\(\dfrac{16+x^2}{x^2-16}\) rút gọn = bnhieu
\(\dfrac{x^2-16}{x}:\dfrac{x^2-8x+16}{x}\)(x≠0,x≠4)
giúp em rút gọn bài này với
\(=\dfrac{\left(x-4\right)\cdot\left(x+4\right)}{x}\cdot\dfrac{x}{\left(x-4\right)^2}=\dfrac{x+4}{x-4}\)
rút gọn
\(\dfrac{x}{x+2}+\dfrac{7x-16}{\left(x+2\right)\left(4x-7\right)}\)
\(\dfrac{x}{x-2}+\dfrac{2}{2-x}\)
\(\dfrac{x}{x+2}+\dfrac{7x-16}{\left(x+2\right)\left(4x-7\right)}\left(dkxd:x\ne-2;x\ne\dfrac{7}{4}\right)\)
\(=\dfrac{x\left(4x-7\right)+7x-16}{\left(x+2\right)\left(4x-7\right)}\)
\(=\dfrac{4x^2-7x+7x-16}{\left(x+2\right)\left(4x-7\right)}\)
\(=\dfrac{4x^2-16}{\left(x+2\right)\left(4x-7\right)}\)
\(=\dfrac{4\left(x^2-4\right)}{\left(x+2\right)\left(4x-7\right)}\)
\(=\dfrac{4\left(x-2\right)\left(x+2\right)}{\left(x+2\right)\left(4x-7\right)}\)
\(=\dfrac{4\left(x-2\right)}{4x-7}\)
\(=\dfrac{4x-8}{4x-7}\)
===============================
\(\dfrac{x}{x-2}+\dfrac{2}{2-x}\left(dkxd:x\ne2\right)\)
\(=\dfrac{x}{x-2}-\dfrac{2}{x-2}\)
\(=\dfrac{x-2}{x-2}\)
\(=1\)
Rút gọn biểu thức N=\(\left(\dfrac{\sqrt{x}}{\sqrt{x}+4}+\dfrac{4}{\sqrt{x}-4}\right):\dfrac{\sqrt{x}+16}{\sqrt{x}+2}\) với x≥0 ; x≠16
\(=\dfrac{x-4\sqrt{x}+4\sqrt{x}+16}{x-16}\cdot\dfrac{\sqrt{x}+2}{\sqrt{x}+16}=\dfrac{\left(x+16\right)\left(\sqrt{x}+2\right)}{\left(x-16\right)\left(\sqrt{x}+16\right)}\)
Cho biểu thức A = \(\left(\dfrac{4x}{x+2}-\dfrac{x^3-8}{x^3+8}.\dfrac{4x^2-8x+16}{x^2-4}\right):\dfrac{16}{x^2-x-6}\)
a) Rút gọn A
b) Tìm x để A < 0
c) Tìm x để A ≥ 5
Rút gọn:
\(A=\dfrac{\sqrt{x}+2}{\sqrt{x}+3}-\dfrac{5}{x+\sqrt{x}-6}+\dfrac{1}{2-\sqrt{x}}\) với x ≠4, x ≠16, x >0
Lời giải:
\(A=\frac{(\sqrt{x}+2)(\sqrt{x}-2)}{(\sqrt{x}+3)(\sqrt{x}-2)}-\frac{5}{(\sqrt{x}+3)(\sqrt{x}-2)}-\frac{\sqrt{x}+3}{(\sqrt{x}-2)(\sqrt{x}+3)}\)
\(=\frac{x-4-5-\sqrt{x}-3}{(\sqrt{x}-2)(\sqrt{x}+3)}=\frac{x-\sqrt{x}-12}{(\sqrt{x}-2)(\sqrt{x}+3)}=\frac{(\sqrt{x}+3)(\sqrt{x}-4)}{(\sqrt{x}-2)(\sqrt{x}+3)}=\frac{\sqrt{x}-4}{\sqrt{x}-2}\)
Rút gọn biểu thức sau :
\(B=\left(\dfrac{\sqrt{x}}{\sqrt{x}+4}+\dfrac{4}{\sqrt{x}-4}\right):\dfrac{x+16}{\sqrt{x}+2}\) ( x ≥ 0 ; x ≠ 16 )
\(P=\dfrac{x}{x-1}+\dfrac{3}{x+1}-\dfrac{6x-4}{x^2-1}\)
\(A=\dfrac{1}{x+\sqrt{x}}+\dfrac{2\sqrt{x}}{x-1}-\dfrac{1}{x-\sqrt{x}}\)
\(B=\left(2-\sqrt{3}\right)\sqrt{26+15\sqrt{3}-\left(2+\sqrt{3}\right)\sqrt{26-15\sqrt{3}}}\)
\(A=\left(\sqrt{10}-\sqrt{2}\right)\sqrt{3+\sqrt{5}}\)
b) ĐKXĐ : \(x\ne\pm1\)
\(P=\dfrac{x}{x-1}+\dfrac{3}{x+1}-\dfrac{6x-4}{x^2-1}\)
\(=\dfrac{x\left(x+1\right)+3\left(x-1\right)-\left(6x-4\right)}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{x^2-2x+1}{\left(x-1\right)\left(x+1\right)}=\dfrac{\left(x-1\right)^2}{\left(x-1\right)\left(x+1\right)}=\dfrac{x-1}{x+1}\)
c) ĐKXĐ : \(\left\{{}\begin{matrix}x>0\\x\ne1\end{matrix}\right.\)
\(A=\dfrac{1}{x+\sqrt{x}}+\dfrac{2\sqrt{x}}{x-1}-\dfrac{1}{x-\sqrt{x}}\)
\(=\dfrac{1}{\sqrt{x}\left(\sqrt{x}+1\right)}+\dfrac{2\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\dfrac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}\)
