\(P=\dfrac{\sqrt{x}-2}{\sqrt{x}-3}+\dfrac{\sqrt{x}+1}{\sqrt{x}+3}+\dfrac{x-4\sqrt{x}-9}{9-x}\)và \(Q=\dfrac{\sqrt{x}+5}{3-\sqrt{x}}\)(với\(x\ge0;x\ne9\))
1, Rút gọn P
2, Tìm x để P=3
3, Tìm \(M=P:Q\). Tìm x để \(\left|M\right|< \dfrac{1}{2}\)
\(A=\dfrac{-3\sqrt{x}+1}{\sqrt{x}-3}\) và \(B=\dfrac{3\sqrt{x}-2}{x-5\sqrt{x}+6}-\dfrac{1}{\sqrt{x}-2}+\dfrac{3\sqrt{x}-2}{3-\sqrt{x}}\) \(\left(x\ge0;x\ne4;x\ne9\right)\). Với \(x>9\), so sánh \(\dfrac{A}{B}\) và 1.
\(B=\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{x+9\sqrt{x}}{9-x};\left(x\ge0;x\ne9;x\ne16\right)\)
\(B=\dfrac{3}{\sqrt{x}-3}+\dfrac{2}{\sqrt{x}+3}+\dfrac{x-5\sqrt{x}-3}{x-9}\)
\(B=\dfrac{\sqrt{x}+1}{\sqrt{x}-1}-\dfrac{x+\sqrt{x}}{x-1};\left(x>0;x\ne1\right)\)
1.
\(A=\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{x+9\sqrt{x}}{9-x}\)
\(=\dfrac{2\sqrt{x}}{\sqrt{x}+3}-\dfrac{x+9\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{2\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}-\dfrac{x+9\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{2x-6\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}-\dfrac{x+9\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{x-15\sqrt{x}}{x-9}\)
2.
\(B=\dfrac{3}{\sqrt{x}-3}+\dfrac{2}{\sqrt{x}+3}+\dfrac{x-5\sqrt{x}-3}{x-9}\)
\(=\dfrac{3\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}+\dfrac{2\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}+\dfrac{x-5\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{3\sqrt{x}+9+2\sqrt{x}-6+x-5\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{x}{x-9}\)
3.
\(C=\dfrac{\sqrt{x}+1}{\sqrt{x}-1}-\dfrac{x+\sqrt{x}}{x-1}\)
\(=\dfrac{\sqrt{x}+1}{\sqrt{x}-1}-\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{\sqrt{x}+1}{\sqrt{x}-1}-\dfrac{\sqrt{x}}{\sqrt{x}-1}\)
\(=\dfrac{1}{\sqrt{x}-1}\)
Rút gọn
\(\dfrac{6}{\sqrt{5}+1}+\sqrt{\dfrac{2}{3-\sqrt{5}}}-\dfrac{10}{\sqrt{5}}\)
B1. Với \(x\ge0,x\ne4.Chobiểuthức\)
\(A=\dfrac{x-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}-\dfrac{1}{2-\sqrt{x}}-\dfrac{\sqrt{x}}{\sqrt{x}+3}\)
\(B=\dfrac{1}{x\sqrt{x}+27}\)
a, tính giá trị biểu thức khi B= 1/4
b, Rút gọn A
c, Tìm giá trị của x để A>1/2
d, Với C= B : A. Tìm GTLN C
a) Ta có: \(\dfrac{6}{\sqrt{5}+1}+\sqrt{\dfrac{2}{3-\sqrt{5}}}-\dfrac{10}{\sqrt{5}}\)
\(=\dfrac{6\left(\sqrt{5}-1\right)}{4}+\sqrt{\dfrac{2\left(3+\sqrt{5}\right)}{4}}-2\sqrt{5}\)
\(=\dfrac{3}{2}\left(\sqrt{5}-1\right)+\dfrac{\sqrt{3+\sqrt{5}}}{\sqrt{2}}-2\sqrt{5}\)
\(=\dfrac{3}{2}\sqrt{5}-\dfrac{3}{2}-2\sqrt{5}+\dfrac{\sqrt{5}+1}{2}\)
\(=-\dfrac{1}{2}\sqrt{5}-\dfrac{3}{2}+\dfrac{1}{2}\sqrt{5}+\dfrac{1}{2}\)
=-1
Bài 1:
a) Thay \(x=\dfrac{1}{4}\)vào B, ta được:
\(B=1:\left(\dfrac{1}{4}\cdot\dfrac{1}{2}+27\right)=1:\left(27+\dfrac{1}{8}\right)=\dfrac{8}{217}\)
