Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Bác google được sinh ra để làm gì, đăng nhiều vc, google có hết mà ;v
Bài 1,2,3,4 đơn giản, tự làm :v
7) \(\dfrac{ab}{c^2}+\dfrac{bc}{a^2}+\dfrac{ca}{b^2}=\dfrac{abc}{c^3}+\dfrac{abc}{a^3}+\dfrac{abc}{b^3}=abc\left(\dfrac{1}{a^3}+\dfrac{1}{b^3}+\dfrac{1}{c^3}\right)=abc.\dfrac{1}{3abc}=\dfrac{1}{3}\)
P/S: \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=0\Rightarrow\dfrac{1}{a^3}+\dfrac{1}{b^3}+\dfrac{1}{c^3}=\dfrac{3}{abc}\)
5) ĐK: a>b>0
\(3a^2+3b^2=10ab\Leftrightarrow\left(a-3b\right)\left(3a-b\right)=0\)
Tự phân tích
Mà a>b>0=> Chọn a=3b
Thay vào
Bài 6 tương tự bài 5
Có bất mãn chỗ nào thì ib nha bạn :))
Bài 2:
a, ĐKXĐ: \(x\ne\pm1;x\ne\dfrac{-1}{2}\)
\(P=\left(\dfrac{x-1}{x+1}-\dfrac{x}{x-1}-\dfrac{3x+1}{1-x^2}\right):\dfrac{2x+1}{x^2-1}\)
\(P=\left(\dfrac{x-1}{x+1}-\dfrac{x}{x-1}+\dfrac{3x+1}{x^2-1}\right).\dfrac{x^2-1}{2x+1}\)
\(P=\dfrac{\left(x-1\right)^2-x\left(x+1\right)+3x+1}{\left(x-1\right)\left(x+1\right)}.\dfrac{\left(x-1\right)\left(x+1\right)}{2x+1}\)
\(P=\dfrac{x^2-2x+1-x^2-x+3x+1}{\left(x-1\right)\left(x+1\right)}.\dfrac{\left(x-1\right)\left(x+1\right)}{2x+1}\)
\(P=\dfrac{2}{2x+1}\)
b, ĐKXĐ: \(x\ne\pm1;x\ne\dfrac{-1}{2}\)
Để \(P=\dfrac{3}{x-1}\Leftrightarrow\dfrac{2}{2x+1}=\dfrac{3}{x-1}\Leftrightarrow2\left(x-1\right)=3\left(2x+1\right)\)
\(\Leftrightarrow2x-2=6x+3\)\(\Leftrightarrow-4x=5\Leftrightarrow x=\dfrac{-5}{4}\)(TMĐK)
c, \(ĐKXĐ:x\ne\pm1;x\ne\dfrac{-1}{2}\)
Để \(P\in Z\Leftrightarrow\dfrac{2}{2x+1}\in Z\Leftrightarrow2x+1\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)
+) Với \(2x+1=1\Leftrightarrow x=0\left(TMĐK\right)\)
+) Với \(2x+1=-1\Leftrightarrow x=-1\left(KTMĐK\right)\)
+) Với \(2x+1=2\Leftrightarrow x=\dfrac{1}{2}\left(TMĐK\right)\)
+) Với \(2x+1=-2\Leftrightarrow x=\dfrac{-3}{2}\left(TMĐK\right)\)
Vậy để \(P\in Z\Leftrightarrow x\in\left\{0;\dfrac{1}{2};\dfrac{-3}{2}\right\}\)
Áp dụng BĐT Cauchy-Schwarz ta có:
\(\left(1^2+1^2+1^2\right)\left(a^2+b^2+c^2\right)\ge\left(a+b+c\right)^2=9^2\)
\(\Rightarrow3\left(a^2+b^2+c^2\right)\ge9\Rightarrow a^2+b^2+c^2\ge3\)
Lại có: \(a^2+b^2+c^2\ge ab+bc+ac\forall a,b,c\)
\(\Rightarrow3\ge ab+bc+ac\Rightarrow ab+bc+ac\le3\)
Bất đẳng thức ban đầu tương đương với:
