Q=(\(\dfrac{1}{x+1}+\dfrac{3.\left(2x+1\right)}{x^3+1}-\dfrac{2}{x^2+1-x}\)):(x+2)
Tìm GTLN
Những câu hỏi liên quan
Chứng minh đẳng thức :
a) left(dfrac{x^2-2x}{2x^2+8}-dfrac{2x^2}{8-4x+2x^2-x^3}right)left(1-dfrac{1}{x}-dfrac{2}{x^2}right)dfrac{x+1}{2x}
b) left[dfrac{2}{3x}-dfrac{2}{x+1}left(dfrac{x+1}{3}-x-1right)right]:dfrac{x-1}{x}dfrac{2x}{x-1}
c) left[dfrac{2}{left(x+1right)^3}.left(dfrac{1}{x}+1right)+dfrac{1}{x^2+2x+1}left(dfrac{1}{x^2}+1right)right]:dfrac{x-1}{x^3}dfrac{x}{x-1}
Đọc tiếp
Chứng minh đẳng thức :
a) \(\left(\dfrac{x^2-2x}{2x^2+8}-\dfrac{2x^2}{8-4x+2x^2-x^3}\right)\left(1-\dfrac{1}{x}-\dfrac{2}{x^2}\right)=\dfrac{x+1}{2x}\)
b) \(\left[\dfrac{2}{3x}-\dfrac{2}{x+1}\left(\dfrac{x+1}{3}-x-1\right)\right]:\dfrac{x-1}{x}=\dfrac{2x}{x-1}\)
c) \(\left[\dfrac{2}{\left(x+1\right)^3}.\left(\dfrac{1}{x}+1\right)+\dfrac{1}{x^2+2x+1}\left(\dfrac{1}{x^2}+1\right)\right]:\dfrac{x-1}{x^3}=\dfrac{x}{x-1}\)
chứng minh rằng :
a) left(dfrac{x^2-2x}{2x^2+8}-dfrac{2x^2}{8-4x+2x^2-x}right)left(1-dfrac{1}{x}-dfrac{2}{x^2}right)dfrac{x+1}{2x}
b)left[dfrac{2}{3x}-dfrac{2}{x+1}left(dfrac{x+1}{3x}-x-1right)right]:dfrac{x+1}{x}dfrac{2x}{x-1}
c)left[dfrac{2}{left(x+1right)^3}left(dfrac{1}{x}+1right)+dfrac{1}{x^2+2x+1}left(dfrac{1}{x^2}+1right)right]:dfrac{x-1}{x^3}dfrac{x}{x-1}
Đọc tiếp
chứng minh rằng :
a) \(\left(\dfrac{x^2-2x}{2x^2+8}-\dfrac{2x^2}{8-4x+2x^2-x}\right)\left(1-\dfrac{1}{x}-\dfrac{2}{x^2}\right)=\dfrac{x+1}{2x}\)
b)\(\left[\dfrac{2}{3x}-\dfrac{2}{x+1}\left(\dfrac{x+1}{3x}-x-1\right)\right]:\dfrac{x+1}{x}=\dfrac{2x}{x-1}\)
c)\(\left[\dfrac{2}{\left(x+1\right)^3}\left(\dfrac{1}{x}+1\right)+\dfrac{1}{x^2+2x+1}\left(\dfrac{1}{x^2}+1\right)\right]:\dfrac{x-1}{x^3}=\dfrac{x}{x-1}\)
b: \(=\left[\dfrac{2}{3x}-\dfrac{2}{x+1}\cdot\dfrac{x+1-3x^2-3x}{3x}\right]\cdot\dfrac{x}{x+1}\)
\(=\left(\dfrac{2}{3x}-\dfrac{2}{x+1}\cdot\dfrac{-3x^2-2x+1}{3x}\right)\cdot\dfrac{x}{x+1}\)
\(=\dfrac{2x+2+6x^2+4x-2}{3x\left(x+1\right)}\cdot\dfrac{x}{x+1}\)
\(=\dfrac{6x^2+6x}{3\left(x+1\right)}\cdot\dfrac{1}{x+1}\)
\(=\dfrac{6x\left(x+1\right)}{3\left(x+1\right)^2}=\dfrac{2x}{x+1}\)
c: \(VT=\left[\dfrac{2}{\left(x+1\right)^3}\cdot\dfrac{x+1}{x}+\dfrac{1}{\left(x+1\right)^2}\cdot\dfrac{1+x^2}{x^2}\right]\cdot\dfrac{x^3}{x-1}\)
