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a) \(A=\left|2-\sqrt{5}\right|+\left|2\sqrt{2}-\sqrt{5}\right|\)
\(=\sqrt{5}-2+2\sqrt{2}-\sqrt{5}=2\sqrt{2}-2\)
b) \(B=\left|\sqrt{7}-2\sqrt{2}\right|+\left|3-2\sqrt{2}\right|\)
\(=2\sqrt{2}-7+3-2\sqrt{2}=-4\)
c) \(C=\sqrt{9+6\sqrt{2}+2}-\sqrt{9-6\sqrt{2}+2}\)
\(=\sqrt{\left(\sqrt{2}+3\right)^2}-\sqrt{\left(3-\sqrt{2}\right)^2}=\left(3+\sqrt{2}\right)-\left(3-\sqrt{2}\right)=2\sqrt{2}\)
d) \(D=\sqrt{9+12\sqrt{2}+8}+\sqrt{9-12\sqrt{2}+8}\)
\(=\sqrt{\left(3+2\sqrt{2}\right)^2}+\sqrt{\left(3-2\sqrt{2}\right)^2}=\left(3+2\sqrt{2}\right)-\left(3-2\sqrt{2}\right)=4\sqrt{2}\)
a, Ta có : \(x=\sqrt{3+2\sqrt{2}}+\sqrt{11-6\sqrt{2}}\)
\(=\sqrt{\left(\sqrt{2}+1\right)^2}+\sqrt{\left(3-\sqrt{2}\right)^2}=4\)
Thay x = 4 => \(\sqrt{x}=2\) vào B ta được :
\(B=\frac{2+5}{2-3}=-7\)
b, Ta có : Với \(x\ge0;x\ne9\)
\(A=\frac{4}{\sqrt{x}+3}+\frac{2x-\sqrt{x}-13}{x-9}-\frac{\sqrt{x}}{\sqrt{x}-3}\)
\(=\frac{4\left(\sqrt{x}-3\right)+2x-\sqrt{x}-13-\sqrt{x}\left(\sqrt{x}+3\right)}{x-9}\)
\(=\frac{4\sqrt{x}-12+2x-\sqrt{x}-13-x-3\sqrt{x}}{x-9}=\frac{x-25}{x-9}\)
Lại có \(P=\frac{A}{B}\Rightarrow P=\frac{\frac{x-25}{x-9}}{\frac{\sqrt{x}+5}{\sqrt{x}-3}}=\frac{\sqrt{x}-5}{\sqrt{x}+3}\)
1.
\(a.\sqrt{6+2\sqrt{5}}+\sqrt{6-2\sqrt{5}}=\sqrt{\left(\sqrt{5}+1\right)^2}+\sqrt{\left(\sqrt{5}-1\right)^2}=\sqrt{5}+1+\sqrt{5}-1=2\sqrt{5}\)
\(b.\sqrt{3+2\sqrt{2}}+\sqrt{6-4\sqrt{2}}=\sqrt{\left(\sqrt{2}+1\right)^2}+\sqrt{\left(2-\sqrt{2}\right)^2}=\sqrt{2}+1+2-\sqrt{2}=3\)\(c.\sqrt{11+6\sqrt{2}}-\sqrt{11-6\sqrt{2}}=\sqrt{\left(3+\sqrt{2}\right)^2}-\sqrt{\left(3-\sqrt{2}\right)^2}=3+\sqrt{2}-3+\sqrt{2}=2\sqrt{2}\)
\(d.\sqrt{3+\sqrt{5}}-\sqrt{3-\sqrt{5}}=\dfrac{\sqrt{6+2\sqrt{5}}-\sqrt{6-2\sqrt{5}}}{\sqrt{2}}=\dfrac{\sqrt{\left(\sqrt{5}+1\right)^2}-\sqrt{\left(\sqrt{5}-1\right)^2}}{\sqrt{2}}=\dfrac{\sqrt{5}+1-\sqrt{5}+1}{\sqrt{2}}=\dfrac{2}{\sqrt{2}}=\sqrt{2}\)
2.
