ta có :

G/s \(n+26=a^3\) và \(n-11=b^3\) với a,b là các STN

\(\Rightarrow a^3-b^3=n+26-n+11\)

\(\Leftrightarrow\left(a-b\right)\left(a^2+ab+b^2\right)=37\)

Vì \(\hept{\begin{cases}a-b>0\\a^2+ab+b^2\ge0\end{cases}\left(\forall a,b\right)}\)

Ta có 2 TH sau:

Nếu \(\hept{\begin{cases}a-b=1\\a^2+ab+b^2=37\end{cases}}\Leftrightarrow\hept{\begin{cases}a=b+1\\a^2+ab+b^2=37\end{cases}}\)

\(\Leftrightarrow\left(b+1\right)^2+\left(b+1\right)b+b^2-37=0\)

\(\Leftrightarrow3b^2+3b-36=0\)

\(\Leftrightarrow\left(b-3\right)\left(b+4\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}b=3\left(tm\right)\\b=-4\left(ktm\right)\end{cases}}\Rightarrow b=3\Rightarrow a=4\)

\(\Rightarrow n=38\)

Nếu \(\hept{\begin{cases}a-b=37\\a^2+ab+b^2=1\end{cases}}\)

\(\Leftrightarrow\left(b+37\right)^2+\left(b+37\right)b+b^2=1\)

\(\Leftrightarrow b^2+74b+1369+b^2+37b+b^2-1=0\)

\(\Leftrightarrow3b^2+111b+1368=0\)

\(\Leftrightarrow b^2+37b+456=0\)

\(\Leftrightarrow\left(b^2+37b+\frac{1369}{4}\right)+\frac{455}{4}=0\)

\(\Leftrightarrow\left(b+\frac{37}{2}\right)^2=-\frac{455}{4}\)

=> vô lý

Vậy n = 38

Ta có: \(\sqrt{x^2-2x+1}+\sqrt{x^2+4x+4}\)

\(=\sqrt{\left(x-1\right)^2}+\sqrt{\left(x+2\right)^2}\)

\(=\left|x-1\right|+\left|x+2\right|\)

\(=\left|1-x\right|+\left|x+2\right|\ge\left|1-x+x+2\right|=\left|3\right|=3\)

Dấu "=" xảy ra khi: \(\left(1-x\right)\left(x+2\right)\ge0\)

\(\Rightarrow-2\le x\le1\)

Vậy \(-2\le x\le1\)

\(\sqrt{x^2-2x+1}+\sqrt{x^2+4x+4}=3\)

\(\Leftrightarrow\sqrt{\left(x-1\right)^2}+\sqrt{\left(x+2\right)^2}=3\)

\(\Leftrightarrow\left|x-1\right|+\left|x+2\right|=3\)(1)

Xét \(\left|x-1\right|+\left|x+2\right|\)

\(=\left|-\left(x-1\right)\right|+\left|x+2\right|\)

\(=\left|1-x\right|+\left|x+2\right|\)

\(\ge\left|1-x+x+2\right|=\left|3\right|=3\)( BĐT \(\left|a\right|+\left|b\right|\ge\left|a+b\right|\))

Dấu "=" xảy ra ( tức (1) ) khi ab ≥ 0

=> \(\left(1-x\right)\left(x+2\right)\ge0\)

=> \(-2\le x\le1\)

Vậy \(-2\le x\le1\)là nghiệm của pt

Ta có: \(\left(x+y+z\right)^2=x^2+y^2+z^2+2\left(xy+yz+zx\right)\)

\(=x^2+y^2+z^2+2.1=x^2+y^2+z^2+2\left(2y^2-3z^2\right)\)\(=x^2+5y^2-5z^2\)

\(\Leftrightarrow\left(x+y+z\right)^2-x^2+5\left(z-y\right)\left(z+y\right)=0\)

\(\Leftrightarrow\left(2x+y+z\right)\left(y+z\right)+5\left(z-y\right)\left(z+y\right)=0\)

\(\Leftrightarrow\left(y+z\right)\left(2x+y+z+5z-5y\right)=0\)

\(\Leftrightarrow\left(y+z\right)\left(2x-4y+6z\right)=0\)

\(\Leftrightarrow\left(y+z\right)\left(x-2y+3z\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}y=-z\\x-2y+3z=0\end{cases}}\)

Với y=-z ta có: \(2y^2-3z^2=1\Rightarrow2y^2-3y^2=1\Leftrightarrow-y^2=1\)( do \(y^2\ge0\)) => pt  vô nghiệm

mong các bạn giúp mk giải bài này

= 324 NHÉ BẠN

đk: \(\hept{\begin{cases}x\ge0\\x\ne1\end{cases}}\)

Ta có:

\(A=\frac{\sqrt{x}-2}{x-1}-\frac{\sqrt{x}+2}{x+2\sqrt{x}+1}\)

\(A=\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)-\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}\)

\(A=\frac{x-\sqrt{x}-2-x-\sqrt{x}+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}\)

\(A=-\frac{2\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}\)

a) ĐK: \(x>2009;y>2010;z>2011\)

\(\Leftrightarrow\frac{\sqrt{x-2009}-1}{x-2009}-\frac{1}{4}+\frac{\sqrt{y-2010}-1}{y-2010}-\frac{1}{4}+\frac{\sqrt{z-2011}-1}{z-2011}-\frac{1}{4}=0\)

\(\Leftrightarrow\frac{-\left(\sqrt{x-2009}-2\right)^2}{4\left(x-2009\right)}+\frac{-\left(\sqrt{y-2010}-2\right)^2}{4\left(y-2010\right)}+\frac{-\left(\sqrt{z-2011}-2\right)^2}{4\left(z-2011\right)}=0\left(1\right)\)

Dễ thấy với đkxđ thì \(VT\left(1\right)\le0\)

Dấu "=" xảy ra khi \(\hept{\begin{cases}\sqrt{x-2009}=2\\\sqrt{y-2010}=2\\\sqrt{z-2011}=2\end{cases}\Leftrightarrow\hept{\begin{cases}x=2013\\y=2014\\z=2015\end{cases}\left(tm\right)}}\)

\(\sqrt{x^2-9}+\sqrt{x^2-6x+9}=0\)(*)

\(ĐK:\orbr{\begin{cases}x\ge3\\x\le-3\end{cases}}\)

(*)\(\Leftrightarrow\sqrt{\left(x+3\right)\left(x-3\right)}+\sqrt{\left(x-3\right)^2}=0\)

\(\Leftrightarrow\sqrt{x-3}\left(\sqrt{x+3}+\sqrt{x-3}\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=3\left(tm\right)\\\sqrt{x+3}+\sqrt{x-3}=0\end{cases}}\)

Xét phương trình\(\sqrt{x+3}+\sqrt{x-3}=0\)(**) có \(\sqrt{x+3}\ge0;\sqrt{x-3}\ge0\)nên (**) xảy ra khi \(\hept{\begin{cases}\sqrt{x+3}=0\\\sqrt{x-3}=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-3\\x=3\end{cases}}\left(L\right)\)

Vậy phương trình có một nghiệm duy nhất là 3