}\le\frac{1}{2}\left(\frac{a}{a+1}+\frac{a}{a+b}+\frac{b}{b+1}+\frac{b}{a+b}\right)\)

\(=\frac{1}{2}\left(\frac{a}{a+1}+\frac{b}{b+1}+1\right)=\frac{1}{2}\left[\frac{2ab+a+b}{\left(a+1\right)\left(b+1\right)}+1\right]\)\(\le\frac{1}{2}\left[\frac{ab+a+b+1}{\left(a+1\right)\left(b+1\right)}+1\right]=1\)
Dấu "=" xảy ra khi \(x=y=1\)

a) \(4\sqrt{x}+\frac{2}{\sqrt{x}}< 2x+\frac{1}{2x}+2\)

hay \(2\sqrt{x}+\frac{1}{\sqrt{x}}< x+\frac{1}{4x}+1\)

\(\Leftrightarrow0< x+\frac{1}{4x}+1-2\sqrt{x}-\frac{1}{\sqrt{x}}\)

\(\Leftrightarrow0< \left(\sqrt{x}\right)^2-2\sqrt{x}-2\sqrt{x}\cdot1+1+\frac{1}{\left(2\sqrt{x}\right)^2}-2\cdot\frac{1}{2\sqrt{x}}\)

\(\Leftrightarrow1< \left(\sqrt{x}-1\right)^2+\left(\frac{1}{2\sqrt{x}}-1\right)^2\)

\(\Rightarrow\hept{\begin{cases}x>0\\\sqrt{x}>1\\2\sqrt{x}>1\end{cases}\Rightarrow\hept{\begin{cases}x>1\\x>\frac{1}{4}\end{cases}\Rightarrow}x>1}\)

b) \(\frac{1}{1-x^2}>\frac{3}{\sqrt{1-x^2}}-1\left(1\right)\left(ĐK:-1< x< 1\right)\)

Ta có (1) <=> \(\frac{1}{1-x^2}-1-\frac{3x}{\sqrt{1-x^2}}+2>0\)\(\Leftrightarrow\frac{x^2}{1-x^2}-\frac{3x}{\sqrt{1-x^2}}+2>0\)

Đặt \(t=\frac{x}{\sqrt{1-x^2}}\)ta được

\(t^2-3t+2>0\Leftrightarrow\orbr{\begin{cases}\frac{x}{\sqrt{1-x^2}}< 1\\\frac{x}{\sqrt{1-x^2}}>2\end{cases}\Leftrightarrow\orbr{\begin{cases}\sqrt{1-x^2}>x\left(a\right)\\2\sqrt{1-x^2}< x\left(b\right)\end{cases}}}\)

(a) <=> \(\hept{\begin{cases}x< 0\\1-x^2>0\end{cases}\Leftrightarrow\hept{\begin{cases}x\ge0\\1-x^2>x^2\end{cases}}}\)

\(\Leftrightarrow-1< x< 0\)hoặc \(\hept{\begin{cases}x\ge0\\x^2< \frac{1}{2}\end{cases}}\)

\(\Leftrightarrow-1< x< 0\)hoặc \(0\le x\le\frac{\sqrt{2}}{2}\Leftrightarrow-1< x< \frac{\sqrt{2}}{2}\)

(b) \(\Leftrightarrow\hept{\begin{cases}1-x^2>0\\x>0\\4\left(1-x^2\right)< x^2\end{cases}\Leftrightarrow\hept{\begin{cases}0< x< 1\\x^2>\frac{4}{5}\end{cases}\Leftrightarrow}\frac{2}{\sqrt{5}}< x< 1}\)

ok đợi nấu ăn xong r làm cho

a) điều kiện x>0

khi đó

\(\left(a\right)\Leftrightarrow4\left(\sqrt{4}+\frac{1}{2\sqrt{x}}\right)< 2\left(\sqrt{x}+\frac{1}{2\sqrt{x}}\right)^2\)

\(\Leftrightarrow\sqrt{x}+\frac{1}{2\sqrt{x}}>2\Leftrightarrow2x-4\sqrt{x}+1>0\)

\(\Leftrightarrow\orbr{\begin{cases}\sqrt{x}< \frac{2-\sqrt{2}}{2}\\\sqrt{x}>\frac{2+\sqrt{2}}{2}\end{cases}}\)

\(\frac{1023}{1024}\)