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cho S=1-3+32-33+...+398-399
a. Chứng minh: S chia hêt cho 20
b. Rút gọn S, từ đó suy ra 3100 chia 4 dư 1
chịu
Mấy bài này dài vật vã ghê =)))))))))))))
1, a, \(\frac{3+4\sqrt{3}}{\sqrt{6}+\sqrt{2}-\sqrt{5}}\)
= \(\frac{\left(3+4\sqrt{3}\right)\left(\sqrt{6}+\sqrt{2}+\sqrt{5}\right)}{\left(\sqrt{6}+\sqrt{2}-\sqrt{5}\right)\left(\sqrt{6}+\sqrt{2}+\sqrt{5}\right)}\)
=\(\frac{\left(3+4\sqrt{3}\right)\left(\sqrt{6}+\sqrt{2}+\sqrt{5}\right)}{\left(\sqrt{6}+\sqrt{2}\right)^2-5}\)
=\(\frac{\left(3+4\sqrt{3}\right)\left(\sqrt{6}+\sqrt{2}+\sqrt{5}\right)}{8+4\sqrt{3}-5}\)
= \(\frac{\left(3+4\sqrt{3}\right)\left(\sqrt{6}+\sqrt{2}+\sqrt{5}\right)}{3+4\sqrt{3}}\)
=\(\sqrt{6}+\sqrt{2}+\sqrt{5}\)
b, M = \(\frac{\sqrt{3}\left(x-1\right)}{\sqrt{x^2}-x+1}\)(ĐKXĐ: \(x\ge0\))
= \(\frac{\sqrt{3}\left(x-1\right)}{x-x+1}\)
= \(\sqrt{3}\left(x-1\right)\)
Thay x = \(2+\sqrt{3}\)(TMĐK) vào M ta có:
M = \(\sqrt{3}\left(2+\sqrt{3}-1\right)=\sqrt{3}\left(1+\sqrt{3}\right)=3+\sqrt{3}\)
Vậy với x = \(2+\sqrt{3}\)thì M = \(3+\sqrt{3}\)
2, Mình chỉ giải câu a thôi nhé:
\(\sqrt{1+b}+\sqrt{1+c}\ge2\sqrt{1+a}\)
\(\Leftrightarrow\left(\sqrt{1+b}+\sqrt{1+c}\right)^2\ge\left(2\sqrt{1+a}\right)^2\)
\(\Leftrightarrow1+b+2\sqrt{\left(1+b\right)\left(1+c\right)}+1+c\ge4\left(1+a\right)\)
\(\Leftrightarrow2+b+c+2\sqrt{\left(1+b\right)\left(1+c\right)}\ge4\left(1+a\right)\left(1\right)\)
Vì \(\left(\sqrt{1+b}-\sqrt{1+c}\right)^2\ge0\)
\(\Rightarrow2+b+c\ge2\sqrt{\left(1+b\right)\left(1+c\right)}\left(2\right)\)
Từ \(\left(1\right),\left(2\right)\Rightarrow4+2\left(b+c\right)+2\sqrt{\left(1+b\right)\left(1+c\right)}\ge4\left(1+a\right)+2\sqrt{\left(1+b\right)\left(1+c\right)}\)
\(\Leftrightarrow4+2\left(b+c\right)\ge4\left(1+a\right)\)
\(\Leftrightarrow4+2\left(b+c\right)\ge4+4a\)
\(\Leftrightarrow2\left(b+c\right)\ge4a\)
\(\Leftrightarrow b+c\ge2a\)
4*. Thật ra cái này mình xài làm trội, làm giảm là được mà
Đặt A = \(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+....+\frac{1}{\sqrt{n}}\)
\(\frac{1}{2}A=\frac{1}{2\sqrt{2}}+\frac{1}{2\sqrt{3}}+....+\frac{1}{2\sqrt{n}}\)
\(\frac{1}{2}A=\frac{1}{\sqrt{2}+\sqrt{2}}+\frac{1}{\sqrt{3}+\sqrt{3}}+....+\frac{1}{\sqrt{n}+\sqrt{n}}\)
Ta có: \(\frac{1}{\sqrt{2}+\sqrt{2}}>\frac{1}{\sqrt{3}+\sqrt{2}}\)
\(\frac{1}{\sqrt{3}+\sqrt{3}}>\frac{1}{\sqrt{4}+\sqrt{3}}\)
+ .........................................................
