Cho biểu thức P=\(\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}+\dfrac{\sqrt{x}}{x-\sqrt{x}}\right):\dfrac{\sqrt{x+1}}{3}\) với x >0; x khác 1
a. Rút gọn biểu thức P
b. Tìm giá trị của x để =1
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Y
25 tháng 7 2023
\(a,P=\dfrac{3\left(x+2\sqrt{x}\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}-\dfrac{\sqrt{x}+2}{\sqrt{x}-1}-\dfrac{\sqrt{x}+1}{\sqrt{x}+2}\left(dk:x\ge0,x\ne1\right)\)
\(=\dfrac{3\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}-\dfrac{\sqrt{x}+2}{\sqrt{x}-1}-\dfrac{\sqrt{x}+1}{\sqrt{x}+2}\\ =\dfrac{3\sqrt{x}}{\sqrt{x}-1}-\dfrac{\sqrt{x}+2}{\sqrt{x}-1}-\dfrac{\sqrt{x}+1}{\sqrt{x}+2}\\ =\dfrac{3\sqrt{x}-\sqrt{x}-2}{\sqrt{x}-1}-\dfrac{\sqrt{x}+1}{\sqrt{x}+2}\\ =\dfrac{2\sqrt{x}-2}{\sqrt{x}-1}-\dfrac{\sqrt{x}+1}{\sqrt{x}+2}\\ =\dfrac{2\left(\sqrt{x}-1\right)}{\sqrt{x}-1}-\dfrac{\sqrt{x}+1}{\sqrt{x}+2}\\ =\dfrac{2\left(\sqrt{x}+2\right)-\left(\sqrt{x}+1\right)}{\sqrt{x}+2}\\ =\dfrac{2\sqrt{x}+4-\sqrt{x}-1}{\sqrt{x}+2}\\ =\dfrac{\sqrt{x}+3}{\sqrt{x}+2}\)
\(b,x=6-2\sqrt{5}=\left(\sqrt{5}-1\right)^2\)
\(\Rightarrow P=\dfrac{\sqrt{\left(\sqrt{5}-1\right)^2}+3}{\sqrt{\left(\sqrt{5}-1\right)^2}+2}=\dfrac{\left|\sqrt{5}-1\right|+3}{\left|\sqrt{5}-1\right|+2}=\dfrac{\sqrt{5}-1+3}{\sqrt{5}-1+2}=\dfrac{\sqrt{5}+2}{\sqrt{5}+1}\)
KV
3 tháng 6 2023
a) \(P=\left(3-\dfrac{3}{\sqrt{x}-1}\right):\left(\dfrac{x+2}{x+\sqrt{x}-2}-\dfrac{\sqrt{x}}{\sqrt{x}+2}\right)\)
\(=\left(\dfrac{3\left(\sqrt{x}-1\right)-3}{\sqrt{x}-1}\right):\left[\dfrac{x+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x+2}\right)}-\dfrac{\sqrt{x}}{\sqrt{x}+2}\right]\)
\(=\dfrac{3\sqrt{x}-3-3}{\sqrt{x}-1}:\dfrac{x+2-\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{3\sqrt{x}-6}{\sqrt{x}-1}:\dfrac{x+2-x+\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{3\sqrt{x}-6}{\sqrt{x}-1}:\dfrac{\sqrt{x}+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{3\sqrt{x}-6}{\sqrt{x}-1}:\dfrac{1}{\sqrt{x}-1}\)
\(=\dfrac{3\sqrt{x}-6}{\sqrt{x}-1}.\left(\sqrt{x}-1\right)\)
\(=3\sqrt{x}-6\)
b) \(P=\dfrac{4\sqrt{x}-1}{\sqrt{x}}\)
\(\Leftrightarrow3\sqrt{x}-6=\dfrac{4\sqrt{x}-1}{\sqrt{x}}\) (1)
ĐKXĐ: \(x>0\)
\(\left(1\right)\Leftrightarrow3x-6\sqrt{x}=4\sqrt{x}-1\)
\(\Leftrightarrow3x-6\sqrt{x}-4\sqrt{x}+1=0\)
\(\Leftrightarrow3x-10\sqrt{x}+1=0\) (2)
Đặt \(t=\sqrt{x}\ge0\)
\(\left(2\right)\Leftrightarrow3t^2-10t+1=0\)
