Bài 6: Biến đối đơn giản biểu thức chứa căn bậc hai

\(\sqrt{25x-25}-\dfrac{15}{2}\cdot\sqrt{\dfrac{x-1}{9}}=6+\sqrt{x-1}\) (1)

\(\Leftrightarrow\sqrt{25\left(x-1\right)}-\dfrac{15}{2}\cdot\dfrac{\sqrt{x-1}}{3}=6+\sqrt{x-1}\)

\(\Leftrightarrow\sqrt{25}\sqrt{x-1}-\dfrac{5}{2}\cdot\sqrt{x-1}=6+\sqrt{x-1}\)

\(\Leftrightarrow5\sqrt{x-1}-\dfrac{5}{2}\cdot\sqrt{x-1}=6+\sqrt{x-1}\)

\(\Leftrightarrow\dfrac{5}{2}\cdot\sqrt{x-1}=6+\sqrt{x-1}\)

\(\Leftrightarrow5\sqrt{x-1}=12+2\sqrt{x-1}\)

\(\Leftrightarrow5\sqrt{x-1}-2\sqrt{x-1}=12\)

\(\Leftrightarrow3\sqrt{x-1}=12\)

\(\Leftrightarrow\sqrt{x-1}=4\)

\(\Leftrightarrow x-1=16\)

\(\Leftrightarrow x=16+1\)

\(\Leftrightarrow x=17\)

Vậy tập nghiệm phương trình (1) là \(S=\left\{17\right\}\)

\(\sqrt{16b}+2\sqrt{40b}-30\sqrt{90b}\)

\(=\sqrt{16}\sqrt{b}+2\sqrt{40}\sqrt{b}-30\sqrt{90}\sqrt{b}\)

\(=\sqrt{b}\left(\sqrt{16}+2\sqrt{40}-30\sqrt{90}\right)\)

\(=\sqrt{b}\left(4+4\sqrt{10}-90\sqrt{10}\right)\)

\(=\sqrt{b}\left(4-86\sqrt{10}\right)\)

\(\left(\sqrt{28}-\sqrt{12}-\sqrt{7}\right)\sqrt{7}+2\sqrt{2}\)

\(=\left(2\sqrt{7}-\sqrt{12}-\sqrt{7}\right)\sqrt{7}+2\sqrt{2}\)

\(=2\sqrt{7}.\sqrt{7}-\sqrt{12}.\sqrt{7}-\sqrt{7}.\sqrt{7}+2\sqrt{2}\)

\(=14-\sqrt{84}-7+2\sqrt{2}\)

\(=7-\sqrt{84}+2\sqrt{2}\)

Chúc bạn học tốt!!!

cách ngắn:

\(\left(\sqrt{28}-\sqrt{12}-\sqrt{7}\right)\sqrt{7}+2\sqrt{2}\)

\(=\left(2\sqrt{7}-2\sqrt{3}-\sqrt{7}\right)\sqrt{7}+2\sqrt{2}\)

\(=\left(\sqrt{7}-2\sqrt{3}\right)\sqrt{7}+2\sqrt{2}\)

\(=7-2\sqrt{21}+2\sqrt{2}\)

đk : \(x\ge0;y\ge0;x\ne y\)

A = \(\dfrac{\sqrt{x}}{\sqrt{x}+\sqrt{y}}+\dfrac{\sqrt{y}}{\sqrt{y}-\sqrt{x}}=\dfrac{2\sqrt{xy}}{x-y}\)

\(\Leftrightarrow\) \(\dfrac{\sqrt{x}}{\sqrt{x}+\sqrt{y}}-\dfrac{\sqrt{y}}{\sqrt{x}-\sqrt{y}}=\dfrac{2\sqrt{xy}}{x-y}\)

\(\Leftrightarrow\) \(\dfrac{\sqrt{x}\left(\sqrt{x}-\sqrt{y}\right)-\sqrt{y}\left(\sqrt{x}+\sqrt{y}\right)}{\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}\right)}=\dfrac{2\sqrt{xy}}{x-y}\)

\(\Leftrightarrow\) \(\dfrac{x-\sqrt{xy}-\sqrt{xy}-y}{x-y}=\dfrac{2\sqrt{xy}}{x-y}\)

\(\Rightarrow\) \(x-2\sqrt{xy}-y=2\sqrt{xy}\) \(\Leftrightarrow\) \(x-y=4\sqrt{xy}\)

\(\Leftrightarrow\) A = \(\dfrac{2\sqrt{xy}}{4\sqrt{xy}}=\dfrac{1}{2}\)

không biết sai chỗ nào ??? sao bài làm lại trái với câu hỏi thế này ???