\(\frac{1}{\sqrt{n}+\sqrt{n}}>\frac{1}{\sqrt{n+1}+\sqrt{n}}\)
Cộng tất cả vào
\(\Rightarrow\frac{1}{\sqrt{2}+\sqrt{2}}+\frac{1}{\sqrt{3}+\sqrt{3}}+...+\frac{1}{\sqrt{n}+\sqrt{n}}>\frac{1}{\sqrt{3}+\sqrt{2}}+\frac{1}{\sqrt{4}+\sqrt{3}}+...+\frac{1}{\sqrt{n+1}+\sqrt{n}}\)\(\frac{1}{2}A>\frac{\sqrt{3}-\sqrt{2}}{3-2}+\frac{\sqrt{4}-\sqrt{3}}{4-3}+...+\frac{\sqrt{n+1}-\sqrt{n}}{n+1-n}\)
\(\frac{1}{2}A>\sqrt{3}-\sqrt{2}+\sqrt{4}-\sqrt{3}+...+\sqrt{n+1}-\sqrt{n}\)
\(\frac{1}{2}A>\sqrt{n+1}-\sqrt{2}\)
\(A>2\sqrt{n+1}-2\sqrt{2}>2\sqrt{n+1}-3\)
\(A+1>2\sqrt{n+1}-3+1\)
\(A+1>2\sqrt{n+1}-2\)
\(A+1>2\left(\sqrt{n+1}-1\right)\)
Vậy ta có điều phải chứng minh.
Rút gọn bt:
Câu 1: a, \(\left(\sqrt{50}+\sqrt{48}-\sqrt{72}\right)2\sqrt{3}\)
b, \(\sqrt{25a}+2\sqrt{45a}-3\sqrt{80a}+2\sqrt{16a}\left(a\ge0\right)\)ư
Câu 2: Cho bt: P =\(\left(1+\frac{\sqrt{a}}{a+1}\right):\left(\frac{1}{\sqrt{a}-1}-\frac{2\sqrt{a}}{a\sqrt{a}+\sqrt{a}-a-1}\right)\)
a, Tìm ĐKXĐ . Rút gọn P
B, Tìm x nguyên để P có gt nguyên
c, Tìm GTNN của P với a >1
Câu 3: Giair các pt
a, \(\sqrt{\left(2x-1\right)^2}=4\)
b, \(\sqrt{4x+4}+\sqrt{9x+9}-8\sqrt{\frac{x+1}{16}}=5\)
\(A=2-x\sqrt{\frac{x\left(x-2\right)}{\left(x-2\right)^2}+\frac{1}{\left(x-2\right)^2}}=2-x\sqrt{\frac{\left(x-1\right)^2}{\left(x-2\right)^2}}\)
\(=2-x\cdot\frac{x-1}{x-2}=\frac{2x-4}{x-2}-\frac{x^2-x}{x-2}=\frac{-x^2+3x-4}{x-2}\)
\(B=\frac{2\sqrt{5}x}{x-2}\cdot\left|x-2\right|+\frac{3\sqrt{5}x^2}{x}=\frac{2\sqrt{5}x}{x-2}\cdot\left|x-2\right|+3\sqrt{5}x\)
Với 0 < x < 2 \(B=-2\sqrt{5}x+3\sqrt{5}x=\sqrt{5}x\)
Với x > 2 \(B=2\sqrt{5}x+3\sqrt{5}x=5\sqrt{5}x\)
\(C=\frac{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}{\sqrt{x}\left(\sqrt{x}+5\right)}+\sqrt{\frac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-5\right)^2}}=\frac{\sqrt{x}-5}{\sqrt{x}}+\left|\frac{\sqrt{x}-1}{\sqrt{x}-5}\right|\)
Với 0 < x < 1 \(C=\frac{\sqrt{x}-5}{\sqrt{x}}+\frac{\sqrt{x}-1}{\sqrt{x}-5}=\frac{x-10\sqrt{x}+25}{x\left(\sqrt{x}-5\right)}+\frac{x-\sqrt{x}}{x\left(\sqrt{x}-5\right)}=\frac{2x-11\sqrt{x}+25}{x\left(\sqrt{x}-5\right)}\)
Với 1 < x < 5 \(C=\frac{\sqrt{x}-5}{\sqrt{x}}-\frac{\sqrt{x}-1}{\sqrt{x}-5}=\frac{x-10\sqrt{x}+25}{x\left(\sqrt{x}-5\right)}-\frac{x-\sqrt{x}}{x\left(\sqrt{x}-5\right)}=\frac{-9\sqrt{x}+25}{x\left(\sqrt{x}-5\right)}\)
Với x > 5 \(C=\frac{\sqrt{x}-5}{\sqrt{x}}+\frac{\sqrt{x}-1}{\sqrt{x}-5}=\frac{x-10\sqrt{x}+25}{x\left(\sqrt{x}-5\right)}+\frac{x-\sqrt{x}}{x\left(\sqrt{x}-5\right)}=\frac{2x-11\sqrt{x}+25}{x\left(\sqrt{x}-5\right)}\)
\(a,\left(\sqrt{50}+\sqrt{48}-\sqrt{72}\right)2\sqrt{3}\)
\(=\left(5\sqrt{2}+4\sqrt{3}-6\sqrt{2}\right)2\sqrt{3}\)
\(=\left(4\sqrt{3}-\sqrt{2}\right)2\sqrt{3}\)
\(=24-2\sqrt{6}\)