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Ta có:\(a^3+b^3+c^3=3abc\Rightarrow a^3+b^3+c^3-3abc=0\)
\(\Rightarrow a^3+3a^2b+3ab^2+b^3+c^3-3a^2b-3ab^2-3abc=0\)
\(\Rightarrow\left(a+b\right)^3+c^3-3ab\left(a+b+c\right)=0\)
\(\Rightarrow\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right).c+c^2\right]-3ab\left(a+b+c\right)=0\)
\(\Rightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)=0\)
\(\Rightarrow\orbr{\begin{cases}a+b+c=0\\a^2+b^2+c^2-ab-bc-ca=0\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}a+b+c=0\left(loai\right)\\a=b=c\end{cases}}\)
\(\Rightarrow P=2007.2007.2007=2007^3\)
Ta có \(a^3+b^3+c^3=3abc\)
=> \(\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)=0\)
Mà \(a+b+c\ne0\)
=> \(a^2+b^2+c^2-ab-bc-ac=0\)
<=> \(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)
Do \(VT\ge0\)
=> a=b=c
Thay vào ta được
P=2018^3
Cần chứng minh BĐT khác
\(\frac{a^3-b^3}{\left(a-b\right)^3}+\frac{b^3-c^3}{\left(b-c\right)^3}+\frac{c^3-a^3}{\left(c-a\right)^3}\ge\frac{9}{4}\)
\(\LeftrightarrowΣ\frac{3\left(a+b\right)^2+\left(a-b\right)^2}{\left(a-b\right)^2}\ge4\)
\(\Leftrightarrow\frac{\left(a+b\right)^2}{\left(a-b\right)^2}+\frac{\left(b+c\right)^2}{\left(b-c\right)^2}+\frac{\left(c+a\right)^2}{\left(c-a\right)^2}\ge2\)
Vậy chứng minh BĐT đầu bài quay ra chứng minh BĐT dòng đầu
\(\Leftrightarrow\frac{\left(a+b\right)^2}{\left(a-b\right)^2}-1+\frac{\left(b+c\right)^2}{\left(b-c\right)^2}-1+\frac{\left(c+a\right)^2}{\left(c-a\right)^2}-1\ge-1\)
\(\Leftrightarrow\frac{4ab}{\left(a-b\right)^2}+\frac{4bc}{\left(b-c\right)^2}+\frac{4ca}{\left(a-c\right)^2}\ge-1\)
\(\Leftrightarrow\frac{3ab}{\left(a-b\right)^2}+\frac{3bc}{\left(b-c\right)^2}+\frac{3ca}{\left(a-c\right)^2}\ge-\frac{3}{4}\)
\(\Leftrightarrow\frac{3ab}{\left(a-b\right)^2}+1+\frac{3bc}{\left(b-c\right)^2}+1+\frac{3ca}{\left(a-c\right)^2}+1\ge3-\frac{3}{4}\)
\(\Leftrightarrow\frac{a^2+ab+b^2}{\left(a-b\right)^2}+\frac{b^2+bc+c^2}{\left(b-c\right)^2}+\frac{c^2+ac+c^2}{\left(a-c\right)^2}\ge\frac{9}{4}\)
\(\Leftrightarrow\frac{a^3-b^3}{\left(a-b\right)^3}+\frac{b^3-c^3}{\left(b-c\right)^3}+\frac{c^3-a^3}{\left(a-c\right)^3}\ge\frac{9}{4}\)
BĐT cuối đúng nên ta có ĐPCM
ko pic
mik pic nhưng giải rất dài dòng
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mik kb hít lun nha
\(B=\frac{9-x}{\sqrt{x}+3}-\frac{x-6\sqrt{x}+9}{\sqrt{x}-3}-6\)(đk: x ≥ 0 và x ≠ 9)
\(B=\frac{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)}{\sqrt{x}+3}-\frac{\left(\sqrt{x}-3\right)^2}{\sqrt{x}-3}-6\)
\(B=\left(3-\sqrt{x}\right)-\left(\sqrt{x}-3\right)-6\)
