Cho A=\(\frac{15\sqrt{x}-11}{x+2\sqrt{x}-3}-\frac{3\sqrt{x}-2}{\sqrt{x}-1}-\frac{2\sqrt{x}+3}{\sqrt{x}+3}\)
a) Rút gọn A
b)Tìm GTLN
Cho \(U=\frac{15\sqrt{x}-11}{x+2\sqrt{x}-3}+\frac{3\sqrt{x}-2}{1-\sqrt{x}}-\frac{2\sqrt{x}+3}{\sqrt{x}+3}\)
a/ Rút gọn U
b/ Tìm gtln của U và giá trị tương ứng của x
Cho \(A=\frac{15\sqrt{x}-11}{x+2\sqrt{x}-3}+\frac{3\sqrt{x}-2}{1-\sqrt{x}}-\frac{2\sqrt{x}+3}{\sqrt{x}+3}\)
với \(x\ge0,x\ne1\)
a. Rút gọn
b.Tìm GTLN
c. Tìm x để A\(=\frac{1}{2}\)
d. CMR \(A\le\frac{2}{3}\)
giúp mình với
a) Với \(x\ge0;x\ne1\)
\(A=\frac{15\sqrt{x}-11}{x+2\sqrt{x}-3}+\frac{3\sqrt{x}-2}{1-\sqrt{x}}-\frac{2\sqrt{x}+3}{\sqrt{x}+3}\)
\(A=\frac{15\sqrt{x}-11}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}-\frac{3\sqrt{x}-2}{\sqrt{x}-1}-\frac{2\sqrt{x}+3}{\sqrt{x}+3}\)
\(A=\frac{15\sqrt{x}-11-\left(3\sqrt{x}-2\right)\left(\sqrt{x}+3\right)-\left(2\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(A=\frac{15\sqrt{x}-11-\left(3x-9\sqrt{x}-2\sqrt{x}-6\right)-\left(2x-2\sqrt{x}+3\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(A=\frac{15\sqrt{x}-11-3x-9\sqrt{x}+2\sqrt{x}+6-2x+2\sqrt{x}-3\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(A=\frac{-5x+7\sqrt{x}-2}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}=\frac{\left(\sqrt{x}-1\right)\left(-5\sqrt{x}+2\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}=\frac{-5\sqrt{x}+2}{\sqrt{x}+3}\)
Vậy : \(A=\frac{-5\sqrt{x}+2}{\sqrt{x}+3}\)
b) \(A=\frac{-5\sqrt{x}+2}{\sqrt{x}+3}=\frac{-5\left(\sqrt{x}+3\right)+17}{\sqrt{x}+3}=-5+\frac{17}{\sqrt{x}+3}\)
\(A_{max}\Leftrightarrow\left(\frac{17}{\sqrt{x}+3}\right)_{max}\)
Vì \(x\ge0;x\ne1\Rightarrow\hept{\begin{cases}\sqrt{x}\ge0\\\frac{17}{\sqrt{x}+3}>0\end{cases}A_{max}\Leftrightarrow}\left(\sqrt{x}+3\right)_{min}\Leftrightarrow\sqrt{x}_{min}\Leftrightarrow x=0\)
Vậy : \(A_{max}=\frac{17}{3}\Leftrightarrow x=0\)
c,d chưa làm được .-.
c) Để \(A=\frac{1}{2}\)
<=> \(\frac{-5\sqrt{x}+2}{\sqrt{x}+3}=\frac{1}{2}\)
<=> \(-10\sqrt{x}+4=\sqrt{x}+3\)
<=> \(-11\sqrt{x}=-1\)
<=> \(\sqrt{x}=\frac{1}{11}\)
<=> \(x=\frac{1}{121}\left(tm\right)\)
Vậy ...
d) \(A\le\frac{2}{3}\)
<=> \(\frac{-5\sqrt{x}+2}{\sqrt{x}+3}\le\frac{2}{3}\)
<=> \(\frac{-5\sqrt{x}+2}{\sqrt{x}+3}-\frac{2}{3}\le0\)
<=> \(\frac{-15\sqrt{x}+6-2\sqrt{x}-6}{3\sqrt{x}+9}\le0\)
<=> \(\frac{-17\sqrt{x}}{3\sqrt{x}+9}\le0\)
Vì \(\hept{\begin{cases}-17\sqrt{x}\le0\\3\sqrt{x}+9>0\end{cases}}\) \(\Rightarrow\frac{-17\sqrt{x}}{3\sqrt{x}+9}\le0\)(luôn đúng)
=> Ta có ĐPCM
cho biểu thức
A=\(\frac{15\sqrt{x}-11}{x+2\sqrt{x}-3}+\frac{3\sqrt{x}-2}{1-\sqrt{x}}-\frac{2\sqrt{x}+3}{3+\sqrt{x}}\)
a, Rút gọn A
b, Tìm x để A= \(\frac{1}{2}\)
Mk làm như này, k biết có sai chỗ nào k. Nếu sai thì bạn sửa nhé.
