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27 tháng 5 2017

Căn bậc hai. Căn bậc ba

19 tháng 6 2017

Bài 1 : Rút gọn biểu thức :

\(\left(2-\sqrt{2}\right)\left(-5\sqrt{2}\right)-\left(3\sqrt{2}-5\right)^2\)

\(=\left(-10\sqrt{2}+10\right)-\left(18-30\sqrt{2}+25\right)\)

\(=\left(-10\sqrt{2}+10\right)-\left(7-30\sqrt{2}\right)\)

\(=-10\sqrt{2}+10-7+30\sqrt{2}\)

\(=20\sqrt{2}+3\)

19 tháng 6 2017

Bài 2:

a) ĐKXĐ : x # 4 ; x # - 4

P = \(\dfrac{\sqrt{x}+1}{\sqrt{x}-2}+\dfrac{2\sqrt{x}}{\sqrt{x}+2}+\dfrac{2+5\sqrt{x}}{4-x}\)

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

P = \(\dfrac{x+2\sqrt{x}+\sqrt{x}+2+2x-4\sqrt{x}-2-5\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)

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

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

b ) Để P = 2 \(\Leftrightarrow\dfrac{3\sqrt{x}}{\sqrt{x}+2}\) = 2

\(\Leftrightarrow3\sqrt{x}=2\sqrt{x}+4\)

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

\(\Leftrightarrow x=16\)

Vậy, để P = 2 thì x = 16.

NV
4 tháng 1 2019

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

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

\(P=\dfrac{\sqrt{a}+1}{\sqrt{a}}\)

b/

\(a=2\sqrt{3-\sqrt{5}}\left(3+\sqrt{5}\right)\left(\sqrt{10}-\sqrt{2}\right)\)

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

\(a=\sqrt{6-2\sqrt{5}}\left(6+2\sqrt{5}\right)\left(\sqrt{5}-1\right)=\sqrt{\left(\sqrt{5}-1\right)^2}\left(\sqrt{5}+1\right)^2\left(\sqrt{5}-1\right)\)

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

\(a=\left[\left(\sqrt{5}+1\right)\left(\sqrt{5}-1\right)\right]^2=4^2=16\)

\(\Rightarrow P=\dfrac{\sqrt{a}+1}{\sqrt{a}}=\dfrac{\sqrt{16}+1}{\sqrt{16}}=\dfrac{4+1}{4}=\dfrac{5}{4}\)

17 tháng 6 2018

ĐKXĐ: a ≥ 0

a) Ta có:

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

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

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

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

Vậy P = \(\sqrt{a}+1\) với a ≥ 0

b) Ta có: a = \(1996-2\sqrt{1995}\) = \(\left(\sqrt{1995}-1\right)^2\) (TMĐK)

\(\sqrt{a}=\sqrt{1995}-1\). Thay vào P ta được

P = \(\sqrt{1995}-1+1=\sqrt{1995}\)

Vậy P = \(\sqrt{1995}\) khi a = \(1996-2\sqrt{1995}\)

24 tháng 6 2017

bài 2 ) a) đk : \(a>0;b>0\)

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

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

P = \(\dfrac{a+2\sqrt{ab}+b}{\sqrt{a}+\sqrt{b}}.\sqrt{a}-\sqrt{b}\) = \(\dfrac{\left(\sqrt{a}+\sqrt{b}\right)^2}{\sqrt{a}+\sqrt{b}}.\sqrt{a}-\sqrt{b}\) = \(\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)\) = \(a-b\)

c) ta có P = \(a-b\) thay \(a=2\sqrt{3};b=\sqrt{3}\) vào ta có

P = \(2\sqrt{3}-\sqrt{3}=\sqrt{3}\) vậy khi \(a=2\sqrt{3};b=\sqrt{3}\) thì P = \(\sqrt{3}\)

24 tháng 6 2017

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

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

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

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

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

P = \(2+\dfrac{2a+2}{\sqrt{a}}\) = \(\dfrac{2a+2\sqrt{a}+2}{\sqrt{a}}\)

b) ta có P = 7 \(\Leftrightarrow\) \(\dfrac{2a+2\sqrt{a}+2}{\sqrt{a}}=7\) \(\Leftrightarrow\) \(2a+2\sqrt{a}+2=7\sqrt{a}\)