Ta thấy: \(\dfrac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\dfrac{1}{\sqrt{n}}-\dfrac{1}{\sqrt{n+1}}\)

\(\Rightarrow\dfrac{1}{2\sqrt{1}+1\sqrt{2}}+\dfrac{1}{3\sqrt{2}+2\sqrt{3}}+...+\dfrac{1}{2016\sqrt{2015}+2015\sqrt{2016}}\)

\(=\dfrac{1}{\sqrt{1}}-\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{3}}+.....+\dfrac{1}{\sqrt{2015}}-\dfrac{1}{\sqrt{2016}}\)

\(=\dfrac{1}{\sqrt{1}}-\dfrac{1}{\sqrt{2016}}=\dfrac{\sqrt{2016}-1}{\sqrt{2016}}\)

\(\frac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\frac{1}{\sqrt{n}.\sqrt{n+1}\left(\sqrt{n+1}+\sqrt{n}\right)}=\frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n}.\sqrt{n+1}\left(\sqrt{n+1}+\sqrt{n}\right)\left(\sqrt{n+1}-\sqrt{n}\right)}=\frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n+1}.\sqrt{n}}=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\)

\(\Rightarrow\frac{1}{2\sqrt{1}+1\sqrt{2}}+.....+\frac{1}{2016\sqrt{2015}+2015\sqrt{2016}}=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+......-\frac{1}{\sqrt{2016}}=1-\frac{1}{\sqrt{2016}}=\frac{\sqrt{2016}-1}{\sqrt{2016}}\)

(bài 1) a) \(\dfrac{1}{5+2\sqrt{6}}-\dfrac{1}{5-2\sqrt{6}}\) = \(\dfrac{5-2\sqrt{6}-5-2\sqrt{6}}{25-24}\)

= \(\dfrac{-4\sqrt{6}}{1}\) = \(-4\sqrt{6}\)

b) \(\sqrt{6+2\sqrt{5}}-\dfrac{\sqrt{15}-\sqrt{3}}{\sqrt{3}}\) = \(\sqrt{\left(\sqrt{5}+1\right)^2}-\dfrac{\sqrt{3}\left(\sqrt{5}-1\right)}{\sqrt{3}}\)

= \(\left(\sqrt{5}+1\right)-\left(\sqrt{5}-1\right)\) = \(\sqrt{5}+1-\sqrt{5}+1\) = \(2\)

c) \(\dfrac{3\sqrt{2}-2\sqrt{3}}{\sqrt{3}-\sqrt{2}}:\dfrac{1}{\sqrt{16}}\) = \(\dfrac{\sqrt{6}\left(\sqrt{3}-\sqrt{2}\right)}{\sqrt{3}-\sqrt{2}}:\dfrac{1}{\sqrt{16}}\)

= \(\sqrt{6}.\sqrt{16}\) = \(4\sqrt{6}\)

d) \(\dfrac{3+2\sqrt{3}}{\sqrt{3}}+\dfrac{2+\sqrt{2}}{1+\sqrt{2}}-\dfrac{1}{2-\sqrt{3}}\)

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

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

= \(\dfrac{2\sqrt{3}-3+4-2\sqrt{3}+2\sqrt{2}-\sqrt{6}-1}{2-\sqrt{3}}\)

= \(\dfrac{2\sqrt{2}-\sqrt{6}}{2-\sqrt{3}}\) = \(\dfrac{\sqrt{2}\left(2-\sqrt{3}\right)}{2-\sqrt{2}}\) = \(\sqrt{2}\)

e) \(\dfrac{4}{1+\sqrt{3}}-\dfrac{\sqrt{15}+\sqrt{3}}{1+\sqrt{5}}\) = \(\dfrac{4}{1+\sqrt{3}}-\dfrac{\sqrt{3}\left(\sqrt{5}+1\right)}{1+\sqrt{5}}\)

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

= \(\dfrac{\left(1-\sqrt{3}\right)\left(1-\sqrt{3}\right)}{1-3}\) = \(\dfrac{1-2\sqrt{3}+3}{-2}\) = \(\dfrac{4-2\sqrt{3}}{-2}\)

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

bài 2)

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

= \(\dfrac{a\sqrt{a}+a\sqrt{b}+b\sqrt{a}+b\sqrt{b}-2a\sqrt{b}-2b\sqrt{a}}{\sqrt{a}-\sqrt{b}}\)

