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\(=\dfrac{\sqrt{a}+2+\sqrt{a}-2}{a-4}:\dfrac{\sqrt{a}+2-2}{\sqrt{a}+2}\)

\(=\dfrac{2\sqrt{a}}{a-4}\cdot\dfrac{\sqrt{a}+2}{\sqrt{a}}=\dfrac{2}{\sqrt{a}-2}\)

\(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=0\) (\(x,y,z\ne0;x\ne y\ne z\)

\(\Leftrightarrow xy+yz+xz=0\)

\(\Leftrightarrow2yz=yz-xy-xz\)

\(\Leftrightarrow x^2+2yz=\left(x-y\right)\left(x-z\right)\)

CMTT : \(\left\{{}\begin{matrix}y^2+2xz=\left(y-z\right)\left(y-x\right)\\z^2+2xy=\left(z-x\right)\left(z-y\right)\end{matrix}\right.\)

\(A=\dfrac{yz\left(y-z\right)-xz\left(x-z\right)+xy\left(x-y\right)}{\left(x-y\right)\left(x-z\right)\left(y-z\right)}\)

\(A=\dfrac{y^2z-yz^2-x^2z+xz^2+xy\left(x-y\right)}{\left(x-y\right)\left(x-z\right)\left(y-z\right)}\)

\(A=\dfrac{z^2\left(x-y\right)-z\left(x-y\right)\left(x+y\right)+xy\left(x-y\right)}{\left(x-y\right)\left(x-z\right)\left(y-z\right)}\)

\(A=\dfrac{z^2-xz-yz+xy}{\left(x-z\right)\left(y-z\right)}=\dfrac{x\left(y-z\right)-z\left(y-z\right)}{\left(x-z\right)\left(y-1\right)}=1\)

Thề, gõ máy mệt gấp đôi viết tay =))

ap dung bdt am gm

\(\sqrt{1+8a^3}=\sqrt{\left(1+2a\right)\left(4a^2-4a+1\right)}\)\(\le\frac{1+2a+4a^2-2a+1}{2}=\frac{4a^2+2}{2}=2a^2+1\)

\(\Rightarrow\frac{1}{\sqrt{1+8a^3}}\ge\frac{1}{2a^2+1}\)

tuongtu ta cung co \(\frac{1}{\sqrt{1+8b^3}}\ge\frac{1}{2b^2+1};\frac{1}{\sqrt{1+8c^3}}\ge\frac{1}{2c^2+1}\)

\(\Rightarrow\)VT\(\ge\frac{1}{2a^2+1}+\frac{1}{2b^2+1}+\frac{1}{2c^2+1}\)

tiep tuc ap dung bat cauchy-schwarz dang engel ta co

\(VT\ge\frac{1}{2a^2+1}+\frac{1}{2b^2+1}+\frac{1}{2c^2+1}\ge\frac{\left(1+1+1\right)^2}{2\left(a^2+b^2+c^2\right)+3}=\frac{3^2}{6+3}=1\)(dpcm)

dau = xay ra \(\Leftrightarrow a=b=c=1\)

#include <bits/stdc++.h>

using namespace std;

long long a,b;

int main()

{

cin>>a>>b;

cout<<fixed<<setprecision(2)<<1/(a*1.0+b*1.0);

return 0;

}

Bài 1:

Vì $a\geq 1$ nên:

\(a+\sqrt{a^2-2a+5}+\sqrt{a-1}=a+\sqrt{(a-1)^2+4}+\sqrt{a-1}\)

\(\geq 1+\sqrt{4}+0=3\)

Ta có đpcm

Dấu "=" xảy ra khi $a=1$

Bài 2:
ĐKXĐ: x\geq -3$

Xét hàm:

\(f(x)=x(x^2-3x+3)+\sqrt{x+3}-3\)

\(f'(x)=3x^2-6x+3+\frac{1}{2\sqrt{x+3}}=3(x-1)^2+\frac{1}{2\sqrt{x+3}}>0, \forall x\geq -3\)

Do đó $f(x)$ đồng biến trên TXĐ

\(\Rightarrow f(x)=0\) có nghiệm duy nhất

Dễ thấy pt có nghiệm $x=1$ nên đây chính là nghiệm duy nhất.

a, ĐKXĐ:\(x\ne-3\)

\(x+1+\dfrac{2}{x+3}=\dfrac{x+5}{x+3}\\ \Leftrightarrow x+1=\dfrac{x+5}{x+3}-\dfrac{2}{x+3}\\ \Leftrightarrow x+1=\dfrac{x+3}{x+3}\\ \Leftrightarrow x+1=1\\ \Leftrightarrow x=0\left(tm\right)\)

b, ĐKXĐ:\(x>2\)

\(\dfrac{x^2-4x-2}{\sqrt{x-2}}=\sqrt{x-2}\\ \Leftrightarrow x^2-4x-2=x-2\\ \Leftrightarrow x^2-5x=0\\ \Leftrightarrow x\left(x-5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\left(ktm\right)\\x=5\left(tm\right)\end{matrix}\right.\)

\(\dfrac{1}{c}+b^2c=ab\left(a+b+c\right)+b^2c=ab\left(a+c\right)+b^2\left(a+c\right)=b\left(a+b\right)\left(a+c\right)\)

\(\dfrac{1}{c}+a^2c=ab\left(a+b+c\right)+a^2c=a\left(a+b\right)\left(b+c\right)\)

\(\Rightarrow\left(\dfrac{1}{c}+b^2c\right)\left(\dfrac{1}{c}+a^2c\right)=ab\left(a+b\right)^2\left(b+c\right)\left(a+c\right)\)

\(\Leftrightarrow\left(1+b^2c^2\right)\left(1+a^2c^2\right)=c^2\left(a+b\right)^2ab\left(ab+bc+ac+c^2\right)\)\(=c^2\left(a+b\right)^2\left(a^2b^2+ab^2c+a^2bc+abc^2\right)\)\(=c^2\left(a+b\right)^2\left[a^2b^2+abc\left(a+b+c\right)\right]=c^2\left(a+b\right)^2\left(a^2b^2+1\right)\)

\(\Rightarrow\dfrac{\left(1+b^2c^2\right)\left(1+a^2c^2\right)}{c^2\left(a^2b^2+1\right)}=\left(a+b\right)^2\)

\(\Leftrightarrow\sqrt{\dfrac{\left(1+b^2c^2\right)\left(1+a^2c^2\right)}{c^2+a^2b^2c^2}}=a+b\) (đpcm)