A=\(\frac{15\sqrt{x}-11}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}-\frac{3\sqrt{x}-2}{\sqrt{x}-1}-\frac{2\sqrt{x}+3}{3+\sqrt{x}}\)
A=\(\frac{15\sqrt{x}-11-\left(3x-2\right)\left(\sqrt{x}+3\right)-\left(2\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
A=\(\frac{15\sqrt{x}-11-3x-7\sqrt{x}+6-2x-\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
A=\(\frac{-5x+7\sqrt{x}-2}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
A=\(\frac{-5x+5\sqrt{x}+2\sqrt{x}-2}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
A=\(\frac{-5\sqrt{x}\left(\sqrt{x}-1\right)+2\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
A=\(\frac{2-5\sqrt{x}}{\sqrt{x}+3}\)
cho C=\(\frac{15\sqrt{x}-11}{x+2\sqrt{x}-3}+\frac{b\sqrt{x}-2}{1-\sqrt{x}}-\frac{2\sqrt{x}+3}{\sqrt{x}+3}\)
a. rút gọn C
b. tìm x để C=\(\frac{1}{2}\)
Cho P = \(\frac{15\sqrt{x}-11}{x+2\sqrt{x}-3}+\frac{3\sqrt{x}-2}{1-\sqrt{x}}-\frac{2\sqrt{x}+3}{\sqrt{x}+3}\)
a. Tìm điều kiện xác định
b. Rút gọn
c. Tìm Pmax
a. ĐKXĐ : \(\orbr{\begin{cases}x\ge0\\1-\sqrt{x}\ne0\end{cases}}\)<=> \(\orbr{\begin{cases}x\ge0\\x\ne1\end{cases}}\)
b. \(P=\frac{15\sqrt{x}-11}{x+2\sqrt{x}-3}+\frac{3\sqrt{x}-2}{1-\sqrt{x}}-\frac{2\sqrt{x}+3}{\sqrt{x}+3}\)
\(\Leftrightarrow P=\frac{15\sqrt{x}-11}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}-\frac{3\sqrt{x}-2}{\sqrt{x}-1}-\frac{2\sqrt{x}+3}{\sqrt{x}+3}\)
\(\Leftrightarrow P=\frac{15\sqrt{x}-11-\left(3\sqrt{x}-2\right)\left(\sqrt{x}+3\right)-\left(2\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
\(\Leftrightarrow P=\frac{15\sqrt{x}-11-3x-7\sqrt{x}+6-2x-\sqrt{x}+3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
\(\Leftrightarrow P=\frac{-5x+7\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
\(\Leftrightarrow P=\frac{\left(\sqrt{x}-1\right)\left(2-5\sqrt{x}\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
\(\Leftrightarrow P=\frac{2-5\sqrt{x}}{\sqrt{x}+3}\)
là bằng 2 phần 3 phải ko
a) ĐKXĐ : \(\hept{\begin{cases}x\ge0\\1-\sqrt{x}\ne0\end{cases}\Leftrightarrow}\hept{\begin{cases}x\ge0\\x\ne1\end{cases}}\)
b) \(P=\frac{15\sqrt{x}-11}{x+2\sqrt{x}-3}+\frac{3\sqrt{x}-2}{1-\sqrt{x}}-\frac{2\sqrt{x}+3}{\sqrt{x}+3}\)
\(P=\frac{15\sqrt{x}-11}{x+3\sqrt{x}-\sqrt{x}-3}+\frac{3\sqrt{x}-2}{1-\sqrt{x}}-\frac{2\sqrt{x}+3}{\sqrt{x}+3}\)
\(P=\frac{15\sqrt{x}-11}{\sqrt{x}\left(\sqrt{x+3}\right)-\left(\sqrt{x}+3\right)}+\frac{3\sqrt{x}-2}{1-\sqrt{x}}-\frac{2\sqrt{x}+3}{\sqrt{x}+3}\)
\(P=\frac{15\sqrt{x}-11+\left(\sqrt{x}+3\right)\left(2-3\sqrt{x}\right)-\left(2\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
\(P=\frac{15\sqrt{x}-11+2\sqrt{x}-3x+6-9\sqrt{x}-\left(2x-2\sqrt{x}+3\sqrt{x}-3\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
