Cho biểu thức A=\(\dfrac{\sqrt{x}+1}{x+4\sqrt{x}+4}:\left(\dfrac{x}{x+2\sqrt{x}}+\dfrac{x}{\sqrt{x}+2}\right)\)( x ≥ 0)
a) Rút gọn
b) Tìm x để A ≥ \(\dfrac{1}{3\sqrt{x}}\)
(mink đag cần gấp)
Cho biểu thức A=\(\dfrac{\sqrt{x}+1}{x+4\sqrt{x}+4}:\left(\dfrac{x}{x+2\sqrt{x}}+\dfrac{x}{\sqrt{x}+2}\right)\)( x ≥ 0)
a) Rút gọn
b) Tìm x để A ≥ \(\dfrac{1}{3\sqrt{x}}\)
(mink đag cần gấp)
Cho biểu thức A = \(\dfrac{x+4}{\sqrt{x}+4}\) ; B = \(\left(\dfrac{x+3\sqrt{x}-3}{x-16}-\dfrac{1}{\sqrt{x}+4}\right):\dfrac{\sqrt{x}+1}{\sqrt{x}-4}\)
x ≥ 0, x ≠ 16
a) Rút gọn
b) Tìm giá trị nhỏ nhất của biểu thức \(\dfrac{A}{B}\)
(mink đag cần gấp)
a) Ta có: \(B=\left(\dfrac{x+3\sqrt{x}-3}{x-16}-\dfrac{1}{\sqrt{x}+4}\right):\dfrac{\sqrt{x}+1}{\sqrt{x}-4}\)
\(=\left(\dfrac{x+3\sqrt{x}-3-\sqrt{x}+4}{\left(\sqrt{x}+4\right)\left(\sqrt{x}-4\right)}\right):\dfrac{\sqrt{x}+1}{\sqrt{x}-4}\)
\(=\dfrac{x+2\sqrt{x}+1}{\left(\sqrt{x}+4\right)\left(\sqrt{x}-4\right)}\cdot\dfrac{\sqrt{x}-4}{\sqrt{x}+1}\)
\(=\dfrac{\sqrt{x}+1}{\sqrt{x}+4}\)
Cho biểu thức A = \(\left(\dfrac{x}{x+3\sqrt{x}}+\dfrac{1}{\sqrt{x}+3}\right):\left(1-\dfrac{2}{\sqrt{x}}+\dfrac{6}{x+3\sqrt{x}}\right)\) x ≥ 0. Rút B
(mink đag cần gấp)
Ta có: \(A=\left(\dfrac{x}{x+3\sqrt{x}}+\dfrac{1}{\sqrt{x}+3}\right):\left(1-\dfrac{2}{\sqrt{x}}+\dfrac{6}{x+3\sqrt{x}}\right)\)
\(=\left(\dfrac{x}{\sqrt{x}\left(\sqrt{x}+3\right)}+\dfrac{1}{\sqrt{x}+3}\right):\left(\dfrac{x+3\sqrt{x}-2\left(\sqrt{x}+3\right)+6}{\sqrt{x}\left(\sqrt{x}+3\right)}\right)\)
\(=\dfrac{\sqrt{x}+1}{\sqrt{x}+3}:\dfrac{x+3\sqrt{x}-2\sqrt{x}-6+6}{\sqrt{x}\left(\sqrt{x}+3\right)}\)
\(=\dfrac{\sqrt{x}+1}{\sqrt{x}+3}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}+3\right)}{x+\sqrt{x}}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}=1\)
\(\left(\dfrac{2\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}+\dfrac{\sqrt{x}}{\sqrt{x}-3}\right):\dfrac{\sqrt{x}}{\sqrt{x}-3}\)
Rút gọn biểu thức
(mink đag cần gấp)
\(\left(\dfrac{2\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}+\dfrac{\sqrt{x}}{\sqrt{x}-3}\right):\dfrac{\sqrt{x}}{\sqrt{x}-3}\) Đk: \(\left\{{}\begin{matrix}x>0\\x\ne9\end{matrix}\right.\)
= \(\dfrac{2\sqrt{x}+x+2\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}.\dfrac{\sqrt{x}-3}{\sqrt{x}}\)
= \(\dfrac{\sqrt{x}\left(\sqrt{x}+4\right)}{\sqrt{x}\left(\sqrt{x}+2\right)}\)
= \(\dfrac{\sqrt{x}+4}{\sqrt{x}+2}\)
Bài 4:
Cho biểu thức \(M=\dfrac{2\sqrt{x}-9}{x-5\sqrt{x}+6}-\dfrac{\sqrt{x}+3}{\sqrt{x}-2}-\dfrac{2\sqrt{x}+1}{3-\sqrt{x}}\)
a) Tìm đkxđ của M và rút gọn
b) Tìm x thuộc Z để M thuộc Z
Em cần gấp :<
a) ĐKXĐ: \(x\ge0;x\ne9;x\ne4\)
\(M=\dfrac{2\sqrt{x}-9}{x-5\sqrt{x}+6}-\dfrac{\sqrt{x}+3}{\sqrt{x}-2}-\dfrac{2\sqrt{x}+1}{3-\sqrt{x}}\)
\(M=\dfrac{2\sqrt{x}-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\dfrac{\sqrt{x}+3}{\sqrt{x}-2}+\dfrac{2\sqrt{x}+1}{\sqrt{x}-3}\)
\(M=\dfrac{2\sqrt{x}-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}+\dfrac{\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(M=\dfrac{2\sqrt{x}-9-x+9+2x-4\sqrt{x}+\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(M=\dfrac{x-\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(M=\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(M=\dfrac{\sqrt{x}+1}{\sqrt{x}-3}\)
b) Ta có M ϵ Z thì \(\dfrac{\sqrt{x}+1}{\sqrt{x}-3}=\dfrac{\sqrt{x}-3+4}{\sqrt{x}-3}=\dfrac{\sqrt{x}-3}{\sqrt{x}-3}+\dfrac{4}{\sqrt{x}-3}=1+\dfrac{4}{\sqrt{x}-3}\)
Phải thuộc Z vậy:
4 ⋮ \(\sqrt{x}-3\)
\(\Rightarrow\sqrt{x}-3\inƯ\left(4\right)=\left\{1;-1;2;-2;4;-4\right\}\)
Mà: \(x\ge0,x\ne4,x\ne9\) nên \(\sqrt{x}-3\in\left\{1;2;-2;4\right\}\)
\(\Rightarrow x\in\left\{16;25;1;49\right\}\)
giúp mik với mik đag cần gấp !
