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a) Thay x=4 vào biểu thức \(B=\dfrac{3}{\sqrt{x}-1}\), ta được:
\(B=\dfrac{3}{\sqrt{4}-1}=\dfrac{3}{2-1}=3\)
Vậy: Khi x=4 thì B=3
b) Ta có: P=A-B
\(\Leftrightarrow P=\dfrac{6}{x-1}+\dfrac{\sqrt{x}}{\sqrt{x}+1}-\dfrac{3}{\sqrt{x}-1}\)
\(\Leftrightarrow P=\dfrac{6}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}-\dfrac{3\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(\Leftrightarrow P=\dfrac{6+x-\sqrt{x}-3\sqrt{x}-3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(\Leftrightarrow P=\dfrac{x-\sqrt{x}-3\sqrt{x}+3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(\Leftrightarrow P=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)-3\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(\Leftrightarrow P=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(\Leftrightarrow P=\dfrac{\sqrt{x}-3}{\sqrt{x}+1}\)
a: Thay x=2 vào B, ta được:
\(B=\dfrac{2}{\sqrt{2}-1}=2\sqrt{2}+2\)
ĐKXĐ : \(x\ne0;x\ne\pm1\)
a) Bạn ghi lại rõ đề.
b) \(B=\dfrac{x-1}{x+1}+\dfrac{3x-x^2}{x^2-1}=\dfrac{x-1}{x+1}+\dfrac{3x-x^2}{\left(x-1\right).\left(x+1\right)}\)
\(=\dfrac{\left(x-1\right)^2+3x-x^2}{\left(x-1\right).\left(x+1\right)}=\dfrac{x+1}{\left(x-1\right).\left(x+1\right)}=\dfrac{1}{x-1}\)
c) \(P=A.B=\dfrac{x^2+x-2}{x.\left(x-1\right)}=\dfrac{\left(x-1\right).\left(x+2\right)}{x\left(x-1\right)}=\dfrac{x+2}{x}=1+\dfrac{2}{x}\)
Không tồn tại Min P \(\forall x\inℝ\)
a) \(A=5x-\sqrt{4x^2-4x+1}\)
\(=5x-\sqrt{\left(2x-1\right)^2}\)
\(=5x-\left|2x-1\right|\)
+) Với x < 1/2
A = 5x - [ -( 2x - 1 ) ] = 5x - ( 1 - 2x ) = 5x - 1 + 2x = 7x - 1
+) Với x ≥ 1/2
A = 5x - ( 2x - 1 ) = 5x - 2x + 1 = 3x + 1
b) Với x = -2 < 1/2
=> A = 7.(-2) - 1 = -14 - 1 = -15
Trả lời:
a. rút gọn biểu thức A.B:
A= 3\(\sqrt{7}\)-2\(\sqrt{7}\)+5\(\sqrt{7}\)-3=-3
B= \(\sqrt{x}\)-1 + \(\sqrt{x}\)=2\(\sqrt{x}\)-1
b. Tìm x để A=3B
ta có:
A=-3= 3 (2\(\sqrt{x}\)-1)
=> -3= 6\(\sqrt{x}\)-3
=> \(\sqrt{x}\)=0
Vậy x=0 thì A=3B
1. \(x=\frac{1}{9}\) thỏa mãn đk: \(x\ge0;x\ne9\)
Thay \(x=\frac{1}{9}\) vào A ta có:
\(A=\frac{\sqrt{\frac{1}{9}}+1}{\sqrt{\frac{1}{9}}-3}=-\frac{1}{2}\)
2. \(B=...\)
\(B=\frac{3\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}+\frac{\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}-\frac{4x+6}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(B=\frac{3x-9\sqrt{x}+x+3\sqrt{x}-4x-6}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(B=\frac{-6\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
3. \(P=A:B=\frac{\sqrt{x}+1}{\sqrt{x}-3}:\frac{-6\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(P=\frac{\sqrt{x}+3}{-6}\)
Vì \(\sqrt{x}+3\ge3\forall x\)\(\Rightarrow\frac{\sqrt{x}+3}{-6}\le\frac{3}{-6}=-\frac{1}{2}\)
hay \(P\le-\frac{1}{2}\)
Dấu "=" xảy ra <=> x=0
a: Khi x=16/9 thì \(A=\left(\dfrac{4}{3}-2\right):\left(\dfrac{4}{3}-3\right)=\dfrac{-2}{3}:\dfrac{-5}{3}=\dfrac{2}{5}\)
b: \(=\dfrac{x+2\sqrt{x}+3\sqrt{x}-6-9\sqrt{x}-10}{x-4}\)
\(=\dfrac{x-4\sqrt{x}-16}{x-4}\)
ĐK: \(\left\{{}\begin{matrix}x-2\sqrt{x}-3\ne0\\\sqrt{x}+1\ne0\\3-\sqrt{x}\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)\ne0\\\sqrt{x}+1\ne0\left(hiển-nhiên\right)\\x\ne\sqrt{3}\end{matrix}\right.\)
\(\Leftrightarrow x\ne\sqrt{3}\)
\(P=\dfrac{x\sqrt{x}-3}{x-2\sqrt[]{x}-3}-\dfrac{2\left(\sqrt{x-3}\right)}{\sqrt{x}+1}+\dfrac{\sqrt{x}+3}{3-\sqrt{x}}\)
\(\Leftrightarrow\dfrac{x\sqrt{x}-3}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}-\dfrac{2\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}+\dfrac{\left(-\sqrt{x}-3\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}\)
\(\Leftrightarrow\dfrac{x\sqrt{x}-3-2\left(x-9\right)-x-\sqrt{x}-3\sqrt{x}-3}{\left(\sqrt{x+1}\right)\left(\sqrt{x}-3\right)}\)
\(\Leftrightarrow\dfrac{\left(x-4\right)\sqrt{x}-3x+12}{\left(\sqrt{x+1}\right)\left(\sqrt{x}-3\right)}\)
Chúc bạn học tốt ^^
Không thấy câu b =))
\(x=14-6\sqrt{5}=\left(3+\sqrt{5}\right)^2\)
\(\Rightarrow\sqrt{x}=3+\sqrt{5}\)
Thay vào ta được
\(\dfrac{14-6\sqrt{5}-3\left(14-6\sqrt{5}\right)+12}{\left(3+\sqrt{5}+1\right)\left(3+\sqrt{5}-3\right)}\)
\(=\dfrac{12\sqrt{5}-16}{\left(4+\sqrt{5}\right)\sqrt{5}}=\dfrac{12\sqrt{5}-16}{4\sqrt{5}+5}\)