A=\(\left(\frac{x-5\sqrt{x}}{x-25}-1\right):\left(\frac{25-x}{x+2\sqrt{x}-15}-\frac{\sqrt{x}+3}{\sqrt{x}+5}+\frac{\sqrt{x}-5}{\sqrt{x-3}}\right)\)
a,Rút gọn A
b,Tính giá trị của A khi x=29-12\(\sqrt{5}\)
Cho biểu thức
A= \(\left(\frac{x-5\sqrt{x}}{x-25}-1\right):\left(\frac{25-x}{x+2\sqrt{x}-15}-\frac{\sqrt{x}+3}{\sqrt{x}+5}+\frac{\sqrt{x}-5}{\sqrt{x}-3}\right)\)
a, Rút gọn A
b, Tìm x để A<1
ĐKXĐ : \(x\ge0,x\ne25,x\ne9\)
a) \(A=\left(\frac{x-5\sqrt{x}}{x-25}-1\right):\left(\frac{25-x}{x+2\sqrt{x}-15}-\frac{\sqrt{x}+3}{\sqrt{x}+5}+\frac{\sqrt{x}-5}{\sqrt{x}-3}\right)\)
\(=\left(\frac{x-5\sqrt{x}-\left(x-25\right)}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}\right):\left(\frac{-\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}-\frac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}+\frac{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}\right)\)
\(=\frac{-5\left(\sqrt{x}-5\right)}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}:\left(\frac{25-x-x+9+x-25}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}\right)\)
\(=-\frac{5}{\sqrt{x}+5}:\frac{-\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+5\right)}=\frac{-5}{\sqrt{x}+5}.\left(\frac{-\left(\sqrt{x}+5\right)}{\sqrt{x}+3}\right)=\frac{5}{\sqrt{x}+3}\)
b) \(A< 1\Rightarrow\frac{5}{\sqrt{x}+3}< 1\Rightarrow\sqrt{x}+3>5\Rightarrow\sqrt{x}>2\Rightarrow x>4\)
Chú ý kết hợp với điều kiện xác định.
Bài 2: Cho A= \(\frac{4\left(\sqrt{x}+1\right)}{25-x}\); B= \(\left(\frac{15-\sqrt{x}}{x-25}+\frac{2}{\sqrt{x}+5}\right):\frac{\sqrt{x}+1}{\sqrt{x}-5}\)
a) Rút gọn B
b) Tìm x để P=A.B đạt giá trị nguyên lớn nhất
bổ sung thêm đề bài là \(x\ge0;x\ne25\) nha
\(a,B=\left(\frac{15-\sqrt{x}}{x-25}+\frac{2}{\sqrt{x}+5}\right):\frac{\sqrt{x}+1}{\sqrt{x}-5}\)
\(B=\left(\frac{15-\sqrt{x}+2\sqrt{x}-10}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}\right).\frac{\sqrt{x}-5}{\sqrt{x}+1}\)
\(B=\frac{5+\sqrt{x}}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}\frac{\sqrt{x}-5}{\sqrt{x}+1}\)
\(B=\frac{1}{\sqrt{x}+1}\)
\(b,P=A.B=\frac{4\left(\sqrt{x}+1\right)}{25-x}.\frac{1}{\sqrt{x}+1}\)
\(P=\frac{4}{25-x}\)
bổ sung điều kiện cho câu b là x nguyên
