x/y=2 =>x=2y

thay vao de bai ta duoc

4y+y/2y-y=5y/y=5

Lời giải:

ĐKXĐ: $x\geq 1$
PT $\Leftrightarrow \sqrt{(x-1)+2\sqrt{x-1}+1}-\sqrt{(x-1)-2\sqrt{x-1}+1}=2$
$\Leftrightarrow \sqrt{(\sqrt{x-1}+1)^2}-\sqrt{(\sqrt{x-1}-1)^2}=2$

$\Leftrightarrow |\sqrt{x-1}+1|-|\sqrt{x-1}-1|=2$
Nếu $2\geq x\geq 1$ thì:

$\sqrt{x-1}+1+(1-\sqrt{x-1})=2$
$\Leftrightarrow 2=2$ (luôn đúng)

Nếu $x>2$ thì: $\sqrt{x-1}+1+(\sqrt{x-1}-1)=2$
$\Leftrightarrow 2\sqrt{x-1}=2$
$\Leftrightarrow x-1=1$

$\Leftrihgtarrow x=2$ (loại)

Vậy $2\geq x\geq 1$

$

\(\dfrac{\sqrt{x-1}}{\sqrt{x+3}}=\dfrac{\sqrt{x-2}}{1}\)(Đk x>2;x≠-3)

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

⇔\(\left(x-2\right)\left(x+3\right)=x-1\)

⇔\(x^2+x-6-x+1=0\)

⇔\(x^2-5=0\)

⇔\(x^2=5\)

⇔x=\(\pm\sqrt{5}\)(thỏa điều kiện)

Vậyx=\(\pm\sqrt{5}\)

ĐKXĐ:x khác -3; x≥2

quy đồng và khử mẩu 2 vế ta đc:

\(\sqrt{x-1}=\sqrt{x-2}\cdot\sqrt{x+3}\)Bình phương 2 vế ta đc:

x-1=(x-2)*(x+3)<=> x-1=x²+x-6 <=>  x²-5=0

<=>\(\left\{{}\begin{matrix}x=\sqrt{5}\left(nhận\right)\\x=-\sqrt{5}\left(loại\right)\end{matrix}\right.\)

vậy x=\(\sqrt{5}\)

b: Thay \(x=7-2\sqrt{6}\) vào A, ta được:

\(A=\dfrac{3\cdot\left(\sqrt{6}-1\right)}{-7+2\sqrt{6}-5\left(\sqrt{6}+1\right)-1}\)

\(=\dfrac{3\cdot\left(\sqrt{6}-1\right)}{-8+2\sqrt{6}-5\sqrt{6}-5}\)

\(=\dfrac{-3\sqrt{6}+3}{13+3\sqrt{6}}=\dfrac{93-48\sqrt{6}}{115}\)

Mình sửa lại đề tí:

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

ĐKXĐ: \(x\ge1\)

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

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

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

\(\Leftrightarrow\sqrt{x-1}-\left|\sqrt{x-1}-1\right|=1\)

TH1: \(x\ge2\Rightarrow\sqrt{x-1}-1\ge0\) pt trở thành:

\(\sqrt{x-1}-\left(\sqrt{x-1}-1\right)=1\) (luôn đúng)

TH2: \(1\le x< 2\)

\(\Rightarrow\sqrt{x-1}-\left(1-\sqrt{x-1}\right)=1\)

\(\Leftrightarrow2\sqrt{x-1}=2\Rightarrow x=2\) (ktm)

Vậy nghiệm của pt là \(x\ge2\)

tích mình với

ai tích mình

mình tích lại

thanks

Tích mình đi mình tích lại

\(a,P=\left[\dfrac{\left(1-\sqrt{x}\right)\left(x+\sqrt{x}+1\right)}{1-\sqrt{x}}+\sqrt{x}\right]\left[\dfrac{\left(1+\sqrt{x}\right)\left(x-\sqrt{x}+1\right)}{1+\sqrt{x}}-\sqrt{x}\right]\\ P=\left(x+2\sqrt{x}+1\right)\left(x-2\sqrt{x}+1\right)\\ P=\left(\sqrt{x}-1\right)^2\left(\sqrt{x}+1\right)^2\\ P=\left(x-1\right)^2\\ b,x=\sqrt{3+2\sqrt{2}}=\sqrt{\left(\sqrt{2}+1\right)^2}=\sqrt{2}+1\\ \Leftrightarrow P=\left(\sqrt{2}+1-1\right)^2=\left(\sqrt{2}\right)^2=2\)

a) \(P=\left(\dfrac{\left(1-\sqrt{x}\right)\left(1+\sqrt{x}+x\right)}{1-\sqrt{x}}+\sqrt{x}\right)\left(\dfrac{\left(1+\sqrt{x}\right)\left(1-\sqrt{x}+x\right)}{1+\sqrt{x}}-\sqrt{x}\right)\)

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

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