hok

tốt 

nha

a) Ta có ${\left(ac+bd\right)}^2+{\left(ad-bc\right)}^2$$=a^2{c}^2+2acbd+b^2{d}^2+a^2{d}^2-2adbc+b^2{c}^2$

$=\left(a^2{c}^2+b^2{c}^2\right)+\left(a^2{d}^2+b^2{d}^2\right)$$=c^2\left(a^2+b^2\right)+d^2\left(a^2+b^2\right)$$=\left(a^2+b^2\right)\left(c^2+d^2\right)$

b) Ta có $0\le {\left(ad-bc\right)}^2$$\iff {\left(ac+bd\right)}^2\le {\left(ac+bd\right)}^2+{\left(ad-bc\right)}^2$

Mà theo câu a, ta có ${\left(ac+bd\right)}^2+{\left(ad-bc\right)}^2=\left(a^2+b^2\right)\left(c^2+d^2\right)$

Nên ${\left(ac+bd\right)}^2\le \left(a^2+b^2\right)\left(c^2+d^2\right)$

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

\(\sqrt{2}A=\sqrt{2x-5+2\sqrt{2x-5}+1}+\sqrt{2x-5+6\sqrt{2x-5}+9}\)

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

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

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

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

 \(\sqrt{46}\)nha

\(\left(\frac{2}{\sqrt{7}-\sqrt{5}}-\frac{3}{\sqrt{7}+\sqrt{5}}\right)\left(5\sqrt{5}+\sqrt{7}\right)\)

\(=\left(\frac{2\sqrt{7}+2\sqrt{5}-3\sqrt{7}+3\sqrt{5}}{2}\right)\left(5\sqrt{5}+\sqrt{7}\right)\)

\(=\left(\frac{-\sqrt{7}+5\sqrt{5}}{2}\right)\left(5\sqrt{5}+\sqrt{7}\right)=\frac{25.5-7}{2}=\frac{118}{2}=59\)

Với x >= -1,5 ; y < 0 

\(\sqrt{\frac{\left(9+2x\right)^2}{y^2}}=\frac{\left|9+2x\right|}{\left|y\right|}=\frac{9+2x}{-y}\)

\(\frac{5\sqrt{2}-2\sqrt{5}}{\sqrt{5}-\sqrt{2}}+\frac{6}{2-\sqrt{10}}\)

\(=\frac{\sqrt{10}\left(\sqrt{5}-\sqrt{2}\right)}{\sqrt{5}-\sqrt{2}}+\frac{6\left(2+\sqrt{10}\right)}{-6}\)

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

Áp dụng bđt bunhiacopxki ta có:

(√b+1+√c+1)2≤(b+1+c+1)(12+12)(b+1+c+1)2≤(b+1+c+1)(12+12)

⇔2(b+c+2)≥4(a+1)⇔2(b+c+2)≥4(a+1)

⇔b+c+2≥2a+2⇔b+c+2≥2a+2

⇔b+c≥2a

bạn viết lại đề bài đi 

Đề: lười ghi

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

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

\(=-2\)

Trục căn thức ở mẫu có công thức đó bạn còn nếu bạn lười thì ấn máy tính là ra

\(\left(\frac{\sqrt{15}-\sqrt{12}}{\sqrt{5}-2}=\sqrt{3}\right)\)

\(\frac{2+\sqrt{3}}{2-\sqrt{3}}+\frac{2-\sqrt{3}}{2+\sqrt{3}}=\frac{\left(2+\sqrt{3}\right)^2+\left(2-\sqrt{3}\right)^2}{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}=\frac{7+4\sqrt{3}+7-4\sqrt{3}}{4-3}=14\)

\(\frac{\sqrt{15}-\sqrt{12}}{\sqrt{5}-\sqrt{2}}+\frac{6}{2+\sqrt{10}}=\frac{\left(\sqrt{15}-\sqrt{12}\right)\left(2+\sqrt{10}\right)+6\left(\sqrt{5}-\sqrt{2}\right)}{\left(\sqrt{5}-\sqrt{2}\right)\left(2+\sqrt{10}\right)}\)

\(=\frac{2\sqrt{15}+5\sqrt{6}-4\sqrt{3}-2\sqrt{30}+6\sqrt{5}-6\sqrt{2}}{5\sqrt{2}-2\sqrt{2}}\)

a. \(\sqrt{\frac{y}{5x^3}}=\sqrt{\frac{5xy}{25x^4}}=\frac{\sqrt{5xy}}{25x^2}\)

b\(\sqrt{\frac{5}{x\left(1-\sqrt{2}\right)}}=\sqrt{\frac{5\times x\left(1+\sqrt{2}\right)}{x^2\left(1-\sqrt{2}\right)\left(1+\sqrt{2}\right)}}=\sqrt{\frac{-5\times x\left(1+\sqrt{2}\right)}{x^2}}=-\frac{\sqrt{-5\times x\left(1+\sqrt{2}\right)}}{x}\)

c.\(\sqrt{\frac{x-1}{2\left(\sqrt{x}-1\right)}}=\sqrt{\frac{\sqrt{x}+1}{2}}=\frac{\sqrt{2\sqrt{x}+2}}{2}\)

d.\(a\sqrt{\frac{4}{a}}=\sqrt{\frac{4a^2}{a}}=\sqrt{4a}=2\sqrt{a}\)

e.\(2\sqrt{\frac{1}{-a}}=2\sqrt{\frac{-a}{a^2}}=-\frac{2}{a}\sqrt{-a}\left(\text{ do a< 0}\right)\)\(2\sqrt{\frac{1}{-a}}=2\sqrt{\frac{-a}{a^2}}=-\frac{2}{a}\sqrt{-a}\)( do a <0)

f.\(\sqrt{\frac{2}{x-1}-\frac{1}{\left(x-1\right)^2}}=\sqrt{\frac{2\left(x-1\right)-1}{\left(x-1\right)^2}}=\frac{\sqrt{2x-3}}{\left|x-1\right|}\)