\(TH1:x>25< =>P< 0\left(KTM\right)\)
\(TH2:x< 25< =>P>0\)mà x nguyên
\(\frac{4}{25-x}\le4\)
dấu "=" xảy ra khi \(x=24\)
\(< =>MAX:P=4\)
Rút gọn A=\(\left(\frac{5\sqrt{x}+50}{x+5\sqrt{x}}+\frac{2\sqrt{x}-10}{\sqrt{x}}+\frac{x}{5\sqrt{x}+25}\right).\frac{7}{15+3\sqrt{x}}\)
1. tìm x thuộc z để \(\frac{\sqrt{x}+1}{\sqrt{x}-3}\)nguyên
2.\(B=\frac{x+2}{x\sqrt{x}-1}+\frac{\sqrt{x}+1}{x+\sqrt{x}+1}+\frac{\sqrt{x}+1}{x-1}\)
a. điều kiện
b. rút gọn
c. chứng minh 3B<1
3.\(C=\left(\frac{x-5\sqrt{x}}{x-25}-1\right):\left(\frac{25-x}{x+2\sqrt{x}-15}-\frac{\sqrt{x}+3}{\sqrt{x+5}}+\frac{\sqrt{x}-5}{\sqrt{x}-3}\right)\)
a. điều kiện
b. rút gọn
c.tìm x thuộc z để C thuộc z
4. tìm giá trị nhỏ nhất của biểu thức
\(A=x+\sqrt{x}+1\)
\(1,\frac{\sqrt{x}+1}{\sqrt{x}-3}=\frac{\sqrt{x}-3+4}{\sqrt{x}-3}=1+\frac{4}{\sqrt{x}-3}\)
Để \(\frac{\sqrt{x}+1}{\sqrt{x}-3}\in Z\Rightarrow\frac{4}{\sqrt{x}-3}\in Z\)
\(\Rightarrow\sqrt{x}-3\in\left(1;4;-1;-4\right)\)
\(\Rightarrow\sqrt{x}\in\left(4;7;2;-1\right)\)
\(\Rightarrow\sqrt{x}=4\Leftrightarrow x=2\)
\(4,A=x+\sqrt{x}+1\)
\(A=\left(\sqrt{x}\right)^2+2.\frac{1}{2}.\sqrt{x}+\left(\frac{1}{2}\right)^2+\frac{3}{4}\)
\(A=\left(\sqrt{x}+\frac{1}{2}\right)^2+\frac{3}{4}\)
\(\Rightarrow A\ge\frac{3}{4}.\left(\sqrt{x}+\frac{1}{2}\right)^2\ge0\)
Dấu "=" xảy ra khi :
\(\sqrt{x}+\frac{1}{2}=0\Leftrightarrow\sqrt{x}=-\frac{1}{2}\)
Vậy Min A = 3/4 khi căn x = -1/2
A = \(\left(\frac{x-5\sqrt{x}}{x-25}-1\right):\left(\frac{25-x}{x+2\sqrt{x}-15}-\frac{\sqrt{x}+3}{\sqrt{x}+5}+\frac{\sqrt{x}-5}{\sqrt{x}-3}\right)\)
a) rút gọn
b) Tìm x nguyên để A nguyên
1) cho biểu thức P=\(\frac{\sqrt{a}+2}{\sqrt{a}+3-}-\frac{5}{a+\sqrt{a}-6}+\frac{1}{2-\sqrt{a}}\)
a/ rút gọn P
b/ tìm giá trị của a để P<1
2) cho biểu thức P=\(\left(1-\frac{\sqrt{x}}{\sqrt{x}+1}\right):\left(\frac{\sqrt{x}+3}{\sqrt{x}-2}+\frac{\sqrt{x}+2}{3-\sqrt{x}}+\frac{\sqrt{x}+2}{x-5\sqrt{x}+6}\right)\)
a/ rút gọn P
b/ tìm giá trị của P<0
Cho biểu thức \(A=\left(\frac{x-5\sqrt{x}}{x-25}-1\right):\left(\frac{25-x}{x+2\sqrt{x}-15}-\frac{\sqrt{x}+3}{\sqrt{x}+5}+\frac{\sqrt{x}-5}{\sqrt{x}-3}\right)\)
a)Rút gọn
b)Với \(x\ge0,x\ne25,x\ne9\).Tìm GTNN của \(B=\frac{A\left(x+16\right)}{5}\)
ĐKXĐ: ...
