a , CM A = 1/1^2+1/2^2+......+1/50^2 < 2
b, Tinhs B = 3+3/2+3/2^2 +...+3/2^9
Những câu hỏi liên quan
cho a,b,c là các số thực dương thỏa mãn a^2+b^2+c^2=3
cm 1/(1+a^2b^2) +1/(1+b^2c^2) +1/(1+c^2a^2) >=9/(2a+2b+2c)
mong các thầy cô giúp em giải bài này với ạ
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Cho M =1/99+2/98+3/97+.........+99/1/1/2+1/3+1/4+.........+1/100 va N = 92-1/9-2/10-3/11-......-92/100/1/45+1/50+1/55+......+1/500. Tinhs M-2.N = ?
Khử mẫu của bthuc lấy căna)√3/2a^2b)√1/600 √11/540 √3/50 √5/98c)√(1-√3)^2/27d)√2/3e)√x^2/5f) √3/xg)√x^2- x^2/7h)ab√a/bi)a/b√a/b √1/b +1/b^2 √9a^3/36b 3ab√2/ab
Đọc tiếp
Khử mẫu của bthuc lấy căn
a)√3/2a^2
b)√1/600
√11/540
√3/50
√5/98
c)√(1-√3)^2/27
d)√2/3
e)√x^2/5
f) √3/x
g)√x^2- x^2/7
h)ab√a/b
i)a/b√a/b
√1/b +1/b^2
√9a^3/36b
3ab√2/ab
a: \(\sqrt{\dfrac{3}{2}a^2}=\left|a\right|\cdot\dfrac{\sqrt{6}}{2}\)
b: \(\sqrt{\dfrac{1}{600}}=\dfrac{1}{10\sqrt{6}}=\dfrac{\sqrt{6}}{60}\)
\(\sqrt{\dfrac{11}{540}}=\dfrac{\sqrt{165}}{90}\)
\(\sqrt{\dfrac{3}{50}}=\sqrt{\dfrac{6}{100}}=\dfrac{\sqrt{6}}{10}\)
\(\sqrt{\dfrac{5}{98}}=\sqrt{\dfrac{10}{196}}=\dfrac{1}{14}\cdot\sqrt{10}\)
c: \(\sqrt{\dfrac{\left(1-\sqrt{3}\right)^2}{27}}=\dfrac{\sqrt{3}-1}{3\sqrt{3}}=\dfrac{3-\sqrt{3}}{9}\)
d: căn 2/3=căn 6/9=1/3*căn 6
e: \(\sqrt{\dfrac{x^2}{5}}=\sqrt{\dfrac{5x^2}{25}}=\pm\dfrac{x\sqrt{5}}{5}\)
f: \(\sqrt{\dfrac{3}{x}}=\sqrt{\dfrac{3x}{x^2}}=\dfrac{\sqrt{3x}}{\left|x\right|}\)
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Cho a+b=1. CM \(\dfrac{a}{b^3-1}+\dfrac{b}{a^3-1}=\dfrac{2.\left(ab-2\right)}{a^2b^2+3}\)
Tính:
a, (3a^2-1/2)^3+(a^3+1/4)^2-(a+1)^3
b,(1/3a^2-1/2b).(1/3a^2-1/2b)-(a+1/2b)-(a+1/2b).(a^2-1/2ab)+1/4b^2
3x^4 + 3x^2y^2 + 6x^3y - 27x^2
x^4 + x^3 - x^2 + x
2x^5 - 6x^4 - 2a^2x^3 - 6ax^3
x^5 + x^4 + x^3 + x^2 + x + 1
x^3 - 1 + 5x^2 - 5 + 3x - 3
1/4.(a + 1)^2 - 4/9.(a - 2)^2
12a^2b^2 - 3.(a^2b^2)^2
4x^2y^2 - (x^2 + y^2 - a^2)^2
(a + b + c)^2 + (a + b - c)^2 - 4c^2
x^3 - 1 + 5x^2 - 5 + 3x - 3
Bài 1: Rút gọn
a)(x+9)(x-9)-x2
b)(10x-1)(10x+1)-(10x-1)2
c)(a+2b+3)(2a-2b-3)+(b-2c)2
d)(x-1)(x-2)-(x-2)(x+2)
a) (x+9)(x-9)-x2=x2-81-x2=-81
b) (10x-1)(10x+1)-(10x-1)2=100x2-1-100x2+20x-1=20x-2
d) (x-1)(x-2)-(x-2)(x+2)=x2-3x+2-x2+4=-3x+6
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Cho a+b=1
CM: \(\frac{a}{b^3-1}+\frac{b}{a^3-1}=\frac{2\left(ab-2\right)}{a^2b^2+3}\)
\(\frac{a}{b^3-1}+\frac{b}{a^3-1}=\frac{a}{\left(b-1\right)\left(b^2+b+1\right)}+\frac{b}{\left(a-1\right)\left(a^2+a+1\right)}\)
\(=\frac{a}{-a\left(b^2+b+1\right)}+\frac{b}{-b\left(a^2+a+1\right)}=-\frac{1}{b^2+b+1}-\frac{1}{a^2+a+1}\)
\(=-\frac{a^2+a+1+b^2+b+1}{\left(b^2+b+1\right)\left(a^2+a+1\right)}=-\frac{a^2+b^2+3}{a^2b^2+b^2a+b^2+ba^2+ab+b+a^2+a+1}\)
\(=-\frac{\left(a+b\right)^2-2ab+3}{a^2b^2+ab\left(a+b\right)+a^2+b^2+ab+\left(a+b\right)+1}\)
\(=\frac{2ab-4}{a^2b^2+2ab+\left(a+b\right)^2-2ab+2}=\frac{2\left(ab-2\right)}{a^2b^2+3}\)
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1) (a+b+c)(1/(a+b)+1/(b+c)+1/(c+a))>=9/2
2) (a+b)(a^2+b^2)/4<=(a^3+b^3)/2
3) a/(2a+b)+b/(2b+a)<=2/3
4) (a^2+2)(b^2+2)(c^2+2)>=abc16√2
Hộ mình nhanh nha
Thiếu \(a,b\ge0\) nhé
\(1)\) Cauchy-Schwarz dạng Engel :
\(\left(a+b+c\right)\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)\ge\frac{9\left(a+b+c\right)}{2\left(a+b+c\right)}=\frac{9}{2}\) ( đpcm )
\(2)\)
\(\frac{\left(a+b\right)\left(a^2+b^2\right)}{4}=\frac{a^3+b^3+ab^2+a^2b}{4}=\frac{a^3+b^3+ab\left(a+b\right)}{4}\)
Cần CM : \(a^3+b^3\ge ab\left(a+b\right)\)
\(\Leftrightarrow\)\(\left(a+b\right)\left(a^2-ab+b^2\right)-ab\left(a+b\right)\ge0\)
\(\Leftrightarrow\)\(\left(a+b\right)\left(a^2-ab+b^2-ab\right)=\left(a+b\right)\left(a-b\right)^2\ge0\) ( đúng )
\(\frac{a^3+b^3+ab\left(a+b\right)}{4}=\frac{2\left(a^3+b^3\right)}{4}=\frac{a^3+b^3}{2}\) ( đpcm )
3,4 làm sau
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