Ôn tập chương 1: Căn bậc hai. Căn bậc ba

Ta có \(A=\dfrac{2\sqrt{x}-3}{\sqrt{x}+1}=\dfrac{2\sqrt{x}+2-5}{\sqrt{x}+1}=2-\dfrac{5}{\sqrt{x}+1}\) có GT nguyên <=> \(\dfrac{5}{\sqrt{x}+1}\) có GT nguyên <=> \(5⋮\sqrt{x}+1\) <=> \(\sqrt{x}+1\in\left\{-5,-1,1,5\right\}\Leftrightarrow\sqrt{x}\in\left\{-6,-2,0,4\right\}\Leftrightarrow x\in\left\{36,4,0,16\right\}\left(tm\right)\)Vậy...

ĐKXĐ: x>=0

\(A=\dfrac{\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}+1\right)}\cdot\dfrac{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{x-\sqrt{x}+1}=\dfrac{\sqrt{x}-1}{\sqrt{x}}\)

ĐKXĐ: x>=0

ĐK: \(x\ge0\)

\(D=\left(\dfrac{1}{\sqrt{x}+1}-\dfrac{1}{x+\sqrt{x}}\right):\dfrac{x-\sqrt{x}+1}{x\sqrt{x}+1}\\ =\dfrac{\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}+1\right)}:\dfrac{x-\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\\ =\dfrac{\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}+1\right)}:\dfrac{1}{\sqrt{x}+1}\\ =\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}=\dfrac{\sqrt{x}-1}{\sqrt{x}}\)

ĐKXĐ: x>=0

Sửa đề: \(A=\left(\dfrac{\sqrt{a}-2}{\sqrt{a}+2}+\dfrac{\sqrt{a}+2}{\sqrt{a}-2}\right)\cdot\left(1-\dfrac{2}{a+1}\right)^2\)

\(=\dfrac{a-4\sqrt{a}+4+a+4\sqrt{a}+4}{a-4}\cdot\left(\dfrac{a+1-2}{a+1}\right)^2\)

\(=\dfrac{2\left(a+4\right)}{a-4}\cdot\dfrac{\left(a-1\right)^2}{\left(a+1\right)^2}\)

\(C=\dfrac{a\sqrt{a}-1}{\sqrt{a}-1}=a+\sqrt{a}+1\)

Bài 1: 

a: ĐKXĐ: a>0

\(A=\sqrt{a}\left(\sqrt{a}+1\right)-2\sqrt{a}-1+1=a-\sqrt{a}\)

b: Để A=2 thì \(a-\sqrt{a}-2=0\)

=>a=4

c: \(A=a-\sqrt{a}+\dfrac{1}{4}-\dfrac{1}{4}=\left(\sqrt{a}-\dfrac{1}{2}\right)^2-\dfrac{1}{4}>=-\dfrac{1}{4}\)

Dấu '=' xảy ra khi a=1/4

a: \(A=\left(\dfrac{\sqrt{a}-2}{\sqrt{a}+2}+\dfrac{\sqrt{a}+2}{\sqrt{a}-2}\right):\left(\dfrac{a+1-2}{a+1}\right)^2\)

\(=\dfrac{a-4\sqrt{a}+4+a+4\sqrt{a}+4}{a-4}\cdot\dfrac{\left(a+1\right)^2}{\left(a-1\right)^2}\)

\(=\dfrac{-8\sqrt{a}\left(a+1\right)^2}{\left(a-1\right)^2\cdot\left(a-4\right)}\)

đk x ≠ 1; x ≥ 0

\(A=\left(\dfrac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}+\dfrac{\sqrt{x}}{x-1}\right):\dfrac{\left(\sqrt{x}\right)^3-1}{\sqrt{x}\left(x-1\right)}\\ =\left(\dfrac{\sqrt{x}+1+x}{\sqrt{x}\left(x-1\right)}\right).\dfrac{\sqrt{x}\left(x-1\right)}{\sqrt{x}^3-1}\\ =\dfrac{1}{\sqrt{x}-1}\)

a: Thay \(x=7-4\sqrt{3}\) vào B, ta được:

\(B=\dfrac{2}{2-\sqrt{3}-2}=\dfrac{-2\sqrt{3}}{3}\)

b: \(P=B:A\)

\(=\dfrac{2}{\sqrt{x}-2}:\dfrac{\sqrt{x}+\sqrt{x}+2}{x-4}\)

\(=\dfrac{2}{\sqrt{x}-2}\cdot\dfrac{x-4}{2\left(\sqrt{x}+1\right)}=\dfrac{\sqrt{x}+2}{\sqrt{x}+1}\)

a) Xem bạn kia nha

b) \(A=\dfrac{\sqrt{x}+\sqrt{x}+2}{x-4}=\dfrac{2\sqrt{x}+2}{x-4}\)

\(B=\dfrac{2\left(\sqrt{x}+2\right)}{x-4}=\dfrac{2\sqrt{x}+4}{x-4}\)

SUy ra: \(\dfrac{B}{A}=\dfrac{2\sqrt{x}+4}{2\sqrt{x}+2}\)

P = \(\dfrac{4}{3}\)

Suy ra \(6\sqrt{x}+12=8\sqrt{x}+8\)(nhân chéo)

tương đương với: -2\(\sqrt{x}\)-4=0 suy ra pt vô nghiệm (hay bạn xem lại đề đi)