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

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

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

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

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

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

đề sai à

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

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

k biết

tốt ghê ha

nếu vậy thì đừng trả lời

99,(9)% sai đề

M = x.√[(2008+y²).(2008+z²)\(2008+x²)] + y.√[(2008+x²).(2008+z²)\(2008+y²)] + z.√[(2008+y²).(2008+x²)\(2008+z²)]

ta có:
2008 + x² = xy + xz + yz + x²
2008 + x² = (x+y).(x+z)
tương tự: 2008 + y² = (x+y).(y+z) và 2008 + z² = (z+y).(x+z)
chỉ việc thay vào rùi rút gọn thui

=> M = x.√[(x+y).(y+z).(x+z).(z+y)\ (x+y).(x+z)] + y.√[(x+y).(x+z).(x+z).(z+y)\(y+x).(y+z)] + z.√[(x+y).(x+z).(y+z).(y+x)\(x+z).(z+y)]

=> M = x.|y+z| + y.|z+x| + z.|x+y|
=> M = 2.2008

Thay \(xy+yz+xz=2018\) ta được:

\(\left\{{}\begin{matrix}2018+x^2=x^2+xy+yz+xz=\left(x+y\right)\left(x+z\right)\\2018+y^2=y^2+xy+yz+xz=\left(y+z\right)\left(x+y\right)\\2018+z^2=z^2+xy+yz+xz=\left(x+z\right)\left(y+z\right)\end{matrix}\right.\)

Sau đó thay vào lần lượt đề bài là được

Câu 1:

\(A=21\left(a+\frac{1}{b}\right)+3\left(b+\frac{1}{a}\right)=21a+\frac{21}{b}+3b+\frac{3}{a}\)

\(=(\frac{a}{3}+\frac{3}{a})+(\frac{7b}{3}+\frac{21}{b})+\frac{62}{3}a+\frac{2b}{3}\)

Áp dụng BĐT Cô-si:
\(\frac{a}{3}+\frac{3}{a}\geq 2\sqrt{\frac{a}{3}.\frac{3}{a}}=2\)

\(\frac{7b}{3}+\frac{21}{b}\geq 2\sqrt{\frac{7b}{3}.\frac{21}{b}}=14\)

Và do $a,b\geq 3$ nên:

\(\frac{62}{3}a\geq \frac{62}{3}.3=62\)

\(\frac{2b}{3}\geq \frac{2.3}{3}=2\)

Cộng tất cả những BĐT trên ta có:

\(A\geq 2+14+62+2=80\) (đpcm)

Dấu "=" xảy ra khi $a=b=3$

Câu 2:

Bình phương 2 vế ta thu được:

\((x^2+6x-1)^2=4(5x^3-3x^2+3x-2)\)

\(\Leftrightarrow x^4+12x^3+34x^2-12x+1=20x^3-12x^2+12x-8\)

\(\Leftrightarrow x^4-8x^3+46x^2-24x+9=0\)

\(\Leftrightarrow (x^2-4x)^2+6x^2+24(x-\frac{1}{2})^2+3=0\) (vô lý)

Do đó pt đã cho vô nghiệm.

a) \(\sqrt{25-x^2}-\sqrt{10-x^2}=3\) (*)

Đk: \(-\sqrt{10}\le x\le\sqrt{10}\)

(*) \(\Leftrightarrow\sqrt{25-x^2}=3+\sqrt{10-x^2}\Leftrightarrow25-x^2=19-x^2+6\sqrt{10-x^2}\)

\(\Leftrightarrow6\sqrt{10-x^2}=6\Leftrightarrow\sqrt{10-x^2}=1\Leftrightarrow\left[{}\begin{matrix}x=-3\left(N\right)\\x=3\left(N\right)\end{matrix}\right.\)

Kl: x = +- 3

b) \(\sqrt{x^2-x-6}+x^2-x-18=0\) (*)

đk: \(\left[{}\begin{matrix}x\le-2\\x\ge3\end{matrix}\right.\)

(*) \(\Leftrightarrow x^2-x-6+\sqrt{x^2-x-6}-12=0\)

Đặt \(t=\sqrt{x^2-x-6}\Rightarrow t^2=x^2-x-6\) (t >/ 0)

phương trình (*) trở thành : \(t^2+t-12=0\Leftrightarrow\left[{}\begin{matrix}t=3\left(N\right)\\t=-4\left(L\right)\end{matrix}\right.\)

Với t=3. ta có: \(\sqrt{x^2-x-6}=3\Leftrightarrow x^2-x-15=0\Leftrightarrow x=\dfrac{1\pm\sqrt{61}}{2}\left(N\right)\)

Kl: \(x=\dfrac{1\pm\sqrt{61}}{2}\)

c) \(\sqrt{x-2009}+\sqrt{y+2008}+\sqrt{z-2}=\dfrac{1}{2}\left(x+y+z\right)\) (*)

Đk: \(\left\{{}\begin{matrix}x\ge2009\\y\ge-2008\\z\ge2\end{matrix}\right.\)

(*) \(\Leftrightarrow2\sqrt{x-2009}+2\sqrt{y+2008}+2\sqrt{z-2}=x+y+z\)

\(\Leftrightarrow\left(x-2009-2\sqrt{x-2009}+1\right)+\left(y+2008-2\sqrt{y+2008}+1\right)+\left(z-2-2\sqrt{z-2}+1\right)=0\)

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

\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x-2009}=1\\\sqrt{y+2008}=1\\\sqrt{z-2}=1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=2010\left(N\right)\\y=-2007\left(N\right)\\z=3\left(N\right)\end{matrix}\right.\)

Kl: x= 2010, y= -2007, z=3

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