\(\Delta'=25-4=22\)
Phương trình có hai nghiệm phân biệt:
\(t_1=\dfrac{5+\sqrt{22}}{3}\) (nhận)
\(t_2=\dfrac{5-\sqrt{22}}{3}\) (nhận)
Với \(t=\dfrac{5+\sqrt{22}}{3}\) \(\Leftrightarrow\sqrt{x}=\dfrac{5+\sqrt{22}}{3}\Leftrightarrow x=\dfrac{47+10\sqrt{22}}{9}\) (nhận)
Với \(t=\dfrac{5-\sqrt{22}}{3}\Leftrightarrow\sqrt{x}=\dfrac{5-\sqrt{22}}{3}\Leftrightarrow x=\dfrac{47-10\sqrt{22}}{9}\) (nhận)
Vậy \(x=\dfrac{47+10\sqrt{22}}{9};x=\dfrac{47-10\sqrt{22}}{9}\) thì \(P=\dfrac{4\sqrt{x}-1}{\sqrt{x}}\)
2 tháng 6 2023
a: \(P=\dfrac{3\sqrt{x}-3-3}{\sqrt{x}-1}:\dfrac{x+2-x+\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{3\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\sqrt{x}+2}=3\sqrt{x}-6\)
b: P=(4căn x-1)/căn x
=>3x-6căn x-4căn x+1=0
=>3x-10căn x+1=0
=>x=(47+10căn 22)/9 hoặc x=(47-10căn 22)/9
XO
31 tháng 8 2023
a) \(M=\left(\dfrac{3}{\sqrt{x}+3}+\dfrac{x+9}{x-9}\right):\left(\dfrac{2\sqrt{x}-5}{x-3\sqrt{x}}-\dfrac{1}{\sqrt{x}}\right)\)
\(=\dfrac{3.\left(\sqrt{x}-3\right)+x+9}{\left(\sqrt{x}-3\right).\left(\sqrt{x}+3\right)}:\dfrac{2\sqrt{x}-5-\left(\sqrt{x}-3\right)}{\sqrt{x}.\left(\sqrt{x}-3\right)}\)
\(=\dfrac{x+3\sqrt{x}}{\left(\sqrt{x}-3\right).\left(\sqrt{x}+3\right)}:\dfrac{\sqrt{x}-2}{\sqrt{x}.\left(\sqrt{x}-3\right)}\)
\(=\dfrac{\sqrt{x}.\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right).\left(\sqrt{x}+3\right)}.\dfrac{\sqrt{x}.\left(\sqrt{x}-3\right)}{\sqrt{x}-2}=\dfrac{x}{\sqrt{x}-2}\)
b) \(M< 0\Leftrightarrow\sqrt{x}-2< 0\Leftrightarrow x< 4\)
Kết hợp điều kiện ta được \(0< x< 4\) thì M < 0
c) Từ câu b ta có M < 0 \(\Leftrightarrow0< x< 4\)
nên \(x\inℤ\) để M nguyên âm <=> \(x\in\left\{1;2;3\right\}\)
Thay lần lượt các giá trị vào M được x = 1 thỏa
d) \(M=\dfrac{x}{\sqrt{x}-2}=\sqrt{x}+2+\dfrac{4}{\sqrt{x}-2}=\left(\sqrt{x}-2+\dfrac{4}{\sqrt{x}-2}\right)+4\)
Vì x > 4 nên \(\sqrt{x}-2>0\)
Áp dụng BĐT Cauchy ta có
\(M=\left(\sqrt{x}-2+\dfrac{4}{\sqrt{x}-2}\right)+4\ge2\sqrt{\left(\sqrt{x}-2\right).\dfrac{4}{\sqrt{x}-2}}+4=8\)
Dấu "=" xảy ra khi \(\sqrt{x}-2=\dfrac{4}{\sqrt{x}-2}\Leftrightarrow x=16\left(tm\right)\)
ND
Nguyễn Đức Trí
VIP
31 tháng 8 2023
1) \(M=\left(\dfrac{3}{\sqrt[]{x}+3}+\dfrac{x+9}{x-9}\right):\left(\dfrac{2\sqrt[]{x}-5}{x-3\sqrt[]{x}}-\dfrac{1}{\sqrt[]{x}}\right)\left(x>0;x\ne9\right)\)
\(\Leftrightarrow M=\left(\dfrac{3\left(\sqrt[]{x}-3\right)}{\left(\sqrt[]{x}+3\right)\left(\sqrt[]{x}-3\right)}+\dfrac{x+9}{x-9}\right):\left(\dfrac{2\sqrt[]{x}-5}{\sqrt[]{x}\left(\sqrt[]{x}-3\right)}-\dfrac{1}{\sqrt[]{x}}\right)\)
