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29: Ta có: \(\dfrac{1}{\sqrt{7}+\sqrt{5}}+\dfrac{2}{1-\sqrt{7}}\)
\(=\dfrac{\sqrt{7}-\sqrt{5}}{2}-\dfrac{2\sqrt{7}-2}{6}\)
\(=\dfrac{3\sqrt{7}-3\sqrt{5}-2\sqrt{7}+2}{6}\)
\(=\dfrac{-3\sqrt{5}-2}{6}\)
30: Ta có: \(\dfrac{4}{1-\sqrt{3}}+\dfrac{\sqrt{3}-1}{\sqrt{3}+1}\)
\(=\dfrac{-4\sqrt{3}-4}{2}+\dfrac{4-2\sqrt{3}}{2}\)
\(=\dfrac{-4\sqrt{3}-4+4-2\sqrt{3}}{2}=-3\sqrt{3}\)
31: Ta có: \(\dfrac{1}{\sqrt{2}-\sqrt{3}}-\dfrac{3}{\sqrt{18}+2\sqrt{3}}\)
\(=-\sqrt{3}-\sqrt{2}-\dfrac{3}{3\sqrt{2}+2\sqrt{3}}\)
\(=-\sqrt{3}-\sqrt{2}-\dfrac{9\sqrt{2}-6\sqrt{3}}{6}\)
\(=\dfrac{-6\sqrt{3}-6\sqrt{2}-9\sqrt{2}+6\sqrt{3}}{6}=\dfrac{-15\sqrt{2}}{6}\)
\(=\dfrac{-5\sqrt{2}}{2}\)
29.
\(=\frac{\sqrt{7}-\sqrt{5}}{(\sqrt{7}-\sqrt{5})(\sqrt{7}+\sqrt{5})}+\frac{2(1+\sqrt{7})}{(1-\sqrt{7})(1+\sqrt{7})}\)
\(=\frac{\sqrt{7}-\sqrt{5}}{7-5}+\frac{2(1+\sqrt{7})}{1-7}=\frac{\sqrt{7}-\sqrt{5}}{2}-\frac{1+\sqrt{7}}{3}=\frac{\sqrt{7}-3\sqrt{5}-2}{6}\)
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c, \(C=\left(2\sqrt{3}-5\sqrt{27}+4\sqrt{12}\right):\sqrt{3}\)
<=> \(C=\left(2\sqrt{3}-15\sqrt{3}+8\sqrt{3}\right):\sqrt{3}\)
<=> \(C=-5\sqrt{3}:\sqrt{3}=-5\)
e. \(\left(\sqrt{3-\sqrt{5}}+\sqrt{3+\sqrt{5}}\right)^2\)
\(=3-\sqrt{5}+3+\sqrt{5}+2\sqrt{\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)}\)
\(=6+2\sqrt{9-5}\)
\(=6+4=10\)
b. \(\left(\sqrt{3}+2\right)^2-\sqrt{75}\)
\(=3+4\sqrt{3}+4-5\sqrt{3}\)
\(=7-\sqrt{3}\)
d. \(\left(1+\sqrt{3}-\sqrt{2}\right)\left(1+\sqrt{3}+\sqrt{2}\right)\)
\(=\left(1+\sqrt{3}\right)^2-2\)
\(=1+2\sqrt{3}+3-2\)
\(=2+2\sqrt{3}\)
f. \(\sqrt{\left(\sqrt{3}+2\right)^2}-\sqrt{\left(\sqrt{3}-2\right)^2}\)
\(=\left|\sqrt{3}+2\right|-\left|\sqrt{3}-2\right|\)
\(=\sqrt{3}+2-2+\sqrt{3}\)
\(=2\sqrt{3}\)
c: Ta có: \(C=\left(2\sqrt{3}-5\sqrt{27}+4\sqrt{12}\right):\sqrt{3}\)
\(=\left(2\sqrt{3}-5\cdot3\sqrt{3}+4\cdot2\sqrt{3}\right):\sqrt{3}\)
\(=2-15+8=-5\)
d: Ta có: \(D=\left(\sqrt{3-\sqrt{5}}+\sqrt{3+\sqrt{5}}\right)^2\)
\(=3-\sqrt{5}+3+\sqrt{5}+2\cdot\sqrt{\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)}\)
