Đặt AH=h

Xong c vô trang này là đc

https://h.vn/hoi-dap/question/92121.html

\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)

\(\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}-\sqrt{2}=0\)

:) trình bày các bước đi bạn :)) ai lại làm thế :v Bấm casio à :)

\(H=\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}\)

\(H^2=4+\sqrt{7}+4-\sqrt{7}+2\sqrt{\left(4+\sqrt{7}\right)\left(4-\sqrt{7}\right)}\)

\(=8-2\sqrt{16-7}=8-6=2\)

\(\Rightarrow H=\sqrt{2}\Rightarrow\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}-2=0\)

Vậy .....................

gọi \(A=\sqrt{3+\sqrt{3}}+\sqrt{3-\sqrt{3}}\)

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

\(< =>A^2=6+2\sqrt{9-3\sqrt{3}+3\sqrt{3}-\sqrt{3^2}}\)

\(< =>A^2=6+2\sqrt{6}\)

\(< =>A=\sqrt{6+2\sqrt{6}}\)

Sửa đề nha :

Đặt 

\(A=\sqrt{1+\sqrt{3+\sqrt{13+4\sqrt{3}}}}+\sqrt{1-\sqrt{3-\sqrt{13-4\sqrt{3}}}}\)

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

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

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

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

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

\(A^2=4+2\sqrt{4}=6\)

\(A=\sqrt{6}\)

Vậy ....

\(\)

Sửa từ dòng 6 :

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

\(A^2=4+2\sqrt{1}=6\)

\(A=6\)

Vậy ...

a) A = \(\left(\frac{x}{x^2-4}+\frac{2}{2-x}+\frac{1}{x+2}\right):\left(x-2+\frac{10-x^2}{x+2}\right)\)

A = \(\left[\frac{x}{\left(x-2\right)\left(x+2\right)}-\frac{2\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{x-2}{x+2}\right]:\left[\frac{\left(x-2\right)\left(x+2\right)}{x+2}+\frac{10-x^2}{x+2}\right]\)

A = \(\left[\frac{x-2x-4+x-2}{\left(x-2\right)\left(x+2\right)}\right]:\left[\frac{x^2-4+10-x^2}{x+2}\right]\)

A = \(-\frac{6}{\left(x-2\right)\left(x+2\right)}:\frac{6}{x+2}\)

A = \(-\frac{6\left(x+2\right)}{6\left(x-2\right)\left(x+2\right)}\)

A = \(-\frac{6}{6\left(x-2\right)}\)

A = \(-\frac{1}{x-2}\)

b) |x| = \(\hept{\begin{cases}x=\frac{1}{2}\\x=-\frac{1}{2}\end{cases}}\)

+) với x = 1/2, ta có: 

A = \(-\frac{1}{\frac{1}{2}-2}=\frac{2}{3}\)

+) với x = -1/2, ta có:

A = \(-\frac{1}{\left(-\frac{1}{2}\right)-2}=\frac{2}{5}\)

a, Đẻ \(P< 1\)thì : 

\(P=\left(\frac{x}{x+2}+\frac{x}{x-2}-\frac{2}{x^2-4}\right).\frac{x-2}{2x+2}< 1\)

\(=\left(\frac{x\left(x-2\right)\left(x^2-4\right)}{\left(x+2\right)\left(x-2\right)\left(x^2-4\right)}+\frac{x\left(x+2\right)\left(x^2-4\right)}{\left(x-2\right)\left(x+2\right)\left(x^2-4\right)}-\frac{2\left(x+2\right)\left(x-2\right)}{\left(x^2+4\right)\left(x+2\right)\left(x-2\right)}\right).\frac{x-2}{2x+2}\)

\(=\left(\frac{x\left(x-2\right)\left(x^2-4\right)+x\left(x+2\right)\left(x^2-4\right)-2\left(x+2\right)\left(x-2\right)}{\left(x+2\right)\left(x-2\right)\left(x^2-4\right)}\right).\frac{x-2}{2x+2}\)

\(=\left(\frac{2x^4-10x^2+8}{x^4-8x^2+16}\right).\frac{x-2}{2x+2}=\left(2x^4-10x^2+8\right)\left(2x+2\right)=\left(x-2\right)\left(x^4-8x^2+16\right)\)

PT tương đương vs : \(\left(2x^4-10x^2+8\right)\left(2x+2\right)-\left(x-2\right)\left(x^4-8x^2+16\right)< 1\)

Khi đó pt trở thành : \(3x^5+6x^4-12x^3-36x^2+48< 1\)

Chắc vại đó ==