1. Cho \(a,b\in Z^+,\left(a^2+b^2\right)⋮\left(ab\right)\) Tính \(A=\frac{a^2+b^2}{ab}\)
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30 tháng 9 2017
Nhiều quá làm 1 bài tiêu biểu thôi nhé:
a/ \(A=\frac{\left(a+b\right)^2\left(b+c\right)^2\left(c+a\right)^2}{\left(1+a^2\right)\left(1+b^2\right)\left(1+c^2\right)}\)
\(=\frac{\left(a+b\right)^2\left(b+c\right)^2\left(c+a\right)^2}{\left(ab+bc+ca+a^2\right)\left(ab+bc+ca+b^2\right)\left(ab+bc+ca+c^2\right)}\)
\(=\frac{\left(a+b\right)^2\left(b+c\right)^2\left(c+a\right)^2}{\left(a+b\right)\left(c+a\right)\left(b+c\right)\left(a+b\right)\left(c+a\right)\left(b+c\right)}=1\)
13 tháng 7 2016
Ta có : \(\frac{a^2+b^2}{2}=ab\Rightarrow a^2+b^2=2ab\)
\(\Rightarrow a^2-ab+b^2=0\Rightarrow\left(a-b\right)^2=0\Rightarrow a=b\)
Tương tự : \(\frac{b^2+c^2}{2}=bc\Rightarrow b=c\)
\(\frac{a^2+c^2}{2}=ac\Rightarrow a=c\)
Áp dụng t/c bắc cầu ta dc : \(a=b=c\)
\(\Rightarrow\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=3a\times3=9a\)
13 tháng 7 2016
=>a2+b2=2ab
=>a2-2ab+b2=0
=>(a-b)2=0=>a=b
tương tự=>b=c
=>a=b=c
\(\Rightarrow\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=3a.3=9a\)
24 tháng 9 2016
\(\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=1+\frac{a}{b}+\frac{b}{c}+\frac{b}{a}+1+\frac{b}{c}+\frac{c}{a}+\frac{c}{b}+1\)
\(=\left(1+1+1\right)+\left(\frac{a}{b}+\frac{b}{a}\right)+\left(\frac{a}{c}+\frac{c}{a}\right)+\left(\frac{b}{c}+\frac{c}{b}\right)\)
\(=3+\frac{a^2+b^2}{ab}+\frac{a^2+c^2}{ac}+\frac{b^2+c^2}{bc}\)
\(=3+\frac{a^2+b^2}{\frac{a^2+b^2}{2}}+\frac{a^2+c^2}{\frac{a^2+c^2}{2}}+\frac{b^2+c^2}{\frac{b^2+c^2}{2}}\)
\(=3+2+2+2=9\)
21 tháng 2 2022
Câu 3:
a: \(G=\dfrac{a^2}{b\left(a+b\right)}-\dfrac{b^2}{a\left(a-b\right)}+\dfrac{-\left(a^2+b^2\right)}{ab}\)
\(=\dfrac{a^3\left(a-b\right)-b^3\left(a+b\right)-\left(a^2+b^2\right)\left(a^2-b^2\right)}{ab\left(a-b\right)\left(a+b\right)}\)
\(=\dfrac{a^4-a^3b-ab^3-b^4-a^4+b^4}{ab\left(a-b\right)\left(a+b\right)}\)
\(=\dfrac{-ab\left(a^2+b^2\right)}{ab\left(a-b\right)\left(a+b\right)}=\dfrac{-a^2-b^2}{a^2-b^2}\)
b: \(\dfrac{a}{b}=\dfrac{a+1}{b+5}\)
nên ab+5a=ab+b
=>5a=b
\(G=\dfrac{-a^2-\left(5a\right)^2}{a^2-\left(5a\right)^2}=\dfrac{-a^2-25a^2}{a^2-25a^2}=\dfrac{-26}{-24}=\dfrac{13}{12}\)
2 tháng 10 2021