\(\dfrac{a^2}{a\left(b^2+1\right)}+\dfrac{b^2}{b\left(c^2+1\right)}+\dfrac{c^2}{c\left(a^2+1\right)}\ge\dfrac{3}{2}\)
Áp dụng BĐT Cauchy-Schwarz dạng Engel ta có:
\(\dfrac{a^2}{a\left(b^2+1\right)}+\dfrac{b^2}{b\left(c^2+1\right)}+\dfrac{c^2}{c\left(a^2+1\right)}\ge\dfrac{\left(a+b+c\right)^2}{a\left(b^2+1\right)+b\left(c^2+1\right)+c\left(a^2+1\right)}\)
Áp dụng BĐT AM-GM ta có:
\(\left\{{}\begin{matrix}a\left(b^2+1\right)\ge a\cdot2\sqrt{b^2}=2ba\\b\left(c^2+1\right)\ge b\cdot2\sqrt{c^2}=2cb\\c\left(a^2+1\right)\ge c\cdot2\sqrt{a^2}=2ac\end{matrix}\right.\)
\(\Rightarrow\dfrac{a^2}{a\left(b^2+1\right)}+\dfrac{b^2}{b\left(c^2+1\right)}+\dfrac{c^2}{c\left(a^2+1\right)}\ge\dfrac{\left(a+b+c\right)^2}{2\left(ab+bc+ca\right)}\)
Mà \(ab+bc+ca\le3\)\(\Rightarrow\dfrac{\left(a+b+c\right)^2}{2\left(ab+bc+ca\right)}\ge\dfrac{\left(a+b+c\right)^2}{2\cdot3}=\dfrac{9}{6}=\dfrac{3}{2}\)
Đẳng thức xảy ra khi \(a=b=c=1\)
\(VT=\dfrac{a}{b^2+1}+\dfrac{b}{c^2+1}+\dfrac{c}{a^2+1}\)
\(VT=a-\dfrac{ab^2}{b^2+1}+b-\dfrac{bc^2}{c^2+1}+c-\dfrac{ca^2}{a^2+1}\)
\(VT=3-\left(\dfrac{ab^2}{b^2+1}+\dfrac{bc^2}{c^2+1}+\dfrac{ca^2}{a^2+1}\right)\)
Áp dụng bất đẳng thức Cauchy - Schwarz
\(\Rightarrow\left\{{}\begin{matrix}b^2+1\ge2\sqrt{b^2}=2b\\c^2+1\ge2\sqrt{c^2}=2c\\a^2+1\ge2\sqrt{a^2}=2a\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{ab^2}{b^2+1}\le\dfrac{ab^2}{2b}=\dfrac{ab}{2}\\\dfrac{bc^2}{c^2+1}\le\dfrac{bc^2}{2c}=\dfrac{bc}{2}\\\dfrac{ca^2}{a^2+1}\le\dfrac{ca^2}{2a}=\dfrac{ca}{2}\end{matrix}\right.\)
\(\Rightarrow\dfrac{ab^2}{b^2+1}+\dfrac{bc^2}{c^2+1}+\dfrac{ca^2}{a^2+1}\le\dfrac{ab+bc+ca}{2}\)
\(\Rightarrow3-\left(\dfrac{ab^2}{b^2+1}+\dfrac{bc^2}{c^2+1}+\dfrac{ca^2}{a^2+1}\right)\ge3-\dfrac{ab+bc+ca}{2}\) (1)
Theo hệ quả của bất đẳng thức Cauchy
\(\Rightarrow\left(a+b+c\right)^2\ge3\left(ab+bc+ca\right)\)
\(\Rightarrow3\ge ab+bc+ca\)
\(\Rightarrow\dfrac{3}{2}\ge\dfrac{ab+bc+ca}{2}\)
\(\Rightarrow\dfrac{3}{2}\le3-\dfrac{ab+bc+ca}{2}\)(2)
Từ (1) và (2)
\(\Rightarrow3-\left(\dfrac{ab^2}{b^2+1}+\dfrac{bc^2}{c^2+1}+\dfrac{ca^2}{a^2+1}\right)\ge\dfrac{3}{2}\)
\(\Leftrightarrow\dfrac{a}{b^2+1}+\dfrac{b}{c^2+1}+\dfrac{c}{a^2+1}\ge\dfrac{3}{2}\) ( đpcm )
Dấu "=" xảy ra khi \(a=b=c=1\)
5)
a)
Có 3x+y = 1
\(\Rightarrow x+x+x+y=1\)
Áp dụng bất đẳng thức bunhiacopxki ta có :
\(\left(x^2+x^2+x^2+y^2\right)\left(1^2+1^2+1^2+1^2\right)\ge\left(x+x+x+y\right)^2\)
