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\(\frac{31}{2}\)\(.\)\(\frac{32}{2}\)\(.\)\(\frac{33}{2}\)\(....\)\(\frac{60}{2}\)
\(=\)\(\left[\left(31.32.33....60\right)\right]\)\(.\)\(\left(\frac{1.2.3....30}{2^{30}}\right)\)\(.\)\(\left(1.2.3....30\right)\)
\(=\)\(\left[\frac{\left(1.3.5....59\right).\left(2.4.6....60\right)}{2.4.6....60}\right]\)\(=\)\(1.3.5....59\)
Vậy \(\frac{31}{2}\)\(.\)\(\frac{32}{2}\)\(.\)\(\frac{33}{2}\)\(....\)\(\frac{60}{2}\)\(=\)\(1.3.5....59\)
ta có:Đặt A= \(1.3.5.....59=\frac{1.2.3.4.....59.60}{2.4.6.....60}\)
=\(\frac{1.2.3.....59.60}{2^{30}.\left(1.2.3.....30\right)}=\frac{31.32.....59.60}{2^{30}}\)
= \(\frac{31}{2}.\frac{32}{2}.....\frac{59}{2}.\frac{60}{2}\)
vì \(\frac{31}{2}.\frac{32}{2}.....\frac{59}{2}.\frac{60}{2}\) = \(\frac{31}{2}.\frac{32}{2}.....\frac{59}{2}.\frac{60}{2}\)
\(\Rightarrow\)A= \(\frac{31}{2}.\frac{32}{2}.....\frac{59}{2}.\frac{60}{2}\)
( Điều phải chứng minh)
toán nâng cao lớp 6 đấy bạn nha
1) \(\left|x-2\right|+2=x\)
\(\Leftrightarrow\left|x-2\right|=x-2\)
\(\Leftrightarrow x-2\ge0\Leftrightarrow x\ge2\)
2) \(x^2+5x+4=0\)
\(\Leftrightarrow x^2+4x+x+4=0\)
\(\Leftrightarrow x\left(x+4\right)+\left(x+4\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x+4\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+1=0\\x+4=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-1\\x=-4\end{cases}}\)
3) \(8\sqrt{x}=x^2\)
Bình phương hai vế, ta được: \(64x=x^4\)
\(\Leftrightarrow x^4-64x=0\)
\(\Leftrightarrow x\left(x^3-64\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x^3-64=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=4\end{cases}}\)
4) \(\frac{x+29}{31}-\frac{x+27}{33}=\frac{x+17}{43}-\frac{x+15}{45}\)
\(\Leftrightarrow\frac{x+29}{31}-\frac{x+27}{33}-\frac{x+17}{43}+\frac{x+15}{45}=0\)
\(\Leftrightarrow\frac{x+29}{31}+1-\frac{x+27}{33}-1-\frac{x+17}{43}-1+\frac{x+15}{45}+1=0\)
\(\Leftrightarrow\frac{x+60}{31}+\frac{x+60}{45}-\frac{x+60}{33}-\frac{x+60}{43}=0\)
\(\Leftrightarrow\left(x+60\right)\left(\frac{1}{31}+\frac{1}{45}-\frac{1}{33}-\frac{1}{43}\right)=0\)
\(\Leftrightarrow x+60=0\Leftrightarrow x=-60\)
5)\(\left|x-1\right|+3x=1\)
\(\Leftrightarrow\left|x-1\right|=1-3x\)(1)
* Nếu \(x\ge1\)thì \(\left(1\right)\Leftrightarrow x-1=1-3x\Leftrightarrow4x=2\Leftrightarrow x=\frac{1}{2}\left(L\right)\)
* Nếu \(x< 1\)thì \(\left(1\right)\Leftrightarrow1-x=1-3x\Leftrightarrow2x=0\Leftrightarrow x=0\left(TM\right)\)
Vậy x = 0
a) \(\frac{x}{-15}=\frac{-60}{x}\)
\(\Rightarrow x^2=900\)
\(\Rightarrow x=30\)
chọn câu nào thì giải thích giúp mik luôn nha