giải phương trình với tham số a:
\(3x+\frac{x}{a}-\frac{3a}{a+1}=\frac{4ax}{\left(a+1\right)^2}+\frac{\left(2a+1\right)x}{a\left(a+1\right)^2}-\frac{3a^2}{\left(a+1\right)^3}\)
giải phương trình với tham số a:
\(3x+\frac{x}{a}-\frac{3a}{a+1}=\frac{4ax}{\left(a+1\right)^2}+\frac{\left(2a+1\right)x}{a\left(a+1\right)^2}-\frac{3a^2}{\left(a+1\right)^3}\)
Cho :\(A=\frac{1}{x\left(x+1\right)}+\frac{1}{\left(x+1\right)\left(x+2\right)}+\frac{1}{\left(x+2\right)\left(x+3\right)}+\frac{1}{x+3};B=\frac{a}{x\left(x+a\right)}+\frac{a}{\left(x+a\right)\left(x+2a\right)}+\frac{a}{\left(x+2a\right)\left(x+3a\right)}+\frac{1}{x+3a}\)CMR : A = B
Giải phương trình với tham số a:
\(3x+\dfrac{x}{a}-\dfrac{3a}{a+1}=\dfrac{4ax}{\left(a+1\right)^2}+\dfrac{\left(2a+1\right)x}{a\left(a+1\right)^2}-\dfrac{3a^2}{\left(a+1\right)^3}\).
1.Giải phương trình: \(\left(1+\frac{1}{x}\right)^3.\left(1+x^3\right)=16\)
2.Cho a,b,c là các số thực dương thỏa mãn abc=1. Chứng minh rằng:
\(\frac{1}{a^3.\left(7b+3c\right)}+\frac{1}{b^3.\left(7c+3a\right)}+\frac{1}{c^3.\left(7a+3b\right)}\ge\frac{1}{10}.\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)
3.Tìm tham số m để phương trình ẩn x sau \(\left(x^2+4x+12\right).\left(x^2+12x+20\right)=m\)có 4 nghiệm phân biệt
GIÚP MÌNH VỚI NHA
Giải các phương trình sau :
a) \(x^4-\left(x^2+2\right)=4\)
b) \(\frac{x+2}{x-2}-\frac{1}{x}=\frac{2}{x\left(x-2\right)}\)
c) \(\frac{2x-10}{4}=5+\frac{2-3x}{6}\)
d) \(\frac{2x}{\left(x-3\right)\left(x+1\right)}+\frac{x}{2\left(x-3\right)}=\frac{x}{2x+2}\)
e) \(\left(\frac{x+2}{x}\right)^2+\left(\frac{x}{x+2}\right)^2=2\)
f) \(\left(x-a\right)\left(x+a\right)+2x+a^2=-1\)
g) \(\frac{x-a}{2a}+\frac{x-2a}{3a}+\frac{x-3a}{4a}+\frac{x-4a}{5a}=-4\)
h) \(\left(x^2-3x+4\right)^2=\left(x^2-2x+3\right)\left(x^2-4x+5\right)\)
i) \(\frac{x^2-4x+12}{x^2-4x+6}=x^2-4x+8\)
ta có
\(\frac{\left(2018-x\right)^2+\left(2018-x\right)\left(x-2019\right)+\left(x-2019\right)^2}{\left(2018-x\right)^2-\left(2018-x\right)\left(x-2019\right)-\left(x-2019\right)^2}=\frac{19}{49}\) ( điều kiện : x khác : 2018;2019 )
đặt a = x - 2019 ( a khác 0 )
ta có hệ thức :
\(\frac{\left(a+1\right)^2-\left(a+1\right)a+a^2}{\left(a+1\right)^2+\left(a+1\right)a+a^2}=\frac{19}{49}\\ \Leftrightarrow\frac{a^2+a+1}{3a^2+3a+1}=\frac{19}{49}\)
