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4)a)\(\dfrac{x+5}{x-5}-\dfrac{x-5}{x+5}=\dfrac{20}{x^2-25}\)(1)
ĐKXĐ:\(\left\{{}\begin{matrix}x-5\ne0\\x+5\ne0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x\ne5\\x\ne-5\end{matrix}\right.\)
(1)\(\Rightarrow\left(x+5\right)\left(x+5\right)-\left(x-5\right)\left(x-5\right)=20\)
\(\Leftrightarrow x^2+10x+25-\left(x^2-10x+25\right)=20\)
\(\Leftrightarrow x^2+10x+25-x^2+10x-25=20\)
\(\Leftrightarrow x^2-x^2+10x+10x=-25+25=20\)
\(\Leftrightarrow20x=20\)
\(\Leftrightarrow x=1\left(nh\text{ậ}n\right)\)
S=\(\left\{1\right\}\)
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) \(\left(x+1\right)\left(x+2\right)+\left(x+2\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x+1+x-2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(2x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{1}{2}\end{matrix}\right.\)
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 \(x\ne7\))
\(\Leftrightarrow8-x-8x+56-1=0\)
\(\Leftrightarrow-9x+63=0\)
\(\Leftrightarrow x=7\)
a: \(=\dfrac{4a^2-4a+1-4a^2-2a+6a+3}{\left(2a-1\right)\left(2a+1\right)}\)
\(=\dfrac{4}{\left(2a-1\right)\left(2a+1\right)}\)
b: \(=\dfrac{x-1-x-1+2x^2}{\left(x-1\right)\left(x+1\right)}=2\)
d: \(=\dfrac{x-5+6x}{x\left(x+3\right)}=\dfrac{7x-5}{x\left(x+3\right)}\)
e: \(=\dfrac{x^2-4+3}{x-2}=\dfrac{x^2-1}{x-2}\)
i: \(=\dfrac{x}{x\left(x-4\right)}-\dfrac{3}{5x}=\dfrac{1}{x-4}-\dfrac{3}{5x}\)
\(=\dfrac{5x-3x+12}{5x\left(x-4\right)}=\dfrac{2x+12}{5x\left(x-4\right)}\)
a , \(16x^2+8x+1=\left(4x\right)^2+2.4x.1+1^2=\left(4x+1\right)^2\)
b , \(x^2-x+\dfrac{1}{4}=x^2-2.x.\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2=\left(x-\dfrac{1}{2}\right)^2\)
a,(4x+1)2 e,\(\left(\dfrac{3}{2}x-\dfrac{2}{5}\right)^2\)
b,(x-\(\dfrac{1}{2}\))2 g,\(\left(xy+1\right)^2\)
c,(\(x+\dfrac{3}{2}\))2 h,\(\left(x+5\right)^2\)
d,\(\left(x-\dfrac{5}{4}\right)^2\) i,\(-\left(x-6\right)^2\)
k,\(-\left(2x+3\right)^2\)
b.\(x^3-16x^2+64x=0\)
=>\(x^3-8x^2-8x^2+64x=0\)
=>\(x^2\left(x-8\right)-8x\left(x-8\right)=0\)
=>\(x\left(x-8\right)\left(x-8\right)=0\)
=>\(x=0\) và \(x-8=0\)
=>x=0 và x= 8
Vậy S={0; 8}
\(|6x-1|=2x+5\)
-Nếu 6x - 1 \(\ge0\Leftrightarrow x\ge\dfrac{1}{6}\)
\(|6x-1|=2x+5\)
\(\Leftrightarrow6x-1=2x+5\)
\(\Leftrightarrow6x-2x=5+1\)
\(\Leftrightarrow4x=6\)
\(\Leftrightarrow x=\dfrac{3}{2}\) (Loại)
-Nếu 6x-1 < 0 \(\Leftrightarrow x< \dfrac{1}{6}\)
\(|6x-1|=2x+5\)
\(\Leftrightarrow-6x+1=2x+5\)
\(\Leftrightarrow-6x-2x=5-1\)
\(\Leftrightarrow-8x=4\)
\(\Leftrightarrow x=-\dfrac{1}{2}\)(Nhận)
Vậy S={\(-\dfrac{1}{2}\)}
a. \(\dfrac{x+3}{x-3}-\dfrac{x-3}{x+3}=\dfrac{9}{x^2-9}\) (ĐKXĐ: \(x\ne\pm3\))
\(\Leftrightarrow\left(x+3\right)^2-\left(x-3\right)^2=9\)
\(\Leftrightarrow x^2+6x+9-x^2+6x-9=9\)
\(\Leftrightarrow12x=9\Leftrightarrow x=\dfrac{3}{4}\left(tm\right)\)
\(\Rightarrow S=\left\{\dfrac{3}{4}\right\}\)
b. \(\dfrac{x+2}{4}-x+3=\dfrac{1-x}{8}\)
\(\Leftrightarrow2\left(x+2\right)-8\left(x-3\right)=1-x\)
\(\Leftrightarrow2x+4-8x+24=1-x\)
\(\Leftrightarrow2x-8x+x=1-4-24\)
\(\Leftrightarrow-3x=-27\Leftrightarrow x=9\)
\(\Rightarrow S=\left\{9\right\}\)
-Mệt -.-
\(\Rightarrow x^2-x+1=5x\Leftrightarrow x^2+1=6x\)
\(\dfrac{x^2}{x^4+x^2+1}=\dfrac{x^2}{\left(x^2+1\right)^2-x^2}=\dfrac{x^2}{\left(6x\right)^2-x^2}=\dfrac{1}{35}\)
\(\dfrac{x}{x^2-x+1}=\dfrac{1}{5}\left(1\right)\)
có x^2 -x+1 > 0 mọi x => x>0
\(\left(1\right)\Leftrightarrow\dfrac{x^2-x+1}{x}=5\)
\(x+\dfrac{1}{x}=6;x^2+\dfrac{1}{x^2}=34\)
\(\left(1\right)\Leftrightarrow\dfrac{1}{E}=\left(\dfrac{x^4+x^2+1}{x^2}\right)=x^2+1+\dfrac{1}{x^2}=34+1=35\)
\(E=\dfrac{1}{35}\)