Ch A= 2n-5/3-n . Tìm giá trị nguyên của n để A là một số nguyên
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Hình tự vẽ! a) Vì AH là đường cao của \(\Delta ABC\Rightarrow\widehat{AHB}=\widehat{AHC}=90^0\)
Xét \(\Delta ABC\) có \(\widehat{AHC}=90^0\Rightarrow AH^2+HC^2=AC^2\) ( ĐL Pytago )
\(\Rightarrow AC^2=12^2+16^2=144+256=400=20^2\Rightarrow AC=20\left(cm\right)\)
b) Xét \(\Delta ABC\) có \(\widehat{AHB}=90^0\Rightarrow AH^2+HB^2=AB^2\) ( ĐL Pytago )
\(\Rightarrow HB^2=AB^2-AH^2=13^2-12^2=169-144=25=5^2\Rightarrow HB=5\left(cm\right)\)
\(BC=HB+HC=16+5=21\left(cm\right)\)
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( a + b ) ( a + b - c ) - ( a + b )2
=a2 + ab - ac + ab + b2 - bc - a2 - 2ab - b2
=( a2 - a2 ) + ( b2 - b2 ) + ( ab +ab - 2ab ) - ac - bc
=-ac - bc
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x2 +2x+2=0
x2+2x=-2
x(x+2)=-2
\(\Rightarrow\orbr{\begin{cases}x=-2\\x+2=-2\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-2\\x=-4\end{cases}}.}\)
Vậy x=-2 hoặc x=-4
\(x^2+2x+2=0\)
\(\Rightarrow\left(x^2+2x+1\right)+1=0\)
\(\Rightarrow\left(x^2+x+x+1\right)+1=0\)
\(\Rightarrow\left[x\left(x+1\right)+\left(x+1\right)\right]+1=0\)
\(\Rightarrow\left(x+1\right)\left(x+1\right)+1=0\)
\(\Rightarrow\left(x+1\right)^2+1=0\left(1\right)\)
Ta có :
\(\left(x+1\right)^2\ge0\forall x\)
\(\Rightarrow\left(x+1\right)^2+1\ge1\forall x\)
\(\Rightarrow\left(x+1\right)^2+1>0\forall x\left(2\right)\)
Từ (1) và (2).
\(\Rightarrow\)Mâu thuẫn nhau.
Do đó \(x=\varnothing\)
Vậy \(x=\varnothing\)
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Sửa lại đề như sau:
\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\)
BÀI LÀM:
\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\)
\(=\frac{1}{1}\times2+\frac{1}{2}\times3+\frac{1}{3}\times4+\frac{1}{4}\times5+\frac{1}{5}\times6+\frac{1}{6}\times7\)
\(=\left(1-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+\left(\frac{1}{4}-\frac{1}{5}\right)+\left(\frac{1}{5}-\frac{1}{6}\right)+\left(\frac{1}{6}-\frac{1}{7}\right)\)
\(=1-\frac{1}{7}\)
\(=\frac{6}{7}\)
Trả lời:
\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\)
\(=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}\)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\)
\(=\frac{1}{1}-\frac{1}{7}\)
\(=\frac{7}{7}-\frac{1}{7}\)
\(=\frac{6}{7}\)
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\(\frac{2n-5}{3-n}=\frac{2n-6+1}{3-n}=-2+\frac{1}{3-n}\inℤ\)
Suy ra \(\frac{1}{3-n}\inℤ\Rightarrow\left(3-n\right)\inƯ\left(1\right)=\left\{-1,1\right\}\Rightarrow n\in\left\{4,2\right\}\)