\(\Leftrightarrow M=\left(\dfrac{3\sqrt[]{x}-9+x+9}{x-9}\right):\left(\dfrac{2\sqrt[]{x}-5-\left(\sqrt[]{x}-3\right)}{\sqrt[]{x}\left(\sqrt[]{x}-3\right)}\right)\)
\(\Leftrightarrow M=\left(\dfrac{3\sqrt[]{x}+x}{x-9}\right):\left(\dfrac{2\sqrt[]{x}-5-\sqrt[]{x}+3}{\sqrt[]{x}\left(\sqrt[]{x}-3\right)}\right)\)
\(\Leftrightarrow M=\left(\dfrac{\sqrt[]{x}\left(\sqrt[]{x}+3\right)}{x-9}\right):\left(\dfrac{\sqrt[]{x}-2}{\sqrt[]{x}\left(\sqrt[]{x}-3\right)}\right)\)
\(\Leftrightarrow M=\left(\dfrac{\sqrt[]{x}}{\sqrt[]{x}-3}\right):\left(\dfrac{\sqrt[]{x}-2}{\sqrt[]{x}\left(\sqrt[]{x}-3\right)}\right)\)
\(\Leftrightarrow M=\dfrac{\sqrt[]{x}}{\sqrt[]{x}-3}.\dfrac{\sqrt[]{x}\left(\sqrt[]{x}-3\right)}{\sqrt[]{x}-2}\)
\(\Leftrightarrow M=\dfrac{x}{\sqrt[]{x}-2}\)
2) Để \(M< 0\) khi và chỉ chi
\(M=\dfrac{x}{\sqrt[]{x}-2}< 0\left(1\right)\)
Nghiệm của tử là \(x=0\)
Nghiệm của mẫu \(\sqrt[]{x}-2=0\Leftrightarrow\sqrt[]{x}=2\Leftrightarrow x=4\)
Lập bảng xét dấu... ta được
\(\left(1\right)\Leftrightarrow0< x< 4\)
26 tháng 10 2021
a) \(M=\dfrac{x+\sqrt{x}+\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(x+1\right)}:\dfrac{x+1-2\sqrt{x}}{\left(x+1\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{\left(\sqrt{x}+1\right)^2}{\left(\sqrt{x}+1\right)\left(x+1\right)}.\dfrac{\left(x+1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)^2}=\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)
b) \(x=\sqrt{7+4\sqrt{3}}+\sqrt{7-4\sqrt{3}}=\sqrt{\left(2+\sqrt{3}\right)^2}+\sqrt{\left(2-\sqrt{3}\right)^2}\)
\(=2+\sqrt{3}+2-\sqrt{3}=4\)
\(M=\dfrac{\sqrt{x}+1}{\sqrt{x}-1}=\dfrac{\sqrt{4}+1}{\sqrt{4}-1}=\dfrac{2+1}{2-1}=3\)
7 tháng 12 2021
\(a,P=\dfrac{\sqrt{x}+2+\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\cdot\dfrac{2-\sqrt{x}}{\sqrt{x}}=\dfrac{-2\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+2\right)}=\dfrac{-2}{\sqrt{x}+2}\\ P=-\dfrac{3}{5}\Leftrightarrow\dfrac{2}{\sqrt{x}+2}=\dfrac{3}{5}\\ \Leftrightarrow3\sqrt{x}+6=10\Leftrightarrow\sqrt{x}=\dfrac{4}{3}\Leftrightarrow x=\dfrac{16}{9}\left(tm\right)\)
TP
8 tháng 12 2021
\(P=-\dfrac{3}{5}\) sao suy ra đc \(\dfrac{2}{\sqrt{x}+2}=\dfrac{3}{5}\) thế
16 tháng 11 2021
a: \(A=\dfrac{2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\cdot\dfrac{\sqrt{x}-1}{\sqrt{x}}=\dfrac{2\sqrt{x}+1}{x+\sqrt{x}}\)