b) Ta có: \(A=\dfrac{x-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}+\dfrac{1}{\sqrt{x}-2}-\dfrac{\sqrt{x}}{\sqrt{x}+3}\)
\(=\dfrac{x-9+\sqrt{x}+3-\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{x+\sqrt{x}-6-x+2\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{3\sqrt{x}-6}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{3}{\sqrt{x}+3}\)
c) Để \(A>\dfrac{1}{2}\) thì \(A-\dfrac{1}{2}>0\)
\(\Leftrightarrow\dfrac{6-\sqrt{x}-3}{2\left(\sqrt{x}+3\right)}>0\)
\(\Leftrightarrow3-\sqrt{x}>0\)
\(\Leftrightarrow\sqrt{x}< 3\)
hay x<9
Kết hợp ĐKXĐ, ta được: \(\left\{{}\begin{matrix}0\le x< 9\\x\ne4\end{matrix}\right.\)
Rút gọn : a) \(\dfrac{a\sqrt{b}-b\sqrt{a}}{\sqrt{a}-\sqrt{b}}-\sqrt{ab}\)
b)\(\dfrac{x+4y-4\sqrt{xy}}{\sqrt{x}-2\sqrt{y}}+\dfrac{y+\sqrt{xy}}{\sqrt{x}+\sqrt{y}}\left(x\ge0;y\ge0;x\ne4y\right)\)
c)\(\dfrac{x+4\sqrt{x}+4}{\sqrt{x}+2}+\dfrac{4-x}{\sqrt{x}-2}\left(x\ge0;x\ne4\right)\)
d)\(\dfrac{9-x}{\sqrt{3x}+3}-\dfrac{9-6\sqrt{x}+x}{\sqrt{x}-3}\)
e)\(\dfrac{\left(\sqrt{x}-\sqrt{y}\right)^2+4\sqrt{xy}}{\sqrt{x}+\sqrt{y}}-\dfrac{x\sqrt{y}+y\sqrt{x}}{\sqrt{xy}}\)
g)\(\left(2-\dfrac{a-3\sqrt{a}}{\sqrt{a}-3}\right)\left(2-\dfrac{5\sqrt{a}-\sqrt{ab}}{\sqrt{b}-5}\right)với\) a, b \(\ge\)0 , a \(\ne\)9; b\(\ne\)25
Mọi người giúp tớ với , cảm ơn nhiều nhiều ạ !!
a: \(=\dfrac{\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}{\sqrt{a}-\sqrt{b}}-\sqrt{ab}=\sqrt{ab}-\sqrt{ab}=0\)
b: \(=\dfrac{\left(\sqrt{x}-2\sqrt{y}\right)^2}{\sqrt{x}-2\sqrt{y}}+\dfrac{\sqrt{y}\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{x}+\sqrt{y}}\)
\(=\sqrt{x}-2\sqrt{y}+\sqrt{y}=\sqrt{x}-\sqrt{y}\)
c: \(=\sqrt{x}+2-\dfrac{x-4}{\sqrt{x}-2}\)
\(=\sqrt{x}+2-\sqrt{x}-2=0\)
a, Tính: \(A=\dfrac{2}{2+\sqrt{5}}-\sqrt{9-2\sqrt{20}}+\sqrt[3]{5\sqrt{5}}\)
b, Cho biểu thức: \(B=\left(\dfrac{2}{2\sqrt{x}-1}+\dfrac{3}{\sqrt{x}+2}-\dfrac{2\sqrt{x}}{2x+3\sqrt{x}-2}\right).\dfrac{2\sqrt{x}-\sqrt{x}}{6\sqrt{x}+4}\) với \(\left\{{}\begin{matrix}x\ge0\\x\ne\dfrac{1}{4}\end{matrix}\right.\)
a: \(=-4+2\sqrt{5}-\sqrt{5}+2+\sqrt{5}=2\sqrt{5}-2\)
b: \(B=\dfrac{2\sqrt{x}+4+6\sqrt{x}-3-2\sqrt{x}}{\left(2\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\cdot\dfrac{\sqrt{x}}{6\sqrt{x}+4}\)
\(=\dfrac{\left(6\sqrt{x}+1\right)\cdot\sqrt{x}}{\left(2\sqrt{x}-1\right)\left(\sqrt{x}+2\right)\left(6\sqrt{x}+4\right)}\)
Bài 1: Rút gọn biểu thức dạng chữ:
3) \(A=\left(\dfrac{\sqrt{x}}{\sqrt{x}+3}.\dfrac{3}{\sqrt{x}-3}\right).\dfrac{\sqrt{x}+3}{x+9}\) với x\(\ge0\), x ≠9
4) \(P=\dfrac{x-2}{x+2\sqrt{x}}-\dfrac{1}{\sqrt{x}}+\dfrac{1}{\sqrt{x}+2}\) với x ≥ 0
5) \(B=\left(1+\dfrac{x+\sqrt{x}}{\sqrt{x}+1}\right).\left(1-\dfrac{x-\sqrt{x}}{\sqrt{x}-1}\right)\)với x > 0, x ≠ 1