\(=\dfrac{\sqrt{x}-1+2x-\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{2\left(x-1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{2\left(x-1\right)}{\sqrt{x}\left(x-1\right)}=\dfrac{2}{\sqrt{x}}\)
a) ĐKXĐ : \(x\ge0;x\ne16\)
\(B=\left(\dfrac{\sqrt{x}}{\sqrt{x}+4}+\dfrac{4}{\sqrt{x-4}}\right):\dfrac{x+16}{\sqrt{x}+2}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}-4\right)+4\left(\sqrt{x}+4\right)}{x-16}:\dfrac{x+16}{\sqrt{x}+2}\)
\(=\dfrac{x+16}{x-16}:\dfrac{x+16}{\sqrt{x}+2}=\dfrac{\sqrt{x}+2}{x-16}\)
\(=\left(\dfrac{\sqrt{x}.\left(\sqrt{x}-4\right)}{x-4}+\dfrac{4.\left(\sqrt{x}+4\right)}{x-4}\right):\dfrac{x+16}{\sqrt{x}+2}\)
\(=\left(\dfrac{x-4\sqrt{x}+4\sqrt{x}+16}{x-4}\right).\dfrac{\sqrt{x}+2}{x+16}\)
\(=\dfrac{x+16}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}.\dfrac{\sqrt{x}+2}{x+16}\)
\(=\dfrac{1}{\sqrt{x}-2}\)
Cho hai biểu thức: A=\(\dfrac{4x-16}{x^2-16}\)và B=\(\dfrac{1}{x}+\dfrac{1}{x+4}+\dfrac{2x-4}{x\left(x+4\right)}\)
a) Rút gọn biểu thức A;
b) Chứng minh: B = A.
\(a,A=\dfrac{4\left(x-4\right)}{\left(x-4\right)\left(x+4\right)}=\dfrac{4}{x+4}\\ b,B=\dfrac{x+4+x+2x-4}{x\left(x+4\right)}=\dfrac{4x}{x\left(x+4\right)}=\dfrac{4}{x+4}=A\)
RÚT GỌN BIỂU THỨC:
16) \(A = \dfrac{\sqrt{x}}{\sqrt{x} + 3} + \dfrac{2\sqrt{x}}{\sqrt{x} - 3} - \dfrac{3x + 9}{x - 9}\)
\(=\dfrac{\sqrt{x}}{\sqrt{x}+3}+\dfrac{2\sqrt{x}}{\sqrt{x}-3}-\dfrac{3x+9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)+2\sqrt{x}\left(\sqrt{x}+3\right)-3x-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{x-3\sqrt{x}+2x+6\sqrt{x}-3x-9}{x-9}=\dfrac{3\sqrt{x}-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{3}{\sqrt{x}+3}\)
\(A=\dfrac{\sqrt{x}}{\sqrt{x}+3}+\dfrac{2\sqrt{x}}{\sqrt{x}-3}-\dfrac{3x+9}{x-9}\) (ĐK: \(x\ge0;x\ne9\))
\(A=\dfrac{\sqrt{x}}{\sqrt{x}+3}+\dfrac{2\sqrt{x}}{\sqrt{x}-3}-\dfrac{3x+9}{\left(\sqrt{x}\right)^2-3^2}\)
\(A=\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}+\dfrac{2\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}-\dfrac{3x+9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(A=\dfrac{x-3\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}+\dfrac{2x+6\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}-\dfrac{3x+9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(A=\dfrac{x-3\sqrt{x}+2x+6\sqrt{x}-3x-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(A=\dfrac{3\sqrt{x}-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(A=\dfrac{3\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(A=\dfrac{3}{\sqrt{x}+3}\)
cho P= \(\dfrac{x^2}{x-8}\)(\(\dfrac{x^2-64}{x}\) -16) +19
a) Rút gọn P
b) Tìm GTNN của P
help tui nha mn
\(A=\left(\dfrac{4x}{x+2}-\dfrac{x^3-8}{x^3+8}\times\dfrac{4x^2-8x+16}{x^2-4}\right)\div\dfrac{16}{x+2}\times\dfrac{x^2+3x+2}{x^2+x+1}\)
\(B=\dfrac{x^2+x-2}{x^3-1}\)
a) Tìm ĐKXĐ của A, B. Rút gọn A, B
b)Tìm GTLN của A+B