\(=\left(\dfrac{2}{x\left(x+1\right)^2}+\dfrac{x^2+1}{x^2\cdot\left(x+1\right)^2}\right)\cdot\dfrac{x^3}{x-1}\)
\(=\dfrac{2x+x^2+1}{x^2\cdot\left(x+1\right)^2}\cdot\dfrac{x^3}{x-1}\)
\(=\dfrac{\left(x+1\right)^2}{\left(x+1\right)^2}\cdot\dfrac{x}{x-1}=\dfrac{x}{x-1}\)
Đúng 0
Bình luận (0)
thuc hien phep tinh
a.left(dfrac{2x+1}{2x-1}-dfrac{2x-1}{2x+1}right):dfrac{4x}{10x-5}
b.left(dfrac{1}{x^2+1}-dfrac{2-x}{x+1}right):left(dfrac{1}{x}+1-2right)
c.dfrac{1}{x-1}-dfrac{x^3-x}{x^2+1}.left(dfrac{1}{x^2-2x+1}+dfrac{1}{1-x^2}right)
d.left(dfrac{x^2+xy}{x^3+x^2y+xy^2+y^3}+dfrac{y}{x^2+y^2}right):left(dfrac{1}{x-y}-dfrac{2xy}{x^3-x^2y+xy^2-y^3}right)
Đọc tiếp
thuc hien phep tinh
a.\(\left(\dfrac{2x+1}{2x-1}-\dfrac{2x-1}{2x+1}\right):\dfrac{4x}{10x-5}\)
b.\(\left(\dfrac{1}{x^2+1}-\dfrac{2-x}{x+1}\right):\left(\dfrac{1}{x}+1-2\right)\)
c.\(\dfrac{1}{x-1}-\dfrac{x^3-x}{x^2+1}.\left(\dfrac{1}{x^2-2x+1}+\dfrac{1}{1-x^2}\right)\)
d.\(\left(\dfrac{x^2+xy}{x^3+x^2y+xy^2+y^3}+\dfrac{y}{x^2+y^2}\right):\left(\dfrac{1}{x-y}-\dfrac{2xy}{x^3-x^2y+xy^2-y^3}\right)\)
a: \(=\dfrac{4x^2+4x+1-\left(4x^2-4x+1\right)}{\left(2x-1\right)\left(2x+1\right)}\cdot\dfrac{5\left(2x-1\right)}{4x}\)
\(=\dfrac{8x}{2x+1}\cdot\dfrac{5}{4x}=\dfrac{10}{2x+1}\)
c: \(=\dfrac{1}{x-1}-\dfrac{x\left(x-1\right)\left(x+1\right)}{x^2+1}\cdot\left(\dfrac{x+1-x+1}{\left(x-1\right)^2\cdot\left(x+1\right)}\right)\)
\(=\dfrac{1}{x-1}-\dfrac{x}{x^2+1}\cdot\dfrac{2}{\left(x-1\right)}=\dfrac{x^2+1-2x}{\left(x-1\right)\left(x^2+1\right)}=\dfrac{x-1}{x^2+1}\)
Đúng 0
Bình luận (0)
a) x-dfrac{dfrac{x}{2}-dfrac{3+x}{4}}{2}dfrac{2x-dfrac{10-7x}{3}}{3}-left(x-1right)
b) x^2-6x-2+dfrac{14}{x^2-6x+7}0
c) dfrac{8x^2}{3left(1-4x^2right)}dfrac{2x}{6x-3}-dfrac{1+8x}{4+8x}
d) dfrac{13}{left(2x+7right)left(x-3right)}+dfrac{1}{left(2x+7right)}dfrac{6}{x^2-9}
e) left(1-dfrac{2x-1}{x+1}right)^3+6left(1-dfrac{2x-1}{x+1}right)^2dfrac{12left(2x-1right)}{x+1}-20
Đọc tiếp
a) \(x-\dfrac{\dfrac{x}{2}-\dfrac{3+x}{4}}{2}=\dfrac{2x-\dfrac{10-7x}{3}}{3}-\left(x-1\right)\)
b) \(x^2-6x-2+\dfrac{14}{x^2-6x+7}=0\)
c) \(\dfrac{8x^2}{3\left(1-4x^2\right)}=\dfrac{2x}{6x-3}-\dfrac{1+8x}{4+8x}\)
d) \(\dfrac{13}{\left(2x+7\right)\left(x-3\right)}+\dfrac{1}{\left(2x+7\right)}=\dfrac{6}{x^2-9}\)