\(P=\left(\dfrac{3\sqrt{a}}{a+\sqrt{ab}+\sqrt{b}}-\dfrac{3a}{a\sqrt{a}-b\sqrt{b}}+\dfrac{1}{\sqrt{a}-\sqrt{b}}\right):\dfrac{\left(a-1\right)\left(\sqrt{a}-\sqrt{b}\right)}{2a+2\sqrt{ab}+2b}\left(đk:a\ne b,a\ge0,b\ge0\right)\)
\(=\dfrac{3a-3\sqrt{ab}-3a+a+\sqrt{ab}+b}{\left(\sqrt{a}-\sqrt{b}\right)\left(a+\sqrt{ab}+\sqrt{b}\right)}.\dfrac{2\left(a+\sqrt{ab}+b\right)}{\left(a-1\right)\left(\sqrt{a}-\sqrt{b}\right)}\)
\(=\dfrac{a-2\sqrt{ab}+b}{\sqrt{a}-\sqrt{b}}.\dfrac{2}{\left(a-1\right)\left(\sqrt{a}-\sqrt{b}\right)}=\dfrac{\left(\sqrt{a}-\sqrt{b}\right)^2.2}{\left(\sqrt{a}-\sqrt{b}\right)^2\left(a-1\right)}=\dfrac{2}{a-1}\in Z\)
\(\Rightarrow a-1\inƯ\left(2\right)=\left\{-2;-1;1;2\right\}\)
Do \(a\ge0\)
\(\Rightarrow a\in\left\{0;2;3\right\}\)
2 tháng 10 2021
Ta có: \(P=\left(\dfrac{3\sqrt{a}}{a+\sqrt{ab}+b}-\dfrac{3a}{a\sqrt{a}-b\sqrt{b}}+\dfrac{1}{\sqrt{a}-\sqrt{b}}\right):\left(\dfrac{\left(a-1\right)\left(\sqrt{a}-\sqrt{b}\right)}{2a+2\sqrt{ab}+2b}\right)\)
\(=\dfrac{3a-3\sqrt{ab}-3a+a+\sqrt{ab}+b}{\left(\sqrt{a}-\sqrt{b}\right)\left(a+\sqrt{ab}+b\right)}\cdot\dfrac{2\left(a+\sqrt{ab}+b\right)}{\left(a-1\right)\left(\sqrt{a}-\sqrt{b}\right)}\)
\(=\dfrac{\left(\sqrt{a}-\sqrt{b}\right)^2}{\left(\sqrt{a}-\sqrt{b}\right)^2}\cdot\dfrac{2}{a-1}\)
\(=\dfrac{2}{a-1}\)
Để P là số nguyên thì \(a-1\in\left\{1;-1;2;-2\right\}\)
hay \(a\in\left\{2;0;3\right\}\)
8 tháng 7 2016
Ta có:
\(a^2+ac-b^2-bc=\left(a^2-b^2\right)+\left(ac-bc\right)\)
\(=\left(a-b\right)\left(a+b\right)+c\left(a-b\right)\)
\(=\left(a-b\right)\left(a+b+c\right)\)(1)
\(b^2+ab-c^2-ac=\left(b^2-c^2\right)+\left(ab-ac\right)\)
\(=\left(b-c\right)\left(b+c\right)+a\left(b-c\right)\)
\(=\left(b-c\right)\left(a+b+c\right)\)(2)
\(c^2+bc-a^2-ab=\left(c^2-a^2\right)+\left(bc-ab\right)\)
\(=\left(c-a\right)\left(a+c\right)+b\left(c-a\right)\)
\(=\left(c-a\right)\left(a+b+c\right)\)(3)
Ta có : \(\frac{1}{\left(b-c\right)\left(a^2+ac-b^2-bc\right)}\)\(+\frac{1}{\left(c-a\right)\left(b^2+ab-c^2-ac\right)}\)\(+\frac{1}{\left(a-b\right)\left(c^2+bc-a^2-ab\right)}\)(*)
Thế (1),(2),(3) vào (*)
=>\(\frac{1}{\left(b-c\right)\left(a-b\right)\left(a+b+c\right)}+\frac{1}{\left(c-a\right)\left(b-c\right)\left(a+b+c\right)}+\frac{1}{\left(a-b\right)\left(c-a\right)\left(a+b+c\right)}\)
\(\Leftrightarrow\frac{\left(c-a\right)+\left(a-b\right)+\left(b-c\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)\left(a+b+c\right)}=0\)
