Bài 8: Rút gọn biểu thức chứa căn bậc hai

a, \(A=\dfrac{\sqrt{x}-1}{x^2-x}:\left(\dfrac{1}{\sqrt{x}}-\dfrac{1}{\sqrt{x}+1}\right)=\dfrac{\sqrt{x}-1}{x\left(\sqrt{x}\pm1\right)}:\left(\dfrac{\sqrt{x}+1-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}\right)\)

\(=\dfrac{1}{x\left(\sqrt{x}+1\right)}.\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{1}=1\)

b, Cho A = 1 rồi còn gì, hay đề lỗi bạn ? 

\(x=4+2\sqrt{3}=\sqrt{3}^2+2\sqrt{3}+1=\left(\sqrt{3}+1\right)^2\)

\(\Rightarrow\sqrt{x}=\sqrt{\left(\sqrt{3}+1\right)^2}=\left|\sqrt{3}+1\right|=\sqrt{3}+1\)

xem là bài mình làm có sai đâu ko nhé nếu rút gọn ra kq khác thì thay bên trên vào nhé 

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

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

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

=1

Gọi chiều dài, chiều rộng hình chữ nhật lần lượt là x ; y > 0, m 

Chu vi hình chữ nhật là : \(P=\left(a+b\right).2=46\)

Nếu tăng chiều dài 5m, giảm chiều rộng 3m thì hình chữ nhật có chiều dài gấp 4 lần chiều rộng : \(a+5=4\left(b-3\right)\)

Ta có hệ phương trình sau : \(\left\{{}\begin{matrix}\left(a+b\right).2=46\\a+5=4\left(b-3\right)\end{matrix}\right.\)

giải hệ ta được a = 15 ; b = 8 

Vậy diện tích hình chữ nhật là : \(a.b=15.8=120\)m²

ĐKXĐ : x > 0 , x khác 1

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

\(=\left[\dfrac{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}-\dfrac{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}\right]\div\left[\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\dfrac{\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right]\)

\(=\dfrac{x+\sqrt{x}+1-x+\sqrt{x}-1}{\sqrt{x}}\div\left[\dfrac{x+\sqrt{x}-x+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right]\)

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

b) Dễ thấy ∀ x ≥ 0 thì \(2\sqrt{x}-2\) nguyên

Kết hợp với ĐKXĐ => Với \(\left\{{}\begin{matrix}x>0\\x\ne1\end{matrix}\right.\)thì A đạt giá trị nguyên 

a) Xét tứ giác OAMC có 

\(\widehat{OAM}\) và \(\widehat{OCM}\) là hai góc đối

\(\widehat{OAM}+\widehat{OCM}=180^0\left(90^0+90^0=180^0\right)\)

Do đó: OAMC là tứ giác nội tiếp(Dấu hiệu nhận biết tứ giác nội tiếp)

Lời giải:

ĐK: $x>0; a\neq 1; a\neq 4$

a) 

$M=\frac{\sqrt{a}-(\sqrt{a}-1)}{\sqrt{a}(\sqrt{a}-1)}:\frac{(\sqrt{a}+1)(\sqrt{a}-1)-(\sqrt{a}-2)(\sqrt{a}+2)}{(\sqrt{a}-2)(\sqrt{a}-1)}$

$=\frac{1}{\sqrt{a}(\sqrt{a}-1)}:\frac{3}{(\sqrt{a}-2)(\sqrt{a}-1)}=\frac{1}{\sqrt{a}(\sqrt{a}-1)}.\frac{(\sqrt{a}-2)(\sqrt{a}-1)}{3}=\frac{\sqrt{a}-2}{3\sqrt{a}}$

b) 

$M>\frac{-1}{2}\Leftrightarrow \frac{\sqrt{a}-2}{3\sqrt{a}}+\frac{1}{2}>0$

$\Leftrightarrow \frac{5\sqrt{a}-4}{6\sqrt{a}}>0$

$\Leftrightarrow 5\sqrt{a}-4>0$

$\Leftrightarrow a>\frac{16}{25}$

Kết hợp với ĐKXĐ thì $a>\frac{16}{25}; a\neq 1; a\neq 4$

b) Ta có: \(B=\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}}{x+\sqrt{x}+1}+\dfrac{1}{1-\sqrt{x}}\)