\(\Leftrightarrow49\left(a^2+a+1\right)=19\left(3a^2+3a+1\right)\)
\(\Leftrightarrow49a^2+49a+49=57a^2+57a+19\)
\(\Leftrightarrow8a^2+8a-30=0\\ \left(2a+1\right)^2-4^2=0\\ \Leftrightarrow\left(2a+3\right)\left(2a+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=\frac{3}{2}\\a=-\frac{5}{2}\end{matrix}\right.\)( thỏa mãn điều kiện )
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{4041}{2}\\x=\frac{4033}{2}\end{matrix}\right.\)( thỏa mãn điều kiện )
vậy \(x\in\left\{\frac{4041}{2};\frac{4033}{2}\right\}\)
bài 1 giải phương trình
a) \(\frac{x+5}{x-1}=\frac{x+1}{x-3}-\frac{8}{x^2-4x+3}\)
B) \(\frac{2}{\left(1-3x\right)\left(3x+11\right)}=\frac{1}{9x^2-6x+1}-\frac{3}{\left(3x+11\right)^2}\)
Bài 2 cho ẩn z
\(\frac{z}{3z+z}-\frac{z}{z-3a}=\frac{a^2}{9a^2-z^2}\)
a) giải phương trình khi a=1
b) tìm cá giá trị a khi z=1
Rút gọn A = \(\left[\frac{\left(a-1\right)^2}{\left(a-1\right)^2+3a}+\frac{2a^2-4a-1}{a^3-1}+\frac{1}{a+1}\right]:\frac{2a}{3}\)
\(=\left[\dfrac{\left(a-1\right)^2}{a^2+a+1}+\dfrac{2a^2-4a-1}{\left(a-1\right)\left(a^2+a+1\right)}+\dfrac{1}{a-1}\right]:\dfrac{2a}{3}\)
\(=\dfrac{a^3-3a^2+3a-1+2a^2-4a-1+a^2+a+1}{\left(a-1\right)\left(a^2+a+1\right)}\cdot\dfrac{3}{2a}\)
\(=\dfrac{a^3-1}{\left(a-1\right)\left(a^2+a+1\right)}\cdot\dfrac{3}{2a}=\dfrac{3}{2a}\)
Giải các phương trình sau :
a) \(x^4-\left(x^2+2\right)=4\)
b) \(\frac{x+2}{x-2}-\frac{1}{x}=\frac{2}{x\left(x-2\right)}\)
c) \(\frac{2x-10}{4}=5+\frac{2-3x}{6}\)
d) \(\frac{2x}{\left(x-3\right)\left(x+1\right)}+\frac{x}{2\left(x-3\right)}=\frac{x}{2x+2}\)
e) \(\left(\frac{x+2}{x}\right)^2+\left(\frac{x}{x+2}\right)^2=2\)
f) \(\left(x-a\right)\left(x+a\right)+2x+a^2=-1\)
g) \(\frac{x-a}{2a}+\frac{x-2a}{3a}+\frac{x-3a}{4a}+\frac{x-4a}{5a}=-4\)
h) \(\left(x^2-3x+4\right)^2=\left(x^2-2x+3\right)\left(x^2-4x+5\right)\)
i ) \(\frac{x^2-4x+12}{x^2-4x+6}=x^2-4x+8\)
a) đặt \(t=x^2\) (t\(\ge0\))
=>\(t^2-t-2=0\)=>\(\orbr{\begin{cases}t=2\\t=-1\left(loại\right)\end{cases}}\)
=>\(x^2=2\)=>\(\orbr{\begin{cases}x=\sqrt{2}\\x=-\sqrt{2}\end{cases}}\)
a) \(\orbr{\begin{cases}x=\sqrt{3}\\x=-\sqrt{3}\end{cases}}\)
b)\(\orbr{\begin{cases}x=1\\x=-3\end{cases}}\)
c)\(x=\frac{47}{6}\)