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\(A^2=\left(2\sqrt{x-4}+\sqrt{8-x}\right)^2\le\left(2^2+1^2\right)\left(x-4+8-x\right)=20..\)
\(A\le2\sqrt{5}..\)
Bài 2 :
Tìm min : Bình phương
Tìm max : Dùng B.C.S ( bunhiacopxki )
Bài 3 : Dùng B.C.S
KP9
nói thế thì đừng làm cho nhanh bạn ạ
Người ta cũng có chút tôn trọng lẫn nhau nhé đừng có vì dăm ba cái tích
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)
\(c.x-4=\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)\)
\(1a.\sqrt{6+2\sqrt{5}}+\sqrt{6-2\sqrt{5}}=\sqrt{5+2\sqrt{5}+1}+\sqrt{5-2\sqrt{5}+1}=\sqrt{5}+1+\sqrt{5}-1=2\sqrt{5}\)
\(b.\sqrt{3+2\sqrt{2}}+\sqrt{6-4\sqrt{2}}=\sqrt{2+2\sqrt{2}+1}+\sqrt{4-2.2\sqrt{2}+2}=\sqrt{2}+1+2-\sqrt{2}=3\)\(c.\sqrt{11+6\sqrt{2}}-\sqrt{11-6\sqrt{2}}=\sqrt{9+2.3\sqrt{2}+2}-\sqrt{9-2.3\sqrt{2}+2}=3+\sqrt{2}-3+\sqrt{2}=2\sqrt{2}\)\(d.\sqrt{3+\sqrt{5}}-\sqrt{3-\sqrt{5}}=\dfrac{\sqrt{6+2\sqrt{5}}-\sqrt{6-2\sqrt{5}}}{\sqrt{2}}=\dfrac{\sqrt{5+2\sqrt{5}+1}-\sqrt{5-2\sqrt{5}+1}}{\sqrt{2}}=\dfrac{\sqrt{5}+1-\sqrt{5}+1}{\sqrt{2}}=\sqrt{2}\)\(2a.x-1=\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)\)
\(b.x+5\sqrt{x}+6=x+2\sqrt{x}+3\sqrt{x}+6=\sqrt{x}\left(\sqrt{x}+2\right)+3\left(\sqrt{x}+2\right)=\left(\sqrt{x}+2\right)\left(\sqrt{x}+3\right)\)
\(c.x-4=\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)\)
Lời giải:
Bạn cứ nhớ công thức $\sqrt{x^2}=|x|$, rồi dùng điều kiện đề bài để phá dấu trị tuyệt đối là được
a)
\(\sqrt{16a^2}-5a=\sqrt{(4a)^2}-5a=|4a|-5a=4a-5a=-a\)
b)
\(3x+2-\sqrt{9x^2+6x+1}=3x+2-\sqrt{(3x)^2+2.3x.1+1^2}\)
\(=3x+2-\sqrt{(3x+1)^2}=3x+2-|3x+1|=3x+2-(3x+1)=1\)
c)
\(\sqrt{8+2\sqrt{7}}-\sqrt{7}=\sqrt{7+1+2.\sqrt{7}.\sqrt{1}}-\sqrt{7}\)
\(=\sqrt{(\sqrt{7}+1)^2}-\sqrt{7}=|\sqrt{7}+1|-\sqrt{7}=\sqrt{7}+1-\sqrt{7}=1\)
d)
\(\sqrt{14-2\sqrt{13}}+\sqrt{14+2\sqrt{13}}=\sqrt{13+1-2\sqrt{13}}+\sqrt{13+1+2\sqrt{13}}\)
\(=\sqrt{(\sqrt{13}-1)^2}+\sqrt{(\sqrt{13}+1)^2}=|\sqrt{13}-1|+|\sqrt{13}+1|\)
\(=\sqrt{13}-1+\sqrt{13}+1=2\sqrt{13}\)
e)
\(2x-\sqrt{4x^2-4x+1}=2x-\sqrt{(2x-1)^2}=2x-|2x-1|=2x-(2x-1)=1\)
g)
\(|x-2|+\frac{\sqrt{x^2-4x+4}}{x-2}=|x-2|+\frac{\sqrt{(x-2)^2}}{x-2}=|x-2|+\frac{|x-2|}{x-2}\)
\(=(x-2)+\frac{(x-2)}{x-2}=x-2+1=x-1\)
1 ) \(A=\sqrt{x-2}+\sqrt{4-x}\)
ĐKXĐ : \(2\le x\le4\)
\(\Rightarrow A^2=x-2+4-x+2\sqrt{\left(x-2\right)\left(4-x\right)}=2+2\sqrt{\left(x-2\right)\left(4-x\right)}\)
Áp dụng bđt AM - GM ta có :
\(2\sqrt{\left(x-2\right)\left(4-x\right)}\le x-2+4-x=2\)
\(\Rightarrow A^2\le2+2=4\Rightarrow-2\le A\le2\)
Mà A > 0 nên ko thể có min = - 2 nên \(2\le x\le4\) ta chọn x = 2
=> A = \(\sqrt{2}\)
Vậy \(\sqrt{2}\le A\le2\)
1.
\(A=-3x+2\sqrt{x}+6\\ =-3\left(x-\frac{2}{3}\sqrt{x}-2\right)\\ =-3\left[\left(\sqrt{x}\right)^2-2\cdot\sqrt{x}\cdot\frac{1}{3}+\left(\frac{1}{3}\right)^2-\frac{19}{9}\right]\\ =-3\left[\left(\sqrt{x}-\frac{1}{3}\right)^2\right]+\frac{19}{3}\le\frac{19}{3}\forall x\ge0\)
Vậy Max A = \(\frac{19}{3}\Leftrightarrow x=\frac{1}{9}\)
2.
\(\sqrt[3]{8}-\sqrt[3]{x}=-2\Leftrightarrow2-\sqrt[3]{x}=-2\\ \Leftrightarrow\sqrt[3]{x}=-4\\ \Leftrightarrow\left(\sqrt[3]{x}\right)^3=\left(-4\right)^3=-64\\ \Leftrightarrow x=-64\left(tm\right)\)