Ta có: \(\sqrt{\left(5-2\sqrt{6}\right)^2}+\sqrt{\left(5+2\sqrt{6}\right)^x}=10\)

\(\Leftrightarrow\sqrt{\left(5+2\sqrt{6}\right)^x}=10-5+2\sqrt{6}=5+2\sqrt{6}\)

\(\Leftrightarrow\left(5+2\sqrt{6}\right)^x=\left(5+2\sqrt{6}\right)^2\)

hay x=2

6:ĐKXĐ: x>=0; x<>1/25

BPT=>\(\dfrac{3\sqrt{x}}{5\sqrt{x}-1}+3< =0\)

=>\(\dfrac{3\sqrt{x}+15\sqrt{x}-5}{5\sqrt{x}-1}< =0\)

=>\(\dfrac{18\sqrt{x}-5}{5\sqrt{x}-1}< =0\)

=>\(\dfrac{1}{5}< \sqrt{x}< =\dfrac{5}{18}\)

=>\(\dfrac{1}{25}< x< =\dfrac{25}{324}\)

7:

ĐKXĐ: x>=0

BPT \(\Leftrightarrow\dfrac{\sqrt{x}+1}{2\sqrt{x}+3}>\dfrac{8}{3}:\dfrac{8}{3}=1\)

=>\(\dfrac{\sqrt{x}+1}{2\sqrt{x}+3}-1>=0\)

=>\(\dfrac{\sqrt{x}+1-2\sqrt{x}-3}{2\sqrt{x}+3}>=0\)

=>\(-\sqrt{x}-2>=0\)(vô lý)

8:

ĐKXĐ: x>=0; x<>9/4

BPT \(\Leftrightarrow\dfrac{\sqrt{x}-2}{2\sqrt{x}-3}+4< 0\)

=>\(\dfrac{\sqrt{x}-2+8\sqrt{x}-12}{2\sqrt{x}-3}< 0\)

=>\(\dfrac{9\sqrt{x}-14}{2\sqrt{x}-3}< 0\)

TH1: 9căn x-14>0 và 2căn x-3<0

=>căn x>14/9 và căn x<3/2

=>14/9<căn x<3/2

=>196/81<x<9/4

TH2: 9căn x-14<0 và 2căn x-3>0

=>căn x>3/2 hoặc căn x<14/9

mà 3/2<14/9

nên trường hợp này Loại

9: 

ĐKXĐ: x>=0

\(BPT\Leftrightarrow\dfrac{2\sqrt{x}+3}{5\sqrt{x}+7}< =-\dfrac{1}{3}\)

=>\(\dfrac{2\sqrt{x}+3}{5\sqrt{x}+7}+\dfrac{1}{3}< =0\)

=>\(\dfrac{6\sqrt{x}+9+5\sqrt{x}+7}{3\left(5\sqrt{x}+7\right)}< =0\)

=>\(\dfrac{11\sqrt{x}+16}{3\left(5\sqrt{x}+7\right)}< =0\)(vô lý)

10: 

ĐKXĐ: x>=0; x<>1/49

\(BPT\Leftrightarrow\dfrac{6\sqrt{x}-2}{7\sqrt{x}-1}+6>0\)

=>\(\dfrac{6\sqrt{x}-2+42\sqrt{x}-6}{7\sqrt{x}-1}>0\)

=>\(\dfrac{48\sqrt{x}-8}{7\sqrt{x}-1}>0\)

=>\(\dfrac{6\sqrt{x}-1}{7\sqrt{x}-1}>0\)

TH1: 6căn x-1>0 và 7căn x-1>0

=>căn x>1/6 và căn x>1/7

=>căn x>1/6

=>x>1/36

TH2: 6căn x-1<0 và 7căn x-1<0

=>căn x<1/6 và căn x<1/7

=>căn x<1/7

=>0<=x<1/49

câu 9 nhầm đề bài r bạn

Đặt a = x - 2 => x - 1 = a + 1; x - 3 = a -1

Khi đó, A = (a+1)⁴ + (a - 1)⁴ + 6.(a + 1)² .(a - 1)²

A = [(a + 1)²  + (a - 1)²]² + 4.(a + 1)² .(a - 1)²

 = (a² + 2a + 1 + a² - 2a + 1)² + 4.(a² - 1)²

= (2a² +2)² + 4.(a⁴ - 2a² + 1)

= 4a⁴ + 8a² + 4 + 4a⁴ - 8a² + 4 = 8a⁴ + 8 \(\ge\) 8 với mọi a

=> min A = 8 khi a = 0 <=> x - 2 = 0 <=>  x= 2

Ta có:

\(\left|x+\frac{1}{2}\right|\ge0;\left|x+\frac{1}{6}\right|\ge0;....;\left|x+\frac{1}{110}\right|\ge0\)

\(\Rightarrow VT\ge0\Rightarrow VP\ge0\)

\(\Rightarrow x+\frac{1}{2}+x+\frac{1}{6}+...+\frac{1}{110}=11x\)

\(\Rightarrow\left(x+...+x\right)+\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{110}\right)=11x\)

\(\Rightarrow10x+\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{10\cdot11}\right)=11x\)