e) \(\left(1-\dfrac{2x-1}{x+1}\right)^3+6\left(1-\dfrac{2x-1}{x+1}\right)^2=\dfrac{12\left(2x-1\right)}{x+1}-20\)
b: Đặt \(x^2-6x-2=a\)
Theo đề, ta có: \(a+\dfrac{14}{a+9}=0\)
=>(a+2)(a+7)=0
\(\Leftrightarrow\left(x^2-6x\right)\left(x^2-6x+5\right)=0\)
=>x(x-6)(x-1)(x-5)=0
hay \(x\in\left\{0;1;6;5\right\}\)
c: \(\Leftrightarrow\dfrac{-8x^2}{3\left(2x-1\right)\left(2x+1\right)}=\dfrac{2x}{3\left(2x-1\right)}-\dfrac{8x+1}{4\left(2x+1\right)}\)
\(\Leftrightarrow-32x^2=8x\left(2x+1\right)-3\left(8x+1\right)\left(2x-1\right)\)
\(\Leftrightarrow-32x^2=16x^2+8x-3\left(16x^2-8x+2x-1\right)\)
\(\Leftrightarrow-48x^2=8x-48x^2+18x+3\)
=>26x=-3
hay x=-3/26
Đúng 0
Bình luận (0)
Tính
a)left(dfrac{left(x-1right)^2}{left(3x+x-1right)^2}-dfrac{1-2x^2+4x}{x^3-1}+dfrac{1}{x-1}right):dfrac{x^2+x}{x^2+1}
b)left(dfrac{3left(x+2right)}{2left(x^3+x^2+x+1right)}+dfrac{2x^2-x+10}{2left(x^3+x^2+x+1right)}right):left(dfrac{5}{x^2+1}+dfrac{3}{2left(x+1right)}-dfrac{3}{2left(x-1right)}right).dfrac{2}{x-1}
c)left(dfrac{x^2}{x^2-5x+6}+dfrac{x^2}{x^2-3x+2}right):dfrac{left(x-1right)left(x-3right)}{x^4+x^2+1}
Đọc tiếp
Tính
a)\(\left(\dfrac{\left(x-1\right)^2}{\left(3x+x-1\right)^2}-\dfrac{1-2x^2+4x}{x^3-1}+\dfrac{1}{x-1}\right):\dfrac{x^2+x}{x^2+1}\)
b)\(\left(\dfrac{3\left(x+2\right)}{2\left(x^3+x^2+x+1\right)}+\dfrac{2x^2-x+10}{2\left(x^3+x^2+x+1\right)}\right):\left(\dfrac{5}{x^2+1}+\dfrac{3}{2\left(x+1\right)}-\dfrac{3}{2\left(x-1\right)}\right).\dfrac{2}{x-1}\)
c)\(\left(\dfrac{x^2}{x^2-5x+6}+\dfrac{x^2}{x^2-3x+2}\right):\dfrac{\left(x-1\right)\left(x-3\right)}{x^4+x^2+1}\)
( \(\dfrac{x+1}{2\left(x-1\right)}+\dfrac{3}{x^2-1}-\dfrac{x+3}{2x+2}\) ). \(\dfrac{4x^2-4}{5}\)
\(\dfrac{x}{x-3}-\dfrac{x^2+3x}{2x+3}.\left(\dfrac{x+3}{x^2-3x}-\dfrac{x}{x^2-9}\right)\)
\(\left(\dfrac{x+1}{x}\right)^2:\left(\dfrac{x^2+3}{x^2}+\dfrac{2}{x+1}.\left(\dfrac{1}{x}+1\right)\right)\)
c/m biểu thức không phụ thuộc vào biến
Đúng 0
Bình luận (0)
a: \(=\dfrac{x^2+2x+1+6-x^2-2x+3}{2\left(x-1\right)\left(x+1\right)}\cdot\dfrac{2\left(x-1\right)\left(x+1\right)}{5}\cdot2\)
\(=\dfrac{10}{5}\cdot2=4\)
b: \(=\dfrac{x}{x-3}-\dfrac{x\left(x+3\right)}{2x+3}\cdot\dfrac{x^2+6x+9-x^2}{x\left(x-3\right)\left(x+3\right)}\)
\(=\dfrac{x}{x-3}-\dfrac{3}{x-3}=1\)
Đúng 0
