khó hiểu làm sao ?

Đề chỉ nhiêu đâu thôi hả

khó hỉu quá trời

1.

ĐKXĐ: $x\geq 1; y\geq 2; z\geq 3$

PT \(\Leftrightarrow x+y+z+8-2\sqrt{x-1}-4\sqrt{y-2}-6\sqrt{z-3}=0\)

\(\Leftrightarrow [(x-1)-2\sqrt{x-1}+1]+[(y-2)-4\sqrt{y-2}+4]+[(z-3)-6\sqrt{z-3}+9]=0\)

\(\Leftrightarrow (\sqrt{x-1}-1)^2+(\sqrt{y-2}-2)^2+(\sqrt{z-3}-3)^2=0\)

\(\Rightarrow \sqrt{x-1}-1=\sqrt{y-2}-2=\sqrt{z-3}-3=0\)

\(\Leftrightarrow \left\{\begin{matrix} x=2\\ y=6\\ z=12\end{matrix}\right.\)

2.

ĐKXĐ: $x\geq 0$

PT $\Leftrightarrow \sqrt{x+1}=1-\sqrt{x}$

$\Rightarrow x+1=(1-\sqrt{x})^2=x+1-2\sqrt{x}$

$\Leftrightarrow 2\sqrt{x}=0$

$\Leftrightarrow x=0$

Thử lại thấy thỏa mãn 

Vậy $x=0$

3.

ĐKXĐ: $x\geq -1$

PT \(\Leftrightarrow (1+\sqrt{x^2+4033}).\frac{(x+2016)-(x+1)}{\sqrt{x+2016}+\sqrt{x+1}}=2015\)

\(\Leftrightarrow 1+\sqrt{x^2+4033}=\sqrt{x+2016}+\sqrt{x+1}\)

\(\Leftrightarrow (1+\sqrt{x^2+4033})^2=(\sqrt{x+2016}+\sqrt{x+1})^2\)

Áp dụng BĐT Bunhiacopxky:

\(\text{VP}\leq 2(x+2016+x+1)=4x+4034\)

\(\text{VP}=x^2+4034+2\sqrt{x^2+4033}\geq x^2+4034+2\sqrt{4033}>x^2+4034+5\)

Mà: $x^2+4034+5-(4x+4034)=(x-2)^2+1> 0$

$\Rightarrow x^2+4034+5> 4x+4034$

$\Rightarrow \text{VP}> \text{VT}$

Do đó pt vô nghiệm.

ai đó làm giúp mình , mình tích cho

nhân 2 vế cho 2

=>2x²+2y²+2z²=2xy+2yz+2zx

=>2x²+2y²+2z²-2xy-2yz-2zx=0

=>(2x²-2xy)+(2y²-2yz)+(2z²-2zx)=0

=>(x-y)²+(y-z)²+(z-x)²=0

mà (x-y)² >= 0 với mọi x,y

(y-z)² >= 0 với mọi y,z

(z-x)² >=0 với mọi z,x

=>(x-y)²+(y-z)²+(z-x)² >= 0

mà theo đề:(x-y)²+(y-z)²+(z-x)²=0

=>(x-y)²=(y-z)²=(z-x)²=0

=>x=y

   y=z

   z=x

hay x=y=z

do đó x²⁰¹⁵+y²⁰¹⁵+z²⁰¹⁵=3²⁰¹⁶

<=>x²⁰¹⁵+x²⁰¹⁵+x²⁰¹⁵=3²⁰¹⁶

<=>3x²⁰¹⁵=3²⁰¹⁶<=>x²⁰¹⁵=3²⁰¹⁶:3=3²⁰¹⁵<=>x=2015

Vậy x=y=z=2015

cau a ban de o hang dang thuc (x-y-z)^2 di

Đặt \(a=\sqrt{x-2015};b=\sqrt{y-2016};c=\sqrt{z-2017}\left(a,b,c>0\right)\)

Khi đó phương trình trở thành: 

