\(\frac{1}{x\left(x+1\right)}+\frac{1}{\left(x+1\right)\left(x+2\right)}+\frac{1}{\left(x+2\right)\left(x+3\right)}-\frac{1}{x}=\frac{1}{2017}\)

\(\Rightarrow\frac{1}{x}-\frac{1}{x+1}+\frac{x}{x+1}-\frac{1}{x+2}+\frac{1}{x+2}-\frac{1}{x+3}-\frac{1}{x}=\frac{1}{2017}\)

\(\Rightarrow-\frac{1}{x+3}=\frac{1}{2017}\)

\(\Rightarrow x+3=-2017\)

\(\Rightarrow x=-2020\)

\(\frac{1}{x\left(x+1\right)}+\frac{1}{\left(x+1\right)\left(x+2\right)}+\frac{1}{\left(x+2\right)\left(x+3\right)}-\frac{1}{x}=\frac{1}{2017}\)

<=>  \(\frac{1}{x}-\frac{1}{x+1}+\frac{1}{x+1}-\frac{1}{x+2}+\frac{1}{x+2}-\frac{1}{x+3}-\frac{1}{x}=\frac{1}{2017}\)

<=>  \(\frac{-1}{x+3}=\frac{1}{2017}\)

=>  \(x+3=-2017\)

<=>  \(x=-2020\)

Vậy...

a/ ĐKXĐ : \(x\ge0;x\ne1\)

\(P=\left(\frac{\sqrt{x}-2}{x-1}-\frac{\sqrt{x}+2}{x+2\sqrt{x}+1}\right):\frac{2}{x^2-2x+1}\)

\(=\left(\frac{\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\frac{\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2}\right):\frac{2}{\left(x-1\right)^2}\)

\(=\left(\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)}-\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)}\right).\frac{\left(x-1\right)^2}{2}\)

\(=\frac{x-2\sqrt{x}+\sqrt{x}-2-x+\sqrt{x}-2\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)}.\frac{\left(x-1\right)^2}{2}\)

\(=\frac{-2\sqrt{x}}{\left(x-1\right)\left(\sqrt{x}+1\right)}.\frac{\left(x-1\right)^2}{2}\)

\(=\frac{-2\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)\left(x-1\right)}{2\left(x-1\right)\left(\sqrt{x}+1\right)}\)

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

Vậy...

b/ Ta có :

\(P>0\)

\(\Leftrightarrow-\sqrt{x}\left(x-1\right)>0\)

\(\Leftrightarrow\sqrt{x}\left(x-1\right)< 0\)

Mà \(\sqrt{x}\ge0\)

\(\Leftrightarrow x-1< 0\Leftrightarrow x< 1\)

Kết hợp ĐKXĐ

Vậy \(0< x< 1\) thì P > 0

c/ Ta có :

\(x=7-4\sqrt{3}=\left(2-\sqrt{3}\right)^2\) thỏa mãn \(\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)

\(\Leftrightarrow\sqrt{x}=\left|2-\sqrt{3}\right|=2-\sqrt{3}\)

Thay vào P rồi bạn tự tính ra nhé :>

Băng Băng 2k6 giúp vs

nhầm đề ak

Xin phép được sủa đề một chút nhé :)

\(\left\{{}\begin{matrix}x+y=z=a\\x^2+y^2+z^2=b\\a^2=b+4034\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x^2+y^2+z^2+2\left(xy+yz+zx\right)=a^2\\x^2+y^2+z^2=b\\a^2-b=4034\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a^2-b=2\left(xy+yz+zx\right)\\a^2-b=4034\end{matrix}\right.\Leftrightarrow xy+yz+zx=2017\)

\(M=x\sqrt{\frac{\left(2017+y^2\right)\left(2017+z^2\right)}{2017+x^2}}+y\sqrt{\frac{\left(2017+x^2\right)\left(2017+z^2\right)}{2017+y^2}}+z\sqrt{\frac{\left(2017+y^2\right)\left(2017+x^2\right)}{2017+z^2}}\)