Bình luận (0)
Thực hiện các phép tính sau:
a).left(dfrac{2x+1}{2x-1}-dfrac{2x-1}{2x+1}right):dfrac{4x}{10x-5}
b). left(dfrac{1}{x^2+1}-dfrac{2-x}{x+1}right):left(dfrac{1}{x}+x-2right)
c). dfrac{1}{x-1}-dfrac{x^3-x}{x^2+1}.left(dfrac{1}{x^2-2x+1}+dfrac{1}{1-x^2}right)
d). left(dfrac{x^2+xy}{x^3+x^2y+xy^2+y^3}+dfrac{y}{x^2+y^2}right):left(dfrac{1}{x-y}-dfrac{2xy}{x^3-x^2y+xy^2-y^3}right)
Đọc tiếp
Thực hiện các phép tính sau:
a).\(\left(\dfrac{2x+1}{2x-1}-\dfrac{2x-1}{2x+1}\right):\dfrac{4x}{10x-5}\)
b). \(\left(\dfrac{1}{x^2+1}-\dfrac{2-x}{x+1}\right):\left(\dfrac{1}{x}+x-2\right)\)
c). \(\dfrac{1}{x-1}-\dfrac{x^3-x}{x^2+1}.\left(\dfrac{1}{x^2-2x+1}+\dfrac{1}{1-x^2}\right)\)
d). \(\left(\dfrac{x^2+xy}{x^3+x^2y+xy^2+y^3}+\dfrac{y}{x^2+y^2}\right):\left(\dfrac{1}{x-y}-\dfrac{2xy}{x^3-x^2y+xy^2-y^3}\right)\)
a) \(\left(\dfrac{2x+1}{2x-1}-\dfrac{2x-1}{2x+1}\right):\dfrac{4x}{10x-5}\)
\(=\left(\dfrac{\left(2x+1\right)\left(2x+1\right)}{2x^2-1}-\dfrac{\left(2x-1\right)\left(2x-1\right)}{2x^2-1}\right):\dfrac{4x}{10x-5}\)
\(=\left(\dfrac{\left(2x+1\right)^2-\left(2x-1\right)^2}{2x^2-1}\right):\dfrac{4x}{10x-5}\)
\(=\left(\dfrac{\left(2x+1-2x-1\right)\left(2x+1+2x-1\right)}{2x^2-1}\right):\dfrac{4x}{10x-5}\)
\(=\dfrac{4x}{2x^2-1}.\dfrac{5\left(2x-1\right)}{4x}\)
\(=\dfrac{5}{2x+1}\)
b) \(\left(\dfrac{1}{x^2+1}-\dfrac{2-x}{x+1}\right):\left(\dfrac{1}{x}+x-2\right)\)
\(=\left(\dfrac{1}{x^2+1}-\dfrac{x\left(2-x\right)}{x\left(x+1\right)}\right):\left(\dfrac{1}{x}+x-2\right)\)
\(=\left(\dfrac{1-2x+x^2}{x^2+1}\right):\left(\dfrac{1}{x}+x-2\right)\)
\(=\left(\dfrac{1-2x+x^2}{x^2+1}\right):\left(\dfrac{1}{x}+\dfrac{x^2}{x}-\dfrac{2x}{x}\right)\)
\(=\left(\dfrac{1-2x+x^2}{x^2+1}\right):\left(\dfrac{x^2-2x+1}{x}\right)\)
\(=\dfrac{\left(x-1\right)^2}{x^2+1}.\dfrac{x}{\left(x-1\right)^2}\)
\(=\dfrac{x}{x^2+1}\)
c) d) Tự làm đi mình làm biếng quass >.< ^^
Đúng 0
Bình luận (0)
a x^2-x0 b x^2-2x0 c (x+1)(x+2)(2-x)(x+2)
d dfrac{1}{x-2}+3dfrac{3-x}{x-2} đ dfrac{8-x}{x-7}-8dfrac{1}{x-7}
e dfrac{x}{2left(x-3right)}+dfrac{x}{2left(x+1right)}dfrac{2x}{left(x+1right)left(x-3right)}
f 5+dfrac{76}{x^2-16}dfrac{2x-1}{x+4}-dfrac{3x-1}{4-x}
g dfrac{90}{x}-dfrac{36}{x-6}2 h dfrac{1}{x}+dfrac{1}{x+10}dfrac{1}{12} i dfrac{x+3}{x-3}-dfrac{1}{...
