\(\sqrt{5x^4}+10x+30+\sqrt{3x^2}+6x+12-8=0\)
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\(VT=\sqrt{3\left(x^2+2x+1\right)+9}+\sqrt{5\left(x^2-2x+1\right)+25}\ge\sqrt{9}+\sqrt{25}=8\)
Do dấu "=" ko đồng thời xảy ra ở hai bđt nên pt vô nghiệm
\(\sqrt{3\left(x+1\right)^2+9}-3+\sqrt{5\left(x^2-1\right)^2+25}-5=0\)
\(\Leftrightarrow\frac{3\left(x+1\right)^2}{\sqrt{3\left(x+2\right)^2+9}+3}+\frac{5\left(x+1\right)^2\left(x-1\right)^2}{\sqrt{5\left(x^2-1\right)^2+25}+5}=0\)
\(\Leftrightarrow\left(x+1\right)^2\left(\frac{3}{\sqrt{3\left(x+2\right)^2+9}+3}+\frac{5\left(x-1\right)^2}{\sqrt{5\left(x^2-1\right)^2+25}+5}\right)=0\)
\(\left(\frac{3}{\sqrt{3\left(x+2\right)^2+9}+3}+\frac{5\left(x-1\right)^2}{\sqrt{5\left(x^2-1\right)^2+25}+5}\right)>0\left(\forall x\right)\)
\(\Rightarrow x=-1\)
Bạn kia làm sai rùi ạ chắc nhìn nhầm đề
6: \(\Leftrightarrow2x^2+3x+9+\sqrt{2x^2+3x+9}-42=0\)
Đặt \(\sqrt{2x^2+3x+9}=a\left(a>=0\right)\)
Phương trình sẽ trở thành là: a^2+a-42=0
=>(a+7)(a-6)=0
=>a=-7(loại) hoặc a=6(nhận)
=>2x^2+3x+9=36
=>2x^2+3x-27=0
=>2x^2+9x-6x-27=0
=>(2x+9)(x-3)=0
=>x=3 hoặc x=-9/2
8: \(\Leftrightarrow x-1-2\sqrt{x-1}+1+y-2-4\sqrt{y-2}+4+z-3-6\sqrt{z-3}+9=0\)
=>\(\left(\sqrt{x-1}-1\right)^2+\left(\sqrt{y-2}-2\right)^2+\left(\sqrt{z-3}-3\right)^2=0\)
=>\(\left\{{}\begin{matrix}\sqrt{x-1}-1=0\\\sqrt{y-2}-2=0\\\sqrt{z-3}-3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-1=1\\y-2=4\\z-3=9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=6\\z=12\end{matrix}\right.\)
Ta có:
\(\sqrt{3x^2+6x+12}+\sqrt{5x^4-10x^2+9}=\sqrt{3\left(x+1\right)^2+9}+\sqrt{5\left(x^2-1\right)^2+4}\ge\sqrt{9}+\sqrt{4}=5\)
\(3-4x-2x^2=5-2\left(x+1\right)^2\le5\)
Đẳng thức xảy ra khi và chỉ khi:
\(\left\{{}\begin{matrix}3\left(x+1\right)^2=0\\5\left(x^2-1\right)^2=0\\2\left(x+1\right)^2=0\end{matrix}\right.\) \(\Rightarrow x=-1\)
Vậy pt có nghiệm duy nhất \(x=-1\)
a) ĐK: \(x\ge5\)
\(\sqrt{4x-20}+\frac{1}{3}\sqrt{9x-45}-\frac{1}{5}\sqrt{16x-80}=0\)
\(\Leftrightarrow\)\(\sqrt{4\left(x-5\right)}+\frac{1}{3}\sqrt{9\left(x-5\right)}-\frac{1}{5}\sqrt{16\left(x-5\right)}=0\)
\(\Leftrightarrow\)\(2\sqrt{x-5}+\sqrt{x-5}-\frac{4}{5}\sqrt{x-5}=0\)
\(\Leftrightarrow\)\(\frac{11}{5}\sqrt{x-5}=0\)
\(\Leftrightarrow\)\(x-5=0\)
\(\Leftrightarrow\)\(x=5\) (t/m)
Vậy
b) \(-5x+7\sqrt{x}=-12\)
\(\Leftrightarrow\)\(5x-7\sqrt{x}-12=0\)
\(\Leftrightarrow\)\(\left(\sqrt{x}+1\right)\left(5\sqrt{x}-12\right)=0\)
đến đây tự làm
c) d) e) bạn bình phương lên
f) \(VT=\sqrt{3\left(x^2+2x+1\right)+9}+\sqrt{5\left(x^4-2x^2+1\right)+25}\)
\(=\sqrt{3\left(x+1\right)^2+9}+\sqrt{5\left(x^2-1\right)^2}\)
\(\ge\sqrt{9}+\sqrt{25}=8\)
Dấu "=" xảy ra \(\Leftrightarrow\)\(\hept{\begin{cases}x+1=0\\x^2-1=0\end{cases}}\)\(\Leftrightarrow\)\(x=-1\)
Vậy...
