help me pls! thankss guys
rút gọn:
B=(\(\frac{1}{1-\sqrt{x}}+\frac{1}{1+\sqrt{x}}\) ):(\(\frac{1}{1-\sqrt{x}}-\frac{1}{1+\sqrt{x}}\) )+\(\frac{1}{2\sqrt{x}}\)
rút gọn:
a)\(\left(\frac{1}{2+2\sqrt{x}}+\frac{1}{2-2\sqrt{x}}-\frac{x^2+1}{1-x^2}\right)\times\left(1+\frac{1}{x}\right)\)
b)\(\left(\frac{2\sqrt{xy}}{x-y}+\frac{\sqrt{x}-\sqrt{y}}{2\sqrt{x}+\sqrt{y}}\right)\times\frac{2\sqrt{x}}{\sqrt{x}+\sqrt{y}}+\frac{\sqrt{y}}{\sqrt{y}-\sqrt{x}}\)
c)\(\left(\frac{x-1}{\sqrt{x}-1}+\frac{x\sqrt{x}-1}{1-x}\right)\div\frac{\left(\sqrt{x}-1\right)^2+\sqrt{x}}{\sqrt{x}+1}\)
a, dk \(x\ge0.x\ne1\)
\(\left(\frac{1+\sqrt{x}+1-\sqrt{x}}{2\left(1-x\right)}-\frac{x^2+1}{1-x^2}\right)\left(\frac{x+1}{x}\right)\)=\(\left(\frac{1}{1-x}-\frac{x^2+1}{1-x^2}\right)\left(\frac{x+1}{x}\right)\)
=\(\left(\frac{1+x-x^2-1}{1-x^2}\right)\left(\frac{x+1}{x}\right)=\frac{x\left(1-x\right)\left(x+1\right)}{x\left(1-x\right)\left(1+x\right)}=1\)
phan b,c ban tu lam not nhe dai lam mk ko lam dau mk co vc ban rui
1. A=\(\frac{2x+2}{\sqrt{x}}+\frac{x\sqrt{x}-1}{x-\sqrt{x}}-\frac{x\sqrt{x}+1}{x+\sqrt{x}}\)
a. Rút gọn A
b. Chứng minh: A>5
c. Tìm x sao cho \(\frac{8}{A}\)có giá trị nguyên.
Help me. Thanks!
\(B=\left(\frac{\sqrt{X}+1}{\sqrt{X}-1}-\frac{\sqrt{X}-1}{\sqrt{X}+1}-\frac{8\sqrt{X}}{X-1}\right)\div\left(\frac{\sqrt{X}-X-3}{X-1}-\frac{1}{\sqrt{X}-1}\right)\)
Rút gọn B
Rút gọn biểu thức :
E=\(\frac{\sqrt{2x+2\sqrt{x^2-4}}}{\sqrt{x^2-4}+x+2}\)
F=\(\frac{\left(\frac{3}{\sqrt{1+a}}+\sqrt{1+a}\right)}{\left(\frac{3}{\sqrt{1-a^2}}+1\right)}\)
HELP ME !