Đọc tiếp
a \(x^2-x=0\) b \(x^2-2x=0\) c (x+1)(x+2)=(2-x)(x+2)
d \(\dfrac{1}{x-2}+3=\dfrac{3-x}{x-2}\) đ \(\dfrac{8-x}{x-7}-8=\dfrac{1}{x-7}\)
e \(\dfrac{x}{2\left(x-3\right)}+\dfrac{x}{2\left(x+1\right)}=\dfrac{2x}{\left(x+1\right)\left(x-3\right)}\)
f \(5+\dfrac{76}{x^2-16}=\dfrac{2x-1}{x+4}-\dfrac{3x-1}{4-x}\)
g \(\dfrac{90}{x}-\dfrac{36}{x-6}=2\) h \(\dfrac{1}{x}+\dfrac{1}{x+10}=\dfrac{1}{12}\) i \(\dfrac{x+3}{x-3}-\dfrac{1}{x}=\dfrac{3}{x\left(x-3\right)}\)
k \(\dfrac{3}{x+2}-\dfrac{2}{x-2}+\dfrac{8}{x^2-4}=0\) l \(\dfrac{3}{x+2}-\dfrac{2}{x-3}=\dfrac{8}{\left(x-3\right)\left(x+2\right)}\)
m\(\dfrac{x}{2x+6}-\dfrac{x}{2x+2}=\dfrac{3x+2}{\left(x+1\right)\left(x+3\right)}\)
n \(\dfrac{x}{x+1}-\dfrac{2x-3}{1-x}=\dfrac{3x^2+5}{x^2-1}\) j \(\dfrac{5}{x+7}+\dfrac{8}{2x+14}=\dfrac{3}{2}\)
q \(\dfrac{x-1}{x}-\dfrac{1}{x+1}=\dfrac{2x-1}{x^2+x}\)
Cần gấp ạ
câu nào cũng ghi lại đề nha
a) \(x\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
b)\(x\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)
c) \(\left(x+1\right)\left(x+2\right)+\left(x+2\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x+1+x-2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(2x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{1}{2}\end{matrix}\right.\)
Đúng 0
Bình luận (0)
d) \(\dfrac{1}{x-2}+3-\dfrac{3-x}{x-2}=0\)
\(\Leftrightarrow\dfrac{1+3\left(x-2\right)-\left(3-x\right)}{x-2}=0\)
\(\Leftrightarrow\dfrac{1+3x-6-3+x}{x-2}=0\) ( đk \(x\ne2\) )
\(\Leftrightarrow4x-8=0\Rightarrow x=2\)
đ) \(\dfrac{8-x}{x-7}-8-\dfrac{1}{x-7}=0\)
\(\Leftrightarrow\dfrac{8-x-8\left(x-7\right)-1}{x-7}=0\) (đk \(x\ne7\))
\(\Leftrightarrow8-x-8x+56-1=0\)
\(\Leftrightarrow-9x+63=0\)
\(\Leftrightarrow x=7\)
Đúng 0
Bình luận (0)
e) \(\dfrac{x}{2\left(x-3\right)}+\dfrac{x}{2\left(x+1\right)}=\dfrac{2x}{\left(x+1\right)\left(x-3\right)}\) (MTC 2(x+1)(x-3) )
\(\Leftrightarrow\dfrac{x\left(x+1\right)+x\left(x-3\right)-2x.2}{2\left(x-3\right)\left(x+1\right)}=0\) (đk \(x\ne3vsx\ne-1\) )
\(\Leftrightarrow x^2+x+x^2-3x-4x=0\)
\(\Leftrightarrow2x^2-6x=0\)
\(\Leftrightarrow2x\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\)
Đúng 0
Bình luận (0)
Giải các phương trình :
a) dfrac{1}{x-1}-dfrac{3x^2}{x^3-1}dfrac{2x}{x^2+x+1}
b) dfrac{3}{left(x-1right)left(x-2right)}+dfrac{2}{left(x-3right)left(x-1right)}dfrac{1}{left(x-2right)left(x-3right)}
c) 1+dfrac{1}{x+2}dfrac{12}{8+x^3}
d) dfrac{13}{left(x-3right)left(2x+7right)}+dfrac{1}{2x+7}dfrac{6}{left(x-3right)left(x+3right)}
Đọc tiếp
Giải các phương trình :
a) \(\dfrac{1}{x-1}-\dfrac{3x^2}{x^3-1}=\dfrac{2x}{x^2+x+1}\)
b) \(\dfrac{3}{\left(x-1\right)\left(x-2\right)}+\dfrac{2}{\left(x-3\right)\left(x-1\right)}=\dfrac{1}{\left(x-2\right)\left(x-3\right)}\)
c) \(1+\dfrac{1}{x+2}=\dfrac{12}{8+x^3}\)
d) \(\dfrac{13}{\left(x-3\right)\left(2x+7\right)}+\dfrac{1}{2x+7}=\dfrac{6}{\left(x-3\right)\left(x+3\right)}\)
a)
Ta có:
cho nên x3 – 1 ≠ 0 khi x – 1 ≠ 0⇔ x ≠ 1
Vậy ĐKXĐ: x ≠ 1
Khử mẫu ta được:
Đúng 0
Bình luận (0)