4) Ta có: \(P=\dfrac{x-2}{x+2\sqrt{x}}-\dfrac{1}{\sqrt{x}}+\dfrac{1}{\sqrt{x}+2}\)
\(=\dfrac{x-2-\sqrt{x}-2+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+2\right)}\)
\(=\dfrac{x-4}{\sqrt{x}\left(\sqrt{x}+2\right)}\)
\(=\dfrac{\sqrt{x}-2}{\sqrt{x}}\)
5) Ta có: \(B=\left(1+\dfrac{x+\sqrt{x}}{\sqrt{x}+1}\right)\left(1-\dfrac{x-\sqrt{x}}{\sqrt{x}-1}\right)\)
\(=\left(1+\sqrt{x}\right)\left(1-\sqrt{x}\right)\)
=1-x
Cho P = \(\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}}{\sqrt{x}-3}+\dfrac{3x+3}{9-x}\) và Q = \(\dfrac{\sqrt{x}+1}{\sqrt{x}-3}\) với \(x\ge0;x\ne9\)
a) Rút gọn biểu thức P. Tính M = P : Q
b) Tìm giá trị nhỏ nhất của biểu thức A = \(x.M+\dfrac{4x+7}{\sqrt{x}+3}\)
a: \(P=\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{x-9}=\dfrac{-3\left(\sqrt{x}+1\right)}{x-9}\)
\(M=\dfrac{-3\left(\sqrt{x}+1\right)}{x-9}\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}+1}=\dfrac{-3}{\sqrt{x}+3}\)
b: \(A=\dfrac{-3x+4x+7}{\sqrt{x}+3}=\dfrac{x+7}{\sqrt{x}+3}=\dfrac{x-9+16}{\sqrt{x}+3}\)
=>\(A=\sqrt{x}-3+\dfrac{16}{\sqrt{x}+3}=\sqrt{x}+3+\dfrac{16}{\sqrt{x}+3}-6>=2\sqrt{16}-6=2\)
Dấu = xảy ra khi x=1
cho biểu thức \(M=\dfrac{3\sqrt{x}-3}{\sqrt{x}-2}-\dfrac{2\sqrt{x}+4}{\sqrt{x}+1}-\dfrac{9}{x-\sqrt{x}-2}\),(với \(x\ge0,x\ne4\))chứng minh A>1
Ta có: \(M=\dfrac{3\sqrt{x}-3}{\sqrt{x}-2}-\dfrac{2\sqrt{x}+4}{\sqrt{x}+1}-\dfrac{9}{x-\sqrt{x}-2}\)
\(=\dfrac{3\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}-\dfrac{2\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}-\dfrac{9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{3x-3-2x+8-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{x-4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{\sqrt{x}+2}{\sqrt{x}+1}\)
Ta có: \(A-1=\dfrac{\sqrt{x}+2}{\sqrt{x}+1}-1\)
\(=\dfrac{\sqrt{x}+2-\sqrt{x}-1}{\sqrt{x}+1}\)
\(=\dfrac{1}{\sqrt{x}+1}>0\forall x\) thỏa mãn ĐKXĐ
hay A>1
\(M=\dfrac{3\sqrt{x}-3}{\sqrt{x}-2}-\dfrac{2\sqrt{x}+4}{\sqrt{x}+1}-\dfrac{9}{x-\sqrt{x}-2}\\ =\dfrac{3\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}-\dfrac{2\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}-\dfrac{9}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\\ =\dfrac{3\left(x-1\right)-2\left(x-4\right)-9}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}=\dfrac{x-4}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\\ =\dfrac{\sqrt{x}+2}{\sqrt{x}+1}=1+\dfrac{1}{\sqrt{x}+1}>1\)
Cho biểu thức D = \(\left(\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}}{\sqrt{x}-3}-\dfrac{3x+3}{x-9}\right):\left(\dfrac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right)\)
với \(x\ne9,x\ge0\)
a) Rút gọn D
b)Tìm x để \(D< \dfrac{-1}{4}\)
