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Lời giải:
$x^2+x+1\vdots x+1$
$\Rightarrow x(x+1)+1\vdots x+1$
$\Rightarrow 1\vdots x+1$
$\Rightarrow x+1\in \left\{1; -1\right\}$
$\Rightarrow x\in \left\{0; -2\right\}$
x\(^2\)+x+1⋮x+1
=x(x+1)+1⋮x+1
=1⋮x+1
=x+1∈{1;−1}
=x∈{0;−2}
\(x-1^{x+2}=x-1^{x+6}\)
→x + 2 = x + 6
→Vậy không có giá trị của x thỏa mãn
`@` `\text {Ans}`
`\downarrow`
\(\left(1-\dfrac{1}{2}\right)\times\left(1-\dfrac{1}{3}\right)\times\left(1-\dfrac{1}{4}\right)\times\left(1-\dfrac{1}{5}\right)...\left(1-\dfrac{1}{1000}\right)\)
`=`\(\left(\dfrac{2}{2}-\dfrac{1}{2}\right)\times\left(\dfrac{3}{3}-\dfrac{1}{3}\right)\times\left(\dfrac{4}{4}-\dfrac{1}{4}\right)...\left(\dfrac{1000}{1000}-\dfrac{1}{1000}\right)\)
`=`\(\dfrac{1}{2}\times\dfrac{2}{3}\times\dfrac{3}{4}\times...\times\dfrac{999}{1000}\)
`=`\(\dfrac{1}{1000}\)
\(=\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}.\dfrac{4}{5}.......\dfrac{99}{100}=\dfrac{1}{100}\)
60-[15*X+4]=15/2:1/2
60-[15*X+4]=15
15*X+4=60-15
15*X+4=45
15*X=45-4
15*X=41
X=41:15
X=41/15
ko ghi đề
\(60-\left(15.x+4\right)=\frac{15}{2}.\frac{2}{1}\)
\(60-\left(15.x+4\right)=15\)
\(15.x+4=60-15\)
\(15.x+4=45\)
\(15.x=45-4\)
\(15.x=41\)
\(x=41:15\)
\(x=\text{2.7333}\)
Ta có : \(\frac{1}{2}+\frac{1}{14}+\frac{1}{35}+...+\frac{1}{x\left(x+3\right):2}=\frac{9}{14}\)
\(\Rightarrow2\left(\frac{1}{4}+\frac{1}{28}+\frac{1}{70}+...+\frac{1}{x\left(x+3\right)}\right)=\frac{9}{14}\)
\(\Rightarrow2.\frac{1}{3}\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{x\left(x+3\right)}\right)=\frac{9}{14}\)
\(\Rightarrow\frac{2}{3}\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{x}-\frac{1}{x+3}\right)=\frac{9}{14}\)
\(\Rightarrow1-\frac{1}{x+3}=\frac{27}{28}\)
\(\Rightarrow\frac{1}{x+3}=\frac{1}{28}\)
=> x + 3 = 28
=> x = 25
\(\Leftrightarrow2^x\cdot\dfrac{1}{8}+2^x\cdot\dfrac{1}{4}+2^x\cdot\dfrac{1}{2}=254\)
\(\Leftrightarrow2^x\cdot\dfrac{7}{8}=254\)
\(\Leftrightarrow2^x=\dfrac{2032}{7}\)
mà x là số tự nhiên
nên \(x\in\varnothing\)
\(C=\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}.....\dfrac{48}{49}.\dfrac{49}{50}=\dfrac{1}{50}\)
\(x-1=\left(x-1\right)^2\)
=> \(\left(x-1\right)-\left(x-1\right)^2=0\)
=> \(\left(x-1\right)\left(1-\left(x-1\right)\right)=0\)
=> \(\orbr{\begin{cases}x-1=0\\1-\left(x-1\right)=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=1\\x-1=1\end{cases}}\Rightarrow\orbr{\begin{cases}x=1\\x=2\end{cases}}\)
\(x-1=\left(x-1\right)^2\)
\(\Leftrightarrow\left(x-1\right)-\left(x-1\right)^2=0\)
\(\Leftrightarrow\left(x-1\right)\left[1-\left(x-1\right)\right]=0\)
\(\Leftrightarrow\left(x-1\right)\left(2-x\right)=0\Leftrightarrow x=1;2\)