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

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

b) \(\left(\dfrac{\sqrt{a}}{2}-\dfrac{1}{2\sqrt{a}}\right).\left(\dfrac{a-\sqrt{a}}{\sqrt{a}+1}-\dfrac{a+\sqrt{a}}{\sqrt{a}-1}\right)\)

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

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

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

= \(\dfrac{2\left(a-1\right)}{4\sqrt{a}}.\dfrac{-4a}{a-1}\) = \(-2\)

1) Đặt \(a=\sqrt{x-1}\ge0\)

\(\Leftrightarrow a^2=x-1\Rightarrow x=a^2+1\)

pt (1) \(\Leftrightarrow a^2+1-5a+3=0\)

\(\Leftrightarrow a^2-5a+4=0\) (trở về dạng quen thuộc).

\(\Rightarrow\left\{{}\begin{matrix}a_1=4\\a_2=1\end{matrix}\right.\left(TMĐK\right)\)

+) \(a=4\Leftrightarrow\sqrt{x-1}=4\Rightarrow x=17\left(TMĐK\right)\)

+) \(a=1\Leftrightarrow\sqrt{x-1}=1\Rightarrow x=2\left(TMĐK\right)\)

Vậy \(S_1=\left\{2;17\right\}\)

2) Đặt \(a=\sqrt{2x-1}\ge0\)

\(\Leftrightarrow a^2=2x-1\Rightarrow2x=a^2+1\)

pt (2) <=> \(a^2+1+5a-7=0\)

\(\Leftrightarrow a^2+5a-6=0\)

\(\Rightarrow\left\{{}\begin{matrix}a_1=1\left(TMĐK\right)\\a_2=-6\left(KTMĐK\right)\end{matrix}\right.\)

+) \(a=1\Leftrightarrow\sqrt{2x-1}=1\Rightarrow x=1\left(TMĐK\right)\)

Vậy \(S_2=\left\{1\right\}\)

ĐKXĐ : \(x\ne1;x\ne0;x>0\)

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

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

\(=\dfrac{4x\sqrt{x}}{\sqrt{x}}=4x\)

b) \(x=\dfrac{\sqrt{6}}{2+\sqrt{6}}=\dfrac{\sqrt{6}\cdot\left(2-\sqrt{6}\right)}{-2}=\dfrac{2\sqrt{6}-6}{-2}=3-\sqrt{6}\)

Suy ra : \(A=4\cdot\left(3-\sqrt{6}\right)=12-4\sqrt{6}\)

c) \(\sqrt{A}>A\Leftrightarrow\sqrt{4x}>4x\)

\(\Leftrightarrow2\sqrt{x}>4x\)

\(\Leftrightarrow2\sqrt{x}-4x>0\)

\(\Leftrightarrow2\sqrt{x}\cdot\left(1-2\sqrt{x}\right)>0\)

\(\Rightarrow1-2\sqrt{x}>0\) (do x > 0 nên \(2\sqrt{x}>0\))

\(\Leftrightarrow1>2\sqrt{x}\)

\(\Leftrightarrow\dfrac{1}{2}>\sqrt{x}\Rightarrow x< \dfrac{1}{4}\)

Theo điều kiện suy ra giá trị của x để \(\sqrt{A}>A\) là \(0< x< \dfrac{1}{4}\)

a) \(A=\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}-\dfrac{\sqrt{x}-1}{\sqrt{x}+1}+4\sqrt{x}\right)\left(\sqrt{x}-\dfrac{1}{\sqrt{x}}\right)\)

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

\(=\left(\dfrac{2\cdot2\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+4\sqrt{x}\right)\cdot\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}}\)

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

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

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

\(=4\left(1+x-1\right)\)

\(=4x\)

b) Thay \(x=\dfrac{\sqrt{6}}{2+\sqrt{6}}\) vào biểu thức A.

Ta có:

\(4\cdot\dfrac{\sqrt{6}}{2+\sqrt{6}}=\dfrac{4\sqrt{6}}{2+\sqrt{6}}=\dfrac{4\sqrt{6}\cdot\left(2-\sqrt{6}\right)}{-2}\\ =-2\sqrt{6}\cdot\left(2-\sqrt{6}\right)=-4\sqrt{6}+12\)

Vậy giá trị biểu thức A tại \(x=\dfrac{\sqrt{6}}{2+\sqrt{6}}\) là \(-4\sqrt{6}+12\)

c) Để \(\sqrt{A}>A\)

\(\Rightarrow\sqrt{4x}>4x\)

\(\Leftrightarrow\sqrt{4x}>4x\left(đk:x\ge0\right)\)

ĐKXĐ \(x\ge0,x\ne4\)

a) \(B=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)-\left(\sqrt{x}+3\right)\cdot\left(\sqrt{x}+1\right)-\left(x+5\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\)

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

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

\(=\dfrac{\sqrt{x}+6}{2-\sqrt{x}}\)

b) B > -1 <=> B + 1 > 0.