MT
8 tháng 7 2016
Dễ thôi bạn chỉ cần quy đồng thôi
\(\frac{1}{\left(b-c\right)\left(a^2+ac-b^2-bc\right)}+\frac{1}{\left(c-a\right)\left(b^2+ab-c^2-ac\right)}+\)\(\frac{1}{\left(a-b\right)\left(c^2+bc-a^2-ab\right)}\)
=\(\frac{1}{\left(b-c\right)\left(a-b\right)\left(a+b+c\right)}+\frac{1}{\left(c-a\right)\left(b-c\right)\left(a+b+c\right)}\)\(+\frac{1}{\left(a-b\right)\left(c-a\right)\left(a+b+c\right)}\)
=\(\frac{c-a+a-b+b-c}{\left(b-c\right)\left(a-b\right)\left(a+b+c\right)}=0\)
8 tháng 7 2016
Ta có :\(\left(a-b\right)\left(c^2+bc-a^2-ab\right)=\left(a-b\right)\left[\left(c^2-a^2\right)+\left(bc-ab\right)\right]\)
\(=\left(a-b\right)\left(c-a\right)\left(a+b+c\right)\)
Tương tự : \(\left(b-c\right)\left(a^2+ac-b^2-bc\right)=\left(b-c\right)\left(a-b\right)\left(a+b+c\right)\)
\(\left(c-a\right)\left(b^2+ab-c^2-ac\right)=\left(c-a\right)\left(b-c\right)\left(a+b+c\right)\)
\(MTC=\left(a-b\right)\left(b-c\right)\left(c-s\right)\left(a+b+c\right)\)
Kí hiệu biểu thức đã cho bởi \(Q\),ta có :
\(Q=\frac{c-a+a-b+b-c}{\left(a-b\right)\left(b-c\right)\left(c-a\right)\left(a+b+c\right)}=0\)
NV
Nguyễn Việt Lâm
Giáo viên
11 tháng 2 2020
Mới nghĩ ra 3 câu:
a/ \(\frac{ab}{\sqrt{\left(1-c\right)^2\left(1+c\right)}}=\frac{ab}{\sqrt{\left(a+b\right)^2\left(1+c\right)}}\le\frac{ab}{2\sqrt{ab\left(1+c\right)}}=\frac{1}{2}\sqrt{\frac{ab}{1+c}}\)
\(\sum\sqrt{\frac{ab}{1+c}}\le\sqrt{2\sum\frac{ab}{1+c}}\)
\(\sum\frac{ab}{1+c}=\sum\frac{ab}{a+c+b+c}\le\frac{1}{4}\sum\left(\frac{ab}{a+c}+\frac{ab}{b+c}\right)=\frac{1}{4}\)
c/ \(ab+bc+ca=2abc\Rightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=2\)
Đặt \(\left(x;y;z\right)=\left(\frac{1}{a};\frac{1}{b};\frac{1}{c}\right)\Rightarrow x+y+z=2\)
\(VT=\sum\frac{x^3}{\left(2-x\right)^2}\)
Ta có đánh giá: \(\frac{x^3}{\left(2-x\right)^2}\ge x-\frac{1}{2}\) \(\forall x\in\left(0;2\right)\)
\(\Leftrightarrow2x^3\ge\left(2x-1\right)\left(x^2-4x+4\right)\)
\(\Leftrightarrow9x^2-12x+4\ge0\Leftrightarrow\left(3x-2\right)^2\ge0\)
d/ Ta có đánh giá: \(\frac{x^4+y^4}{x^3+y^3}\ge\frac{x+y}{2}\)
\(\Leftrightarrow\left(x-y\right)^2\left(x^2+xy+y^2\right)\ge0\)
11 tháng 2 2020
Akai Haruma, Nguyễn Ngọc Lộc , @tth_new, @Băng Băng 2k6, @Trần Thanh Phương, @Nguyễn Việt Lâm
Mn giúp e vs ạ! Thanks!
a = b = 1
hình như kết quả là 2 nhưng mình ko biết cach làm