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Ta có :
\(\left(2x-10\right)\cdot3+40=100\)
=> \(6x-30+40-100=0\)
=> \(x=15\)
Gọi số cần tìm là a
Theo bài ra ta có :
3( 2a - 10 ) + 40 = 100
3( 2a - 10 ) = 60
2x - 10 = 20
2x = 30
x = 15
Vậy x = 15 thỏa mãn đề bài
\(\dfrac{2}{\left(x+1\right)^2}-\dfrac{1}{x^2-1}\)
\(=\dfrac{2}{\left(x+1\right)^2}-\dfrac{1}{\left(x+1\right)\left(x-1\right)}\)
\(=\dfrac{2\left(x-1\right)}{\left(x+1\right)^2\left(x-1\right)}-\dfrac{x+1}{\left(x+1\right)^2\left(x-1\right)}\)
\(=\dfrac{2\left(x-1\right)-x-1}{\left(x+1\right)^2\left(x-1\right)}\)
\(=\dfrac{2x-2-x-1}{\left(x+1\right)^2\left(x-1\right)}\)
\(=\dfrac{x-3}{\left(x+1\right)^2\left(x-1\right)}\)
⇒Chọn B
\(\dfrac{2}{\left(x+1\right)^2}-\dfrac{1}{x^2-1}\\ =\dfrac{2}{\left(x+1\right)^2}-\dfrac{1}{\left(x-1\right)\left(x+1\right)}\\ =\dfrac{2.\left(x-1\right)-\left(x+1\right)}{\left(x+1\right)^2.\left(x-1\right)}\\ =\dfrac{2x-2-x-1}{\left(x+1\right)^2.\left(x-1\right)}\\ =\dfrac{x-3}{\left(x+1\right)^2\left(x-1\right)}\\ =>B\)
\(\dfrac{1}{x-3}-\dfrac{1}{x}=\dfrac{x-\left(x-3\right)}{x\left(x-3\right)}=\dfrac{x-x+3}{x\left(x-3\right)}=\dfrac{3}{x\left(x-3\right)}\)
\(B=\dfrac{1}{x^2-3x}+\dfrac{1}{x^2-9x+18}+\dfrac{1}{x^2-15x+54}+\dfrac{1}{x^2-21x+108}\)
\(=\dfrac{1}{x\left(x-3\right)}+\dfrac{1}{\left(x-3\right)\left(x-6\right)}+\dfrac{1}{\left(x-6\right)\left(x-9\right)}+\dfrac{1}{\left(x-9\right)\left(x-12\right)}\)
\(=\dfrac{1}{3}\left(\dfrac{3}{x\left(x-3\right)}+\dfrac{3}{\left(x-3\right)\left(x-6\right)}+\dfrac{3}{\left(x-6\right)\left(x-9\right)}+\dfrac{3}{\left(x-9\right)\left(x-12\right)}\right)\)
\(=\dfrac{1}{3}\left(-\dfrac{1}{x}+\dfrac{1}{x-3}-\dfrac{1}{x-3}+\dfrac{1}{x-6}-\dfrac{1}{x-6}+\dfrac{1}{x-9}-\dfrac{1}{x-9}+\dfrac{1}{x-12}\right)\)
\(=\dfrac{1}{3}\left(-\dfrac{1}{x}+\dfrac{1}{x-12}\right)\)
\(=\dfrac{1}{3}\cdot\dfrac{-\left(x-12\right)+x}{x\left(x-12\right)}\)
\(=\dfrac{4}{x\left(x-12\right)}\)
a: ĐKXĐ: x<>0; x<>1
\(P=\dfrac{x\left(x+1\right)}{\left(x-1\right)^2}:\dfrac{x^2-1+x+2-x^2}{x\left(x-1\right)}\)
\(=\dfrac{x\left(x+1\right)}{\left(x-1\right)^2}\cdot\dfrac{x\left(x-1\right)}{x+1}=\dfrac{x^2}{x-1}\)
b: |2x+1|=3
=>x=1(loại); x=-2(nhận)
Khi x=-2 thì P=4/-3=-4/3
c: P=-1/2
=>x^2/x-1=-1/2
=>2x^2=-x+1
=>2x^2+x-1=0
=>2x^2+2x-x-1=0
=>(x+1)(2x-1)=0
=>x=1/2; x=-1
a: \(\dfrac{x^2-3x+2}{x^2-1}=\dfrac{\left(x-2\right)\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}=\dfrac{x-2}{x+1}\)
\(a,VP=\dfrac{\left(x-1\right)\left(x-2\right)}{\left(x-1\right)\left(x+1\right)}=\dfrac{x-2}{x+1}=VP\\ b,VT=\dfrac{u\left(4u^2-1\right)}{5\left(1-2u\right)}=\dfrac{-u\left(1-2u\right)\left(1+2u\right)}{5\left(1-2u\right)}=\dfrac{-u\left(1+2u\right)}{5}=-\dfrac{2u^2+u}{5}=VP\)
Khi lấy \(x\) trừ đi \(\dfrac{1}{2}\) ta được số \(x - \dfrac{1}{2}\), sau đó nhân với \(\dfrac{1}{2}\) ta được số \(\left( {x - \dfrac{1}{2}} \right).\dfrac{1}{2}\).
Vì kết quả thu được là \(\dfrac{1}{8}\) nên ta có phương trình:
\(\left( {x - \dfrac{1}{2}} \right).\dfrac{1}{2} = \dfrac{1}{8}\)
\(x - \dfrac{1}{2} = \dfrac{1}{8}:\dfrac{1}{2}\)
\(x - \dfrac{1}{2} = \dfrac{1}{4}\)
\(x = \dfrac{1}{4} + \dfrac{1}{2}\)
\(x = \dfrac{3}{4}\).
Vậy \(x = \dfrac{3}{4}\).
\(\left(x-\dfrac{1}{2}\right)\times\dfrac{1}{2}=\dfrac{1}{8}\\ \Leftrightarrow\left(x-\dfrac{1}{2}\right)=\dfrac{1}{8}:\dfrac{1}{2}=\dfrac{1}{4}\\ \Leftrightarrow x=\dfrac{1}{4}+\dfrac{1}{2}=\dfrac{3}{4}\\ Vậy:x=\dfrac{3}{4}\)