\(B=3-\sqrt{x}-\sqrt{x}+3-6\)
\(B=-2\sqrt{x}\)
\(A=\frac{\sqrt{x}}{\sqrt{x}-6}-\frac{3}{\sqrt{x}+6}+\frac{x}{36-x}\)(đk: x ≥ 0 và x ≠ 36)
\(=\frac{\sqrt{x}}{\sqrt{x}-6}-\frac{3}{\sqrt{x}+6}-\frac{x}{x-36}\)
\(=\frac{\sqrt{x}}{\sqrt{x}-6}-\frac{3}{\sqrt{x}+6}-\frac{x}{x-36}\)
\(=\frac{\sqrt{x}\left(\sqrt{x}+6\right)-3\left(\sqrt{x-6}\right)-x}{(\sqrt{x}-6)\left(\sqrt{x}+6\right)}\)
\(=\frac{x+6\sqrt{x}-3\sqrt{x}+18-x}{(\sqrt{x}-6)\left(\sqrt{x}+6\right)}\)
\(=\frac{3\sqrt{x}+18}{(\sqrt{x}-6)\left(\sqrt{x}+6\right)}\)
\(=\frac{3(\sqrt{x}+6)}{(\sqrt{x}-6)\left(\sqrt{x}+6\right)}\)
\(=\frac{3}{\sqrt{x}-6}\)
a) Ta có: \(\frac{a-b}{\sqrt{a}-\sqrt{b}}-\frac{\sqrt{a^3}-\sqrt{b^3}}{a-b}\)
\(=\frac{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{a}-\sqrt{b}}-\frac{\left(\sqrt{a}-\sqrt{b}\right)\left(a+\sqrt{ab}+b\right)}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\)
\(=\frac{\left(\sqrt{a}+\sqrt{b}\right)^2}{\sqrt{a}+\sqrt{b}}-\frac{a+\sqrt{ab}+b}{\sqrt{a}+\sqrt{b}}\)
\(=\frac{a+2\sqrt{ab}+b-a-\sqrt{ab}-b}{\sqrt{a}+\sqrt{b}}\)
\(=\frac{\sqrt{ab}}{\sqrt{a}+\sqrt{b}}\)
b)Sửa đề: \(\frac{\left(\sqrt{a}+\sqrt{b}\right)^2-4\sqrt{ab}}{\sqrt{a}-\sqrt{b}}-\frac{a\sqrt{b}+b\sqrt{a}}{\sqrt{ab}}\)
Ta có: \(\frac{\left(\sqrt{a}+\sqrt{b}\right)^2-4\sqrt{ab}}{\sqrt{a}-\sqrt{b}}-\frac{a\sqrt{b}+b\sqrt{a}}{\sqrt{ab}}\)
\(=\frac{\left(\sqrt{a}-\sqrt{b}\right)^2}{\left(\sqrt{a}-\sqrt{b}\right)}-\frac{\sqrt{ab}\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{ab}}\)
\(=\sqrt{a}-\sqrt{b}-\sqrt{a}-\sqrt{b}\)
\(=-2\sqrt{b}\)
c) Ta có: \(\left(\frac{1}{\sqrt{a}-1}-\frac{1}{\sqrt{a}}\right):\left(\frac{\sqrt{a}+1}{\sqrt{a}-2}-\frac{\sqrt{a}+2}{\sqrt{a}-1}\right)\)
\(=\left(\frac{\sqrt{a}}{\sqrt{a}\left(\sqrt{a}-1\right)}-\frac{\sqrt{a}-1}{\sqrt{a}\left(\sqrt{a}-1\right)}\right):\left(\frac{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}-\frac{\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\right)\)
\(=\frac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\frac{a-1-a+4}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\)
\(=\frac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\frac{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}{3}\)
\(=\frac{\sqrt{a}-2}{3\sqrt{a}}\)
d) Ta có: \(\left(\frac{a\sqrt{a}+b\sqrt{b}}{\sqrt{a}+\sqrt{b}}-\sqrt{ab}\right)\left(\frac{\sqrt{a}+\sqrt{b}}{a-b}\right)^2\)
\(=\left(\frac{\left(\sqrt{a}+\sqrt{b}\right)\left(a-\sqrt{ab}+b\right)}{\left(\sqrt{a}+\sqrt{b}\right)}-\sqrt{ab}\right)\left(\frac{\left(\sqrt{a}+\sqrt{b}\right)}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}\right)^2\)