\(P=\frac{7\sqrt{x}-5x-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
\(P=\frac{2-5\sqrt{x}}{\sqrt{x}+3}\)
c) Ta có :
\(P=\frac{-5\sqrt{x}+2}{\sqrt{x}+3}\)
+)Với \(x\ge0,x\ne1\)ta có : \(\sqrt{x}+3\ge3\left(1\right)\)
+) \(5\sqrt{x}\ge0\Rightarrow-5\sqrt{x}\le0\Rightarrow-5\sqrt{x}+2\le2\left(2\right)\)
Từ ( 1 ) và ( 2 ) \(\Rightarrow P\le\frac{2}{3}\)
Vậy max \(P=\frac{2}{3}\Leftrightarrow\hept{\begin{cases}x\ge0\\x\ne1\end{cases}}\)
Câu 1 :A= \(\left(\frac{x}{x\sqrt{x}-4\sqrt{x}}-\frac{6}{3\sqrt{x}-6}+\frac{1}{\sqrt{x}+2}\right):\left(\sqrt{x}-2+\frac{10-x}{x+2}\right)\)
a, rút gọn A
b, Tìm X sao cho A<2
Câu 2 \(A=\frac{15\sqrt{x}-11}{x+2\sqrt{x}-3}+\frac{3\sqrt{x}-2}{1-\sqrt{x}}-\frac{2\sqrt{x}+3}{\sqrt{x}+3}\)
a, rút gọn A \(\left(với\right)x\ge0,x\ne1\)
b, chúng minh rằng A\(\le\)\(\frac{2}{3}\)
Câu 3 \(\left(\frac{2}{\sqrt{x}}+\frac{\sqrt{x}}{\sqrt{x}+2}\right):\frac{2\sqrt{x}}{x+2\sqrt{x}}\left(vớix>0\right)\)
a, Rút gọn P
b, tìm giá trị của x để P=3
Câu 3 :
\(ĐKXĐ:x>0\)
\(P=\left(\frac{2}{\sqrt{x}}+\frac{\sqrt{x}}{\sqrt{x}+2}\right):\frac{2\sqrt{x}}{x+2\sqrt{x}}\)
\(\Leftrightarrow P=\frac{2\sqrt{x}+4+x}{x+2\sqrt{x}}\cdot\frac{x+2\sqrt{x}}{2\sqrt{x}}\)
\(\Leftrightarrow P=\frac{2\sqrt{x}+4+x}{2\sqrt{x}}\)
b) Để P = 3
\(\Leftrightarrow\frac{2\sqrt{x}+4+x}{x+2\sqrt{x}}=3\)
\(\Leftrightarrow2\sqrt{x}+4+x=6\sqrt{x}\)
\(\Leftrightarrow x-4\sqrt{x}+4=0\)
\(\Leftrightarrow\left(\sqrt{x}-2\right)^2=0\)
\(\Leftrightarrow\sqrt{x}-2=0\)
\(\Leftrightarrow\sqrt{x}=2\)
\(\Leftrightarrow x=4\)(tm)
Vậy để \(P=3\Leftrightarrow x=4\)
Câu 1 : Hình như sai đề !! Mik sửa :
\(ĐKXĐ:\hept{\begin{cases}x\ge0\\x\ne4\end{cases}}\)
\(A=\left(\frac{x}{x\sqrt{x}-4\sqrt{x}}-\frac{6}{3\sqrt{x}-6}+\frac{1}{\sqrt{x}+2}\right):\left(\sqrt{x}-2+\frac{10-x}{\sqrt{x}+2}\right)\)
\(\Leftrightarrow A=\left(\frac{\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-\frac{2}{\sqrt{x}-2}+\frac{1}{\sqrt{x}+2}\right):\left(\frac{x-4+10-x}{\sqrt{x}+2}\right)\)
\(\Leftrightarrow A=\frac{\sqrt{x}-2\sqrt{x}-4+\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}:\frac{6}{\sqrt{x}+2}\)
\(\Leftrightarrow A=\frac{-6\left(\sqrt{x}+2\right)}{6\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(\Leftrightarrow A=-\frac{1}{\sqrt{x}-2}\)
b) Để A < 2
\(\Leftrightarrow-\frac{1}{\sqrt{x}-2}< 2\)
\(\Leftrightarrow-1< 2\sqrt{x}-4\)
\(\Leftrightarrow2\sqrt{x}>3\)
\(\Leftrightarrow\sqrt{x}>1,5\)
\(\Leftrightarrow x>2,25\)
Vậy để \(A< 2\Leftrightarrow x>2,25\)
Câu 2 :
\(ĐKXĐ:\hept{\begin{cases}x\ne\frac{4}{25}\\x\ne9\\x\ne1\end{cases}}\)
\(A=\frac{15\sqrt{x}-11}{x+2\sqrt{x}-3}+\frac{3\sqrt{x}-2}{1-\sqrt{x}}-\frac{2\sqrt{x}+3}{\sqrt{x}+3}\)
