Ta có pt <=> \(2\sqrt{x-2}+2\sqrt{y+2009}+2\sqrt{z-2010}=x+y+z\)

<=> \(x-2-2\sqrt{x-2}+1+y+2009-2\sqrt{y+2009}+1+z-2010-2\sqrt{z-2010}+1=0\)

<=> \(\left(\sqrt{x-2}-1\right)^2+\left(\sqrt{y+2009}-1\right)^2+\left(\sqrt{z-2010}-1\right)^2=0\)

...

^_^

nát cả óc!

-kindle (sử dụng ở ứng dụng Android

\(x-2008=X;y-2009=Y;z-2010=Z\)

\(\sqrt{X}+\sqrt{Y}+\sqrt{Z}+3012=\frac{1}{2}\left(X+Y+Z+2008+2009+2010\right)\)

\(2.\sqrt{X}+2\sqrt{Y}+2\sqrt{Z}+2.3012=X+Y+Z+2009\cdot3\)

\(\left(X-2\sqrt{X}+1\right)+\left(Y-2\sqrt{Y}+1\right)+\left(Z-2\sqrt{Z}+1\right)+3.2008=2.3012\)

\(\left(\sqrt{X}-1\right)^2+\left(\sqrt{Y}-1\right)^2+\left(\sqrt{Z}-1\right)^2=2.3012-3.2008=0\)

\(X=1;Y=1;Z=1\Rightarrow x=2009;y=2010;z=2011\)

Điều kiện : \(x\ge2;y\ge-2009;z\ge2010;x+y+z\ge0\)

PT <=> \(2.\sqrt{x-2}+2.\sqrt{y+2009}+2.\sqrt{z-2010}=x+y+z\)

Áp dụng B ĐT Cô- si với 2 số dương a; b : \(2\sqrt{ab}\le a+b\) ta có:

\(2.\sqrt{x-2}\le x-2+1=x-1\)

\(2.\sqrt{y+2009}\le y+2009+1=y+2010\)

\(2.\sqrt{z-1010}\le z-2010+1=z-2009\)

=> \(2.\sqrt{x-2}+2.\sqrt{y+2009}+2.\sqrt{z-2010}\le x-1+y+2010+z-2009=x+y+z\)

Dấu "=" xảy ra <=> x - 2 = 1 ; y + 2009 = 1; z - 2010 = 1

=> x = 3; y = -2008; z = 2011 là nghiệm của PT

Điều kiện \(x\ge2\) vs \(y\ge-2009\) vs \(z\ge2010\)  Khi đó

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

nên => x=3 ; y=-2008 vs z=2011

\(x^2+2x\sqrt{x+\frac{1}{x}}=8x-1\)(đk;x>0)

\(\Leftrightarrow x^2+2\sqrt{x}\cdot\sqrt{x^2+1}=8x-1\)

\(\Leftrightarrow\left(x^2+1\right)+2\sqrt{x}\cdot\sqrt{x^2+1}+x=9x\)

\(\Leftrightarrow\left(\sqrt{x^2+1}+\sqrt{x}\right)^2-9x=0\)

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

\(\Leftrightarrow\left(\sqrt{x^2+1}+4\sqrt{x}\right)\left(\sqrt{x^2+1}-2\sqrt{x}\right)=0\)

\(\Leftrightarrow\sqrt{x^2+1}-2\sqrt{x}=0\)(vì \(\sqrt{x^2+1}+4\sqrt{x}>0\))

\(\Leftrightarrow x^2-4x+1=0\)

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

\(\Leftrightarrow\orbr{\begin{cases}x=2-\sqrt{3}\\x=2+\sqrt{3}\end{cases}}\)(thõa mãn điều kiện)

\(\sqrt{x-2009}-\sqrt{y-2008}-\sqrt{z-2}=\frac{1}{2}\left(x+y+z\right)\)(đk:x>2009,y>2008,z>2)