\(=6+2\cdot2=10\)
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Bài 3:
Xét ΔABD và ΔAEB có
\(\widehat{ABD}=\widehat{AEB}\)
\(\widehat{BAD}\) chung
Do đó: ΔABD\(\sim\)ΔAEB
Suy ra: AB/AE=AD/AB
hay \(AB^2=AD\cdot AE\left(1\right)\)
Xét ΔOBA vuông tại B có BH là đường cao
nên \(AB^2=AH\cdot AO\left(2\right)\)
Từ (1) và (2) suy ra \(AD\cdot AE=AH\cdot AO\)
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Gọi tỉ lệ cắt giảm mỗi năm của công ty là \(x\left(\%\right)\) hay \(\dfrac{x}{100}\) với \(0\le x\le100\)
Số công nhân còn lại sau năm thứ nhất là:
\(10000\left(1-\dfrac{x}{100}\right)\)
Số công nhân còn lại sau năm thứ hai là:
\(10000.\left(1-\dfrac{x}{100}\right)\left(1-\dfrac{x}{100}\right)=10000.\left(1-\dfrac{x}{100}\right)^2\)
Do sau 2 năm công ty còn lại 9000 công nhân nên:
\(10000\left(1-\dfrac{x}{100}\right)^2=9000\)
\(\Rightarrow1-\dfrac{x}{100}=\dfrac{3}{\sqrt{10}}\Rightarrow x=100-30\sqrt{10}\) (%)\(\approx5,13\left(\%\right)\)
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1) Vì x=25 thỏa mãn ĐKXĐ nên Thay x=25 vào biểu thức \(A=\dfrac{\sqrt{x}-2}{x+1}\), ta được:
\(A=\dfrac{\sqrt{25}-2}{25+1}=\dfrac{5-2}{25+1}=\dfrac{3}{26}\)
Vậy: Khi x=25 thì \(A=\dfrac{3}{26}\)
2) Ta có: \(B=\dfrac{\sqrt{x}-3}{\sqrt{x}+1}+\dfrac{2x+8\sqrt{x}-6}{x-\sqrt{x}-2}\)
\(=\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}+\dfrac{2x+8\sqrt{x}-6}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{x-5\sqrt{x}+6+2x+8\sqrt{x}-6}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{3x+3\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{3\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{3\sqrt{x}}{\sqrt{x}-2}\)
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\(A=\dfrac{2\sqrt{x}+17}{\sqrt{x+5}}=\dfrac{2\sqrt{x}+10}{\sqrt{x}+5}+\dfrac{7}{\sqrt{x}+5}=2+\dfrac{7}{\sqrt{x}+5}\)
Để \(A\) ∈ \(Z\) thì \(\dfrac{7}{\sqrt{x}+5}\) phải ∈ \(Z\)
=> \(\sqrt{x}+5\) ∈ \(Ư\left(7\right)=\left\{-7;-1;1;7\right\}\)
# Với \(\sqrt{x}+5=-7=>\sqrt{x}=-12\)(Loại)
#Với \(\sqrt{x}+5=-1=>\sqrt{x}=-6\)(Loại)