\(\dfrac{a-1}{a^2}+\dfrac{b-1}{b^2}+\dfrac{c-1}{c^2}=\dfrac{3}{4}\\ \Leftrightarrow\left(\dfrac{1}{4}-\dfrac{1}{a}+\dfrac{1}{a^2}\right)+\left(\dfrac{1}{4}-\dfrac{1}{b}+\dfrac{1}{b^2}\right)+\left(\dfrac{1}{4}-\dfrac{1}{c}+\dfrac{1}{c^2}\right)=0\\ \Leftrightarrow\left(\dfrac{1}{2}-\dfrac{1}{a}\right)^2+\left(\dfrac{1}{2}-\dfrac{1}{b}\right)^2+\left(\dfrac{1}{2}-\dfrac{1}{c}\right)^2=0\\ \Leftrightarrow a=b=c=2\\ \Leftrightarrow x=2019;y=2020;z=2021\)

Tick plz

Có: \(z^2\ge0\forall z\Rightarrow z^2+4\ge4\forall z\Rightarrow\sqrt{z^2+4}\ge\sqrt{4}=2\forall z\)

Mà \(x^{2016}+\left|y-2015\right|+\sqrt{z^2+4}=2\)

\(\Rightarrow\sqrt{z^2+4}=2\)\(\Rightarrow z^2+4=4\Rightarrow z^2=0\Rightarrow z=0\)

Lúc này ta có: x²⁰¹⁶ + |y - 2015| = 0

Mà \(x^{2016}\ge0;\left|y-2015\right|\ge0\forall x;y\)

nên \(\begin{cases}x^{2016}=0\\\left|y-2015\right|=0\end{cases}\)\(\Rightarrow\begin{cases}x=0\\y-2015=0\end{cases}\)\(\Rightarrow\begin{cases}x=0\\y=2015\end{cases}\)

Vậy phương trình trên có nghiệm x = 0; y = 2015; z = 0

Nghiệm nguyên nha

\(x^2+y^2+z^2=xy+yz+xz\)

\(2x^2+2y^2+2z^2=2xy+2yz+2xz\)

\(2x^2+2y^2+2z^2-2xy-2yz-2xz=0\)

\(\left(x^2-2xy+y^2\right)+\left(y^2-2yz+z^2\right)+\left(z^2-2xz+x^2\right)=0\)

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

Vì mũ chẵn luôn lớn hơn hoặc bằng 0

\(\Rightarrow\hept{\begin{cases}x-y=0\\y-z=0\\z-x=0\end{cases}\Rightarrow\hept{\begin{cases}x=y\\y=z\\z=x\end{cases}\Rightarrow}}x=y=z\)

\(\Rightarrow x^{2015}+y^{2015}+z^{2015}=x^{2015}+x^{2015}+x^{2015}=3x^{2015}\)

\(\Rightarrow3x^{2015}=3^{2016}\)

\(\Rightarrow x^{2015}=3^{2015}\)

\(\Rightarrow x=3\)

Vậy \(x=y=z=3\)

Có: \(x^2+y^2+z^2=xy+yz+xz\)

\(\Leftrightarrow2x^2+2y^2+2z^2=2xy+2yz+2xz\)

\(\Leftrightarrow\left(x^2-2xy+y^2\right)+\left(y^2-2yz+z^2\right)+\left(x^2-2xz+z^2\right)=0\)

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

\(\Leftrightarrow\begin{cases}x-y=0\\y-z=0\\x-z=0\end{cases}\)\(\Leftrightarrow x=y=z\)

Lại có: \(x^{2015}+y^{2015}+z^{2015}=3^{2016}\)

\(\Leftrightarrow x^{2015}+x^{2015}+x^{2015}=3^{2016}\)

\(\Leftrightarrow3x^{2015}=3^{2016}\)

\(\Leftrightarrow x=3\)

Vậy \(x=y=z=3\)

Đặt \(\sqrt{x-2014}=a;\sqrt{y-2015}=b;\sqrt{z=2016}=c\)(với a,b,c>0). Khi đó pt trở thành: 

