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ĐKXĐ: \(x\ne\left\{-3;-2;-1;0\right\}\)
\(\dfrac{1}{x\left(x+1\right)}+\dfrac{1}{\left(x+1\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+3\right)}=\dfrac{x}{x\left(x+3\right)}\)
\(\Leftrightarrow\dfrac{1}{x}-\dfrac{1}{x+1}+\dfrac{1}{x+1}-\dfrac{1}{x+2}+\dfrac{1}{x+2}-\dfrac{1}{x+3}=\dfrac{x}{x\left(x+3\right)}\)
\(\Leftrightarrow\dfrac{1}{x}-\dfrac{1}{x+3}=\dfrac{x}{x\left(x+3\right)}\)
\(\Leftrightarrow\dfrac{3}{x\left(x+3\right)}=\dfrac{x}{x\left(x+3\right)}\)
\(\Leftrightarrow x=3\)
Dấu ngoặc và cuối là sai nhé bạn. Phải là ngoặc vuông (x=0 hoặc x=-8) mới đúng, vì x không thể nhận 2 giá trị khác nhau cùng lúc.
=>8(x+1/x)^2+4[(x+1/x)^2-2]^2-4[(x+1/x)^2-2](x+1/x)^2=(x+4)^2
Đặt x+1/x=a(a>=2)
=>8a^2+4[a^2-2]^2-4[a^2-2]*a^2=(x+4)^2
=>8a^2+4a^4-16a^2+16-4a^4+8a^2=(x+4)^2
=>(x+4)^2=16
=>x+4=4 hoặc x+4=-4
=>x=-8;x=0
\(\Leftrightarrow\dfrac{2}{-x^2+6x-8}=\dfrac{x-1}{x-2}+\dfrac{x+3}{x-4}\\ \Leftrightarrow\left\{{}\begin{matrix}2=\left(-x^2+6x-8\right)\left(\dfrac{x-1}{x-2}+\dfrac{x+3}{x-4}\right)\\-x^2+6x-8\ne0\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}2=-2x^2+4x+2\\-x^2+6x-8\ne0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\x=2\\-x^2+6x-8\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left\{{}\begin{matrix}x=0\\-x^2+6x-8\ne0\end{matrix}\right.\\\left\{{}\begin{matrix}x=2\\-x^2+6x-8\ne\end{matrix}\right.\end{matrix}\right.\\\Rightarrow x=0\)
b: \(\Leftrightarrow\dfrac{20}{x}-\dfrac{20}{x+20}=\dfrac{1}{6}\)
=>\(\dfrac{20x+400-20x}{x\left(x+20\right)}=\dfrac{1}{6}\)
=>x*(x+20)=400*6=2400
=>x^2+20x-2400=0
=>(x+60)(x-40)=0
=>x=-60 hoặc x=40
c: \(\dfrac{2x+1}{2x-1}-\dfrac{2x-1}{2x+1}=\dfrac{8}{4x^2-1}\)
=>(2x+1)^2-(2x-1)^2=8
=>4x^2+4x+1-4x^2+4x-1=8
=>8x=8
=>x=1(nhận)
\(\Leftrightarrow56\left(x+1\right)+63\left(x+2\right)=72\left(x+3\right)+84\left(x+4\right)\)
\(\Leftrightarrow56\left(x+1\right)+63\left(x+2\right)-72\left(x+3\right)-84\left(x+4\right)=0\)
\(\Leftrightarrow-37x-370=0\Leftrightarrow x=-10\)
\(\frac{x+1}{9}+\frac{x+2}{8}=\frac{x+3}{7}+\frac{x+4}{6}\)
\(\Rightarrow\left(\frac{x+1}{9}+1\right)+\left(\frac{x+2}{8}+2\right)=\left(\frac{x+3}{7}+1\right)+\left(\frac{x+4}{6}+1\right)\)
\(\Rightarrow\frac{x+10}{9}+\frac{x+10}{8}-\frac{x+10}{7}-\frac{x+10}{6}=0\)
\(\Rightarrow\left(x+10\right)\left(\frac{1}{9}+\frac{1}{8}-\frac{1}{7}-\frac{1}{6}\right)=0\)
Mà \(\frac{1}{9}+\frac{1}{8}-\frac{1}{7}-\frac{1}{6}\ne0\)
\(\Rightarrow x+10=0\)
\(\Rightarrow x=-10\)
Vậy $x = -10$
a: \(\Leftrightarrow\left(1-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{17}+...+\dfrac{1}{49}-\dfrac{1}{57}\right)+2x-2=\dfrac{2}{3}x+\dfrac{7}{3}+\dfrac{5}{4}x-2\)
\(\Leftrightarrow\dfrac{56}{57}+2x-2=\dfrac{23}{12}x+\dfrac{1}{3}\)
=>1/12x=77/57
=>x=308/19
b: =>(x^2-4)(x^2-10)=72
=>x^4-14x^2+40-72=0
=>x^4-14x^2-32=0
=>(x^2-16)(x^2+2)=0
=>x^2-16=0
=>x^2=16
=>x=4 hoặc x=-4
a) \(x^2-10x=-25\)
\(\Leftrightarrow x^2-10x+25=0\)
\(\Leftrightarrow\left(x-5\right)^2=0\)
\(\Leftrightarrow x=5\)
