Bài 7: Biến đối đơn giản biểu thức chứa căn bậc hai (Tiếp theo)

a/ ĐKXĐ: \(x\ge-\dfrac{3}{2}\)

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

\(\Leftrightarrow2x+3=1+2\sqrt{2}+2\)

\(\Leftrightarrow2x=2\sqrt{2}\)

\(\Leftrightarrow x=\sqrt{2}\) ( tmđk)

Vậy phương trình có nghiệm \(x=\sqrt{2}\)

b/ ĐKXĐ: \(x\ge-\dfrac{10}{\sqrt{3}}\)

\(\sqrt{10+\sqrt{3}x}=2+\sqrt{6}\)

\(\Leftrightarrow10+\sqrt{3}x=4+4\sqrt{6}+6\)

\(\Leftrightarrow\sqrt{3}x=4\sqrt{6}\)

\(\Leftrightarrow x=4\sqrt{2}\) ( tmđk)

Vậy phương trình có nghiệm \(x=4\sqrt{2}\)

c/ \(x\ge\dfrac{2}{3}\)

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

\(\Leftrightarrow3x-2=4-4\sqrt{3}+3\)

\(\Leftrightarrow3x=9-4\sqrt{3}\)

\(\Leftrightarrow x=\dfrac{9-4\sqrt{3}}{3}\) ( tmđk)

Vậy phương trình có nghiệm \(x=\dfrac{9-4\sqrt{3}}{3}\)

d/ Xét vế trái phương trình: \(\sqrt{x+1}\ge0\) (1)

vế phải phương trình: \(\sqrt{5}-3< 0\left(2\right)\)

Ta thấy (1) và (2) mâu thuẫn \(\Rightarrow\) Phương trình đã cho vô nghiệm

a) \(\sqrt{2x+3}\)= 1 + \(\sqrt{2}\)( ĐK : x \(\ge\dfrac{-3}{2}\))

<=> 2x + 3 = ( 1 + \(\sqrt{2}\))²

<=> 2x + 3 = 1 + 2\(\sqrt{2}\)+2

<=> 2x = 2\(\sqrt{2}\)

<=> x = \(\sqrt{2}\)( TMĐK )

Vậy x = \(\sqrt{2}\)

sai đề rồi bạn ơi, sửa đề

\(B=\dfrac{1}{\sqrt{1}-\sqrt{2}}-\dfrac{1}{\sqrt{2}-\sqrt{3}}-\dfrac{1}{\sqrt{3}-\sqrt{4}}-...-\dfrac{1}{\sqrt{24}-\sqrt{25}}\)

ta có: \(\dfrac{1}{\sqrt{n}-\sqrt{n+1}}=\dfrac{\sqrt{n}+\sqrt{n+1}}{n-n-1}=-\sqrt{n}-\sqrt{n+1}\)

áp dụng vào B, ta có:

\(B=-\sqrt{1}-\sqrt{2}+\sqrt{2}+\sqrt{3}-\sqrt{3}-\sqrt{4}+...+\sqrt{24}+\sqrt{25}\)

\(B=\sqrt{25}-\sqrt{1}=4\)

\(\dfrac{\sqrt{5}-\sqrt{3}}{\sqrt{2}}=\dfrac{\sqrt{10}-\sqrt{6}}{2}\)

\(\dfrac{\sqrt{5}-\sqrt{3}}{\sqrt{2}}\) \(=\dfrac{\sqrt{10}-\sqrt{6}}{2}\)

yêu cầu là gì

liên hợp ra, áp dụng \(a^2-b^2=\left(a-b\right)\left(a+b\right)\Rightarrow\dfrac{a^2-b^2}{a+b}=a-b\)

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

\(=\dfrac{\left(1+\sqrt{2}-\sqrt{3}\right)\left(2\sqrt{2}-2\right)}{6}\approx0,0941870216...\)

trong khi đó : \(\dfrac{1}{1+\sqrt{2}+\sqrt{3}}=0,2411809549\)

ý tớ là thế này đây....

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

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

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

\(=\dfrac{\sqrt{a}-\sqrt{b}+2\sqrt{b}}{\sqrt{a}+\sqrt{b}}=1=VP\)

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

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

= \(\dfrac{2\sqrt{3}-4}{\sqrt{3}-1}.\dfrac{2}{\sqrt{3}+1}\) = \(\dfrac{4\sqrt{3}-8}{2}\) = \(2\sqrt{3}-4\)

a) \(\sqrt{12-2\sqrt{35}}=\sqrt{\left(\sqrt{5}-\sqrt{7}\right)^2}=\sqrt{7}-\sqrt{5}\)

b) \(\sqrt{4+\sqrt{15}}=...\)

c) \(\left(3-\sqrt{2}\right)\sqrt{11+6\sqrt{2}}=\left(3\sqrt{2}\right)\sqrt{\left(3+\sqrt{2}\right)^2}\\ =\left(3-\sqrt{2}\right)\left(3+\sqrt{2}\right)=9-2=7\)

d) \(\left(\sqrt{5}+\sqrt{7}\right)\sqrt{12-2\sqrt{35}}=\left(\sqrt{7}+\sqrt{5}\right)\sqrt{\left(\sqrt{5}-\sqrt{7}\right)^2}\\ =\left(\sqrt{7}+\sqrt{5}\right)\left(\sqrt{7}-\sqrt{5}\right)=7-5=2\)

e) \(\sqrt{7-2\sqrt{10}-\sqrt{7+2\sqrt{10}}}=\sqrt{7-2\sqrt{10}-\sqrt{\left(\sqrt{2}+\sqrt{5}\right)^2}}\\ =\sqrt{7-2\sqrt{10}-\left(\sqrt{2}+\sqrt{5}\right)}=\sqrt{7-2\sqrt{10}-\sqrt{2}-\sqrt{5}}\\ =\sqrt{7-2\sqrt{10}-\sqrt{2}-\sqrt{5}}\)

f) \(\sqrt{13-\sqrt{160}+\sqrt{53}+4\sqrt{90}}=\sqrt{13-4\sqrt{10}+\sqrt{53}+12\sqrt{10}}\\ =\sqrt{13+8\sqrt{10}+\sqrt{53}}\)

dài quá nhác làm

a) \(\sqrt{14-6\sqrt{5}}=\sqrt{\left(3-\sqrt{5}\right)^2}=3-\sqrt{5}\)

b, c) tương tự câu a.

d) \(\left(3-\sqrt{2}\right)\sqrt{11+6\sqrt{2}}\)

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

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

\(=9-2\)

\(=7\)

e) \(\sqrt{11-6\sqrt{2}+\sqrt{3-2\sqrt{2}}}\)

\(=\sqrt{11-6\sqrt{2}+\sqrt{\left(1-\sqrt{2}\right)^2}}\)

\(=\sqrt{11-6\sqrt{2}+\sqrt{2}-1}\)

\(=\sqrt{10-5\sqrt{2}}\)