a, Để A có nghĩa \(x^2-1\ge0\Leftrightarrow\left(x-1\right)\left(x+1\right)\ge0\Leftrightarrow x\le-1;x\ge1\)

b,  \(A=\sqrt{x^2+2\sqrt{x^2-1}}-\sqrt{x^2-2\sqrt{x^2-1}}\)với \(x\ge\sqrt{2}\)

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

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

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

\(ĐKXĐ:x\ge2\)

\(\sqrt{x-2}+\sqrt{4x-8}-\frac{1}{2}\sqrt{9x-18}=2\)

\(\sqrt{x-2}+2\sqrt{x-2}-\frac{3}{2}\sqrt{x-2}=2\)

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

\(\frac{3}{2}\sqrt{x-2}=2\)

\(\sqrt{x-2}=\frac{4}{3}\)

\(x-2=\frac{16}{9}\)

\(x=\frac{34}{9} \left(TM\right)\)

ĐK : \(\sqrt{x-1}\ge0;\sqrt{x+2\sqrt{x-1}}\ne0\)

\(\Leftrightarrow x-1\ge0;x-1+2\sqrt{x-1}+1\ne0\)

\(\Leftrightarrow x\ge1;\left(\sqrt{x-1}+1\right)^2\ne0\left(luondung\right)\Rightarrow x\ge1\)

đk : x > 2/3

\(\frac{x^2}{3x-2}-\sqrt{3x-2}=1-x\)

đặt \(\sqrt{3x-2}=a\left(a>\right)\)

pt trở thành \(\frac{x^2}{a^2}-a=1-x\) vi a^2 > 0

\(\Leftrightarrow x^2-a^3=a^2-a^2x\)

\(\Leftrightarrow a^3-a^2x+a^2-x^2=0\)

\(\Leftrightarrow a^2\left(a-x\right)+\left(a-x\right)\left(a+x\right)=0\)

\(\Leftrightarrow\left(a^2+a+x\right)\left(a-x\right)=0\)

th1 : \(a-x=0\Leftrightarrow a=x\) nên :

\(\sqrt{3x-2}=x\)

\(\Leftrightarrow x^2=3x-2\)

\(\Leftrightarrow x^2-3x+2=0\)

\(\Leftrightarrow\left(x-1\right)\left(x-2\right)=0\Leftrightarrow\orbr{\begin{cases}x=1\left(tm\right)\\x=2\left(tm\right)\end{cases}}\)

th2 : \(a^2+a+x=0\Leftrightarrow3x-2+\sqrt{3x-2}+x=0\)

có x > 2/3 => 3x -2 > 0 => 3x - 2 + x > 2/3 

=> pt vô nghiệm

vậy x = 1 hoặc x = 2

cứ thay vào thôi bạn 

a, Thay x = 64 vào A ta được : \(\frac{\sqrt{64}}{\sqrt{64}-2}=\frac{8}{8-2}=\frac{8}{6}=\frac{4}{3}\)

b, Thay x = 36 vào A ta được : \(\frac{\sqrt{36}+4}{\sqrt{36}+2}=\frac{6+4}{6+2}=\frac{10}{8}=\frac{5}{4}\)

c, Thay x = 9 vào A ta được \(\frac{\sqrt{9}-5}{\sqrt{9}+5}=\frac{3-5}{3+5}=-\frac{2}{8}=-\frac{1}{4}\)

\(\left(3\sqrt{x^2y}-4\sqrt{xy^2}+5xy\right)\div\sqrt{xy}=3\sqrt{x}-4\sqrt{y}+5\sqrt{xy}\)

Đáp án :

= \(\sqrt{2}\sqrt{5}\left|x\right|-3\sqrt{2^5}\sqrt{5x+36}\)

# Hok tốt !

Gọi số lớn là a, số bé là b

Ta có: a= bx4+1

{(a-6).(b+2)=a.b

suy ra { a=4b+1; ab+2a-6b-12=ab

suy ra {a-4b=1; 2a-6b=12

Vậy {a=21; b=5

a, Với x > 0 

\(B=\frac{\sqrt{x}-1}{\sqrt{x}}+\frac{1}{x+\sqrt{x}}=\frac{x-1+1}{x+\sqrt{x}}=\frac{x}{\sqrt{x}\left(\sqrt{x}+1\right)}=\frac{\sqrt{x}}{\sqrt{x}+1}\)

b, Ta có : \(A>\frac{2}{3}\Rightarrow\frac{\sqrt{x}}{\sqrt{x}+1}-\frac{2}{3}>0\Leftrightarrow\frac{3\sqrt{x}-2\sqrt{x}-2}{3\left(\sqrt{x}+1\right)}>0\)

\(\Rightarrow\sqrt{x}-2>0\Leftrightarrow x>4\)

c, \(\frac{A}{B}=\frac{\sqrt{x}}{\sqrt{x}+1}.\frac{\sqrt{x}+3}{2\sqrt{x}}=\frac{\sqrt{x}+3}{2\sqrt{x}+2}=\frac{2\sqrt{x}+6}{2\sqrt{x}+2}=1+\frac{4}{2\sqrt{x}+2}=1+\frac{2}{\sqrt{x}+1}\)

\(\Rightarrow\sqrt{x}+1\inƯ\left(2\right)=\left\{1;2\right\}\)

Để ptđt trên là hàm bậc nhất khi \(3-m\ne0\Leftrightarrow m\ne3\)

Thay x = 0 ; y = 5 vào ptđt y = (3-m)x + m-4

\(5=m-4\Leftrightarrow m=9\)(tm)