a) \(D=\left(\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}}{\sqrt{x}-3}-\dfrac{3x+3}{x-9}\right):\left(\dfrac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right)\)
\(=\dfrac{2\sqrt{x}\left(\sqrt{x}-3\right)+\sqrt{x}\left(\sqrt{x}+3\right)-3x+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}:\dfrac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-3}\)
\(=\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3x+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}.\dfrac{\sqrt{x}-3}{\sqrt{x}-1}\)
\(=\dfrac{-3\sqrt{x}+3}{\sqrt{x}+3}.\dfrac{1}{\sqrt{x}-1}=\dfrac{-3}{\sqrt{x}+3}\)
b) \(D=-\dfrac{3}{\sqrt{x}+3}< -\dfrac{1}{4}\)
\(\Leftrightarrow12>\sqrt{x}+3\Leftrightarrow\sqrt{x}< 9\)
\(\Leftrightarrow0\le x< 81\) và \(x\ne9\)
a) D=\(\left(\dfrac{2\sqrt{x}.\left(\sqrt{x}-3\right)+\sqrt{x}.\left(\sqrt{x}+3\right)-3x-3}{\left(\sqrt{x}+3\right).\left(\sqrt{x}-3\right)}\right)\) \(:\dfrac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-3}\)
\(\Leftrightarrow D=\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{\left(\sqrt{x}-3\right).\left(\sqrt{x}+3\right)}\) \(.\dfrac{\sqrt{x}-3}{\sqrt{x}+1}\)
\(\Leftrightarrow D=\dfrac{-3-3\sqrt{x}}{\sqrt{x}+3}.\dfrac{1}{\sqrt{x}+1}\)
\(\Leftrightarrow D=\dfrac{-3.\left(\sqrt{x}+1\right)}{\sqrt{x}+3}.\dfrac{1}{\sqrt{x}+1}\)
\(\Leftrightarrow D=\dfrac{-3}{\sqrt{x}+3}\)
b) Để D\(< \dfrac{-1}{4}\) \(\Leftrightarrow\dfrac{-3}{\sqrt{x}+3}< \dfrac{-1}{4}\)
\(\Leftrightarrow12>\sqrt{x}+3\Leftrightarrow9>\sqrt{x}\Leftrightarrow81>x\ge0\)
Tính GTLN của biểu thức A.
\(A=\dfrac{1-\sqrt{x}}{\sqrt{x}+2}\)(đk: \(x\ge0,x\ne1,x\ne4\))
B2. Giải pt
\(\sqrt{x-3}+\sqrt{y-5}+\sqrt{z-4}=20-\dfrac{4}{\sqrt{x-3}}-\dfrac{9}{\sqrt{y-5}}-\dfrac{25}{\sqrt{z-4}}\)
\(A=\dfrac{1-\sqrt{x}}{\sqrt{x}+2}=\dfrac{3-\left(\sqrt{x}+2\right)}{\sqrt{x}+2}=\dfrac{3}{\sqrt{x}+2}-1\)
Có \(\sqrt{x}\ge0\Leftrightarrow\sqrt{x}+2\ge2\Leftrightarrow\dfrac{3}{\sqrt{x}+2}\le\dfrac{3}{2}\)\(\Leftrightarrow\dfrac{3}{\sqrt{x}+2}-1\le\dfrac{1}{2}\)\(\Leftrightarrow A\le\dfrac{1}{2}\)
Dấu "=" xảy ra khi x=0 (tm)
Vậy \(A_{max}=\dfrac{1}{2}\)
Bài 2:
Đk: \(x\ge3;y\ge5;z\ge4\)
Pt\(\Leftrightarrow\sqrt{x-3}+\dfrac{4}{\sqrt{x-3}}+\sqrt{y-5}+\dfrac{9}{\sqrt{y-5}}+\sqrt{z-4}+\dfrac{25}{\sqrt{z-4}}=20\)
Áp dụng AM-GM có:
\(\sqrt{x-3}+\dfrac{4}{\sqrt{x-3}}\ge2\sqrt{\sqrt{x-3}.\dfrac{4}{\sqrt{x-3}}}=4\)
\(\sqrt{y-5}+\dfrac{9}{\sqrt{y-5}}\ge6\)
\(\sqrt{z-4}+\dfrac{25}{\sqrt{z-4}}\ge10\)
Cộng vế với vế \(\Rightarrow VT\ge20\)
Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}\sqrt{x-3}=\dfrac{4}{\sqrt{x-3}}\\\sqrt{y-5}=\dfrac{9}{\sqrt{y-5}}\\\sqrt{z-4}=\dfrac{25}{\sqrt{z-4}}\end{matrix}\right.\)\(\Leftrightarrow x=7;y=14;z=29\) (tm)
Vậy...