giúp với 

1.A^2= 1-x+8+x+2\(\sqrt{\left(1-x\right).\left(8+x\right)}\)= 9+2\(\sqrt{\left(x-1\right).\left(8+x\right)}\)

ta thấy \(\sqrt{\left(x-1\right).\left(8+x\right)}\)>= 0 =>A^2>= 9

KL: A min= 9 khi x=1 hoặc x=-8

1. ĐKXĐ: \(-4\le x\le6\)

\(\Leftrightarrow-x^2+2x+24+\sqrt{-x^2+2x+24}-12=0\)

Đặt \(\sqrt{-x^2+2x+24}=t\ge0\)

\(t^2+t-12=0\Rightarrow\left[{}\begin{matrix}t=3\\t=-4\left(l\right)\end{matrix}\right.\)

\(\Leftrightarrow\sqrt{-x^2+2x+24}=3\)

\(\Leftrightarrow-x^2+2x+15=0\) (casio)

2. ĐKXĐ: \(x\ge1\)

\(\Leftrightarrow3x^2-18=8\sqrt{x^3-1}-24\)

\(\Leftrightarrow3\left(x^2+2\right)=8\sqrt{\left(x-1\right)\left(x^2+x+1\right)}\)

Đặt \(\left\{{}\begin{matrix}\sqrt{x^2+x+1}=a>0\\\sqrt{x-1}=b\ge0\end{matrix}\right.\)

\(\Rightarrow3\left(a^2-b^2\right)=8ab\)

\(\Leftrightarrow3a^2-8ab-3b^2=0\)

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

\(\Leftrightarrow a=3b\) (do \(3a+b>0\))

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

\(\Leftrightarrow x^2+x+1=9\left(x-1\right)\) (casio)

\(\left(x-1\right)\left(\sqrt{3x+4}-1\right)=3\left(x+1\right)\)

\(\Leftrightarrow x=7\)

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 (x−1)(√3x+4−1)=3(x+1)  ⇔x=7

tk mk nha

1)chia x^2 cho 2 vế

ra được hđt giải bt

chịu thua

b/ ĐKXĐ:...

\(\Leftrightarrow x-19-2\sqrt{x-19}+1+y-7-4\sqrt{y-7}+4+z-1997-6\sqrt{z-1997}+9=0\)

\(\Leftrightarrow\left(\sqrt{x-19}-1\right)^2+\left(\sqrt{y-7}-2\right)^2+\left(\sqrt{z-1997}-3\right)^2=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x-19}=1\\\sqrt{y-7}=2\\\sqrt{z-1997}=3\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=20\\y=11\\z=2006\end{matrix}\right.\)

c/ ĐKXĐ: \(x\ge-1\)

\(\Leftrightarrow10\sqrt{\left(x+1\right)\left(x^2-x+1\right)}=3\left(x^2+2\right)\)

Đặt \(\left\{{}\begin{matrix}\sqrt{x+1}=a\\\sqrt{x^2-x+1}=b\end{matrix}\right.\) \(\Rightarrow a^2+b^2=x^2+2\)

Pt tương đương:

\(10ab=3\left(a^2+b^2\right)\Leftrightarrow3a^2-10ab+3b^2=0\)

\(\Leftrightarrow\left(3a-b\right)\left(a-3b\right)=0\Rightarrow\left[{}\begin{matrix}3a=b\\a=3b\end{matrix}\right.\)

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

\(\Leftrightarrow\left[{}\begin{matrix}9\left(x+1\right)=x^2-x+1\\x+1=9\left(x^2-x+1\right)\end{matrix}\right.\) \(\Leftrightarrow...\)

a/ ĐKXĐ; \(-1\le x\le8\)

Đặt \(\sqrt{1+x}+\sqrt{8-x}=t>0\Rightarrow\sqrt{\left(1+x\right)\left(8-x\right)}=\frac{t^2-9}{2}\)

\(\Rightarrow t+\frac{t^2-9}{2}=3\)

\(\Leftrightarrow t^2+2t-15=0\Rightarrow\left[{}\begin{matrix}t=3\\t=-5\left(l\right)\end{matrix}\right.\)

\(\Rightarrow\sqrt{1+x}+\sqrt{8-x}=3\)

\(\Leftrightarrow9+2\sqrt{\left(1+x\right)\left(8-x\right)}=9\)

\(\Leftrightarrow\left(1+x\right)\left(8-x\right)=0\Rightarrow\left[{}\begin{matrix}x=-1\\x=8\end{matrix}\right.\)

7.  \(S=9y^2-12\left(x+4\right)y+\left(5x^2+24x+2016\right)\)

\(=9y^2-12\left(x+4\right)y+4\left(x+4\right)^2+\left(x^2+8x+16\right)+1936\)

\(=\left[3y-2\left(x+4\right)\right]^2+\left(x-4\right)^2+1936\ge1936\)

Vậy   \(S_{min}=1936\)    \(\Leftrightarrow\)    \(\hept{\begin{cases}3y-2\left(x+4\right)=0\\x-4=0\end{cases}}\)    \(\Leftrightarrow\)    \(\hept{\begin{cases}x=4\\y=\frac{16}{3}\end{cases}}\)

8. \(x^2-5x+14-4\sqrt{x+1}=0\)       (ĐK: x > = -1).

\(\Leftrightarrow\)   \(\left(x+1\right)-4\sqrt{x+1}+4+\left(x^2-6x+9\right)=0\)

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

Với mọi x thực ta luôn có:   \(\left(\sqrt{x+1}-2\right)^2\ge0\)   và   \(\left(x-3\right)^2\ge0\) 

Suy ra   \(\left(\sqrt{x+1}-2\right)^2+\left(x-3\right)^2\ge0\)

Đẳng thức xảy ra   \(\Leftrightarrow\)   \(\hept{\begin{cases}\left(\sqrt{x+1}-2\right)^2=0\\\left(x-3\right)^2=0\end{cases}}\)    \(\Leftrightarrow\)    x = 3 (Nhận)