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

+) câu này không làm được :

nếu muốn phân tích thì ta phải đưa nó về dạng \(\left(a+b\right)^2=a^2+b^2+2ab\)

trong đó \(a^2+b^2=\) phần nguyên \(\ge\) phần có căn = 2ab

mà ta có : \(\sqrt{48}>3\) \(\Rightarrow\) không thể phân tích đc ...

+) ta có : \(B=\dfrac{\sqrt{8-\sqrt{15}}}{\sqrt{30}-\sqrt{2}}=\dfrac{\sqrt{\left(\sqrt{15}-1\right)^2}}{2\sqrt{15}-2}=\dfrac{\sqrt{15}-1}{2\left(\sqrt{15}-1\right)}=\dfrac{1}{2}\)

+) ta có : \(C=\dfrac{1}{\sqrt{3}+1}-\dfrac{1}{\sqrt{3}-\sqrt{2}}=\dfrac{\sqrt{3}-1}{2}-\dfrac{\sqrt{3}+\sqrt{2}}{1}\)

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

Mysterious Person giúp mk

a)Đkxđ : x#1 , x > 0

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

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

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

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

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

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

b)Thay x = 2\(\sqrt{2}\)+3 vào phương trình ta được :

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

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

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

Q= 2

Mysterious Person giup mk

a) điều kiện xác định : \(a\ge0;a\ne1\)

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

b) ta có : \(A\in Z\Leftrightarrow\dfrac{2\sqrt{a}}{\sqrt{a}-1}\in Z\Leftrightarrow2+\dfrac{2}{\sqrt{a}-1}\in Z\)

\(\Leftrightarrow\sqrt{a}-1\) là ước của \(2\) \(\Leftrightarrow\left(\sqrt{a}-1\right)\in\left\{\pm1;\pm2\right\}\)

\(\Rightarrow\) \(a\in\left\{0;4;9\right\}\)

c) ta có : \(A-2=\dfrac{2\sqrt{a}}{\sqrt{a}-1}-2=\dfrac{2\sqrt{a}-2\sqrt{a}+2}{\sqrt{a}-1}=\dfrac{2}{\sqrt{a}-1}\)

(không so sánh được ) --> kiểm tra để

Mysterious Person giup mk

a) điều kiện xác định : \(x>0;x\ne1\)

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

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

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

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

b) để \(A>-6\Leftrightarrow-2\sqrt{x}>-6\Leftrightarrow\sqrt{x}< 3\Leftrightarrow0< x< 9\) và \(x\ne1\)

vậy ....

Mysterious Person giúp mk

Đk: x >0 ; x khác 1

sau khi rút gọn ra -2

b, 9>x>0

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

Mysterious Person giups mk nha

a) ta có \(2-\sqrt{6}=\sqrt{2}\left(\sqrt{2}-\sqrt{3}\right)>1-\sqrt{3}\)

b) ta có : \(2-\sqrt{2}=\dfrac{1}{2}\left(4-2\sqrt{2}\right)\)

mà ta có : \(2\sqrt{2}< 3\) (vì \(8< 9\))

\(\Rightarrow4-2\sqrt{2}>4-3>1\) \(\Rightarrow2-\sqrt{2}>\dfrac{1}{2}\)

c) ta có : \(\left(\sqrt{2003}+\sqrt{2005}\right)^2=4008+2\sqrt{2003.2005}\)

\(\left(2\sqrt{2004}\right)^2=8016=4008+4008=4008+2\sqrt{2004.2004}\)

mà ta có : \(x^2\ge x^2-1\Rightarrow x^2>\left(x-1\right)\left(x+1\right)\)

\(\Rightarrow4008+2\sqrt{2004.2004}>4008+2\sqrt{2003.2005}\)

\(\Rightarrow2\sqrt{2004}>\sqrt{2003}+\sqrt{2005}\)

Mysterious Person giúp mk

Giải:

\(\sqrt{42-10\sqrt{17}}+\sqrt{33-8\sqrt{17}}\)

\(=\sqrt{\left(5-\sqrt{17}\right)^2}+\sqrt{\left(4-\sqrt{17}\right)^2}\)

\(=\left|5-\sqrt{17}\right|+\left|4-\sqrt{17}\right|\)

\(=5-\sqrt{17}+\sqrt{17}-4\)

\(=1\)

Vậy ...