5 tháng 2 2022
\(=\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)=3-1=2\)
b: \(=\dfrac{\sqrt{x}-\sqrt{x}-2}{\sqrt{x}\left(\sqrt{x}+2\right)}\cdot\dfrac{2}{\sqrt{x}+1}=\dfrac{-4}{\sqrt{x}\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)}\)
5 tháng 2 2022
a, \(=\left(\dfrac{\sqrt{3}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}+1\right)\left(\sqrt{3}-1\right)=\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)=2\)
b, với x > 0
\(=\left(\dfrac{\sqrt{x}-\sqrt{x}-2}{\sqrt{x}\left(\sqrt{x}+2\right)}\right)\left(\dfrac{2}{\sqrt{x+1}}\right)\)
\(=-\dfrac{-4}{\sqrt{x}\left(\sqrt{x}+2\right)\sqrt{x+1}}=\dfrac{4}{\left(\sqrt{x}+2\right)\sqrt{x^2+x}}\)
27 tháng 7 2021
a) Ta có: \(B=\left(\dfrac{3}{\sqrt{x}+2}-\dfrac{1}{\sqrt{x}-2}\right):\left(\dfrac{\sqrt{x}-6}{x-2\sqrt{x}}+\dfrac{1}{\sqrt{x}}\right)\)
\(=\dfrac{3\sqrt{x}-6-\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}:\dfrac{\sqrt{x}-6+\sqrt{x}-2}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(=\dfrac{2\sqrt{x}-8}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)}{2\sqrt{x}-8}\)
\(=\dfrac{\sqrt{x}}{\sqrt{x}+2}\)
b) Để \(B=\dfrac{1}{3}\) thì \(\dfrac{\sqrt{x}}{\sqrt{x}+2}=\dfrac{1}{3}\)
\(\Leftrightarrow3\sqrt{x}=\sqrt{x}+2\)
\(\Leftrightarrow2\sqrt{x}=2\)
\(\Leftrightarrow x=1\)(thỏa ĐK)
LD
27 tháng 7 2021
a) B= \(\left(\dfrac{3\left(\sqrt{x}-2\right)-1\left(\sqrt{x}+2\right)}{x-4}\right):\left(\dfrac{\sqrt{x}-6+1\left(\sqrt{x}-2\right)}{x-2\sqrt{x}}\right)\)
\(=\dfrac{2\sqrt{x}-8}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}.\dfrac{\left(\sqrt{x}-2\right)\sqrt{x}}{2\sqrt{x}-8}\)=\(\dfrac{\sqrt{x}}{\sqrt{x}+2}\)
b) Để B=\(\dfrac{1}{3}\Rightarrow\dfrac{\sqrt{x}}{\sqrt{x}+2}=\dfrac{1}{3}\Leftrightarrow\sqrt{x}+2=3\sqrt{x}\Rightarrow x=1\)
a: \(P=\dfrac{x+\sqrt{x}}{x-\sqrt{x}}\cdot\dfrac{3}{\sqrt{x}+1}=\dfrac{3}{\sqrt{x}-1}\)
b: Để P=1 thì \(\sqrt{x}-1=3\)
hay x=16
\(P=\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}+\dfrac{\sqrt{x}}{x-\sqrt{x}}\right):\dfrac{\sqrt{x}+1}{3}\)
\(P=\left(\dfrac{x+\sqrt{x}}{x\left(\sqrt{x}-1\right)}\right):\dfrac{\sqrt{x}+1}{3}\)
\(P=\left(\dfrac{x\left(\sqrt{x}+1\right)}{x\left(\sqrt{x}-1\right)}\right).\dfrac{3}{\sqrt{x}+1}\)
\(P=\dfrac{3}{\sqrt{x}-1}\)
\(P=1\)
\(\Leftrightarrow1=\dfrac{3}{\sqrt{x}-1}\)
\(\Leftrightarrow\sqrt{x}-1=3\)
\(\Leftrightarrow\sqrt{x}=4\)
\(\Leftrightarrow x=16\left(tm\right)\)