#Với \(\sqrt{x}+5=1=>\sqrt{x}=-4\left(Loại\right)\)
#Với \(\sqrt{x}+5=7=>\sqrt{x}=2< =>x=4\left(Nhận\right)\)
Vậy \(x=4\) thì \(A\)∈\(Z\)
\(\sqrt[3]{\dfrac{a^4}{b^2\left(a^2-ab+b^2\right)}}+\sqrt[3]{\dfrac{b^4}{c^2\left(b^2-bc+c^2\right)}}\sqrt[3]{\dfrac{c^4}{a^2\left(c^2-ac+b^2\right)}}\) \(\text{≥}3\)
\(Ta\) \(Có\) : \(\sqrt[3]{\dfrac{a^4}{b^2\left(a^2-ab+b^2\right)}}=\sqrt[3]{\dfrac{a^6}{ab.ab\left(a^2-ab+b^2\right)}}=\dfrac{a^2}{\sqrt[3]{ab.ab.\left(a^2-ab+b^2\right)}}\)
\(Áp\) \(dụng\) \(bđt\) \(AM-GM\)
\(\sqrt[3]{ab.ab\left(a^2-ab+b^2\right)}\text{≤}\) \(\dfrac{ab+ab+a^2-ab+b^2}{3}\)
\(=>\dfrac{a^2}{\sqrt[3]{ab.ab\left(a^2-ab+b^2\right)}}\) \(\text{≥}\) \(\dfrac{3a^2}{a^2+ab+b^2}\) \(Hay\) \(\sqrt[3]{\dfrac{a^4}{b^2\left(a^2-ab+b^2\right)}}\text{≥}\dfrac{3a^2}{a^2+ab+b^2}\)
Tương tự ta cũng có :
\(\sqrt[3]{\dfrac{b^4}{c^2\left(b^2-bc+c^2\right)}}\text{≥}\dfrac{3b^2}{b^2+bc+c^2}\)
\(\sqrt[3]{\dfrac{c^4}{a^2\left(c^2-ac+a^2\right)}}\text{≥}\dfrac{3c^2}{a^2+ac+c^2}\)
\(=>\text{}\text{}\)\(\sqrt[3]{\dfrac{a^4}{b^2\left(a^2-ab+b^2\right)}}+\sqrt[3]{\dfrac{b^4}{c^2\left(b^2-bc+c^2\right)}}\sqrt[3]{\dfrac{c^4}{a^2\left(c^2-ac+b^2\right)}}\) \(\text{≥}\) \(3\left(\dfrac{a^2}{a^2+ab+b^2}+\dfrac{b^2}{b^2+bc+c^2}+\dfrac{c^2}{a^2+ac+c^2}\right)\)
Cần c/m \(\left(\dfrac{a^2}{a^2+ab+b^2}+\dfrac{b^2}{b^2+bc+c^2}+\dfrac{c^2}{a^2+ac+c^2}\right)\) ≥ \(1\)
Ta có : \(\dfrac{a^2}{a^2+ab+b^2}\text{≥}\dfrac{1}{3}\)
\(< =>3a^2\text{≥}a^2+ab+b^2\) \(< =>2a^2-b\left(a+b\right)\text{≥}0\) (1)
Lại có : \(a^2\text{≥}-b\left(a+b\right)\) (2)
Từ (1) và (2) => \(\dfrac{a^2}{a^2+ab+b^2}\text{≥}\dfrac{1}{3}\)
Tương tự ta cũng có :
\(\dfrac{b^2}{b^2+bc+c^2}\text{≥}\dfrac{1}{3}\)
\(\dfrac{c^2}{a^2+ac+c^2}\text{≥}\dfrac{1}{3}\)
Do đó \(\dfrac{a^2}{a^2+ab+b^2}+\dfrac{b^2}{b^2+bc+c^2}+\dfrac{c^2}{a^2+ac+c^2}\text{≥}1\)
Suy ra : \(\sqrt[3]{\dfrac{a^4}{b^2\left(a^2-ab+b^2\right)}}+\sqrt[3]{\dfrac{b^4}{c^2\left(b^2-bc+c^2\right)}}\sqrt[3]{\dfrac{c^4}{a^2\left(c^2-ac+b^2\right)}}\) \(\text{≥}\) \(3\)
Đẳng thức xảy ra <=> \(a=b=c=1\)
Chỗ nào khó hiểu, cứ hỏi mình nhe