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

\(=2\left(xy+yz+zx\right)=4034\)

a, Xét : 3 - E = 3x^3-3xy-3y^3-x^3-xy-y^2/x^2-xy+y^2

= 2x^2-4xy+2y^2/x^2-xy+y^2

= 2.(x^2-2xy+y^2)/x^2-xy+y^2

= 2.(x-y)^2/x^2-xy+y^2 

>= 0 ( vì x^2-xy+y^2 > 0 )

Dấu "=" xảy ra <=> x-y=0 <=> x=y

Vậy ..........

b, Có : (x+1995)^2 = x^2+3990+1995^2 = (x^2-3990x+1995^2)+7980x

= (x-1995)^2 + 7980x >= 7980x

=> M < = x/7980x = 1/7980 ( vì x > 0 )

Dấu "=" xảy ra <=> x-1995=0 <=> x=1995

Vậy ...............

Bài 1:

Ta có : \(\left(x^2-1\right)\left(x^2-4\right)\left(x^2-7\right)\left(x^2-10\right)< 0\)

\(\Leftrightarrow\left[\left(x^2-1\right)\left(x^2-10\right)\right].\left[\left(x^2-4\right)\left(x^2-7\right)\right]< 0\)

\(\Leftrightarrow\left(x^4-11x^2+10\right)\left(x^4-11x^2+28\right)< 0\)

Đặt \(y=x^4-11x^2+19\), ta có : \(\left(y-9\right)\left(y+9\right)< 0\)

\(\Leftrightarrow y^2< 81\Leftrightarrow-9< y< 9\) \(\Leftrightarrow\hept{\begin{cases}y>-9\left(1\right)\\y< 9\left(2\right)\end{cases}}\)

Giải (1) được : \(x^4-11x^2+28>0\) \(\Leftrightarrow\left(x^2-7\right)\left(x^2-4\right)>0\)

\(\Leftrightarrow\orbr{\begin{cases}x^2>7\\x^2< 4\end{cases}}\) \(\Leftrightarrow\orbr{\begin{cases}x>\sqrt{7}\\x< -\sqrt{7}\end{cases}}\)hoặc  \(-2< x< 2\)

Giải (2) được : 

\(\Leftrightarrow\hept{\begin{cases}x^2< 1\\x^2>10\end{cases}}\)(loại)  hoặc \(1< x^2< 10\)(nhận)

\(\Rightarrow\hept{\begin{cases}x^2>1\\x^2< 10\end{cases}}\) \(\Leftrightarrow\orbr{\begin{cases}x< -1\\x>1\end{cases}}\)và \(-\sqrt{10}< x< \sqrt{10}\)

\(\Rightarrow\orbr{\begin{cases}-\sqrt{10}< x< -1\\1< x< \sqrt{10}\end{cases}}\)

Kết hợp (1) và (2) : \(-2< x< -1\);;\(1< x< 2\); \(\sqrt{7}< x< \sqrt{10}\); \(-\sqrt{10}< x< -\sqrt{7}\)

Suy ra các giá trị nguyên của x là : \(x\in\left\{-3;3\right\}\)

Bài 1: 

Có: \(x^2-10< x^2-7< x^2-4< x^2-1\)

Để tích trên < 0

: \(\left(x^2-1\right);\left(x^2-4\right);\left(x^2-7\right)\)cùng dương và \(\left(x^2-10\right)\)âm

\(\Rightarrow x^2-10< 0\)và\(x^2-7>0\)

\(\Rightarrow x^2< 10\)và \(x^2>7\)

\(\Rightarrow7< x^2< 10\)

\(\Rightarrow x^2=9\Rightarrow x=+;-3\)

Câu hỏi của Bui Cam Lan Bui - Toán lớp 7 - Học toán với OnlineMath