= \(\dfrac{2x^3+5-x^3-4}{x^2-x+1}\) = \(\dfrac{x^3-1}{x^2-x+1}\)

\(\dfrac{2x^3+5 -x^3-4}{x^2-x+1}=\dfrac{x^3+1 }{x+1}\)

câu này luôn \(\dfrac{2x^2y^5}{xy^2z}-\dfrac{4x^2y^3}{xy^2z}\)

Tính M =  \(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{100}}\)

a/ \(\dfrac{x-1}{5}=\dfrac{1-2x}{3}\)

\(\Leftrightarrow3\left(x-1\right)=5\left(1-2x\right)\)

\(\Leftrightarrow3x-3=5-10x\)

\(\Leftrightarrow3x+10x=5+3\)

\(\Leftrightarrow13x=8\)

\(\Leftrightarrow x=\dfrac{8}{13}\)

Vậy ...

b/ \(\dfrac{3-\left|x\right|}{5}=1\dfrac{1}{2}:\dfrac{-6}{5}\)

\(\Leftrightarrow\dfrac{3-\left|x\right|}{5}=\dfrac{-5}{4}\)

\(\Leftrightarrow\left(3-\left|x\right|\right)4=5.\left(-5\right)\)

\(\Leftrightarrow\left(3-\left|x\right|\right).4=-25\)

\(\Leftrightarrow3-\left|x\right|=-6,25\)

\(\Leftrightarrow\left|x\right|=-3,25\)

\(\Leftrightarrow x\in\varnothing\)

\(\dfrac{x-1}{5}=\dfrac{1-2x}{3}\Rightarrow3x-3=5-10x\)

Áp dụng tính chất chuyển quế đổi giấu

3x+10x=5+3=8

13x=8

\(\Rightarrow\dfrac{8}{13}\)

b)\(\dfrac{3-|x|}{5}=1\dfrac{1}{2}chia\dfrac{-6}{5}=\dfrac{-5}{4}\)

3-/x/=5chia\(\dfrac{-5}{4}\)=-4

/x/=-4+3=-1

Mà /x/\(\ge0\Rightarrow x\in\varnothing\)

Tick em nha

a: \(=\dfrac{4xy+x^2-2xy+y^2}{2\left(x+y\right)\left(x-y\right)}\cdot\dfrac{2x}{x+y}-\dfrac{y}{x-y}\)

\(=\dfrac{x}{x-y}-\dfrac{y}{x-y}=1\)

b: \(=\dfrac{x^2+x-2-2x^2-2x}{\left(x-1\right)\left(x+1\right)}\cdot\dfrac{3\left(x+1\right)}{x}+\dfrac{4x^2+x+7}{x\left(x-1\right)}\)

\(=\dfrac{-x^2-x-2}{\left(x-1\right)}\cdot\dfrac{3}{x}+\dfrac{4x^2+x+7}{x\left(x-1\right)}\)

\(=\dfrac{4x^2+x+7-3x^2-3x-6}{x\left(x-1\right)}=\dfrac{x^2-2x+1}{x\left(x-1\right)}=\dfrac{x-1}{x}\)

c: \(=\dfrac{1}{x+4}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+6}+\dfrac{1}{x+6}-\dfrac{1}{x+7}\)

\(=\dfrac{x+7-x-4}{\left(x+7\right)\left(x+4\right)}=\dfrac{3}{\left(x+4\right)\left(x+7\right)}\)

a) \(\dfrac{x-4}{15}=\dfrac{5}{3}\)

\(\Leftrightarrow x-4=15.\dfrac{5}{3}\)

\(\Leftrightarrow x-4=25\)

\(\Leftrightarrow x=29\) thỏa \(x\inℤ\)

b) \(\dfrac{x}{4}=\dfrac{18}{x+1}\left(x\ne-1\right)\)

\(\Leftrightarrow x\left(x+1\right)=18.4\)

\(\Leftrightarrow x\left(x+1\right)=72\)

vì \(72=8.9=\left(-8\right).\left(-9\right)\)

\(\Leftrightarrow x\in\left\{8;-9\right\}\left(x\inℤ\right)\)

c) \(2x+3⋮x+4\) \(\left(x\ne-4;x\inℤ\right)\)

\(\Leftrightarrow2x+3-2\left(x+4\right)⋮x+4\)

\(\Leftrightarrow2x+3-2x-8⋮x+4\)

\(\Leftrightarrow-5⋮x+4\)

\(\Leftrightarrow x+4\in\left\{-1;1;-5;5\right\}\)

\(\Leftrightarrow x\in\left\{-5;-3;-9;1\right\}\)

ÁP dụng cái bất đẳng thức j j đó

mk có xem làm ở đâu rùi nhưng chưa học nên ko bt giải

\(1+\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+...+\dfrac{2}{x\left(x+1\right)}=1\dfrac{1989}{1991}\)

\(\Rightarrow2\left(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+...+\dfrac{1}{x\left(x+1\right)}\right)=\dfrac{3980}{1991}\)

\(\Rightarrow2\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{x\left(x+1\right)}\right)=\dfrac{3980}{1991}\)

\(\Rightarrow2\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{x}-\dfrac{1}{x+1}\right)=\dfrac{3980}{1991}\)

\(\Rightarrow2\left(1-\dfrac{1}{x+1}\right)=\dfrac{3980}{1991}\)

\(\Rightarrow1-\dfrac{1}{x+1}=\dfrac{3980}{1991}.\dfrac{1}{2}\)

\(\Rightarrow1-\dfrac{1}{x+1}=\dfrac{1990}{1991}\)

\(\Rightarrow\dfrac{1}{x+1}=1-\dfrac{1990}{1991}\)

\(\Rightarrow\dfrac{1}{x+1}=\dfrac{1}{1991}\)

\(\Rightarrow x+1=1991\)

\(\Rightarrow x=1990\)

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