b) \(\dfrac{x+4}{2000}+\dfrac{x+8}{1996}=\dfrac{x+12}{1992}+\dfrac{x+16}{1988}\)
\(\Leftrightarrow\dfrac{x+4}{2000}+1+\dfrac{x+8}{1996}+1=\dfrac{x+12}{1992}+1+\dfrac{x+16}{1988}+1\)
\(\Leftrightarrow\dfrac{x+2004}{2000}+\dfrac{x+2004}{1996}-\dfrac{x+2004}{1992}-\dfrac{x+2004}{1988}=0\)
\(\Leftrightarrow\left(x+2004\right)\left(\dfrac{1}{2000}+\dfrac{1}{1996}-\dfrac{1}{1992}-\dfrac{1}{1988}\right)=0\)
\(\Leftrightarrow x+2004=0\)(vì \(\dfrac{1}{2000}+\dfrac{1}{1996}-\dfrac{1}{1992}-\dfrac{1}{1988}\ne0\))
\(\Leftrightarrow x=-2004\)
điều kiện xác định : \(x\ne-2\)
ta có : \(\dfrac{2}{x+2}-\dfrac{2x^2+16}{x^3+8}=\dfrac{5}{x^2-2x+4}\)
\(\Leftrightarrow\dfrac{2}{x+2}-\dfrac{2x^2+16}{x^3+8}-\dfrac{5}{x^2-2x+4}=0\)
\(\Leftrightarrow\dfrac{2}{x+2}-\dfrac{2x^2+16}{\left(x+2\right)\left(x^2-2x+4\right)}-\dfrac{5}{x^2-2x+4}=0\)
\(\Leftrightarrow\dfrac{2\left(x^2-2x+4\right)}{\left(x+2\right)\left(x^2-2x+4\right)}-\dfrac{2x^2+16}{\left(x+2\right)\left(x^2-2x+4\right)}-\dfrac{5\left(x+2\right)}{\left(x+2\right)\left(x^2-2x+4\right)}=0\)
\(\Leftrightarrow\dfrac{2x^2-4x+8}{\left(x+2\right)\left(x^2-2x+4\right)}-\dfrac{2x^2+16}{\left(x+2\right)\left(x^2-2x+4\right)}-\dfrac{5x+10}{\left(x+2\right)\left(x^2-2x+4\right)}=0\)\(\Leftrightarrow\dfrac{2x^2-4x+8-2x^2-16-5x-10}{\left(x+2\right)\left(x^2-2x+4\right)}=0\)
\(\Leftrightarrow\dfrac{-9x-18}{\left(x+2\right)\left(x^2-2x+4\right)}=0\Leftrightarrow-9x-18=0\)
\(\Leftrightarrow-9x=18\Leftrightarrow x=-2\left(loại\right)\)
vậy phương trình vô nghiệm
Giải:
\(\dfrac{2}{x+2}-\dfrac{2x^2+16}{x^3+8}=\dfrac{5}{x^2-2x+4}\) (1)
ĐKXĐ: \(x\ne-2\)
\(\left(1\right)\Leftrightarrow\dfrac{2\left(x^2-2x+4\right)}{x^3+8}-\dfrac{2x^2+16}{x^3+8}=\dfrac{5\left(x+2\right)}{x^3+8}\)
\(\Rightarrow2\left(x^2-2x+4\right)-2x^2+16=5\left(x+2\right)\)
\(\Rightarrow2x^2-4x+8-2x^2+16=5x+10\)
\(\Rightarrow-4x-5x=10-8-16\)
\(\Rightarrow-9x=-14\)
\(\Rightarrow x=-\dfrac{14}{-9}=\dfrac{14}{9}\) (thoả mãn ĐKXĐ)
Vậy ...
Câu 1:
\(\left(x-2\right)\left(x^2+2x+4\right)+25x=x\left(x+5\right)\left(x-5\right)+8\)
\(\Leftrightarrow x^3-8+25x=x\left(x^2-25\right)+8\)
\(\Leftrightarrow x^3-8+25x=x^3-25x+8\)
\(\Leftrightarrow x^3-8+25x-x^3+25x-8=0\)
\(\Leftrightarrow50x-16=0\)
\(\Leftrightarrow50x=16\)
\(\Leftrightarrow x=\dfrac{8}{25}\)
Câu 2 :
\(\dfrac{x+5}{4}+\dfrac{3+2x}{3}=\dfrac{6x-1}{3}-\dfrac{1-2x}{12}\)
<=> \(\dfrac{3\left(x+5\right)}{12}+\dfrac{4\left(3+2x\right)}{12}=\dfrac{4\left(6x-1\right)}{12}-\dfrac{1-2x}{12}\)
<=>\(\dfrac{3x+15+12+8x}{12}=\dfrac{24x-4-1+2x}{12}\)
<=> 3x + 15 + 12 + 8x = 24x - 4 - 1 +2x
<=> 11x+27 = 26x -5
<=> ( 26x - 5 ) - ( 11x + 27 ) = 0
<=> 15x - 32 = 0
<=> 15x = 32
<=> x = \(\dfrac{32}{15}\)
- Đề lỗi phải khum ạ ?
Lời giải:
ĐKXĐ: $x\neq \pm 4$
PT $\Leftrightarrow \frac{8(x-4)+8(x+4)}{x^2-16}=\frac{25}{3}$
$\Leftrightarrow \frac{16x}{x^2-16}=\frac{25}{3}$
$\Rightarrow 48x=25x^2-400$
$\Leftrightarrow 25x^2-48x-400=0$
$\Leftrightarrow (5x-\frac{24}{5})^2=\frac{10576}{25}$
$\Rightarrow x=\frac{24\pm 4\sqrt{661}}{25}$ (đều thỏa mãn)