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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\(\Leftrightarrow\left\{{}\begin{matrix}x=-1\\x^2=1\end{matrix}\right.\Leftrightarrow x=-1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\sqrt{3\left(x^2+2x+1\right)+9}\ge3\) (1)
\(\sqrt{5\left(x^4-2x^2+1\right)+4}\ge2\) (2)
\(\Rightarrow\left(1\right)+\left(2\right)\ge5\)
![](https://rs.olm.vn/images/avt/0.png?1311)
tách trong căn thành hđt thôi
căn thứ 1 >=3
căn thứ 2 >=2
=> đpcm
= \(\sqrt{3\left(x^2+2x+4\right)}+\sqrt{5x^2\left(x^2-2\right)+9}\)
=\(\sqrt{3\left(x^2+2x+1+3\right)}+\sqrt{5x^2\left(x^2-2\right)+9}\)
= \(\sqrt{3\left[\left(x+1\right)^2+3\right]}+\sqrt{5x^2\left(x^2-2\right)+9}\)
=\(3\left(x+1\right)+\sqrt{5}.x.x.\left(-\sqrt{2}\right)+3\)
=\(3\left(x+1\right)-\sqrt{10}.x^2+3\)
P/s: Mình mới học lớp 8 nên chỉ có thể khai triển như thế thôi, phần chứng minh bạn làm tiếp nhé.
![](https://rs.olm.vn/images/avt/0.png?1311)
2) năm mới chúc nhau niềm vui ( cho bài dễ thôi )
Vt >/ 3 + 2 = 5
VP </ 5
dấu = xảy ra khi x =-1
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\sqrt{3x^2+6x+12}+\sqrt{5x^2-10x^2+9}=\sqrt{3\left(x^2+2x+1\right)+9}+\sqrt{5\left(x^2-2x+1\right)+4}\)
\(\ge\sqrt{9}+\sqrt{4}=3+2=5\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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 \(x=-1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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: \(\Leftrightarrow x-1-2\sqrt{x-1}+1+y-2-4\sqrt{y-2}+4+z-3-6\sqrt{z-3}+9=0\)
=>\(\left(\sqrt{x-1}-1\right)^2+\left(\sqrt{y-2}-2\right)^2+\left(\sqrt{z-3}-3\right)^2=0\)
=>\(\left\{{}\begin{matrix}\sqrt{x-1}-1=0\\\sqrt{y-2}-2=0\\\sqrt{z-3}-3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-1=1\\y-2=4\\z-3=9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=6\\z=12\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
<=>\(\sqrt{3\left(x+1\right)^2+9}+\sqrt{5\left(x^2-1\right)^2+4}+2\left(x+1\right)^2=5\)
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
![](https://rs.olm.vn/images/avt/0.png?1311)
a/ ĐKXĐ: ...
\(\sqrt{x-7}-\frac{1}{2}+\sqrt{x-5}-\frac{3}{2}=0\)
\(\Leftrightarrow\frac{x-\frac{29}{4}}{\sqrt{x-7}+\frac{1}{2}}+\frac{x-\frac{29}{4}}{\sqrt{x-5}+\frac{3}{2}}=0\)
\(\Leftrightarrow\left(x-\frac{29}{4}\right)\left(\frac{1}{\sqrt{x-7}+\frac{1}{2}}+\frac{1}{\sqrt{x-5}+\frac{3}{2}}\right)=0\)
\(\Leftrightarrow x=\frac{29}{4}\)
b/ \(\Leftrightarrow\sqrt{x^2-6x+9}=3x+2\left(x\ge-\frac{2}{3}\right)\)
\(\Leftrightarrow x^2-6x+9=9x^2+12x+4\)
\(\Leftrightarrow8x^2-18x-5=0\Rightarrow\left[{}\begin{matrix}x=\frac{5}{2}\\x=-\frac{1}{4}\end{matrix}\right.\)
c/
\(\sqrt{3\left(x+1\right)^2+9}+\sqrt{5x^2\left(x^2+2\right)+9}=5-2\left(x+1\right)^2\)
Do \(\left\{{}\begin{matrix}3\left(x+1\right)^2+9\ge9\\5x^2\left(x^2+2\right)\ge9\end{matrix}\right.\) \(\Rightarrow VT\ge\sqrt{9}+\sqrt{9}=6\)
\(VP=5-2\left(x+1\right)^2\le5< VP\)
Pt luôn vô nghiệm
hẽhe kĩckDễ z sao đăg hả bn
\(\sqrt{3x^2+6x+12}+\sqrt{5x^4-10x^2+9}\)
\(=\sqrt{3\left(x+1\right)^2+9}+\sqrt{5\left(x^2-1\right)^2+4}\ge3+2=5\)