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: \(\Leftrightarrow x-1-2\sqrt{x-1}+1+y-2-4\sqrt{y-2}+4+z-3-6\sqrt{z-3}+9=0\)
=>\(\left(\sqrt{x-1}-1\right)^2+\left(\sqrt{y-2}-2\right)^2+\left(\sqrt{z-3}-3\right)^2=0\)

=>\(\left\{{}\begin{matrix}\sqrt{x-1}-1=0\\\sqrt{y-2}-2=0\\\sqrt{z-3}-3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-1=1\\y-2=4\\z-3=9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=6\\z=12\end{matrix}\right.\)

28. \(x^2+\frac{9x^2}{\left(x-3\right)^2}=40\) DK: \(x\ne3\)

PT\(\Leftrightarrow\left(x+\frac{3x}{x-3}\right)^2-6\frac{x^2}{x-3}-40=0\)\(\Leftrightarrow\frac{x^4}{\left(x-3\right)^2}-6\frac{x^2}{x-3}-40=0\)

Dat \(\frac{x^2}{x-3}=a\). PTTT \(a^2-6a-40=0\)\(\Leftrightarrow\left(a-10\right)\left(a+4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}a=10\\a=-4\end{matrix}\right.\)

giai tiep

14. \(\frac{1}{\sqrt{x}+1}+\frac{1}{\sqrt{x}-1}=1\) DK: \(\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)

PT\(\Leftrightarrow\frac{\sqrt{x}-1+\sqrt{x}+1}{x-1}=1\Leftrightarrow2\sqrt{x}=x-1\)\(\Leftrightarrow x-2\sqrt{x}+1=2\Leftrightarrow\left(\sqrt{x}-1\right)^2=2\)

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

Bạn ơi lần sau bạn đăng bài thì cố gắng đăng giãn giãn bớt bớt/ chia nhỏ bài ra chứ một cục bài như thế này nhìn rất đáng sợ và gây tâm lý ngại đọc nhé.

1. ĐKXĐ: $x\geq 1$

Đặt $x\sqrt{x-1}=a\Rightarrow x^3-x^2=x^2(x-1)=a^2$. PT đã cho trở thành:

$a^2+12a+20=0(*)$

Lại thấy rằng vì $x\geq 1$ nên $a\geq 0$

$\Rightarrow a^2+12a+20\geq 20>0$. Do đó $(*)$ vô nghiệm. Kéo theo PT ban đầu vô nghiệm.

2. ĐK: $x\geq -1$

$x^3+\sqrt{(x+1)^3}=9x+8$

$\Leftrightarrow x^3-9x-8+\sqrt{(x+1)^3}=0$

$\Leftrightarrow (x^2-x-8)(x+1)+(x+1)\sqrt{x+1}=0$

$\Leftrightarrow (x+1)(x^2-x-8+\sqrt{x+1})=0$

Nếu $x+1=0\Rightarrow x=-1$ (thỏa mãn)

Nếu $x^2-x-8+\sqrt{x+1}=0$

$\Leftrightarrow (x^2-9)-(x-3)+(\sqrt{x+1}-2)=0$

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

Dễ thấy với $x\geq -1$ thì biểu thức trong ngoặc lớn luôn lớn hơn $0$

Do đó $x-3=0\Rightarrow x=3$

Vậy $x=-1$ hoặc $x=3$

\(3\sqrt{9a^6}-6a^3=3\left|3a^3\right|-6a^3\)

Xét \(a\ge0\Rightarrow\) biểu thức \(=9a^3-6a^3=3a^3\)

Xét \(a< 0\Rightarrow\) biểu thức \(=-9a^3-6a^3=-15a^3\)

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

\(=1-x+3x-1\left(\dfrac{1}{3}< x\le1\right)=2x\)

\(\sqrt{2-\sqrt{3}}\left(\sqrt{6}+\sqrt{2}\right)=\sqrt{2-\sqrt{3}}.\sqrt{2}\left(\sqrt{3}+1\right)=\sqrt{4-2\sqrt{3}}\left(\sqrt{3}+1\right)\)

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

\(\left(\sqrt{10}+\sqrt{2}\right)\left(6-2\sqrt{5}\right)\sqrt{3+\sqrt{5}}=\left(\sqrt{5}+1\right)\left(\sqrt{5}-1\right)^2\sqrt{2}.\sqrt{3+\sqrt{5}}\)

\(=\left(\sqrt{5}+1\right)\left(\sqrt{5}-1\right)^2\sqrt{6+2\sqrt{5}}=\left(\sqrt{5}+1\right)\left(\sqrt{5}-1\right)^2\sqrt{\left(\sqrt{5}+1\right)^2}\)

\(=\left(\sqrt{5}+1\right)^2\left(\sqrt{5}-1\right)^2=4^2=16\)

\(\sqrt{23-8\sqrt{7}}+\sqrt{8-2\sqrt{7}}=\sqrt{\left(2\sqrt{7}-4\right)^2}+\sqrt{\left(\sqrt{7}-1\right)^2}\)

\(=2\sqrt{7}-4+\sqrt{7}-1=3\sqrt{7}-5\)

\(\sqrt{x+2\sqrt{x-1}}+\sqrt{x-2\sqrt{x-1}}\)

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

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

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

\(\sqrt{x+4\sqrt{x-4}}+\sqrt{x-4\sqrt{x-4}}\)

\(=\sqrt{x-4+4\sqrt{x-4}+4}+\sqrt{x-4-4\sqrt{x-4}+4}\)

\(=\sqrt{\left(\sqrt{x-4}+2\right)^2}+\sqrt{\left(\sqrt{x-4}-2\right)^2}=\left|\sqrt{x-4}+2\right|+\left|\sqrt{x-4}-2\right|\)

Xét \(x\ge8\Rightarrow\sqrt{x-4}\ge2\Rightarrow\)biểu thức \(=\sqrt{x-4}+2+\sqrt{x-4}-2\)

\(=2\sqrt{x-4}\)

Xét \(x< 8\Rightarrow\sqrt{x-4}< 2\Rightarrow\) biểu thức \(=\sqrt{x-4}+2+2-\sqrt{x-4}=4\)

b) đặt \(\sqrt{3x+1}=a\)(\(a\ge0\))

\(PT\Leftrightarrow\dfrac{a^2-1}{\sqrt{a^2+9}}+1=a\)

\(\Leftrightarrow\left(a-1\right)\left(1-\dfrac{a+1}{\sqrt{a^2+9}}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}a=1\\a+1=\sqrt{a^2+9}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}a=1\\a=4\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=5\end{matrix}\right.\)(tm)

c) bunyalovsky:

\(VT^2\le2\left(7-x+x-5\right)=4\)

\(\Leftrightarrow VT\le2\)

\(VF=\left(x-6\right)^2+2\ge2\)

Dấu = xảy ra khi x=6