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

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

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

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

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

a) Thay x=4 vào biểu thức \(B=\dfrac{3}{\sqrt{x}-1}\), ta được:

\(B=\dfrac{3}{\sqrt{4}-1}=\dfrac{3}{2-1}=3\)

Vậy: Khi x=4 thì B=3

b) Ta có: P=A-B

\(\Leftrightarrow P=\dfrac{6}{x-1}+\dfrac{\sqrt{x}}{\sqrt{x}+1}-\dfrac{3}{\sqrt{x}-1}\)

\(\Leftrightarrow P=\dfrac{6}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}-\dfrac{3\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

\(\Leftrightarrow P=\dfrac{6+x-\sqrt{x}-3\sqrt{x}-3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

\(\Leftrightarrow P=\dfrac{x-\sqrt{x}-3\sqrt{x}+3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

\(\Leftrightarrow P=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)-3\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

\(\Leftrightarrow P=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

\(\Leftrightarrow P=\dfrac{\sqrt{x}-3}{\sqrt{x}+1}\)

Lời giải:ĐK: $a\geq 0; a\neq 9; a\neq 4$

a) 

\(A=\frac{2\sqrt{a}-9}{(\sqrt{a}-2)(\sqrt{a}-3)}-\frac{\sqrt{a}+3}{\sqrt{a}-2}+\frac{2\sqrt{a}+1}{\sqrt{a}-3}\)

\(\frac{2\sqrt{a}-9}{(\sqrt{a}-2)(\sqrt{a}-3)}-\frac{(\sqrt{a}+3)(\sqrt{a}-3)}{(\sqrt{a}-2)(\sqrt{a}-3)}+\frac{(2\sqrt{a}+1)(\ \sqrt{a}-2)}{(\sqrt{a}-3)(\sqrt{a}-2)}\)

\(=\frac{2\sqrt{a}-9-(a-9)+(2a-3\sqrt{a}-2)}{(\sqrt{a}-3)(\sqrt{a}-2)}=\frac{a-\sqrt{a}-2}{(\sqrt{a}-3)(\sqrt{a}-2)}=\frac{(\sqrt{a}-2)(\sqrt{a}+1)}{(\sqrt{a}-3)(\sqrt{a}-2)}=\frac{\sqrt{a}+1}{\sqrt{a}-3}\)

b) Để \(A< 1\Leftrightarrow \frac{\sqrt{a}+1}{\sqrt{a}-3}<1\Leftrightarrow 1+\frac{4}{\sqrt{a}-3}<1\)

\(\Leftrightarrow \frac{4}{\sqrt{a}-3}< 0\Leftrightarrow \sqrt{a}-3< 0\Leftrightarrow 0\leq a< 9\)

Kết hợp ĐKXĐ: suy ra $0\leq a< 9; a\neq 4$

c) Với $a$ nguyên,  \(A=1+\frac{4}{\sqrt{a}-3}\in\mathbb{Z}\Leftrightarrow 4\vdots \sqrt{a}-3\)

$\Rightarrow \sqrt{a}-3\in\left\{\pm 1; \pm 2;\pm 4\right\}$

$\Rightarrow a\in\left\{4;16; 1;25; 49\right\}$

Kết hợp ĐKXĐ suy ra $a\in\left\{16;1;25;49\right\}$

ĐKXĐ: \(\left\{{}\begin{matrix}a\ge0\\a\notin\left\{4;9\right\}\end{matrix}\right.\)

a) Ta có: \(A=\dfrac{2\sqrt{a}-9}{a-5\sqrt{a}+6}-\dfrac{\sqrt{a}+3}{\sqrt{a}-2}-\dfrac{2\sqrt{a}+1}{3-\sqrt{a}}\)

\(=\dfrac{\left(2\sqrt{a}-9\right)}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-3\right)}-\dfrac{\left(\sqrt{a}+3\right)\left(\sqrt{a}-3\right)}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-3\right)}+\dfrac{\left(2\sqrt{a}+1\right)\left(\sqrt{a}-2\right)}{\left(\sqrt{a}-3\right)\left(\sqrt{a}-2\right)}\)

