Tính \(M=\sqrt{11111115555556}\)
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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Lời giải:
Nếu .... là vô hạn thì:
$M=\sqrt{15-2M}$
$\Rightarrow M^2=15-2M$
$\Leftrightarrow M^2+2M-15=0$
$\Leftrightarrow (M-3)(M+5)=0$
$\Leftrightarrow M=3$ (do $M>0$)
![](https://rs.olm.vn/images/avt/0.png?1311)
Nhân tử và mẫu của biểu thức với \(\sqrt{m}+\sqrt{n}-\sqrt{m+n}.\)
\(\Rightarrow\frac{2\sqrt{mn}\left(\sqrt{m}+\sqrt{n}-\sqrt{m+n}\right)}{\left(\sqrt{m}+\sqrt{n}+\sqrt{m+n}\right)\left(\sqrt{m}+\sqrt{n}-\sqrt{m+n}\right)}\)
\(=\frac{2\sqrt{mn}\left(\sqrt{m}+\sqrt{n}-\sqrt{m+n}\right)}{\left(\sqrt{m}+\sqrt{n}\right)^2-\left(\sqrt{m+n}\right)^2}\)
\(=\frac{2\sqrt{mn}\left(\sqrt{m}+\sqrt{n}-\sqrt{m+n}\right)}{m+n+2\sqrt{mn}-m-n}=\sqrt{m}+\sqrt{n}-\sqrt{m+n}\)
Ta có: \(\frac{2\sqrt{mn}}{\sqrt{m}+\sqrt{n}+\sqrt{m+n}}=\frac{2\sqrt{mn}.\left(\sqrt{m}+\sqrt{n}-\sqrt{m+n}\right)}{(\sqrt{m}+\sqrt{n}+\sqrt{m+n})\left(\sqrt{m}+\sqrt{n}-\sqrt{m+n}\right)}\)
\(=\frac{2\sqrt{mn}.\left(\sqrt{m}+\sqrt{n}-\sqrt{m+n}\right)}{\left(\sqrt{m}+\sqrt{n}\right)^2-\left(\sqrt{m+n}\right)^2}=\frac{2\sqrt{mn}.\left(\sqrt{m}+\sqrt{n}-\sqrt{m+n}\right)}{m+2\sqrt{mn}+n-m-n}\)
\(=\frac{2\sqrt{mn}\left(\sqrt{m}+\sqrt{n}-\sqrt{m+n}\right)}{2\sqrt{mn}}=\sqrt{m}+\sqrt{n}-\sqrt{m+n}\)( đpcm )
Áp dụng: Với \(m=2\)và \(n=5\)và \(mn=10\); \(m+n=7\)ta có:
\(\frac{2\sqrt{10}}{\sqrt{2}+\sqrt{5}+\sqrt{7}}=\sqrt{2}+\sqrt{5}-\sqrt{2+5}=\sqrt{2}+\sqrt{5}-\sqrt{7}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
ĐKXĐ: \(x>0;x\ne1\)
\(M=\left(\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}-1}-\dfrac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}\right):\dfrac{\sqrt{x}+1}{x}\)
\(=\left(\sqrt{x}-\dfrac{1}{\sqrt{x}}\right).\dfrac{x}{\sqrt{x}+1}=\dfrac{\left(x-1\right)}{\sqrt{x}}.\dfrac{x}{\sqrt{x}+1}\)
\(=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)x}{\sqrt{x}\left(\sqrt{x}+1\right)}=\sqrt{x}\left(\sqrt{x}-1\right)\)
\(=x-\sqrt{x}=\left(\sqrt{x}-\dfrac{1}{2}\right)^2-\dfrac{1}{4}\ge-\dfrac{1}{4}\)
\(M_{min}=-\dfrac{1}{4}\) khi \(x=\dfrac{1}{4}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(=>M^2=4-\sqrt{10-2\sqrt{5}}+2\sqrt{\left(4-\sqrt{10-2\sqrt{5}}\right)\left(4+\sqrt{10-2\sqrt{5}}\right)}\)
\(+4+\sqrt{10-2\sqrt{5}}\)
\(M^2=8+2\)\(\sqrt{16-\left(\sqrt{10-2\sqrt{5}}\right)^2}\)\(=8+2\sqrt{16-10+2\sqrt{5}}\)
\(=>M^2=8+2\sqrt{6+2\sqrt{5}}=8+2\sqrt{\left(\sqrt{5}+1\right)^2}=8+2\sqrt{5}+2\)
\(=10+2\sqrt{5}\)
\(=>M=\sqrt{10+2\sqrt{5}}\)
thực hiện phép tính
\(\frac{\sqrt{m^3}+4\sqrt{mn^2}-4\sqrt{m^2n}}{\sqrt{m^2n}-2\sqrt{mn^2}},m>0,n>o\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{\sqrt{m^3}+4\sqrt{mn^2}-4\sqrt{m^2n}}{\sqrt{m^2n}-2\sqrt{mn^2}}=\frac{\sqrt{m}\left(m+4n-4\sqrt{m}\sqrt{n}\right)}{\sqrt{m}\left(\sqrt{mn}-2n\right)}=\frac{\left(\sqrt{m}-2\sqrt{n}\right)^2}{\sqrt{n}\left(\sqrt{m}-2\sqrt{n}\right)}=\frac{\sqrt{m}-2\sqrt{n}}{\sqrt{n}}\)
thực hiện phép tính
\(\frac{\sqrt{m^3}+4\sqrt{mn^2}-4\sqrt{m^2n}}{\sqrt{m^2n}-2\sqrt{mn^2}},m>0,n>0\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có :
\(\left(\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}\right)^2=4+\sqrt{7}-2\sqrt{\left(4+\sqrt{7}\right)\left(4-\sqrt{7}\right)}+4-\sqrt{7}\)
\(8-2\sqrt{16-7}=8-2\sqrt{9}=8-2.3=8-6=2\)
\(\Rightarrow\)\(\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}=\sqrt{2}\) ( vì \(\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}>0\) )
\(\Rightarrow\)\(M=\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}-\sqrt{2}=\sqrt{2}-\sqrt{2}=0\)
Vậy \(M=0\)
Chúc bạn học tốt ~
đề bài hỏi j vậy bạn . hình như có kết quả rồi mà
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