\(=\left(a-\sqrt{ab}+b-\sqrt{ab}\right)\cdot\left(\frac{1}{\sqrt{a}-\sqrt{b}}\right)^2\)
\(=\left(a-2\sqrt{ab}+b\right)\cdot\frac{1}{\left(\sqrt{a}-\sqrt{b}\right)^2}\)
\(=\frac{\left(\sqrt{a}-\sqrt{b}\right)^2}{\left(\sqrt{a}-\sqrt{b}\right)^2}=1\)
e) Ta có: \(\left(\frac{\sqrt{x}}{3+\sqrt{x}}+\frac{x+9}{9-x}\right):\left(\frac{3\sqrt{x}+1}{x-3\sqrt{x}}-\frac{1}{\sqrt{x}}\right)\)
\(=\left(\frac{\sqrt{x}\left(3-\sqrt{x}\right)}{\left(3+\sqrt{x}\right)\left(3-\sqrt{x}\right)}+\frac{x+9}{\left(3+\sqrt{x}\right)\left(3-\sqrt{x}\right)}\right):\left(\frac{3\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-3\right)}-\frac{\sqrt{x}-3}{\sqrt{x}\left(\sqrt{x}-3\right)}\right)\)
\(=\frac{3\sqrt{x}+9}{\left(3+\sqrt{x}\right)\left(3-\sqrt{x}\right)}:\frac{3\sqrt{x}+1-\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}-3\right)}\)
\(=\frac{3\left(\sqrt{x}+3\right)}{-\left(\sqrt{x}-3\right)\cdot\left(\sqrt{x}+3\right)}\cdot\frac{\sqrt{x}\left(\sqrt{x}-3\right)}{2\left(\sqrt{x}+2\right)}\)
\(=\frac{-3\sqrt{x}}{2\sqrt{x}+4}\)
5/ĐK: \(\left[{}\begin{matrix}x\le-1\\x\ge5\end{matrix}\right.\)
PT \(\Leftrightarrow2\left(x^2-4x-6\right)+\sqrt{x^2-4x-5}-1=0\)
\(\Leftrightarrow\left(x^2-4x-6\right)\left(2+\frac{1}{\sqrt{x^2-4x-5}+1}\right)=0\)
\(\Leftrightarrow x^2-4x-6=0\Leftrightarrow\left[{}\begin{matrix}x=2+\sqrt{10}\\x=2-\sqrt{10}\end{matrix}\right.\)
Vậy..
Bạn tham khảo :
Ta có :
\(\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=1\)
\(\Rightarrow\frac{a}{b}+\frac{a}{c}+\frac{b}{a}+\frac{b}{c}+\frac{c}{a}+\frac{c}{b}+3=1\)
\(\Rightarrow\frac{a}{b}+\frac{a}{c}+\frac{b}{a}+\frac{b}{c}+\frac{c}{a}+\frac{c}{b}+2=0\)
\(\Rightarrow abc\left(\frac{a}{b}+\frac{a}{c}+\frac{b}{a}+\frac{b}{c}+\frac{c}{a}+\frac{c}{b}+2\right)=abc.0\)
\(\Rightarrow a^2b+b^2c+a^2c+b^2a+c^2a+c^2b+2abc=0\)
\(\Rightarrow\left(a^2b+ab^2\right)+\left(b^2c+abc\right)+\left(a^2c+abc\right)+\left(c^2a+c^2b\right)=0\)
\(\Rightarrow ab\left(a+b\right)+bc\left(a+b\right)+ac\left(a+b\right)+c^2\left(a+b\right)=0\)
\(\Rightarrow\left(ab+bc+ac+c^2\right)\left(a+b\right)=0\)
\(\Rightarrow\left[\left(ab+bc\right)+\left(ac+c^2\right)\right]\left(a+b\right)=0\)
\(\Rightarrow\left[b\left(a+c\right)+c\left(a+c\right)\right]\left(a+b\right)=0\)
\(\Rightarrow\left(a+c\right)\left(b+c\right)\left(a+b\right)=0\)
TH1 : \(a+c=0\)
\(\Rightarrow a=-c\)
\(\Rightarrow c^{2006}=a^{2006}\)
\(\Rightarrow P=\left(a^{2004}-b^{2004}\right)\left(b^{2005}+c^{2005}\right)\left(c^{2006}-a^{2006}\right)\)
\(=\left(a^{2004}-b^{2004}\right)\left(b^{2005}+c^{2005}\right)0\)
\(=0\)
CMTT đều có \(P=0\)
Vậy ...