\(A=\left(\frac{\sqrt{x}\left(\sqrt{x}-5\right)}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}-1\right):\left(\frac{25-x+\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+5\right)}-\frac{\sqrt{x}+3}{\sqrt{x}+5}\right)\)
\(=\left(\frac{\sqrt{x}}{\sqrt{x}+5}-\frac{\sqrt{x}+5}{\sqrt{x}+5}\right):\left(\frac{25-x+x-25}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+5\right)}-\frac{\sqrt{x}+3}{\sqrt{x}+5}\right)\)
\(=\frac{-5}{\left(\sqrt{x}+5\right)}.\frac{\left(\sqrt{x}+5\right)}{-\left(\sqrt{x}+3\right)}=\frac{5}{\sqrt{x}+3}\)
b/ \(B=\frac{x+16}{\sqrt{x}+3}=\sqrt{x}-3+\frac{25}{\sqrt{x}+3}=\sqrt{x}+3+\frac{25}{\sqrt{x}+3}-6\)
\(\Rightarrow B\ge2\sqrt{\frac{\left(\sqrt{x}+3\right).25}{\sqrt{x}+3}}-6=4\)
\(B_{min}=4\) khi \(\left(\sqrt{x}+3\right)^2=25\Rightarrow x=4\)
Cho biểu thức A= \(\frac{1-\sqrt{x}}{\sqrt{x}+1}\) và \(B=\left(\frac{15-\sqrt{x}}{x-25}+\frac{2}{\sqrt{x}+5}\right):\frac{\sqrt{x}+1}{\sqrt{x}-5}\)
a, Tính giá trị của A khi x=9
b, Rút gọn B
c, Với các biểu thức A và B nói trên hãy tìm x để A-B =a có nghiệm
1,rút gọn
\(\frac{5+\sqrt{5}}{5-\sqrt{5}}+\frac{5-\sqrt{5}}{5+\sqrt{5}}\)
2.giải pt
\(a,\sqrt{45x}-2\sqrt{20x}+2\sqrt{80x}=21\) \(b,\sqrt{x^2-10x+25}=4\)
3,cho biểu thức :A=\(\left(\frac{1}{\sqrt{x}-1}-\frac{1}{\sqrt{x}}\right).\left(\frac{\sqrt{x}+1}{\sqrt{x}-2}-\frac{\sqrt{x}+2}{\sqrt{x}-1}\right)\) (x>0;x#0)
a, rút gọn biểu thức
b, tính giá trị của x khi A>\(\frac{1}{6}\)
1.\(\frac{5+\sqrt{5}}{5-\sqrt{5}}+\frac{5-\sqrt{5}}{5+\sqrt{5}}=\frac{\left(5+\sqrt{5}\right)\left(5+\sqrt{5}\right)}{\left(5-\sqrt{5}\right)\left(5+\sqrt{5}\right)}+\frac{\left(5-\sqrt{5}\right)\left(5-\sqrt{5}\right)}{\left(5-\sqrt{5}\right)\left(5+\sqrt{5}\right)}\)
\(=\frac{25+10\sqrt{5}+5}{25-5}+\frac{25-10\sqrt{5}+5}{25-5}\)
\(=\frac{25+10\sqrt{5}+5+25-10\sqrt{5}+5}{20}\)
\(=\frac{60}{20}=3\)
2.
a) \(\sqrt{45x}-2\sqrt{20x}+2\sqrt{80x}=21\)
ĐK : x ≥ 0
<=> \(\sqrt{5x\cdot9}-2\sqrt{5x\cdot4}+2\sqrt{5x\cdot16}=21\)
<=> \(\sqrt{5x\cdot3^2}-2\sqrt{2^2\cdot5x}+2\sqrt{5x\cdot4^2}=21\)
<=> \(\left|3\right|\sqrt{5x}-2\cdot\left|2\right|\sqrt{5x}+2\cdot\left|4\right|\sqrt{5x}=21\)
<=> \(\sqrt{5x}\cdot\left(3-4+8\right)=21\)
<=> \(\sqrt{5x}\cdot7=21\)
<=> \(\sqrt{5x}=3\)
<=> \(5x=9\)
<=> \(x=\frac{9}{5}\left(tm\right)\)
ơ đang làm lại bấm " Gửi trả lời " ._.
