Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(x-2008=X;y-2009=Y;z-2010=Z\)
\(\sqrt{X}+\sqrt{Y}+\sqrt{Z}+3012=\frac{1}{2}\left(X+Y+Z+2008+2009+2010\right)\)
\(2.\sqrt{X}+2\sqrt{Y}+2\sqrt{Z}+2.3012=X+Y+Z+2009\cdot3\)
\(\left(X-2\sqrt{X}+1\right)+\left(Y-2\sqrt{Y}+1\right)+\left(Z-2\sqrt{Z}+1\right)+3.2008=2.3012\)
\(\left(\sqrt{X}-1\right)^2+\left(\sqrt{Y}-1\right)^2+\left(\sqrt{Z}-1\right)^2=2.3012-3.2008=0\)
\(X=1;Y=1;Z=1\Rightarrow x=2009;y=2010;z=2011\)
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
\(x^2+2x\sqrt{x+\frac{1}{x}}=8x-1\)(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
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
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.
\(ĐKXĐ:x\ge1;y\ge2;z\ge3\)
\(\Leftrightarrow x+y+z+8-2\sqrt{x-1}-4\sqrt{y-2}-6\sqrt{z-3}=0\)
\(\Leftrightarrow\left[\left(x-1\right)-2\sqrt{x-1}\cdot1+1\right]+\left[\left(y-2\right)-2\cdot\sqrt{y-2}\cdot2+4\right]+\left[\left(z-3\right)-2\cdot\sqrt{z-3}.3+9\right]=0\)
\(\Leftrightarrow\left(\sqrt{x-1}-1\right)^2+\left(\sqrt{y-2}-2\right)^2+\left(\sqrt{z-3}-3\right)^2=0\)
\(\Rightarrow\left\{{}\begin{matrix}\sqrt{x-1}-1=0\\\sqrt{y-2}-2=0\\\sqrt{z-3}-3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=6\\z=12\end{matrix}\right.\)