cho M= \(\frac{2\sqrt{x}}{\sqrt{x+3}}+\frac{\sqrt{x}+1}{\sqrt{x}-3}+\frac{11\sqrt{x}-3}{x-9}\) với x>0;x#9
a. Rút gọn M
b. Tìm x sao cho \(\frac{1}{M}\)< \(\frac{1}{6}\)
1/ Cho D=\(\frac{x^2-\sqrt{x}}{x+\sqrt{x}+1}-\frac{x^2+\sqrt{x}}{x-\sqrt{x}+1}\)với 0≤x≤1
a) Rút gọn
b) CMinh 1\(-\sqrt{D+x+1}=\sqrt{x}\)
2/Cho E=\(\frac{15\sqrt{x}-11}{x+2\sqrt{x}-3}+\frac{3\sqrt{x}-2}{1-\sqrt{x}}-\frac{2\sqrt{x}+3}{\sqrt{x}+3}\)với x≥0 và x≠1
a) Rút gọn
b) Tìm giá trị của x để E = \(\frac{1}{2}\)
c) So sánh E với \(\frac{2}{3}\)
3/Cho G=\(\left(\frac{x-3\sqrt{x}}{x-9}-1\right):\left(\frac{9-x}{x+\sqrt{x}-6}+\frac{\sqrt{x}-3}{\sqrt{x}-2}-\frac{\sqrt{x}+2}{\sqrt{x}+3}\right)\)với x≥0,x≠4,x≠9
a) Rút gọn
b) Tìm x để G<1
Cho M =\(\frac{2\sqrt{y}}{x-y}+\frac{\sqrt{x}+1}{\sqrt{x}-3}+\frac{11\sqrt{x}-3}{x-9}\)với\(x>0,x\ne9\)
1.Rút gọc biểu thức M
2.Tìm x sao cho \(\frac{1}{M}< \frac{1}{6}\)
Bạn xem lại đề. Biểu thức M có chứa y không?
Cho biểu thức: \(M=\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}+1}{\sqrt{x}-3}+\frac{11\sqrt{x}-3}{x-9}\) với x>0 , x≠9
a)Rút gọn M
b)Tính giá trị của M tại =81
c)Tìm x sao cho \(\frac{1}{M}\)<\(\frac{1}{6}\)
rút gọn
P=\(\frac{15\sqrt{x}-11}{x+2\sqrt{x}-3}+\frac{3\sqrt{x}-2}{1-\sqrt{x}}-\frac{2\sqrt{x}+3}{\sqrt{x}+3}\)
N= \(\left(\frac{x-3\sqrt{x}}{x-9}-1\right):\left(\frac{9-x}{x+\sqrt{x}-6}-\frac{\sqrt{x}-3}{2-\sqrt{x}}-\frac{\sqrt{x}-2}{\sqrt{x}+3}\right)\)
\(P=\frac{15\sqrt{x}-11}{x+2\sqrt{x}-3}+\frac{3\sqrt{x}-2}{1-\sqrt{x}}-\frac{2\sqrt{x}+3}{\sqrt{x}+3}\)
\(=\frac{15\sqrt{x}-11}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}+\frac{3\sqrt{x}-2}{1-\sqrt{x}}-\frac{2\sqrt{x}+3}{\sqrt{x}+3}\)
\(=\frac{15\sqrt{x}-11-\left(3\sqrt{x}-2\right)\left(\sqrt{x}+3\right)-\left(2\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
\(=\frac{-5x+7\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}=\frac{\left(1-\sqrt{x}\right)\left(5\sqrt{x}-2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}=\frac{2-5\sqrt{x}}{3+\sqrt{x}}\)
\(N=\left(\frac{x-3\sqrt{x}}{x-9}-1\right):\left(\frac{9-x}{x+\sqrt{x}-6}-\frac{\sqrt{x}-3}{2-\sqrt{x}}-\frac{\sqrt{x}-2}{\sqrt{x}+3}\right)\)
\(=\left(\frac{\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}-1\right):\left(\frac{9-x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}-\frac{\sqrt{x}-3}{2-\sqrt{x}}-\frac{\sqrt{x}-2}{\sqrt{x}+3}\right)\)
\(=\left(\frac{\sqrt{x}}{\left(\sqrt{x}+3\right)}-1\right):\left(\frac{9-x+\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)-\left(\sqrt{x}-2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\right)\)
\(=-3:\left(\frac{-\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+3\right)}\right)\)
