\(a,\sqrt{4x^4}+6x^2=2x^2+6x^2=8x^2\)

\(b,\sqrt{25a^4}-2a^2=5a^2-2a^2=3a^2\)

\(c,\sqrt{36a^4}+8a=6a^2+8a\)

\(d,\sqrt{\left(x-3\right)^4}-x^2+3x-1=\left(x-3\right)^2-x^2+3x-1=x^2-6x+9-x^2+3x-1=-3x+8\)

A = \(x^2+3x-7=x^2+2x\frac{3}{2}+\frac{9}{4}-\frac{37}{4}\)

\(=\left(x+\frac{3}{2}\right)^2-\frac{37}{4}\ge-\frac{37}{4}\)

\(\Rightarrow\)min A = \(-\frac{37}{4}\Leftrightarrow x=-\frac{3}{2}\)

B = \(x-5\sqrt{x}-1\) ĐKXĐ: \(x\ge0\)

\(=x-2\sqrt{x}\frac{5}{2}+\frac{25}{4}-\frac{29}{4}=\left(\sqrt{x}-\frac{5}{2}\right)^2-\frac{29}{4}\ge-\frac{29}{4}\)

\(\Rightarrow\)min B = \(-\frac{29}{4}\Leftrightarrow x=\frac{25}{4}\)( thỏa mãn)

C = \(\frac{-4}{\sqrt{x}+7}\) ĐKXĐ:\(x\ge0\)

Ta có: \(\sqrt{x}+7\ge7\Rightarrow\frac{4}{\sqrt{x}+7}\le\frac{4}{7}\)\(\Leftrightarrow\frac{-4}{\sqrt{x}+7}\ge-\frac{4}{7}\)

\(\Rightarrow\)min C = \(-\frac{4}{7}\Leftrightarrow x=0\)

D = \(\frac{\sqrt{x}+1}{\sqrt{x}+3}\) ĐKXĐ:\(x\ge0\)

\(=1-\frac{2}{\sqrt{x}+3}\ge1-\frac{2}{3}=\frac{1}{3}\)

\(\Rightarrow\)min D = \(\frac{1}{3}\Leftrightarrow x=0\)

E = \(\frac{x+7}{\sqrt{x}+3}\) ĐKXĐ:\(x\ge0\)

\(=\frac{x-9+16}{\sqrt{x}+3}=\frac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)+16}{\sqrt{x}+3}=\sqrt{x}-3+\frac{16}{\sqrt{x}+3}=\sqrt{x}+3+\frac{16}{\sqrt{x}+3}-6\ge2\sqrt{16}-6=2\)

\(\Rightarrow\)min E = \(2\Leftrightarrow x=1\)(thỏa mãn)

F = \(\frac{x^2+3x+5}{x^2}\) ĐKXĐ: \(x\ne0\)

\(\Leftrightarrow\)\(x^2\left(F-1\right)-3x-5=0\)

△ = \(3^2+20\left(F-1\right)\ge0\)\(\Leftrightarrow F\ge\frac{11}{20}\)

\(\Rightarrow\)min F = \(\frac{11}{20}\Leftrightarrow x=-\frac{10}{3}\)( thỏa mãn)

Bài 1:

a) Để căn thức \(\sqrt{\frac{2}{9-x}}\) có nghĩa thì \(\left\{{}\begin{matrix}\frac{2}{9-x}\ge0\\9-x\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}9-x>0\\x\ne9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x< 9\\x\ne9\end{matrix}\right.\Leftrightarrow x< 9\)

b) Ta có: \(x^2+2x+1\)

\(=\left(x+1\right)^2\)

mà \(\left(x+1\right)^2\ge0\forall x\)

nên \(x^2+2x+1\ge0\forall x\)

Do đó: Căn thức \(\sqrt{x^2+2x+1}\) xác được với mọi x

c) Để căn thức \(\sqrt{x^2-4x}\) có nghĩa thì \(x^2-4x\ge0\)

\(\Leftrightarrow x\left(x-4\right)\ge0\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge0\\x-4\ge0\end{matrix}\right.\\\left\{{}\begin{matrix}x< 0\\x-4< 0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge0\\x\ge4\end{matrix}\right.\\\left\{{}\begin{matrix}x< 0\\x< 4\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x\ge4\\x< 0\end{matrix}\right.\)

Bài 3:

a) Ta có: \(\sqrt{\left(3-\sqrt{10}\right)^2}\)