\(\Rightarrow10x+\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{10}-\frac{1}{11}\right)=11x\)

\(\Rightarrow10x+\left(1-\frac{1}{11}\right)=11x\)

\(\Rightarrow10x+\frac{10}{11}=11x\)

\(\Rightarrow x=\frac{10}{11}\)

mk viết vội nên nhầm dòng thứ 4 từ trên xuống, bn sửa 1/3 thành 1/6 nhé

kq vẫn đúng đấy

a

d) Ta có: \(D=\left(\dfrac{5\sqrt{x}-6}{x-9}-\dfrac{2}{\sqrt{x}+3}\right):\left(1+\dfrac{6}{x-9}\right)\)

\(=\dfrac{5\sqrt{x}-6-2\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}:\dfrac{x-9+6}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)

\(=\dfrac{5\sqrt{x}-6-2\sqrt{x}+6}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\cdot\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}{x-3}\)

\(=\dfrac{3\sqrt{x}}{x-3}\)

f) Ta có: \(\left(\dfrac{3}{\sqrt{1+x}}+\sqrt{1-x}\right):\left(\dfrac{3}{\sqrt{1-x^2}}+1\right)\)

\(=\dfrac{3+\sqrt{1-x^2}}{\sqrt{1+x}}:\dfrac{3+\sqrt{1-x^2}}{\sqrt{1-x^2}}\)

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

\(< =>\dfrac{13\left(x+3\right)}{\left(2x+7\right)\left(x-3\right)\left(x+3\right)}+\dfrac{x^2-9}{\left(2x+7\right)\left(x-3\right)\left(x+3\right)}=\dfrac{6\left(2x+7\right)}{\left(2x+7\right)\left(x-3\right)\left(x+3\right)}\left(ĐK:x\ne\left\{-\dfrac{7}{2};3;-3\right\}\right)\\ =>13x+39+x^2-9=12x+42\\ < =>x^2+x-12=0\\ < =>\left(x+4\right)\left(x-3\right)=0\\ =>\left[{}\begin{matrix}x=-4\left(TM\right)\\x=3\left(KTM\right)\end{matrix}\right.\\ =>S=\left\{-4\right\}\)

\(ĐKXĐ:x\ne\dfrac{7}{2}\) và \(x\ne\pm3\)

mẫu chung : \(\left(2x+7\right)\left(x+3\right)\left(x-3\right)\)

Khử mẫu ta được :

\(13\left(x+3\right)+\left(x+3\right)\left(x-3\right)=6\left(2x+7\right)\)

\(\Leftrightarrow x^2+x-12=0\)

\(\Leftrightarrow\left(x+4\right)\left(x-3\right)=0\)

\(x=\left\{{}\begin{matrix}-4\\3\end{matrix}\right.\)

do \(x=3\) không thỏa mãn điều kiện thích hợp nên pt có nghiệm duy nhất là : \(-4\)

\(Vậy...\)

a/ \(\left(x+3\right)\left(3\left(x^2+1\right)^2+2\left(x+3\right)^2\right)=5\left(x^2+1\right)^3\)

\(\Leftrightarrow3\left(x+3\right)\left(x^2+1\right)^2+2\left(x+3\right)^3-5\left(x^2+1\right)^3=0\)

\(\Leftrightarrow3\left(x+3\right)\left(x^2+1\right)^2-3\left(x^2+1\right)^3+2\left(x+3\right)^3-2\left(x^2+1\right)^3=0\)

\(\Leftrightarrow3\left(x^2+1\right)^2\left(-x^2+x+2\right)+2\left(-x^2+x+2\right)\left(\left(x+3\right)^2+\left(x+3\right)\left(x^2+1\right)+\left(x^2+1\right)^2\right)=0\)

\(\Leftrightarrow\left(-x^2+x+2\right)\left[3\left(x^2+1\right)^2+2\left(x+3+\dfrac{x^2+1}{2}\right)^2+\dfrac{3\left(x^2+1\right)^2}{4}\right]=0\)

\(\Leftrightarrow-x^2+x+2=0\) (phần ngoặc phía sau luôn dương)

\(\Rightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)

b/ \(3\left(x^2+2x-1\right)^2-2\left(x^2+3x-1\right)^2+5\left(x^2+3x-1-\left(x^2+2x-1\right)\right)^2=0\)

Đặt \(\left\{{}\begin{matrix}a=x^2+2x-1\\b=x^2+3x-1\end{matrix}\right.\)

\(3a^2-2b^2+5\left(b-a\right)^2=0\Leftrightarrow8a^2+3b^2-10ab=0\)

\(\Leftrightarrow\left(4a-3b\right)\left(2a-b\right)=0\Leftrightarrow\left[{}\begin{matrix}4a=3b\\2a=b\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}4\left(x^2+2x-1\right)=3\left(x^2+3x-1\right)\\2\left(x^2+2x-1\right)=x^2+3x-1\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x^2-x-1=0\\x^2+x-1=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{1+\sqrt{5}}{2}\\x=\dfrac{1-\sqrt{5}}{2}\\x=\dfrac{-1+\sqrt{5}}{2}\\x=\dfrac{-1-\sqrt{5}}{2}\end{matrix}\right.\)