\(\frac{a-1}{a^2}+\frac{b-1}{b^2}+\frac{c-1}{c^2}=\frac{3}{4}\)\(\Leftrightarrow\left(\frac{1}{4}-\frac{1}{a}+\frac{1}{a^2}\right)+\left(\frac{1}{4}-\frac{1}{b}+\frac{1}{b^2}\right)+\left(\frac{1}{4}-\frac{1}{c}+\frac{1}{c^2}\right)=0\)

\(\Leftrightarrow\left(\frac{1}{2}-\frac{1}{a}\right)^2+\left(\frac{1}{2}-\frac{1}{b}\right)^2+\left(\frac{1}{2}-\frac{1}{c}\right)^2=0\Leftrightarrow a=b=c=2\)

\(\Rightarrow x=2018;y=2019;z=2020\)

\(\frac{\sqrt{x-2014}-1}{x-2014}+\frac{\sqrt{y-2015}-1}{y-2015}+\frac{\sqrt{z-2016}-1}{z-2016}=\frac{3}{4}\)

\(\frac{\sqrt{x-2014}}{x-2014}+\frac{\sqrt{y-2015}}{y-2015}+\frac{\sqrt{z-2016}}{z-2016}-\left(\frac{1}{x-2014+y-2015+z-2016}\right)=\frac{3}{4}\)

\(\frac{\sqrt{x-2014}}{x-2014}+\frac{\sqrt{y-2015}}{y-2015}+\frac{\sqrt{z-2016}}{z-2016}+0=\frac{3}{4}\)

\(\frac{\sqrt{x}-\sqrt{2014}}{x-2014}+\frac{\sqrt{y}-\sqrt{2015}}{y-2015}+\frac{\sqrt{z}-\sqrt{2016}}{z-2016}=\frac{3}{4}\)

\(x=2018,y=2019,z=2020\)

ĐK : \(\hept{\begin{cases}x>2014\\y>2015\\z>2016\end{cases}}\)

\(\frac{\sqrt{x-2014}-1}{x-2014}+\frac{\sqrt{y-2015}-1}{y-2015}+\frac{\sqrt{z-2016}-1}{z-2016}=\frac{3}{4}\)

\(\Leftrightarrow\frac{1}{4}-\frac{\sqrt{x-2014}-1}{x-2014}+\frac{1}{4}-\frac{\sqrt{y-2015}-1}{y-2015}+\frac{1}{4}-\frac{\sqrt{z-2016}-1}{z-2016}=0\)

\(\Leftrightarrow\frac{x-2010-4\sqrt{x-2014}}{4\left(x-2014\right)}+\frac{y-2011-4\sqrt{y-2015}}{4\left(y-2015\right)}+\frac{z-2012-4\sqrt{z-2016}}{4\left(x-2014\right)}=0\)

\(\Leftrightarrow\frac{\left(2-\sqrt{x-2014}\right)^2}{4\left(x-2014\right)}+\frac{\left(2-\sqrt{y-2015}\right)^2}{4\left(y-2015\right)}+\frac{\left(2-\sqrt{z-2016}\right)^2}{4\left(z-2016\right)}=0\)( 1 )

Mà \(\hept{\begin{cases}\frac{\left(2-\sqrt{x-2014}\right)^2}{4\left(x-2014\right)}\ge0\forall x>2014\\\frac{\left(2-\sqrt{y-2015}\right)^2}{4\left(y-2015\right)}\ge0\forall y>2015\\\frac{\left(2-\sqrt{z-2016}\right)^2}{4\left(z-2016\right)}\ge0\forall z>2016\end{cases}}\)( 2 )

Từ ( 1 ) và ( 2 ) => \(\hept{\begin{cases}\left(2-\sqrt{x-2014}\right)^2=0\\\left(2-\sqrt{y-2015}\right)^2=0\\\left(2-\sqrt{z-2016}\right)^2=0\end{cases}}\)

<=> \(\hept{\begin{cases}\sqrt{x-2014}=2\\\sqrt{y-2015}=2\\\sqrt{z-2016}=2\end{cases}}\)<=>\(\hept{\begin{cases}x=2018\\y=2019\\z=2020\end{cases}}\)( tmđk )

Vậy ( x ; y ; z ) = ( 2018 ; 2019 ; 2020 )