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a)\(\sqrt{3x^2+6x+7}+\sqrt{5x^2+10x+14}=4-2x-x^2\)
\(pt\Leftrightarrow\sqrt{3x^2+6x+3+4}+\sqrt{5x^2+10x+5+9}=-x^2-2x+4\)
\(\Leftrightarrow\sqrt{3\left(x^2+2x+1\right)+4}+\sqrt{5\left(x^2+2x+1\right)+9}=-x^2-2x+4\)
\(\Leftrightarrow\sqrt{3\left(x+1\right)^2+4}+\sqrt{5\left(x+1\right)^2+9}=-x^2-2x+4\)
Dễ thấy: \(\hept{\begin{cases}3\left(x+1\right)^2\ge0\\5\left(x+1\right)^2\ge0\end{cases}}\)\(\Rightarrow\hept{\begin{cases}3\left(x+1\right)^2+4\ge4\\5\left(x+1\right)^2+9\ge9\end{cases}}\)\(\Rightarrow\hept{\begin{cases}\sqrt{3\left(x+1\right)^2+4}\ge2\\\sqrt{5\left(x+1\right)^2+9}\ge3\end{cases}}\)
\(\Rightarrow VT=\sqrt{3\left(x+1\right)^2+4}+\sqrt{5\left(x+1\right)^2+9}\ge2+3=5\)
Và \(VP=-x^2-2x+4=-x^2-2x-1+5\)
\(=-\left(x^2+2x+1\right)+5=-\left(x+1\right)^2+5\le5\)
SUy ra \(VT\ge VP=5\Leftrightarrow x=-1\)
b)\(\sqrt{x-2\sqrt{x-1}}-\sqrt{x-1}=1\)
\(pt\Leftrightarrow\sqrt{x-1-2\sqrt{x-1}+1}-\sqrt{x-1}=1\)
\(\Leftrightarrow\left(\sqrt{x-1}-1\right)^2-\sqrt{x-1}=1\)
..... giải nốt tiếp ra x=1
c)Sửa đề \(\sqrt{x-7}+\sqrt{9-x}=x^2-16x+66\)
ĐK:....
Áp dụng BĐT Cauchy-Schwarz ta có:
\(VT^2=\left(\sqrt{x-7}+\sqrt{9-x}\right)^2\)
\(\le\left(1+1\right)\left(x-7+9-x\right)=4\)
\(\Rightarrow VT^2\le4\Rightarrow VT\le2\)
Lại có: \(VP=x^2-16x+66=x^2-16x+64+2\)
\(=\left(x-8\right)^2+2\ge2\)
Suy ra \(VT\ge VP=2\) khi \(VT=VP=2\)
\(\Rightarrow\left(x-8\right)^2+2=2\Rightarrow x-8=0\Rightarrow x=8\)
a) \(x^2-9\ge0\Leftrightarrow x^2\ge9\Leftrightarrow\orbr{\begin{cases}x\ge3\\x\ge-3\end{cases}}\)
b) \(-x-2\ge0\Leftrightarrow-x\ge2\Leftrightarrow x\ge-2\)
c) \(x^2+2x+1=\left(x+1\right)^2\)
\(\Rightarrow\left(x+1\right)^2\ge0\Leftrightarrow x+1\ge0\Leftrightarrow x\ge-1\)
\(\sqrt{9x^2-6x+5}=1-x^2\)
\(\Leftrightarrow9x^2-6x+5=\left(1-x^2\right)^2\)
\(\Leftrightarrow9x^2-6x+5=1-2x^2+x^4\)
\(\Leftrightarrow9x^2-6x+5-1+2x^2-x^4=0\)
\(\Leftrightarrow-x^4+11x^2-6x+4=0\)
\(\Leftrightarrow x^4-11x^2+6x-4=0\)
<=>\(\sqrt{9x^2-6x+5}=1-x^2\)
<=>\(\sqrt{\left(9x^2-6x+1\right)+4}=1-x^2\)
<=>\(\sqrt{\left(3x-1\right)^2+4}=1-x^2\)
<=> 3x - 1 + 2 = 1 - x2
<=> 3x + x2 = 1 +1 - 2
<=> x(3+x) = 0
<=> x = o hoặc 3+x =0 <=> x = -3
Vậy S= {0;-3}
a) ĐKXĐ: 1\(\le x\le7\)
