x+1/3=x+2/4
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\(\Leftrightarrow\dfrac{\left(x+2\right)+5}{2-x}=\dfrac{2x-3}{\left(x-2\right)\left(x+2\right)}\\ \Leftrightarrow-\left(x+2\right)+5\left(x+2\right)=2x-3\\ \Leftrightarrow6x+12-2x+3=0\\ \Leftrightarrow4x+15=0\\ \Leftrightarrow x=\dfrac{-15}{4}\)
\(\dfrac{1}{x+2}+\dfrac{5}{2-x}=\dfrac{2x-3}{x^2-4}\)
\(\Leftrightarrow\dfrac{1}{x+2}-\dfrac{5}{x-2}-\dfrac{2x-3}{\left(x-2\right)\left(x+2\right)}=0\left(đk:x\ne\pm2\right)\)
\(\Leftrightarrow\dfrac{x-2-5\left(x+2\right)-2x-3}{\left(x-2\right)\left(x+2\right)}=0\)
\(\Leftrightarrow x-2-5x-10-2x-3=0\)
\(\Leftrightarrow-6x-15=0\)
\(\Leftrightarrow-6x=15\)
\(\Leftrightarrow x=-\dfrac{15}{6}\left(n\right)\)
Vậy \(S=\left\{-\dfrac{15}{6}\right\}\)
\(\dfrac{1}{x+2}+\dfrac{5}{2-x}=\dfrac{2x-3}{x^2-4}\) đkxđ : x khác 2 , x khác -2.
<=> \(\dfrac{1}{x+2}-\dfrac{5}{x-2}-\dfrac{2x-3}{x^2-4}=0\)
<=> \(\dfrac{1.\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}-\dfrac{5.\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\dfrac{2x-3}{\left(x-2\right)\left(x+2\right)}=0\)
<=> \(\dfrac{x-2}{\left(x+2\right)\left(x-2\right)}-\dfrac{5x+10}{\left(x-2\right)\left(x+2\right)}-\dfrac{2x-3}{\left(x-2\right)\left(x+2\right)}=0\)
<=>\(x-2-5x-10-2x+3=0\)
<=> \(-6x-9=0\)
<=> \(x=-\dfrac{9}{6}=-\dfrac{3}{2}\left(nhận\right)\)
Vậy pt có nghiệm \(S=\left\{-\dfrac{3}{2}\right\}\)
\(a)\)\(\left(50-6.x\right).18=2^3.3^2.5\)
\(\Leftrightarrow\)\(\left(50-6.x\right).18=8.9.5\)
\(\Leftrightarrow\)\(\left(50-6.x\right).18=360\)
\(\Leftrightarrow\)\(\left(50-6.x\right)=360\div18\)
\(\Leftrightarrow\)\(50-6.x=20\)
\(\Leftrightarrow\)\(6.x=50-20\)
\(\Leftrightarrow\)\(6.x=30\)
\(\Leftrightarrow\)\(x=5\)
\(b)\)\(\left(x+1\right)+\left(x+2\right)+\left(x+3\right)+...+\left(x+100\right)=7450\)
\(\Leftrightarrow\)\(100x+\left(1+2+3+...+100\right)=7450\)
\(\Leftrightarrow\)\(100x+5050=7450\)
\(\Leftrightarrow\)\(100x=7450-5050\)
\(\Leftrightarrow\)\(100x=2400\)
\(\Leftrightarrow\)\(x=24\)
b.
(x+1)+(x+2)+...+(x+100)=7450
=> 100x + (1+2+3+...+100)=7450
=>100x + (100+1).50=7450
=>100x=2400
=>x=24
\(x^4-3x^3-6x^2+3x+1\)
\(=x^4-2x^2+1-3x^3+3x-4x^2\)
\(=\left(x^2-1\right)^2-3x\left(x^2-1\right)-4x^2\)
đặt \(a=x^2-1\) khi đó biểu thức trở thành
\(a^2-3ax-4x^2\)
\(=a^2+ax-4ax-4x^2\)
\(=\left(a+x\right)\left(a-4x\right)\)
\(=\left(x^2+x-1\right)\left(x^2-4x+1\right)\)
a: \(\left|3x-2\right|=4\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-2=4\\3x-2=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-\dfrac{2}{3}\end{matrix}\right.\)
b: Ta có: \(\left|5x-3\right|=\left|x-7\right|\)
\(\Leftrightarrow\left[{}\begin{matrix}5x-3=x-7\\5x-3=7-x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}4x=-4\\6x=10\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=\dfrac{5}{3}\end{matrix}\right.\)
2/5 x 1/2 : 1/3 = 2/5 x 1/2 x 3 = 3/5
1/2 x 1/3 + 1/4= 1/6 + 1/4 = 4/24 + 6/24 = 10/24 =5/12
\(\left(x-1\right)^2=\left(x-1\right)^4\)
\(\Rightarrow\left(x-1\right)^4-\left(x-1\right)^2=0\)
\(\Rightarrow\left(x-1\right)^2\left[\left(x-1\right)^2-1\right]=0\)
\(\Rightarrow\left[{}\begin{matrix}x-1=0\\\left(x-1\right)^2=1\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=1\\x-1=1\\x-1=-1\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=1\\x=2\\x=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
x+1/3=x+2/4
\(\Rightarrow\)x-x=2/4-1/3\(\)
\(\Rightarrow\)0=2/4-1/3 ( Vô lí )
Vậy ko có x,y thỏa mãn yêu cấu đề bài