\(\Leftrightarrow\dfrac{\sqrt{x}+6}{2-\sqrt{x}}+1>0\Leftrightarrow\dfrac{8}{2-\sqrt{x}}>0\)

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

Vậy \(0\le x< 4\) thì B > -1.

c) \(B=\dfrac{\sqrt{x}+6}{2-\sqrt{x}}=-1-\dfrac{8}{2-\sqrt{x}}\in Z\)

\(\Rightarrow2-\sqrt{x}\inƯ_{\left(8\right)}=\left\{\pm1;\pm2;\pm4;\pm8\right\}\)

\(\Rightarrow\sqrt{x}\in\left\{1;3;0;4;-2;6;-6;10\right\}\)

\(\Rightarrow x\in\left\{1;9;0;16;36;100\right\}\)thì \(B\in Z\)

a) đk : \(x\ne4;x\ge0\)

B = \(\dfrac{\sqrt{x}-1}{\sqrt{x}+1}-\dfrac{\sqrt{x}+3}{\sqrt{x}-2}-\dfrac{x+5}{x-\sqrt{x}-2}\)

B = \(\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)-\left(\sqrt{x}+3\right)\left(\sqrt{x}+1\right)-\left(x+5\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\)

B = \(\dfrac{x-2\sqrt{x}-\sqrt{x}+2-\left(x+\sqrt{x}+3\sqrt{x}+3\right)-x-5}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\)

B = \(\dfrac{x-2\sqrt{x}-\sqrt{x}+2-x-\sqrt{x}-3\sqrt{x}-3-x-5}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\)

B = \(\dfrac{-x-7\sqrt{x}-6}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\) = \(\dfrac{\left(-\sqrt{x}-6\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\)

B = \(\dfrac{-\sqrt{x}-6}{\sqrt{x}-2}\)

a) B = \(\dfrac{\sqrt{x}-1}{\sqrt{x}+1}-\dfrac{\sqrt{x}+3}{\sqrt{x}-2}-\dfrac{x+5}{x-\sqrt{x}+2}\) ( đk: x \(\ge\) 0; x\(\ne\)4)

<=> B = \(\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)-\left(\sqrt{x}+3\right)\left(\sqrt{x}+1\right)-x-5}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\)

<=> B = \(\dfrac{x-3\sqrt{x}+2-x-4\sqrt{x}-3-x-5}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\)

<=> B = \(\dfrac{x+7\sqrt{x}+6}{-\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}=\dfrac{-\left(\sqrt{x}+1\right)\left(\sqrt{x}+6\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\)

=> B = \(\dfrac{-\sqrt{x}-6}{\sqrt{x}-2}\)

b) Để B > -1 => \(\dfrac{-\sqrt{x}-6}{\sqrt{x}-2}>-1\)

=> \(\dfrac{-\sqrt{x}-6}{\sqrt{x}-2}+1>0\)

<=> \(\dfrac{-8}{\sqrt{x}-2}\) > 0 => \(\sqrt{x}-2< 0\) => \(x< 4\)

Đối chiếu với điều kiện ta được: \(0\le x< 4\)

c) Để B \(\in\) Z => \(\dfrac{-\sqrt{x}-6}{\sqrt{x}-2}\) \(\in\) Z

Mà \(\dfrac{-\sqrt{x}-6}{\sqrt{x}-2}\) = \(\dfrac{-\left(\sqrt{x}-2\right)-8}{\sqrt{x}-2}=-1-\dfrac{8}{\sqrt{x}-2}\)

=> \(\dfrac{8}{\sqrt{x}-2}\in Z\) => 8 \(⋮\) \(\sqrt{x}-2\)

=> \(\sqrt{x}\inƯ\left(8\right)=\left\{\pm1;\pm2;\pm4;\pm8\right\}\)

=> \(\sqrt{x}\in\left\{3;1;4;0;6;-4;10;-6\right\}\)

Mà \(x\ge0\) => \(x\in\left\{9;0;1;16;36;100\right\}\)

Vậy .......................................................