\(=3.\left(\frac{\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-2\right)}\right)\)
Cái đó chỗ số 3 phải có mẫu nữa mà mình quên mất mẫu
\(\frac{3}{\sqrt{x}+3}:\frac{\sqrt{x}-2}{\sqrt{x}+3}=\frac{3}{\sqrt{x}-2}\)
Cho biểu thức A = \(\frac{\sqrt{4+2\sqrt{3}}}{\sqrt{3}+1}+\frac{5+3\sqrt{5}}{\sqrt{5}}-\left(\sqrt{5}+3\right)\)
B = \(\frac{1}{3-\sqrt{x}}+\frac{\sqrt{x}}{3+\sqrt{x}}-\frac{x+9}{x-9}\) với x ≠ 9, x ≥ 0
a, Rút gọn biểu thức A
b, Tìm các giá trị của x để B > A
a) \(A=\frac{\sqrt{4+2\sqrt{3}}}{\sqrt{3}+1}+\frac{5+3\sqrt{5}}{\sqrt{5}}-\left(\sqrt{5}+3\right)\)
\(A=\frac{\sqrt{\left(\sqrt{3}+1\right)^2}}{\sqrt{3}+1}+\frac{5+3\sqrt{5}}{\sqrt{5}}-\frac{\sqrt{5}\left(\sqrt{5}+3\right)}{\sqrt{5}}\)
\(A=\frac{\sqrt{3}+1}{\sqrt{3}+1}+\frac{5+3\sqrt{5}}{\sqrt{5}}-\frac{5+3\sqrt{5}}{\sqrt{5}}\)
\(A=1\)
b) Ta có:
\(B=\frac{1}{3-\sqrt{x}}+\frac{\sqrt{x}}{3+\sqrt{x}}-\frac{x+9}{x-9}\) ( x >= 0, x khác 9 )
\(B=\frac{3+\sqrt{x}}{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)}+\frac{\sqrt{x}\left(3-\sqrt{x}\right)}{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)}+\frac{x+9}{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)}\)
\(B=\frac{3+\sqrt{x}+3\sqrt{x}-x+x+9}{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)}\)
\(B=\frac{3+\sqrt{x}+3\sqrt{x}+9}{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)}\)
\(B=\frac{\left(3+\sqrt{x}\right)+3\left(\sqrt{x}+3\right)}{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)}\)
\(B=\frac{4\left(3+\sqrt{x}\right)}{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)}\)
\(B=\frac{4}{3-\sqrt{x}}\)
Để B > A
\(\Rightarrow\frac{4}{3-\sqrt{x}}>1\)
\(\Rightarrow4>3-\sqrt{x}\)
\(\Rightarrow4-3+\sqrt{x}>0\)
\(\Rightarrow1+\sqrt{x}>0\)
\(\Rightarrow\sqrt{x}>-1\)
\(\Rightarrow x>1\)
a) A=\(\frac{\sqrt{4+2\sqrt{3}}}{\sqrt{3}+1}+\frac{5+3\sqrt{5}}{\sqrt{5}}-\left(\sqrt{5}+3\right)\)
\(=\frac{\sqrt{3+2\sqrt{3}+1}}{\sqrt{3}+1}+\frac{\sqrt{5}\cdot\left(\sqrt{5}+3\right)}{\sqrt{5}}\)
\(=\frac{\sqrt{\left(\sqrt{3}+1\right)^2}}{\sqrt{3}+1}+\left(\sqrt{5}+3\right)-\left(\sqrt{5}+3\right)\)
\(=\frac{\sqrt{3}+1}{\sqrt{3}+1}+0=1\)
b) B=\(\frac{1}{3-\sqrt{x}}+\frac{\sqrt{x}}{3+\sqrt{x}}-\frac{x+9}{x-9}\)
\(=\frac{3+\sqrt{x}+\sqrt{x}\left(3-\sqrt{x}\right)}{\left(3-\sqrt{x}\right)\cdot\left(3+\sqrt{x}\right)}+\frac{x+9}{9-x}\)
\(=\frac{3+\sqrt{x}+3\sqrt{x}-x}{\left(3-\sqrt{x}\right)\cdot\left(3+\sqrt{x}\right)}+\frac{x+9}{\left(3-\sqrt{x}\right)\cdot\left(3+\sqrt{x}\right)}\)
\(=\frac{4\text{}\sqrt{x}+12}{\left(3-\sqrt{x}\right)\cdot\left(3+\sqrt{x}\right)}\)