Câu 1:Cho biểu thức P=\(\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{2}{x-\sqrt{x}}\right)\):\(\left(\dfrac{1}{\sqrt{x+1}}-\dfrac{2}{1-x}\right)\)
a) Rút gọn biểu thức P
b) Tính giá trị của biểu thức P khi x=4+2\(\sqrt{3}\)
\(a,P=\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{2}{x-\sqrt{x}}\right):\left(\dfrac{1}{\sqrt{x}+1}-\dfrac{2}{1-x}\right)\left(dkxd:x\ge0,x\ne1\right)\)
\(=\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{2}{\sqrt{x}\left(\sqrt{x}-1\right)}\right):\left(\dfrac{1}{\sqrt{x}+1}+\dfrac{2}{x-1}\right)\)
\(=\dfrac{\sqrt{x}.\sqrt{x}-2}{\sqrt{x}\left(\sqrt{x}-1\right)}:\dfrac{\sqrt{x}-1+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{x-2}{\sqrt{x}\left(\sqrt{x}-1\right)}.\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}+1}\)
\(=\dfrac{x-2}{\sqrt{x}}\)
\(b,x=4+2\sqrt{3}\Rightarrow P=\dfrac{\left(4+2\sqrt{3}\right)-2}{\sqrt{4+2\sqrt{3}}}\)
\(=\dfrac{2\sqrt{3}+4-2}{\sqrt{\sqrt{3}^2+2\sqrt{3}+1}}\)
\(=\dfrac{2\sqrt{3}+2}{\sqrt{\left(\sqrt{3}+1\right)^2}}\)
\(=\dfrac{2\left(\sqrt{3}+1\right)}{\left|\sqrt{3}+1\right|}\)
\(=\dfrac{2\left(\sqrt{3}+1\right)}{\sqrt{3}+1}=2\)
a: \(P=\dfrac{x-2}{\sqrt{x}\left(\sqrt{x}-1\right)}:\dfrac{\sqrt{x}-1+2}{x-1}\)
\(=\dfrac{x-2}{\sqrt{x}\left(\sqrt{x}-1\right)}\cdot\dfrac{x-1}{\sqrt{x}+1}=\dfrac{x-2}{\sqrt{x}}\)
b: Khi x=4+2căn 3 thì \(P=\dfrac{2+2\sqrt{3}}{\sqrt{3}+1}=2\)
cho biểu thức
P=\(\left(\dfrac{1}{\sqrt{x}-x}+\dfrac{1}{1-\sqrt{x}}\right)\):\(\dfrac{\sqrt{x}+1}{\left(1-\sqrt{x}\right)^2}\)
a) tìm đk và rút gọn
b) Tìm x để P>0
cho biểu thức:
\(A=\left(\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}}{x+\sqrt{x}+1}+\dfrac{1}{1-\sqrt{x}}\right):\dfrac{\sqrt{x}-1}{2}\)
a, rút gọn
b, chứng minh: A > 0 với mọi x ≥ 0, x ≠ 1
\(\left(\dfrac{\sqrt{x}}{\sqrt{x}-2}-\dfrac{2}{\sqrt{x}+2}\right):\dfrac{x+4}{x+2\sqrt{x}}\)
a/rút gọn
b/tìm x sao cho biểu thức trên bằng -x
\(a,\left(\dfrac{\sqrt{x}}{\sqrt{x}-2}-\dfrac{2}{\sqrt{x}+2}\right):\dfrac{x+4}{x+2\sqrt{x}}\left(dkxd:x>0;x\ne4\right)\)
\(=\left[\dfrac{\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-\dfrac{2\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right]\cdot\dfrac{x+2\sqrt{x}}{x+4}\)
\(=\dfrac{x+2\sqrt{x}-2\sqrt{x}+4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}+2\right)}{x+4}\)
\(=\dfrac{x+4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}+2\right)}{x+4}\)
\(=\dfrac{\sqrt{x}}{\sqrt{x}-2}\)
\(---\)
\(b,\) Để biểu thức trên bằng $-x$
thì \(\dfrac{\sqrt{x}}{\sqrt{x}-2}=-x\)
\(\Leftrightarrow\sqrt{x}=-x\sqrt{x}+2x\)
\(\Leftrightarrow x\sqrt{x}-2x+\sqrt{x}=0\)
\(\Leftrightarrow\sqrt{x}\left(x-2\sqrt{x}+1\right)=0\)
\(\Leftrightarrow\sqrt{x}\left(\sqrt{x}-1\right)^2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=0\\\sqrt{x}-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
Kết hợp với ĐKXĐ của $x$, ta được:
\(x=1\)
Vậy biểu thức bằng $-x$ khi $x=1$
\(\text{#}Toru\)