Tim x để M =\(\frac{x-5}{|5-x|}=1\)
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\(\left[\frac{-2}{5}x^3.\left(2x-1\right)^m+\frac{2}{5}x^{m+3}\right]:\left(\frac{-2}{5}x^3\right)\)
\(=\left[\frac{2}{5}x^3\left(2x+1\right)^m+\frac{2}{5}x^3.\left(\frac{2}{5}\right)^m\right]:\left(\frac{-2}{5}x^3\right)\)
\(=\left\{\frac{2}{5}x^3.\left[\left(2x+1\right)^m+\left(\frac{2}{5}\right)^m\right]\right\}:\left(\frac{-2}{5}x^3\right)\)
\(=\left\{\frac{2}{5}x^3.\left[2x+\frac{7}{5}\right]^m\right\}:\frac{-2}{5}x^3\)
\(=-\left(2x+\frac{7}{5}\right)^m\)
đến đây thì mình chịu
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\(\frac{x+2015}{5}+\frac{x+2016}{4}=\frac{x+2017}{3}+\frac{x+2018}{2}\)
\(\Leftrightarrow\left(\frac{x+2015}{5}+1\right)+\left(\frac{x+2016}{4}+1\right)=\left(\frac{x+2017}{3}+1\right)+\left(\frac{x+2018}{2}+1\right)\)
\(\Leftrightarrow\frac{x+2020}{5}+\frac{x+2020}{4}-\frac{x+2020}{3}-\frac{x+2020}{2}=0\)
\(\Leftrightarrow\left(x+2020\right)\left(\frac{1}{5}+\frac{1}{4}-\frac{1}{3}-\frac{1}{2}\right)=0\)
\(\Leftrightarrow x+2020=0\)vì \(\frac{1}{5}+\frac{1}{4}+\frac{1}{3}+\frac{1}{2}\ne0\)
\(\Leftrightarrow x=-2020\)
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3.(x-1/2) -5(x+3/5)=-x+1/5
3x - 3/2 -5x +3 = -x+1/5
3x-5x+x= 3/2-3+1/5
x.(3-5+1)=15/10 + (-30/10)+2/10
x.(-1)= -13/10
x = -13/10 : (-1)
x=13/10
vậy x=13/10
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-5.(x+1/5) -1/2.(x-2/3)=3/2x-5/6
-5x + (-1) -1/2x -1/3=3/2x-5/6
-5x-1/2x-3/2x=1+1/3-5/6
x.(-5-1/2-3/2)= 6/6+2/6+(-5/6)
x.(-10/2+(-1/2)+(-3/2))=3/6
x.6/2=1/2
x=1/2:6/2
x=1/6
Vậy x = 1/6
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a) ĐKXĐ: x∉{0;-5;5}
Ta có: \(M=\left(\frac{x}{x+5}-\frac{5}{5-x}+\frac{10x}{x^2-25}\right)\cdot\left(1-\frac{5}{x}\right)\)
\(=\left(\frac{x}{x+5}+\frac{5}{x-5}+\frac{10x}{x^2-25}\right)\cdot\frac{x-5}{x}\)
\(=\left(\frac{x\left(x-5\right)}{\left(x+5\right)\left(x-5\right)}+\frac{5\left(x+5\right)}{\left(x-5\right)\left(x+5\right)}+\frac{10x}{\left(x-5\right)\left(x+5\right)}\right)\cdot\frac{x-5}{x}\)
\(=\frac{\left(x+5\right)^2}{\left(x+5\right)\left(x-5\right)}\cdot\frac{x-5}{x}\)
\(=\frac{x+5}{x-5}\cdot\frac{x-5}{x}=\frac{x+5}{x}\)
b) Đặt \(M=\frac{1}{20}x+1\)
⇒\(\frac{x+5}{x}=\frac{x}{20}+1\)
⇒\(\frac{20\left(x+5\right)}{20x}-\frac{x^2}{20x}-\frac{20x}{20x}=0\)
\(\Leftrightarrow20x+100-x^2-20x=0\)
⇔\(100-x^2=0\)
⇔\(\left(10-x\right)\left(10+x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}10-x=0\\10+x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=10\left(tm\right)\\x=-10\left(tm\right)\end{matrix}\right.\)
Vậy: x∈{10;-10}
c) Để M là số nguyên thì x+5⋮x
mà x⋮x
nên 5⋮x
⇔x∈Ư(5)
⇔x∈{1;-1;5;-5}
mà x∉{5;-5}
nên x∈{1;-1}
Vậy: Khi x∈{1;-1} thì M là số nguyên
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có :
\(\frac{x+1}{49}+\frac{x+2}{48}+\frac{x+3}{47}+\frac{x+4}{46}+\frac{x+5}{45}=-5\)
\(\Leftrightarrow\)\(\left(\frac{x+1}{49}+1\right)+\left(\frac{x+2}{48}+1\right)+\left(\frac{x+3}{47}+1\right)+\left(\frac{x+4}{46}+1\right)+\left(\frac{x+5}{45}+1\right)=-5+5\)
\(\Leftrightarrow\)\(\frac{x+50}{49}+\frac{x+50}{48}+\frac{x+50}{47}+\frac{x+50}{46}+\frac{x+50}{45}=0\)
\(\Leftrightarrow\)\(\left(x+50\right)\left(\frac{1}{49}+\frac{1}{48}+\frac{1}{47}+\frac{1}{46}+\frac{1}{45}\right)=0\)
Vì \(\frac{1}{49}+\frac{1}{48}+\frac{1}{47}+\frac{1}{46}+\frac{1}{45}\ne0\)
Nên \(x+50=0\)
\(\Rightarrow\)\(x=-50\)
Vậy \(x=-50\)
Chúc bạn học tốt ~
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a) Để \(\frac{11}{\sqrt{x}-5}\)nhận giá trị nguyên thì \(\sqrt{\text{x}}-5\inƯ\left(11\right)\)(DK : \(0\le x\ne25\))
Vì \(\sqrt{\text{x}}-5\ge-5\)nên ta có :
\(\sqrt{x}-5\in\left\{-1;1;11\right\}\)\(\Rightarrow\sqrt{x}\in\left\{4;6;16\right\}\Rightarrow x\in\left\{16;36;256\right\}\)
b) \(B=\frac{\sqrt{x}+1}{\sqrt{x}-3}=\frac{\sqrt{x}-3+4}{\sqrt{x}-3}=1+\frac{4}{\sqrt{x}-3}\)(DK : \(0\le x\ne9\))
Để B nhận giá trị nguyên thì \(\sqrt{x}-3\inƯ\left(4\right)\)
Vì \(\sqrt{\text{x}}-3\ge-3\)nên ta có :
\(\sqrt{\text{x}}-3\in\left\{-2;-1;1;2;4\right\}\)\(\Rightarrow\sqrt{x}\in\left\{1;2;4;5;7\right\}\Rightarrow x\in\left\{1;4;16;25;49\right\}\)
Ta có : \(M=\frac{x-5}{\left|5-x\right|}=1\)
\(\Rightarrow x-5=\left|5-x\right|\)
Vì x - 5 là số đối của 5 - x
\(\Rightarrow5-x\le0\Rightarrow x\ge5\)
\(\Rightarrow x-5=5-x\)
\(\Leftrightarrow x+x=5+5\)
\(\Leftrightarrow x=5\)