\(=\dfrac{2\sqrt{a}-9-\left(a-9\right)+2a-4\sqrt{a}+\sqrt{a}-2}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-3\right)}\)

\(=\dfrac{2a-\sqrt{a}-11-a+9}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-3\right)}\)

\(=\dfrac{a-\sqrt{a}-2}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-3\right)}\)

\(=\dfrac{a-2\sqrt{a}+\sqrt{a}-2}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-3\right)}\)

\(=\dfrac{\sqrt{a}\left(\sqrt{a}-2\right)+\left(\sqrt{a}-2\right)}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-3\right)}\)

\(=\dfrac{\left(\sqrt{a}-2\right)\left(\sqrt{a}+1\right)}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-3\right)}\)

\(=\dfrac{\sqrt{a}+1}{\sqrt{a}-3}\)

b) Để A<1 thì A-1<0

\(\Leftrightarrow\dfrac{\sqrt{a}+1}{\sqrt{a}-3}-1< 0\)

\(\Leftrightarrow\dfrac{\sqrt{a}+1}{\sqrt{a}-3}-\dfrac{\sqrt{a}-3}{\sqrt{a}-3}< 0\)

\(\Leftrightarrow\dfrac{\sqrt{a}+1-\sqrt{a}+3}{\sqrt{a}-3}< 0\)

\(\Leftrightarrow\dfrac{4}{\sqrt{a}-3}< 0\)

mà 4>0

nên \(\sqrt{a}-3< 0\)

\(\Leftrightarrow\sqrt{a}< 3\)

hay a<9

Kết hợp ĐKXĐ, ta được:

\(\left\{{}\begin{matrix}0\le a< 9\\a\ne4\end{matrix}\right.\)

Vậy: Để A<1 thì \(\left\{{}\begin{matrix}0\le a< 9\\a\ne4\end{matrix}\right.\)

c) Để A nguyên thì \(\sqrt{a}+1⋮\sqrt{a}-3\)

\(\Leftrightarrow\sqrt{a}-3+4⋮\sqrt{a}-3\)

mà \(\sqrt{a}-3⋮\sqrt{a}-3\)

nên \(4⋮\sqrt{a}-3\)

\(\Leftrightarrow\sqrt{a}-3\inƯ\left(4\right)\)

\(\Leftrightarrow\sqrt{a}-3\in\left\{1;-1;2;-2;4;-4\right\}\)

mà \(\sqrt{a}-3\ge-3\forall a\) thỏa mãn ĐKXĐ

nên \(\sqrt{a}-3\in\left\{1;-1;2;-2;4\right\}\)

\(\Leftrightarrow\sqrt{a}\in\left\{4;2;5;1;7\right\}\)

\(\Leftrightarrow a\in\left\{16;4;25;1;49\right\}\)

Kết hợp ĐKXĐ, ta được: \(a\in\left\{1;16;25;49\right\}\)

Vậy: Để A nguyên thì \(a\in\left\{1;16;25;49\right\}\)

a) Ta có: \(A=\left(\dfrac{x-5\sqrt{x}}{x-25}-1\right):\left(\dfrac{25-x}{x+2\sqrt{x}-15}-\dfrac{\sqrt{x}+3}{\sqrt{x}+5}+\dfrac{\sqrt{x}-5}{\sqrt{x}-3}\right)\)

\(=\left(\dfrac{\sqrt{x}\left(\sqrt{x}-5\right)}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}-1\right):\left(\dfrac{25-x}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}-\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}+\dfrac{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}\right)\)

\(=\left(\dfrac{\sqrt{x}}{\sqrt{x}+5}-1\right):\left(\dfrac{25-x-\left(x-9\right)+x-25}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}\right)\)

\(=\left(\dfrac{\sqrt{x}}{\sqrt{x}+5}-\dfrac{\sqrt{x}+5}{\sqrt{x}+5}\right):\left(\dfrac{25-x-x+9+x-25}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}\right)\)

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

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

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