\(\sqrt{3x^2+6x+12}+\sqrt{5x^2-10x^2+9}=\sqrt{3\left(x^2+2x+1\right)+9}+\sqrt{5\left(x^2-2x+1\right)+4}\)
\(\ge\sqrt{9}+\sqrt{4}=3+2=5\)
1)
ĐK: \(x\geq 5\)
PT \(\Leftrightarrow \sqrt{4(x-5)}+3\sqrt{\frac{x-5}{9}}-\frac{1}{3}\sqrt{9(x-5)}=6\)
\(\Leftrightarrow \sqrt{4}.\sqrt{x-5}+3\sqrt{\frac{1}{9}}.\sqrt{x-5}-\frac{1}{3}.\sqrt{9}.\sqrt{x-5}=6\)
\(\Leftrightarrow 2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}=6\)
\(\Leftrightarrow 2\sqrt{x-5}=6\Rightarrow \sqrt{x-5}=3\Rightarrow x=3^2+5=14\)
2)
ĐK: \(x\geq -1\)
\(\sqrt{x+1}+\sqrt{x+6}=5\)
\(\Leftrightarrow (\sqrt{x+1}-2)+(\sqrt{x+6}-3)=0\)
\(\Leftrightarrow \frac{x+1-2^2}{\sqrt{x+1}+2}+\frac{x+6-3^2}{\sqrt{x+6}+3}=0\)
\(\Leftrightarrow \frac{x-3}{\sqrt{x+1}+2}+\frac{x-3}{\sqrt{x+6}+3}=0\)
\(\Leftrightarrow (x-3)\left(\frac{1}{\sqrt{x+1}+2}+\frac{1}{\sqrt{x+6}+3}\right)=0\)
Vì \(\frac{1}{\sqrt{x+1}+2}+\frac{1}{\sqrt{x+6}+3}>0, \forall x\geq -1\) nên $x-3=0$
\(\Rightarrow x=3\) (thỏa mãn)
Vậy .............
Ta có: Vế trái = \(\sqrt{3\left(x+1\right)^2+9}+\sqrt{5\left(x^2-1\right)^24}\ge\sqrt{9}+\sqrt{4}=5\)
Dấu "=" xảy ra\(\Leftrightarrow\left\{{}\begin{matrix}x=-1\\x^2=1\end{matrix}\right.\Leftrightarrow x=-1\)
\(\sqrt{3\left(x^2+2x+1\right)+9}\ge3\) (1)
\(\sqrt{5\left(x^4-2x^2+1\right)+4}\ge2\) (2)
\(\Rightarrow\left(1\right)+\left(2\right)\ge5\)
<=>\(\sqrt{3\left(x+1\right)^2+9}+\sqrt{5\left(x^2-1\right)^2+4}+2\left(x+1\right)^2=5\)
mà \(\sqrt{3\left(x+1\right)^2+9}\ge3\), \(\sqrt{5\left(x^2-1\right)^2+4}\ge4\), \(2\left(x+1\right)^2\ge0\)với mọi x
=>\(\sqrt{3\left(x+1\right)^2+9}+\sqrt{5\left(x^2-1\right)^2+4}+2\left(x+1\right)^2\ge3+2+0=5\)
'=" xảy ra<=> x+1=0<=> x=-1