phương trình <=> \(x-1-2\sqrt{x-1}+2\sqrt{7-x}-\sqrt{\left(7-x\right)\left(x-1\right)}=0\\ \Leftrightarrow\sqrt{x-1}\left(\sqrt{x-1}-2\right)-\sqrt{7-x}\left(\sqrt{x-1}-2\right)=0\\ \Leftrightarrow\left(\sqrt{x-1}-2\right)\left(\sqrt{x-1}-\sqrt{7-x}\right)=0\\\Leftrightarrow\left[{}\begin{matrix}x-1=4\\x-1=7-x\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=5\\x=4\end{matrix}\right.\left(thoả.mãn\right) \)
Vậy S={5,4} là tập nghiệm của phương trình
b) PT <=> \(2x^2-6x+4=\sqrt[2]{\left(x+2\right)\left(x^2-2x+4\right)}\)
Đặt \(\sqrt[2]{x+2}=y,\sqrt[2]{x^2-2x+4}=z\) (y,z>=0)
=> z^2-y^2=x^2-3x+2
pt<=> 2z^2-2y^2=3yz <=> (2z+y)(z-2y)=0
đến đó tự làm tự đặt dkxd
\(a,\sqrt{x-2\sqrt{x}-1}-\sqrt{x-1}=1.\)
\(\Rightarrow\sqrt{\left(\sqrt{x}-1\right)^2}-\sqrt{x-1}=1\)
\(\Rightarrow x-1-\sqrt{x-1}=1\)
\(\Rightarrow\sqrt{x-1}=x-1+1\)
\(\Rightarrow x-1=x^2\Rightarrow x^2-x+1=0\) ( vô nghiệm vì nó luôn lớn hơn 0 )
\(đkxđ\Leftrightarrow2x-1\ge0\Rightarrow x\ge\frac{1}{2}\)
\(c,\sqrt{x+\sqrt{2x-1}}+\sqrt{x-\sqrt{2x-1}}=\sqrt{2}.\)
\(\Rightarrow\sqrt{2x+2\sqrt{2x-1}}+\sqrt{2x-2\sqrt{2x-1}}=2\)
\(\Rightarrow\sqrt{2x-1+2\sqrt{2x-1}+1}+\sqrt{2x-1-2\sqrt{2x-1}+1}=2\)
\(\Rightarrow\sqrt{\left(\sqrt{2x-1}+1\right)^2}+\sqrt{\left(\sqrt{2x-1}-1\right)^2}=2\)
\(\Rightarrow\sqrt{2x-1}+1+\sqrt{2x-1}-1=2\)
\(\Rightarrow\sqrt{2x-1}+\sqrt{2x-1}=2\)
\(\Rightarrow\sqrt{2x-1}=1\Rightarrow\sqrt{2x-1}^2=1\)
\(\Rightarrow2x-1=1\Rightarrow2x=2\Leftrightarrow x=1\)\(\left(tm\right)\)
d tương tự nha , nhân thêm 2 vế với \(\sqrt{6}\)là ra
cần gấp thì mình làm cho
\(\sqrt{x^2+2x+1}=\sqrt{x+1}\left(đk:x\ge1\right)\)
\(< =>\sqrt{\left(x+1\right)^2}=\sqrt{x+1}\)
\(< =>x+1=\sqrt{x+1}\)
\(< =>\frac{x+1}{\sqrt{x+1}}=1\)
\(< =>\sqrt{x+1}=1< =>x=0\left(ktm\right)\)
ĐKXĐ : \(x\ge-1\)
Bình phương 2 vế , ta có :
\(x^2+2x+1=x+1\)
\(\Leftrightarrow x^2+2x+1-x-1=0\)
\(\Leftrightarrow x^2+x=0\)
\(\Leftrightarrow x\left(x+1\right)=0\Leftrightarrow\orbr{\begin{cases}x=0\\x=-1\end{cases}\left(TM\right)}\)\
Vậy ...............................
Ta có
\(\sqrt{-x^2+2x+2}=\sqrt{-x^2+2x-1+3}=\sqrt{-\left(x-1\right)^2+3}\le\sqrt{3}\)
\(\sqrt{-x^2-6x-8}=\sqrt{-x^2-6x-9+1}=\sqrt{-\left(x+3\right)^2+1}\le1\)
\(\Rightarrow\sqrt{-x^2+2x+2}+\sqrt{-x^2-6x-8}\le1+\sqrt{3}\)
Dấu "=" xảy ra khi x-1=0 và x+3=0 nên x=1 và x=-3(VL). Phương trình vô nghiệm