2b) \(\sqrt{x^2-10x+25}=4\)
<=> \(\sqrt{\left(x-5\right)^2}=4\)
<=> \(\left|x-5\right|=4\)
<=> \(\orbr{\begin{cases}x-5=4\\x-5=-4\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=9\\x=1\end{cases}}\)
3. \(A=\left(\frac{1}{\sqrt{x}-1}-\frac{1}{\sqrt{x}}\right)\div\left(\frac{\sqrt{x}+1}{\sqrt{x}-2}-\frac{\sqrt{x}+2}{\sqrt{x}-1}\right)\)
ĐK : \(\hept{\begin{cases}x>0\\x\ne1\\x\ne4\end{cases}}\)
\(=\left(\frac{\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}-\frac{\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x-1}\right)}\right)\div\left(\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}-\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\right)\)
\(=\left(\frac{\sqrt{x}-\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}\right)\div\left(\frac{x-1}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}-\frac{x-4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\right)\)
\(=\frac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}\div\left(\frac{x-1-x+4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\right)\)
\(=\frac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}\div\frac{3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\)
\(=\frac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}\times\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}{3}\)
\(=\frac{\sqrt{x}-2}{3\sqrt{x}}\)
1/ \(\frac{5+\sqrt{5}}{5-\sqrt{5}}+\frac{5-\sqrt{5}}{5+\sqrt{5}}=\frac{\sqrt{5}+1}{\sqrt{5}-1}+\frac{\sqrt{5}-1}{\sqrt{5}+1}=\frac{\left(\sqrt{5}+1\right)^2}{5-1}+\frac{\left(\sqrt{5}-1\right)^2}{5-1}\)
\(=\frac{5+2\sqrt{5}+1}{4}+\frac{5-2\sqrt{5}+1}{4}=\frac{12}{4}=3\)
2/ a) \(\sqrt{45x}-2\sqrt{20x}+2\sqrt{80x}=21\Leftrightarrow3\sqrt{5}.\sqrt{x}-4\sqrt{5}.\sqrt{x}+8\sqrt{5}.\sqrt{x}=21\)
\(\Leftrightarrow7\sqrt{5}.\sqrt{x}=21\Leftrightarrow\sqrt{x}=\frac{3}{\sqrt{5}}\Leftrightarrow x=\frac{3}{5}\)
(Bài giải gồm toàn dấu tương đương nên khỏi cần ĐKXĐ ha!! )
b) \(\sqrt{x^2-10x+25}=4\Leftrightarrow\sqrt{\left(x-5\right)^2}=4\Leftrightarrow\left|x-5\right|=4\)
\(\Leftrightarrow\orbr{\begin{cases}x-5=4\\x-5=-4\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=9\\x=1\end{cases}}\)
3/ a) \(A=\left(\frac{1}{\sqrt{x}-1}-\frac{1}{\sqrt{x}}\right)\left(\frac{\sqrt{x}+1}{\sqrt{x}-2}-\frac{\sqrt{x}+2}{\sqrt{x}-1}\right)\)
\(=\frac{\sqrt{x}-\left(\sqrt{x}-1\right)}{\sqrt{x}.\left(\sqrt{x}-1\right)}.\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)-\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\)
\(=\frac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}.\frac{x-1-\left(x-4\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}=\frac{3}{\sqrt{x}\left(\sqrt{x}-1\right)^2\left(\sqrt{x}-2\right)}\)
???? Đề khó hiểu vậy ?? Phải là dấu chia ở giữa mới đúng chứ ??
b) \(A>\frac{1}{6}\Leftrightarrow\sqrt{x}\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)^2< 18\Leftrightarrow\left(x-2\sqrt{x}\right)\left(x-2\sqrt{x}+1\right)< 18\)
\(\Leftrightarrow\frac{-1-\sqrt{73}}{2}< x-2\sqrt{x}< \frac{-1+\sqrt{73}}{2}\)
??? Bó tay, đề kinh quá ???