\(\Leftrightarrow\left(\sqrt{x-2009}-1\right)^2+\left(\sqrt{x-2008}+1\right)^2+\left(\sqrt{z-2}+1\right)^2+4014=0\)(không thõa mãn)

Lý do có kết quả trên là vì chuyển 1\2 qua vế trái và tách theo hằng đẳng thức

Bài tiếp theo cũng làm tương tự

k biết

tốt ghê ha

nếu vậy thì đừng trả lời

99,(9)% sai đề

gt pt nó thành nhân tử thay vào P tính

mk nhớ lm bài tương tự thế này r` bn chịu khó mở ra xem lại ở đây olm.vn/?g=page.display.showtrack&id=424601&limit=260, ấn vào chữ Trang tiếp theo để tìm thêm nhé

thôi mk lm luôn bn chờ tí

đề sai à

ráng làm nốt rồi đi ngủ thoyy

1.

a) ĐK: \(x\ge2\)

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

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

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-1}=1\\\sqrt{x-2}=\sqrt{x+3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x-1=1\\x-2=x+3\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\varnothing\end{matrix}\right.\)

Vậy...

b) \(\left(4x+2\right)\sqrt{x+8}=3x^2+7x+8\)

\(\Leftrightarrow2\left(2x+1\right)\sqrt{x+8}=4x^2+4x+1+x+8-x^2+2x-1\)

\(\Leftrightarrow2\left(2x+1\right)\sqrt{x+8}=\left(2x+1\right)^2+\left(x+8\right)-\left(x-1\right)^2\)

\(\Leftrightarrow\left(2x+1\right)^2-2\left(2x-1\right)\sqrt{x+8}+\left(x+8\right)-\left(x-1\right)^2=0\)

\(\Leftrightarrow\left(2x+1-\sqrt{x+8}\right)^2-\left(x-1\right)^2=0\)

\(\Leftrightarrow\left(2x+1-\sqrt{x+8}-x+1\right)\left(2x+1-\sqrt{x+8}+x-1\right)=0\)

\(\Leftrightarrow\left(x-\sqrt{x+8}+2\right)\left(3x-\sqrt{x+8}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+2=\sqrt{x+8}\\3x=\sqrt{x+8}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=1\end{matrix}\right.\)\(\Leftrightarrow x=1\)

Vậy...

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

Nhân cả 2 vế với \(\sqrt{2}\) ta được :

\(pt\Leftrightarrow\sqrt{2x+2\sqrt{2x-1}}+\sqrt{2x-2\sqrt{2x-1}}=2\)

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

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

Ta có : \(\left|\sqrt{2x-1}+1\right|+\left|\sqrt{2x-1}-1\right|\)

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

Dấu "=" xảy ra \(\Leftrightarrow\left(\sqrt{2x-1}+1\right)\left(1-\sqrt{2x-1}\right)\ge0\Leftrightarrow\frac{1}{2}\le x\le1\)

2) \(\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right):\frac{1}{x+y+z}=1\)

\(\Leftrightarrow\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{x+y+z}\)

\(\Leftrightarrow\frac{1}{x}+\frac{1}{y}=\frac{1}{x+y+z}-\frac{1}{z}\)

\(\Leftrightarrow\frac{x+y}{xy}=\frac{z-x-y-z}{z\left(x+y+z\right)}\)

\(\Leftrightarrow\frac{x+y}{xy}=\frac{-\left(x+y\right)}{z\left(x+y+z\right)}\)

\(\Leftrightarrow z\left(x+y\right)\left(x+y+z\right)=-xy\cdot\left(x+y\right)\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x+y=0\\y+z=0\\z+x=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=-y\\y=-z\\z=-x\end{matrix}\right.\)

TH1: \(x=-y\Leftrightarrow x^{29}=-y^{29}\Leftrightarrow x^{29}+y^{29}=0\)

Khi đó \(B=0\cdot\left(x^{11}+y^{11}\right)\cdot\left(x^{2013}+y^{2013}\right)=0\)

Tương tự 2 trường hợp còn lại ta đều được \(B=0\)

Vậy \(B=0\)