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

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

ĐK: \(x\ge-\dfrac{5}{2}\)

\(\Leftrightarrow3x^2-4x-4=2x+5\)

\(\Leftrightarrow3x^2-6x-9=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=3\end{matrix}\right.\) (thỏa mãn)

b.

ĐKXĐ: \(3\le x\le8\)

\(\Leftrightarrow-x^2+11x-24-\sqrt{-x^2+11x-24}-2=0\)

Đặt \(\sqrt{-x^2+11x-24}=t\ge0\)

\(\Rightarrow t^2-t-2=0\Rightarrow\left[{}\begin{matrix}t=-1\left(loại\right)\\t=2\end{matrix}\right.\)

\(\Leftrightarrow\sqrt{-x^2+11x-24}=2\)

\(\Leftrightarrow-x^2+11x-28=0\Rightarrow\left[{}\begin{matrix}x=7\\x=4\end{matrix}\right.\)

Nhìn không đủ chán rồi không dám động vào

Viết đề kiểu gì v @@

À do nãy máy lag sr :) Chứ bài đặt ẩn phụ mệt lắm :)

ĐK: x \(\ge\)\(\frac{8}{3}\)

pt <=> \(4.\left(x-3\right)+9-3.\sqrt{5x-6}=\sqrt{3x-8}-1\)

<=> \(4.\left(x-3\right)+3.\left(3-\sqrt{5x-6}\right)=\sqrt{3x-8}-1\)

<=>  \(4.\left(x-3\right)+3.\frac{\left(3-\sqrt{5x-6}\right)\left(3+\sqrt{5x-6}\right)}{3+\sqrt{5x-6}}=\frac{\left(\sqrt{3x-8}-1\right)\left(\sqrt{3x-8}+1\right)}{\sqrt{3x-8}+1}\)

<=> \(4.\left(x-3\right)+3.\frac{9-5x+6}{3+\sqrt{5x-6}}=\frac{3x-8-1}{\sqrt{3x-8}+1}\)

<=> \(4.\left(x-3\right)+15.\frac{3-x}{3+\sqrt{5x-6}}-3.\frac{x-3}{\sqrt{3x-8}+1}=0\)

<=> \(\left(x-3\right)\left(4-\frac{15}{3+\sqrt{5x-6}}-\frac{3}{\sqrt{3x-8}+1}\right)=0\)

<=> x = 3 (thoả mãn) hoặc \(4-\frac{15}{3+\sqrt{5x-6}}-\frac{3}{\sqrt{3x-8}+1}=0\) (2)

Giải (2):  (2) <=> \(\frac{15}{6}-\frac{15}{3+\sqrt{5x-6}}+\frac{3}{2}-\frac{3}{\sqrt{3x-8}+1}=0\)

<=> \(15\left(\frac{1}{6}-\frac{1}{3+\sqrt{5x-6}}\right)+3.\left(\frac{1}{2}-\frac{1}{\sqrt{3x-8}+1}\right)=0\)

<=>  \(15.\frac{\sqrt{5x-6}-3}{6.\left(3+\sqrt{5x-6}\right)}+3.\frac{\sqrt{3x-8}-1}{2.\left(\sqrt{3x-8}+1\right)}=0\)

<=> \(15.\frac{5.\left(x-3\right)}{6.\left(3+\sqrt{5x-6}\right)^2}+3.\frac{3.\left(x-3\right)}{2.\left(\sqrt{3x-8}+1\right)^2}=0\)

<=> \(\left(x-3\right).\left(\frac{75}{6.\left(3+\sqrt{5x-6}\right)^2}+\frac{9}{2.\left(\sqrt{3x-8}+1\right)^2}\right)=0\)

<=> x = 3 Vì \(\frac{75}{6.\left(3+\sqrt{5x-6}\right)^2}+\frac{9}{2.\left(\sqrt{3x-8}+1\right)^2}>0\) với mọi x \(\ge\frac{8}{3}\)

Vậy pt có 1 nghiệm duy nhất x = 3

Bài 1:

Ta có: \(\left(3\sqrt{50}-5\sqrt{18}+3\sqrt{8}\right)\cdot\sqrt{2}\)

\(=\left(15\sqrt{2}-15\sqrt{2}+6\sqrt{2}\right)\cdot\sqrt{2}\)

\(=6\sqrt{2}\cdot\sqrt{2}\)

=12

Bài 2: 

1) ĐKXĐ: \(x\le0\)

2) ĐKXĐ: \(x\le2\)

3) ĐKXĐ: \(x>\dfrac{-3}{2}\)

4) ĐKXĐ: x>0

5) ĐKXĐ: x<3