\(=\frac{4\left(\sqrt{x}+3\right)}{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)}\)
\(=\frac{4}{3-\sqrt{x}}\)
\(B>A \Leftrightarrow\frac{4}{3-\sqrt{x}}>1\)
các giá trị của x là \(\left\{x\in R\backslash0\le x\le9\right\}\)
Cho P=\(\left(1-\frac{x-3\sqrt{x}}{x-9}\right):\left(\frac{9-x}{x+\sqrt{x}-6}-\frac{\sqrt{x}-3}{2-\sqrt{x}}-\frac{\sqrt{x}-2}{\sqrt{x}+3}\right)\)
a) Rút gọn P
b) Tìm x để P>0
c) Với x=4, x khác 9, tìm GTLN của Q = P(x+1)
a) Rút gọn biểu thức sau A=\(\sqrt{3+2\sqrt{2}}-\frac{1}{1+\sqrt{2}}\)
b)Chứng minh rằng:\(\left(\frac{\sqrt{x}}{\sqrt{x}+3}+\frac{3}{\sqrt{x}-3}\right).\frac{\sqrt{x}+3}{x+9}=\frac{1}{\sqrt{x}-3}\)với x≥0 và x ≠ 9
a) Ta có: \(A=\sqrt{3+2\sqrt{2}}-\frac{1}{1+\sqrt{2}}\)
\(=\sqrt{1+2\cdot1\cdot\sqrt{2}+2}-\frac{1}{1+\sqrt{2}}\)
\(=\sqrt{\left(1+\sqrt{2}\right)^2}-\frac{1}{1+\sqrt{2}}\)
\(=1+\sqrt{2}-\frac{1}{1+\sqrt{2}}\)
\(=\frac{\left(1+\sqrt{2}\right)^2}{1+\sqrt{2}}-\frac{1}{1+\sqrt{2}}\)
\(=\frac{1+2\sqrt{2}+2-1}{1+\sqrt{2}}\)
\(=\frac{2\sqrt{2}+2}{1+\sqrt{2}}\)
\(=\frac{2\left(\sqrt{2}+1\right)}{\sqrt{2}+1}=2\)
b) Ta có: \(\left(\frac{\sqrt{x}}{\sqrt{x}+3}+\frac{3}{\sqrt{x}-3}\right)\cdot\frac{\sqrt{x}+3}{x+9}\)
\(=\left(\frac{\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}+\frac{3\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\right)\cdot\frac{1}{\sqrt{x}-3}\)
\(=\frac{x-3\sqrt{x}+3\sqrt{x}+9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\cdot\frac{1}{\sqrt{x}-3}\)
\(=\frac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\cdot\frac{1}{\sqrt{x}-3}\)
\(=\frac{1}{\sqrt{x}-3}\)(đpcm)
cho bt với x≥0 , x≠9
P=\(\left(\frac{2\sqrt{2}}{\sqrt{x}+3}+\frac{\sqrt{x}}{\sqrt{x}-3}-\frac{3x+3}{x-9}\right):\left(\frac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right)\)
a) rút gọn
b)tìm x để P<\(-\frac{1}{3}\)
c)tìm GTNN của P
a) \(P=\left(\frac{2\sqrt{2}}{\sqrt{x}+3}+\frac{\sqrt{x}}{\sqrt{x}-3}-\frac{3x+3}{x-9}\right):\left(\frac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right)\)
\(=\left(\frac{2\sqrt{x}\left(\sqrt{x}-3\right)+\sqrt{x}\left(\sqrt{x}+3\right)-\left(3x+3\right)}{x-9}\right):\left(\frac{2\sqrt{x}-2}{\sqrt{x}-3}-\frac{\sqrt{x}-3}{\sqrt{x}-3}\right)\)\(=\frac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{\left(\sqrt{x}+3\right).\left(\sqrt{x}+1\right)}\)
\(=\frac{-3\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{-3}{\sqrt{x}+3}\)
cho bt với x≥0 , x≠9
P=\(\left(\frac{2\sqrt{2}}{\sqrt{x}+3}\frac{\sqrt{x}}{\sqrt{x}-3}-\frac{3x+3}{x-9}\right):\left(\frac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right)\)
a) rút gọn
b) tìm x để P<\(-\frac{1}{3}\)
c) tìm GTNN của P