Question

bài 1,giải các phương trình sau

a,\(\sqrt{5x-2}=7\)

b,\(\sqrt{9x-27}+\sqrt{25x-75}=24\)

c,\(x^2-5x+8=2\sqrt{x-2}\)

bài 2,cho A=\(\left\{\dfrac{\sqrt{x}}{\sqrt{x}+2}-\dfrac{\sqrt{x}}{\sqrt{x}-2}\right\}\div\dfrac{2}{\sqrt{x}+2}\)

NÊU ĐKXĐ VÀ RÚT GỌN A

bài 3,cho B=\(\left\{\dfrac{1}{\sqrt{x}-1}+\dfrac{\sqrt{x}}{x-1}\right\}\times\dfrac{x-\sqrt{x}}{2\sqrt{x}+1}\)

NÊU ĐKXĐ VÀ RÚT GỌN B

bài4,cho C=\(\left(\dfrac{1}{\sqrt{x}-3}-\dfrac{1}{\sqrt{x}+3}\right)\times\left(1-\dfrac{3}{\sqrt{x}}\right)\)

NÊU ĐKXĐ VÀ RÚT GỌN C

Answer 1

Bài 1:
a. ĐKXĐ: $x\geq \frac{2}{5}$

PT $\Leftrightarrow 5x-2=7^2=49$

$\Leftrightarrow 5x=51$

$\Leftrightarrow x=\frac{51}{5}=10,2$

b. ĐKXĐ: $x\geq 3$

PT $\Leftrightarrow \sqrt{9(x-3)}+\sqrt{25(x-3)}=24$

$\Leftrightarrow 3\sqrt{x-3}+5\sqrt{x-3}=24$

$\Leftrightarrow 8\sqrt{x-3}=24$

$\Leftrightarrow \sqrt{x-3}=3$

$\Leftrightarrow x-3=9$

$\Leftrightarrow x=12$ (tm)

Answer 2

Bài 1:

c. ĐKXĐ: $x\geq 2$

PT $\Leftrightarrow x^2-5x+6-2(\sqrt{x-2}-1)=0$

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

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

$x-3=0$ hoặc $x-2=\frac{2}{\sqrt{x-2}+1}$

Nếu $x-3=0$

$\Leftrightarrow x=3$ (tm) 

Nếu $x-2=\frac{2}{\sqrt{x-2}+1}$

$\Leftrightarrow a^2=\frac{2}{a+1}$ (đặt $\sqrt{x-2}=a$)

$\Leftrightarrow a^3+a^2-2=0$

$\Leftrightarrow a^2(a-1)+2a(a-1)+2(a-1)=0$

$\Leftrightarrow (a-1)(a^2+2a+2)=0$

Hiển nhiên $a^2+2a+2=(a+1)^2+1>0$ với mọi $a$ nên $a-1=0$

$\Leftrightarrow a=1\Leftrightarrow \sqrt{x-2}=1\Leftrightarrow x=3$ (tm)

Vậy pt có nghiệm duy nhất $x=3$.

Answer 3

Bài 2:

ĐKXĐ: $x\geq 0; x\neq 4$

\(A=\frac{\sqrt{x}(\sqrt{x}-2)-\sqrt{x}(\sqrt{x}+2)}{(\sqrt{x}+2)\sqrt{x}-2)}.\frac{\sqrt{x}+2}{2}\\ =\frac{-4\sqrt{x}}{(\sqrt{x}-2)(\sqrt{x}+2)}.\frac{\sqrt{x}+2}{2}\\ =\frac{-2\sqrt{x}}{\sqrt{x}-2}=\frac{2\sqrt{x}}{2-\sqrt{x}}\)

Answer 4

Bài 3:

ĐKXĐ: $x\geq 0; x\neq 1$

\(B=\left[\frac{\sqrt{x}+1}{(\sqrt{x}-1)(\sqrt{x}+1)}+\frac{\sqrt{x}}{(\sqrt{x}-1)(\sqrt{x}+1)}\right].\frac{\sqrt{x}(\sqrt{x}-1)}{2\sqrt{x}+1}\\ =\frac{2\sqrt{x}+1}{(\sqrt{x}+1)(\sqrt{x}-1)}.\frac{\sqrt{x}(\sqrt{x}-1)}{2\sqrt{x}+1}\\ =\frac{\sqrt{x}}{\sqrt{x}+1}\)

Answer 5

Bài 4:

ĐKXĐ: $x> 0; x\neq 9$

\(C=\frac{\sqrt{x}+3-(\sqrt{x}-3)}{(\sqrt{x}-3)(\sqrt{x}+3)}.\frac{\sqrt{x}-3}{\sqrt{x}}\\ =\frac{6}{(\sqrt{x}-3)(\sqrt{x}+3)}.\frac{\sqrt{x}-3}{\sqrt{x}}=\frac{6}{\sqrt{x}(\sqrt{x}+3)}\)