\(a.x-1=\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)\)
\(b.x+5\sqrt{x}+6=x+2\sqrt{x}+3\sqrt{x}+6=\sqrt{x}\left(\sqrt{x}+2\right)+3\left(\sqrt{x}+2\right)=\left(\sqrt{x}+2\right)\left(\sqrt{x}+3\right)\)( mạo danh sửa đề)
\(c.x-4=\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)\)
\(1a.\sqrt{6+2\sqrt{5}}+\sqrt{6-2\sqrt{5}}=\sqrt{5+2\sqrt{5}+1}+\sqrt{5-2\sqrt{5}+1}=\sqrt{5}+1+\sqrt{5}-1=2\sqrt{5}\)
\(b.\sqrt{3+2\sqrt{2}}+\sqrt{6-4\sqrt{2}}=\sqrt{2+2\sqrt{2}+1}+\sqrt{4-2.2\sqrt{2}+2}=\sqrt{2}+1+2-\sqrt{2}=3\)\(c.\sqrt{11+6\sqrt{2}}-\sqrt{11-6\sqrt{2}}=\sqrt{9+2.3\sqrt{2}+2}-\sqrt{9-2.3\sqrt{2}+2}=3+\sqrt{2}-3+\sqrt{2}=2\sqrt{2}\)\(d.\sqrt{3+\sqrt{5}}-\sqrt{3-\sqrt{5}}=\dfrac{\sqrt{6+2\sqrt{5}}-\sqrt{6-2\sqrt{5}}}{\sqrt{2}}=\dfrac{\sqrt{5+2\sqrt{5}+1}-\sqrt{5-2\sqrt{5}+1}}{\sqrt{2}}=\dfrac{\sqrt{5}+1-\sqrt{5}+1}{\sqrt{2}}=\sqrt{2}\)\(2a.x-1=\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)\)
\(b.x+5\sqrt{x}+6=x+2\sqrt{x}+3\sqrt{x}+6=\sqrt{x}\left(\sqrt{x}+2\right)+3\left(\sqrt{x}+2\right)=\left(\sqrt{x}+2\right)\left(\sqrt{x}+3\right)\)
\(c.x-4=\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)\)
Bài 1:
a) \(=5.|2a|-5a^2\)
b) \(=7\left(a-1\right)+5a=12a-7\)
c) \(|a-2|-5\sqrt{a+2}\)
Bài 2:
a) \(=3-\sqrt{2}+5-\sqrt{2}=8-2\sqrt{2}\)
b) \(=3+\sqrt{2}-\left(3-\sqrt{2}\right)\)
\(=2\sqrt{2}\)
c) \(=6-\sqrt{5}-\left(6+\sqrt{5}\right)\)
\(=-2\sqrt{5}\)
a) \(5\sqrt{4a^2}-5a^2\)
\(=5.|2a|-5a^2\)
b) \(7\sqrt{\left(a-1\right)^2}+5a\)
\(=7\left(a-1\right)+5a\)
\(=12a-7\)
c) \(\sqrt{\left(2-a\right)^2}-5\sqrt{a+2}\)
\(=|a-2|-5\sqrt{a+2}\)
bài 2:
a)\(\sqrt{\left(3-\sqrt{2}\right)^2}+\sqrt{\left(\sqrt{2}-5\right)^2}\)
\(=3-\sqrt{2}+5-\sqrt{2}\)
\(=8-2\sqrt{2}\)
b) \(\sqrt{11+6\sqrt{2}}-\sqrt{11-6\sqrt{2}}\)
\(=3+\sqrt{2}-\left(3-\sqrt{2}\right)\)
\(=2\sqrt{2}\)
c)\(\sqrt{41-12\sqrt{5}}-\sqrt{41+12\sqrt{5}}\)
\(=6-\sqrt{5}-\left(6+\sqrt{5}\right)\)
\(=-2\sqrt{5}\)
\(A=\left(2-\sqrt{3}\right)\sqrt{4+2.2.\sqrt{3}+3}=\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)=1\)
các câu còn lại làm tương tự nhé bạn !
a) x = 16 (tm) => A = \(\frac{\sqrt{16}-2}{\sqrt{16}+1}=\frac{4-2}{4+1}=\frac{2}{5}\)
b) B = \(\left(\frac{1}{\sqrt{x}+5}-\frac{x+2\sqrt{x}-5}{25-x}\right):\frac{\sqrt{x}+2}{\sqrt{x}-5}\)
B = \(\frac{\sqrt{x}-5+x+2\sqrt{x}-5}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}\cdot\frac{\sqrt{x}-5}{\sqrt{x}+2}\)
B = \(\frac{x+3\sqrt{x}-10}{\left(\sqrt{x}+5\right)\left(\sqrt{x}+2\right)}\)
B = \(\frac{x+5\sqrt{x}-2\sqrt{x}-10}{\left(\sqrt{x}+5\right)\left(\sqrt{x}+2\right)}\)
B = \(\frac{\left(\sqrt{x}+5\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+5\right)\left(\sqrt{x}+2\right)}=\frac{\sqrt{x}-2}{\sqrt{x}+2}\)
c) P = \(\frac{B}{A}=\frac{\sqrt{x}-2}{\sqrt{x}+2}:\frac{\sqrt{x}-2}{\sqrt{x}+1}=\frac{\sqrt{x}+1}{\sqrt{x}+2}\)
=> \(P\left(\sqrt{x}+2\right)\ge x+6\sqrt{x}-13\)
<=> \(\frac{\sqrt{x}+1}{\sqrt{x}+2}.\left(\sqrt{x}+2\right)-x-6\sqrt{x}+13\ge0\)
<=> \(-x-6\sqrt{x}+13+\sqrt{x}+1\ge0\)
<=> \(-x-5\sqrt{x}+14\ge0\)
<=> \(x+5\sqrt{x}-14\le0\)
<=> \(x+7\sqrt{x}-2\sqrt{x}-14\le0\)
<=> \(\left(\sqrt{x}+7\right)\left(\sqrt{x}-2\right)\le0\)
Do \(\sqrt{x}+7>0\) với mọi x => \(\sqrt{x}-2\le0\)
<=> \(\sqrt{x}\le2\) <=> \(x\le4\)
Kết hợp với Đk: x \(\ge\)0; x \(\ne\)4; x \(\ne\)25
và x thuộc Z => x = {0; 1; 2; 3}
d) M = \(3P\cdot\frac{\sqrt{x}+2}{x+\sqrt{x}+4}\) <=>M = \(3\cdot\frac{\sqrt{x}+1}{\sqrt{x}+2}\cdot\frac{\sqrt{x}+2}{x+\sqrt{x}+4}\)
M = \(\frac{3\sqrt{x}+3}{x+\sqrt{x}+4}=\frac{x+\sqrt{x}+4-x+2\sqrt{x}-1}{\left(x+\sqrt{x}+\frac{1}{4}\right)+\frac{15}{4}}=1-\frac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}+\frac{1}{2}\right)^2+\frac{15}{4}}\le1\)(Do \(\left(\sqrt{x}-1\right)^2\ge0\) và \(\left(\sqrt{x}+\frac{1}{2}\right)^2+\frac{15}{4}>0\))
Dấu "=" xảy ra <=> \(\sqrt{x}-1=0\) <=> \(x=1\)
Vậy MaxM = 1 khi x = 1
Đáp án đúng : C