Rút gọn B=\(\left(\frac{3x +3\sqrt{x}-3}{x+\sqrt{x}-2}+\frac{1}{\sqrt{x}+1}+\frac{1}{\sqrt{x}+2}-2\right):\frac{1}{x-1}\)
Rút gọn
\(1.A=\left(\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}}{\sqrt{x}-3}-\frac{3x+3}{x-9}\right):\left(\frac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right)\)
\(2.B=\left(\frac{\sqrt{a}+1}{\sqrt{ab}+1}+\frac{\sqrt{ab}+\sqrt{a}}{\sqrt{ab}-1}-1\right):\left(\frac{\sqrt{a}+1}{\sqrt{ab}+1}-\frac{\sqrt{ab}+\sqrt{a}}{\sqrt{ab}-1}+1\right)\)
\(3.C=\left(\frac{2x-1+\sqrt{x}}{1-x}+\frac{2x\sqrt{x}+x-\sqrt{x}}{1+x\sqrt{x}}\right).\left(\frac{\left(x-\sqrt{x}\right)\left(1-\sqrt{x}\right)}{2\sqrt{x}-1}\right)\)
Rút gọn biểu thức A=\(\left(\frac{x+2}{x\sqrt{x}-1}+\frac{\sqrt{x}+2}{x+\sqrt{x}+1}+\frac{1}{1-\sqrt{x}}\right):\frac{\sqrt{x}+1}{x+\sqrt{x}+1}\)
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Ta có : A = \(\left(\frac{x+2}{x.\sqrt{x}-1}+\frac{\sqrt{x}+2}{x+\sqrt{x}+1}+\frac{1}{1-\sqrt{x}}\right):\frac{\sqrt{x}+1}{x+\sqrt{x}+1}\)
= \(\frac{x+2+x+\sqrt{x}-2-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right).\left(x+\sqrt{x}+1\right)}.\frac{x+\sqrt{x}+1}{\sqrt{x}+1}\)
= \(\frac{x-1}{\left(\sqrt{x}-1\right).\left(x+\sqrt{x}+1\right)}.\frac{x+\sqrt{x}+1}{\sqrt{x}+1}=1\)
Vậy A = 1
các bạn giải chi tiết giúp mk nhé. Cảm ơn
1. a> Rút gọn biểu thức sau : A= \(5\left(\frac{1}{\sqrt{2-\sqrt{3}}}+\sqrt{3-\sqrt{5}}-\frac{\sqrt{10}}{2}\right)^2\)+ \(\left(\frac{1}{\sqrt{2+\sqrt{3}}}+\sqrt{3-\sqrt{5}}-\frac{\sqrt{6}}{2}\right)^2\)
b) Cho biểu thức B= \(\left(\frac{\sqrt{x}+1}{\sqrt{x}-1}-\frac{\sqrt{x}-1}{\sqrt{x+1}}-\frac{8\sqrt{x}}{x-1}\right):\left(\frac{\sqrt{x}-x-3}{x-1}-\frac{1}{\sqrt{x}-1}\right)\)
Rút gọn biểu thức B và chứng minh B nhỏ hơn hoặc bằng 1 với mọi x lớn hơn hoặc bằng 0 và x khác 1
\(B=\frac{3x+\sqrt{9x}-3}{x+\sqrt{x}-2}-\frac{\sqrt{x}+1}{\sqrt{x}-2}+\frac{\sqrt{x}-2}{\sqrt{x}}\left(\frac{1}{1-\sqrt{x}}-1\right)\) (x>0, x khác 1)
a)rút gọn B
b)tìm giá trị của x để B âm
Help me!!!
Lời giải:
Sửa lại đề, chỗ $\sqrt{x}-2}$ chuyển thành $\sqrt{x}+2$ mới đúng.
a)
\(B=\frac{3(x+\sqrt{x}-1)}{(\sqrt{x}-1)(\sqrt{x}+2)}-\frac{\sqrt{x}+1}{\sqrt{x}+2}+\frac{\sqrt{x}-2}{\sqrt{x}}.\frac{\sqrt{x}}{1-\sqrt{x}}=\frac{3(x+\sqrt{x}-1)}{(\sqrt{x}-1)(\sqrt{x}+2)}-\frac{\sqrt{x}+1}{\sqrt{x}+2}-\frac{\sqrt{x}-2}{\sqrt{x}-1}\)
\(=\frac{3(x+\sqrt{x}-1)}{(\sqrt{x}-1)(\sqrt{x}+2)}-\frac{(\sqrt{x}+1)(\sqrt{x}-1)}{(\sqrt{x}+2)(\sqrt{x}-1)}-\frac{(\sqrt{x}-2)(\sqrt{x}+2)}{(\sqrt{x}-1)(\sqrt{x}+2)}\)
\(=\frac{x+3\sqrt{x}+2}{(\sqrt{x}-1)(\sqrt{x}+2)}=\frac{(\sqrt{x}+1)(\sqrt{x}+2)}{(\sqrt{x}-1)(\sqrt{x}+2)}=\frac{\sqrt{x}-1}{\sqrt{x}+1}\)
b)
Để $B$ âm thì $\frac{\sqrt{x}-1}{\sqrt{x}+1}< 0$
Mà $\sqrt{x}+1>0$ nên $\sqrt{x}-1< 0$
$\Leftrightarrow 0\leq x< 1$
Kết hợp với đkxđ suy ra $0< x< 1$ thì $B$ âm.