\(\Leftrightarrow\) \(2a-5\sqrt{a}+2=0\) (1)

đặc \(\sqrt{a}=u\) \(\left(u\ge0\right)\) (1) \(\Leftrightarrow\) \(2u^2-5u+2\)

\(\Delta=\left(-5\right)^2-4.2.2\) = \(25-16=9>0\)

\(\Rightarrow\) phương trình có 2 nghiệm phân biệt

\(u_1=\dfrac{5+3}{4}=\dfrac{8}{4}=2\left(tmđk\right)\)

\(u_2=\dfrac{5-3}{4}=\dfrac{2}{4}=\dfrac{1}{2}\left(tmđk\right)\)

ta có : \(u=\sqrt{a}=2\Leftrightarrow x=4\)

\(u=\sqrt{a}=\dfrac{1}{2}\Leftrightarrow a=\dfrac{1}{4}\)

vậy \(a=4;a=\dfrac{1}{4}\) thì P = 7

10 tháng 9 2017

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

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

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

\(=2\sqrt{2}\)

10 tháng 9 2017

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

ĐKXĐ \(x>0,x\ne1\)

pt (1) <=> \(\left(\dfrac{\sqrt{x}+1+\sqrt{x}-1}{\left(\sqrt{x}-1\right)\cdot\left(\sqrt{x}+1\right)}\right)\cdot\left(\dfrac{\sqrt{x}+1}{\sqrt{x}}\right)\)

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

\(\Leftrightarrow\dfrac{2\sqrt{x}}{x-\sqrt{x}}\)

\(\Leftrightarrow\dfrac{\sqrt{x}\cdot2}{\sqrt{x}\cdot\left(\sqrt{x}-1\right)}\)

\(\Leftrightarrow\dfrac{2}{\sqrt{x}-1}\)

b) Để \(\sqrt{A}>A\Leftrightarrow\sqrt{\dfrac{2}{\sqrt{x}-1}}>\dfrac{2}{\sqrt{x}-1}\)

\(\Leftrightarrow\dfrac{2}{\sqrt{x}-1}>\dfrac{4}{x-2\sqrt{x}+1}\)

\(\Leftrightarrow\dfrac{2}{\sqrt{x}-1}-\dfrac{4}{x-2\sqrt{x}+1}>0\)

\(\Leftrightarrow\dfrac{2\cdot\left(\sqrt{x}-1\right)-4}{x-2\sqrt{x}+1}>0\)

\(\Leftrightarrow\dfrac{2\sqrt{2}-2-4}{x-2\sqrt{x}+1}>0\)

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

\(2\sqrt{2}-6< 0\Rightarrow x-2\sqrt{x}+1< 0\)

\(x-2\sqrt{x}+1=\left(\sqrt{x}-1\right)^2\ge0\forall x\)

Vậy không có giá trị nào của x thỏa mãn \(\sqrt{A}>A\)

(P/s Đề câu b bị sai hay sao vậy, chả có số nào mà \(\sqrt{A}>A\) cả, check lại đề giùm với nhé)

Câu 1: 

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

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

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

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

b: Để P>0 thì căn a-2>0

=>a>4

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

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

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

\(\Leftrightarrow a\left(P-\dfrac{1}{2}\right)< 0\)

20 tháng 7 2018

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

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

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

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

\(=\dfrac{1}{\sqrt{a}}.\dfrac{\sqrt{a}-2}{3}\)

\(=\dfrac{\sqrt{a}-2}{3\sqrt{a}}\) ĐKXĐ: \(x>0\) \(a\ne4\) \(a\ne1\)

b) \(Q>0\)

\(\Leftrightarrow\dfrac{\sqrt{a}-2}{3\sqrt{a}}>0\)

\(3\sqrt{a}>0\) (Kết hợp ĐKXĐ \(a>0\))

\(\Leftrightarrow\sqrt{a}-2>0\)

\(\Leftrightarrow\sqrt{a}>2\)

\(\Leftrightarrow a>4\) (Thỏa mãn ĐKXĐ)

Vậy \(a>4\) thì \(Q>0\)

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