\(\Rightarrow3x^2+y^{2^{ }}.4\ge\left(3x+y\right)^2\)
\(\Rightarrow3x^2+y^2\ge\dfrac{1}{4}\)
b)
Áp dụng bất đẳng thức AM - GM ta có :
\(\left[{}\begin{matrix}a^2+1^2\ge2a\\b^2+1^2\ge2b\\c^2+1^2\ge2c\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\left(a+1\right)^2\ge4a^{ }\\\left(b+1\right)^2\ge4b^{ }\\\left(c+1\right)^2\ge4c^{ }\end{matrix}\right.\)
\(\Rightarrow\left(a+1\right)^2\left(b+1\right)^2\left(c+1\right)^2\ge4a^{ }.4b.4c^{ }\)
\(\Rightarrow\left(a+1\right)^2\left(b+1\right)^2\left(c+1\right)^2\ge64a^{ }bc^{ }\)
\(\Rightarrow\left(a+1\right)^2\left(b+1\right)^2\left(c+1\right)^2\ge64abc\)
\(\Rightarrow\left(a+1\right)^2\left(b+1\right)^2\left(c+1\right)^2\ge64\)
\(\Rightarrow\left(a+1\right)^{ }\left(b+1\right)^{ }\left(c+1\right)^{ }\ge8\) \(\left(đpcm\right)\)
3)
Sửa đề \(A=\dfrac{a}{b+c-a}+\dfrac{b}{a+c-b}+\dfrac{c}{a+b-c}\)
Đặt b + c - a = x , a+c-b = y , a+b-c= z
\(\Rightarrow\left[{}\begin{matrix}2a=y+z\\2b=x+z\\2c=x+y\end{matrix}\right.\)
Có :
\(\dfrac{a}{b+c-a}+\dfrac{b}{a+c-b}+\dfrac{c}{a+b-c}\)
\(\Rightarrow\dfrac{2a}{b+c-a}+\dfrac{2b}{a+c-b}+\dfrac{2c}{a+b-c}\)
\(\Rightarrow\dfrac{y+z}{x}+\dfrac{x+z}{y}+\dfrac{x+y}{z}\)
\(\Rightarrow\left(\dfrac{y}{x}+\dfrac{x}{y}\right)+\left(\dfrac{z}{x}+\dfrac{x}{z}\right)+\left(\dfrac{z}{y}+\dfrac{y}{z}\right)\)
Áp dụng bất đẳng thức \(\dfrac{a}{b}+\dfrac{b}{a}\ge2\forall a,b>0\)
\(\Rightarrow\) \(\left(\dfrac{y}{x}+\dfrac{x}{y}\right)+\left(\dfrac{z}{x}+\dfrac{x}{z}\right)+\left(\dfrac{z}{y}+\dfrac{y}{z}\right)\ge6\)
\(\Rightarrow\dfrac{2a}{b+c-a}+\dfrac{2b}{a+c-b}+\dfrac{2c}{a+b-c}\ge6\)
\(\Rightarrow2\left(\dfrac{a}{b+c-a}+\dfrac{b}{a+c-b}+\dfrac{c}{a+b-c}\right)\ge6\)
\(\Rightarrow\dfrac{a}{b+c-a}+\dfrac{b}{a+c-b}+\dfrac{c}{a+b-c}\ge3\) \(\left(đpcm\right)\)
Bài 2:
Bài 1:
\(a^2+b^2+c^2=14\Rightarrow\left(a+b+c\right)^2-2ab-2bc-2ac=14\)\(\Leftrightarrow-2\left(ab+bc+ac\right)=14\Rightarrow ab+bc+ac=-7\)\(\Rightarrow\left(ab+bc+ac\right)^2=49\)
\(\Leftrightarrow a^2b^2+b^2c^2+c^2a^2+2a^2bc+2ab^2c+2abc^2=49\)\(\Leftrightarrow a^2b^2+b^2c^2+a^2c^2+2abc\left(a+b+c\right)=49\)
\(\Rightarrow a^2b^2+b^2c^2+a^2c^2=49\)
Ta có:
\(a^4+b^4+c^4=\left(a^2+b^2+c^2\right)^2-2a^2b^2-2b^2c^2-2a^2c^2\)\(=14^2-2\left(a^2b^2+b^2c^2+a^2c^2\right)=196-2.49=98\)