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\(\sqrt{14+\sqrt{16900}}-\sqrt{19+\sqrt{900}}+\sqrt{45+\sqrt{3025}}\)
\(=\sqrt{14+\sqrt{130^2}}-\sqrt{19+\sqrt{30^2}}+\sqrt{45+\sqrt{55^2}}\)
\(=\sqrt{14+130}-\sqrt{19+30}+\sqrt{45+55}\)
\(=\sqrt{144}-\sqrt{49}+\sqrt{100}\)
\(=\sqrt{12^2}-\sqrt{7^2}+\sqrt{10^2}\)
\(=12-7+10\)
\(=5+10\)
\(=15\)
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\(=\left(100+900\right)+\left(800+200\right)+...+\left(900+100\right)\)(9 cặp)
\(=1000+1000+...+1000\)(9 số 1000)
\(=1000\times9\)
\(=9000\)
k mình nha
\(\sqrt{300+900.x}=2007\Rightarrow300+900.x=4028049\)
\(\Rightarrow900.x=4027749\Rightarrow x=4475,276667\)
Mk thật sự k chắc đâu đó! chúc bn hok tot~!
\(\sqrt{300+900.x}=2007\Rightarrow300+900.x=4028049\)
\(\Rightarrow900.x=4027749\Rightarrow x=4475,276667\)
Khai triển :
\(A=\frac{\sqrt{x}+1}{\sqrt{x}-3}=\frac{\sqrt{x}-3+4}{\sqrt{x}-3}=1+\frac{4}{\sqrt{x}-3}\)
Ta có :
A nguyên
<=> 1+\(\frac{4}{\sqrt{x}-3}\) nguyên
<=> \(\frac{4}{\sqrt{x}-3}\) nguyên
<=> \(\sqrt{x}-3\inƯ_{\left(4\right)}\)
<=> \(\sqrt{x}-3\in\left\{1;2;4;-1;-2;-4\right\}\)
<=> \(\sqrt{x}\in\left\{4;5;7;2;1;-1\right\}\)
Mà \(\sqrt{x}\ge0\forall x\)
=> \(\sqrt{x}\in\left\{4;5;7;2;1\right\}\)
=> \(x\in\left\{16;25;49;4;1\right\}\)
Vậy \(x\in\left\{16;25;49;4;1\right\}\)
\(A=\frac{\sqrt{x}+1}{\sqrt{x}-3}=\frac{\sqrt{x}-3+4}{\sqrt{x}-3}=\frac{\sqrt{x}-3}{\sqrt{x}-3}+\frac{4}{\sqrt{x}-3}=1+\frac{4}{\sqrt{x}-3}\in Z\)
\(\Rightarrow4⋮\sqrt{x}-3\)
\(\Rightarrow\sqrt{x}-3\inƯ\left(4\right)=\left\{\pm1;\pm2;\pm4\right\}\)
a) Có: \(\sqrt{9\cdot4}=\sqrt{36}=6\)
\(\sqrt{9}\cdot\sqrt{4}=3\cdot2=6\)
=> \(\sqrt{9\cdot4}=\sqrt{9}\cdot\sqrt{4}\)
b) \(\sqrt{16\cdot25}=\sqrt{400}=20\)
\(\sqrt{16}\cdot\sqrt{25}=4\cdot5=20\)
=> \(\sqrt{16\cdot25}=\sqrt{16}\cdot\sqrt{25}\)
c,d tương tự
ai đó giúp mk cái @Trần Việt Linh VS @Mai Phương aNH ơi
mn ơi, giúp vs
\(\sqrt{8100=90}\) \(\sqrt{3136=56}\)
\(\sqrt{3364=58}\) \(\sqrt{722500=850}\)
\(\sqrt{900=30}\)
\(\sqrt{3969=63}\)
\(\sqrt{900}:\sqrt{800}=30:20\sqrt{2}=\frac{3\sqrt{2}}{2}\)