\(\Leftrightarrow A=\frac{15\sqrt{x}-11-\left(3\sqrt{x}-2\right)\left(\sqrt{x}+3\right)-\left(2\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
\(\Leftrightarrow A=\frac{15\sqrt{x}-11-3x-7\sqrt{x}+6-2x-\sqrt{x}+3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
\(\Leftrightarrow A=\frac{-5x+7\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
\(\Leftrightarrow A=\frac{\left(2-5\sqrt{x}\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
\(\Leftrightarrow A=\frac{2-5\sqrt{x}}{\sqrt{x}+3}\)
b) Mik quên mất cách chứng minh rồi :((
Chỉ biết : Dấu " = " xảy ra : \(\Leftrightarrow x=0\)
\(\)
\(B=\frac{15\sqrt{x}-11}{x+2\sqrt{x}-3}-\frac{3\sqrt{x}-2}{\sqrt{x}-1}-\frac{2\sqrt{x}+3}{\sqrt{x}+3}\)
a/ Rút gọn B
b/ Tìm x để B = 1/2
c/ so sánh B và 2/3
a) \(B=\frac{15\sqrt{x}-11}{x+2\sqrt{x}-3}-\frac{3\sqrt{x}-2}{\sqrt{x}-1}-\frac{2\sqrt{x}+3}{\sqrt{x}+3}\)
ĐKXĐ: \(x\ge0,x\ne1\)
\(B=\frac{15\sqrt{x}-11}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}-\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}-\frac{\left(2\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
\(B=\frac{15\sqrt{x}-11-\left(3x+7\sqrt{x}-6\right)-\left(2x+\sqrt{2}-3\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
\(B=\frac{-5x+7\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
\(B=\frac{\left(\sqrt{x}-1\right)\left(2-5\sqrt{x}\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}=\frac{2-5\sqrt{x}}{\sqrt{x+3}}\)
b) Để \(B=\frac{1}{2}\Rightarrow\frac{2-5\sqrt{x}}{\sqrt{x}+3}=\frac{1}{2}\)\(\Rightarrow\sqrt{x}+3=4-10\sqrt{x}\Rightarrow11\sqrt{x}=1\Rightarrow\sqrt{x}=\frac{1}{12}\Rightarrow x=\frac{1}{121}\)(Thoả mãn ĐKXĐ)
Vậy x=1/121 thì B =1/2
Rút gọn biểu thức:
\(A=\frac{15\sqrt{x}-11}{x+2\sqrt{x}-3}-\frac{3\sqrt{x}-2}{\sqrt{x}-1}-\frac{2\sqrt{x}+3}{\sqrt{x}+3}\)
Cho A = \(\frac{15\sqrt{x}-11}{x+2\sqrt{x}-3}+\frac{3\sqrt{x}-2}{1-\sqrt{x}}-\frac{2\sqrt{x}+3}{\sqrt{x}+3}\) với \(x\ge0,x\ne1\)
a, Rút gọn A
b, Tìm GTLN của A
a.
\(A=\frac{15\sqrt{x}-11-\left(3\sqrt{x}-2\right)\left(\sqrt{x}+3\right)-\left(2\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
\(=\frac{15\sqrt{x}-11-3x-9\sqrt{x}+2\sqrt{x}+6-2x+2\sqrt{x}-3\sqrt{x}+3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
\(=\frac{-5x+7\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
\(=\frac{\left(\sqrt{x}-1\right)\left(2-5\sqrt{x}\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
\(=\frac{2-5\sqrt{x}}{\sqrt{x}+3}\)
b.Ta co:
\(A=\frac{2-5\sqrt{x}}{\sqrt{x}+3}=\frac{-5\left(\sqrt{x}+3\right)+17}{\sqrt{x}+3}=-5+\frac{17}{\sqrt{x}+3}\le-5+\frac{17}{3}=\frac{2}{3}\)
Dau '=' xay ra khi \(x=0\)
Vay \(A_{max}=\frac{2}{3}\)khi \(x=0\)