\(\sqrt{42-10\sqrt{17}}+\sqrt{33-8\sqrt{17}}=\sqrt{25-2.5.\sqrt{17}+17}+\sqrt{16-2.4.\sqrt{17}+17}=\sqrt{\left(5-\sqrt{17}\right)^2}+\sqrt{\left(4-\sqrt{17}\right)^2}=\left|5-\sqrt{17}\right|+\left|4-\sqrt{17}\right|=5-\sqrt{17}+\sqrt{17}-4=1\)

3 căn 56 nha

\(3\sqrt{56}-2\sqrt{98}-5\sqrt{18}-\sqrt{63}+2\sqrt{28}=3\sqrt{4.14}-2\sqrt{49.2}-5\sqrt{9.2}-\sqrt{9.7}+2\sqrt{4.7}=3.2\sqrt{14}-2.7\sqrt{2}-5.3\sqrt{2}-3\sqrt{7}+2.2\sqrt{7}=6\sqrt{14}-14\sqrt{2}-15\sqrt{2}-3\sqrt{7}+4\sqrt{7}=6\sqrt{14}-29\sqrt{2}+\sqrt{7}\)

\(\left(5+4\sqrt{2}\right)\left(3+2\sqrt{1+\sqrt{2}}\right)\left(3-2\sqrt{1+\sqrt{2}}\right)=\left(5+4\sqrt{2}\right)\left[3^2-\left(2\sqrt{1+\sqrt{2}}\right)^2\right]=\left(5+4\sqrt{2}\right)\left[9-4\left(1+\sqrt{2}\right)\right]=\left(5+4\sqrt{2}\right)\left(9-4-4\sqrt{2}\right)=\left(5+4\sqrt{2}\right)\left(5-4\sqrt{2}\right)=5^2-\left(4\sqrt{2}\right)^2=25-16.2=25-32=-7\)

Câu 1:

ĐK: \(x\geq \frac{-3}{2}\)

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

\(\Rightarrow 2x+3=(3-\sqrt{5})^2=14-6\sqrt{5}\)

\(\Rightarrow x=\frac{11-6\sqrt{5}}{2}\)

Câu 2: ĐK: \(x\geq 0\)

\(\sqrt{5+\sqrt{7x}}=2+\sqrt{7}\)

\(\Rightarrow 5+\sqrt{7x}=(2+\sqrt{7})^2=11+4\sqrt{7}\)

\(\Rightarrow \sqrt{7x}=6+4\sqrt{7}\)

\(\Rightarrow 7x=(6+4\sqrt{7})^2\Rightarrow x=\frac{(6+4\sqrt{7})^2}{7}\)

Câu 3: ĐK: \(x\geq 0\)

\((\sqrt{x}-2)(5-\sqrt{x})=4-x\)

\(\Leftrightarrow 5\sqrt{x}-x-10+2\sqrt{x}=4-x\)

\(\Leftrightarrow 7\sqrt{x}=14\Rightarrow \sqrt{x}=2\Rightarrow x=4\)

Câu 4: ĐK: \(x\ge 1\)

Sửa đề \(\frac{1}{2}\sqrt{x-1}-\frac{3}{2}\sqrt{9x-9}+24\sqrt{\frac{x-1}{64}}=-17\)

\(\Leftrightarrow \frac{\sqrt{x-1}}{2}-\frac{3}{2}\sqrt{9}.\sqrt{x-1}+24\sqrt{\frac{1}{64}}\sqrt{x-1}=-17\)

\(\Leftrightarrow \frac{\sqrt{x-1}}{2}-\frac{9\sqrt{x-1}}{2}+3\sqrt{x-1}=-17\)

\(\Leftrightarrow \sqrt{x-1}(\frac{1}{2}-\frac{9}{2}+3)=-17\)

\(\Leftrightarrow -\sqrt{x-1}=-17\Rightarrow \sqrt{x-1}=17\Rightarrow x=17^2+1=290\)