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Gọi phân thức cần tìm là \(A\)
Ta có:
\(\dfrac{1}{x}.\dfrac{x}{x+1}.\dfrac{x+1}{x+2}.\dfrac{x+2}{x+3}.\dfrac{x+3}{x+4}.\dfrac{x+4}{x+5}.\dfrac{x+5}{x+6}.\dfrac{x+6}{x+7}.\dfrac{x+7}{x+8}.\dfrac{x+8}{x+9}.\dfrac{x+9}{x+10}\)
\(=\dfrac{x\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)\left(x+6\right)\left(x+7\right)\left(x+8\right)\left(x+9\right)}{\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)\left(x+6\right)\left(x+7\right)\left(x+8\right)\left(x+9\right)\left(x+10\right)}\)\(=\dfrac{x}{x+10}\)
Suy ra:
\(\dfrac{1}{x}.\dfrac{x}{x+1}.\dfrac{x+1}{x+2}.\dfrac{x+2}{x+3}.\dfrac{x+3}{x+4}.\dfrac{x+4}{x+5}.\dfrac{x+5}{x+6}.\dfrac{x+6}{x+7}.\dfrac{x+7}{x+8}.\dfrac{x+8}{x+9}.\dfrac{x+9}{x+10}.A=1\)
\(\Leftrightarrow\dfrac{x}{x+10}.A=1\)
\(\Leftrightarrow A=\dfrac{x+10}{x}\)
Vậy phân thức cần điền vào chỗ trống là \(\dfrac{x+10}{x}\)
\(\begin{array}{l} n) \Leftrightarrow \dfrac{{x + 1}}{7} + 1 + \dfrac{{x + 2}}{6} + 1 = \dfrac{{x + 3}}{5} + 1 + \dfrac{{x + 4}}{4} + 1\\ \Leftrightarrow \dfrac{{x + 8}}{7} + \dfrac{{x + 8}}{6} - \dfrac{{x + 8}}{5} - \dfrac{{x + 8}}{4} = 0\\ \Leftrightarrow \left( {x + 8} \right)\underbrace {\left( {\dfrac{1}{7} + \dfrac{1}{8} - \dfrac{1}{5} - \dfrac{1}{6}} \right)}_{ < 0} = 0\\ \Leftrightarrow x + 8 = 0\\ \Leftrightarrow x = - 8 \end{array}\)
k/
\(8-\dfrac{x-2}{3}=\dfrac{x}{4}\)
\(\Leftrightarrow\dfrac{96}{12}-\dfrac{4\left(x-2\right)}{12}=\dfrac{3x}{12}\)
\(\Leftrightarrow96-4x+8=3x\)
\(\Leftrightarrow96-4x+8-3x=0\)
\(\Leftrightarrow104-7x=0\)
\(\Leftrightarrow7x=104\)
\(\Leftrightarrow x=104:7\)
\(\Leftrightarrow x=\dfrac{104}{7}\)
Vậy tập nghiệm của phương trình là \(S=\left\{\dfrac{104}{7}\right\}\)
m/
\(\dfrac{3x+2}{2}-\dfrac{3x+1}{6}=2x+\dfrac{5}{3}\)
\(\Leftrightarrow\dfrac{3\left(3x+2\right)}{6}-\dfrac{3x+1}{6}=\dfrac{12x}{6}+\dfrac{10}{6}\)
\(\Leftrightarrow9x+6-3x-1-12x-10=0\)
\(\Leftrightarrow-6x-5=0\)
\(\Leftrightarrow-6x=5\)
\(\Leftrightarrow x=-\dfrac{5}{6}\)
Vậy tập nghiệm của phương trình là \(S=\left\{-\dfrac{5}{6}\right\}\)
\(a,A=\dfrac{2x\left(x-3\right)+8\left(x+3\right)-2x-12}{\left(x-3\right)\left(x+3\right)}\cdot\dfrac{x+3}{x^2+6}\\ A=\dfrac{2x^2-6x+8x+24-2x-12}{\left(x-3\right)}\cdot\dfrac{1}{x^2+6}\\ A=\dfrac{2x^2+12}{\left(x-3\right)\left(x^2+6\right)}=\dfrac{2\left(x^2+6\right)}{\left(x-3\right)\left(x^2+6\right)}=\dfrac{2}{x-3}\)
\(b,A=5\Leftrightarrow\dfrac{2}{x-3}=5\Leftrightarrow5x-15=2\Leftrightarrow x=\dfrac{17}{5}\)
a: =>10x-4=15-9x
=>19x=19
hay x=1
b: \(\Leftrightarrow3\left(10x+3\right)=36+4\left(8x+6\right)\)
=>30x+9=36+32x+24
=>30x-32x=60-9
=>-2x=51
hay x=-51/2
c: \(\Leftrightarrow2x+\dfrac{6}{5}=5-\dfrac{13}{5}-x\)
=>3x=6/5
hay x=2/5
d: \(\Leftrightarrow\dfrac{7x}{8}-\dfrac{5\left(x-9\right)}{1}=\dfrac{20x+1.5}{6}\)
\(\Leftrightarrow21x-120\left(x-9\right)=4\left(20x+1.5\right)\)
=>21x-120x+1080=80x+60
=>-179x=-1020
hay x=1020/179
e: \(\Leftrightarrow5\left(7x-1\right)+60x=6\left(16-x\right)\)
=>35x-5+60x=96-6x
=>95x+6x=96+5
=>x=1
f: \(\Leftrightarrow6\left(x+4\right)+30\left(-x+4\right)=10x-15\left(x-2\right)\)
=>6x+24-30x+120=10x-15x+30
=>-24x+96=-5x+30
=>-19x=-66
hay x=66/19