\(=\left|3-\sqrt{10}\right|\)

\(=\sqrt{10}-3\)(Vì \(3< \sqrt{10}\))

b) Ta có: \(\sqrt{9-4\sqrt{5}}\)

\(=\sqrt{5-2\cdot\sqrt{5}\cdot2+4}\)

\(=\sqrt{\left(\sqrt{5}-2\right)^2}\)

\(=\left|\sqrt{5}-2\right|\)

\(=\sqrt{5}-2\)(Vì \(\sqrt{5}>2\))

c) Ta có: \(3x-\sqrt{x^2-2x+1}\)

\(=3x-\sqrt{\left(x-1\right)^2}\)

\(=3x-\left|x-1\right|\)

\(=\left[{}\begin{matrix}3x-\left(x-1\right)\left(x\ge1\right)\\3x-\left(1-x\right)\left(x< 1\right)\end{matrix}\right.\)

\(=\left[{}\begin{matrix}3x-x+1\\3x-1+x\end{matrix}\right.=\left[{}\begin{matrix}2x+1\\4x-1\end{matrix}\right.\)

\(K=\left(\frac{a}{\sqrt{a}\left(\sqrt{a}-1\right)}-\frac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\right):\left(\frac{\sqrt{a}-1}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}+\frac{2}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\right)\)

\(=\left(\frac{a-1}{\sqrt{a}\left(\sqrt{a}-1\right)}\right):\left(\frac{\sqrt{a}+1}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\right)\)

\(=\left(\frac{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}{\sqrt{a}\left(\sqrt{a}-1\right)}\right).\left(\sqrt{a}-1\right)\)

\(=\frac{a-1}{\sqrt{a}}\Rightarrow\left\{{}\begin{matrix}m=1\\n=-1\end{matrix}\right.\Rightarrow m^2+n^2=2\)

\(A=\frac{x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\frac{\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\frac{\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)

\(=\frac{x+2\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\frac{\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)

\(=\frac{\sqrt{x}}{\sqrt{x}-2}\Rightarrow\left\{{}\begin{matrix}m=0\\n=-2\end{matrix}\right.\Rightarrow m-n=2\)

Cảm ơn bạn nha ;)

Lời giải:

a)

\(\sqrt{1-4a+4a^2}-2a=\sqrt{1-2.2a+(2a)^2}-2a\)

\(=\sqrt{(2a-1)^2}-2a=|2a-1|-2a=(2a-1)-2a=-1\)

(do $a\geq \frac{1}{2}$ nên $|2a-1|=2a-1$)

b)

\(x-2y-\sqrt{x^2-4xy+4y^2}=x-2y-\sqrt{(x-2y)^2}=x-2y-|x-2y|\)

\(=x-2y-(2y-x)=2(x-2y)\)

(do $x< 2y$ nên $|x-2y|=-(x-2y)=2y-x$)

c)

\(x^2+\sqrt{x^4-8x^2+16}=x^2+\sqrt{(x^2)^2-2.4.x^2+4^2}\)

\(=x^2+\sqrt{(x^2-4)^2}=x^2+|x^2-4|=x^2+(4-x^2)=4\)

(do $x^2< 4$ nên $|x^2-4|=4-x^2$)

Câu 3:

bạn cứ áp dụng cái \(a^3+b^3+c^3=\left(a+b+c\right)^3-3\left(a+b\right)\left(a+c\right)\left(b+c\right)\)

Câu 4:

từ giả thiết :\(a+b+c+\sqrt{abc}=4\Leftrightarrow\sqrt{abc}=4-a-b-c\Leftrightarrow abc=\left(4-a-b-c\right)^2\)

ta có: \(a\left(4-b\right)\left(4-c\right)=a\left(16-4c-4b+bc\right)=16a-4ac-4ab+abc\)

\(=16a-4ab-4ac+\left[4-\left(a+b+c\right)\right]^2=16a-4ab-4ac+16-8\left(a+b+c\right)+\left(a+b+c\right)^2\)

\(=a^2+b^2+c^2-2ab-2ac+2bc+8a-8b-8c+16\)

\(=\left(a-b-c\right)^2+8\left(a-b-c\right)+16=\left(a-b-c+4\right)^2\)

\(\Rightarrow\sqrt{a\left(4-b\right)\left(4-c\right)}=a-b-c+4\)(vì \(a-b-c+4=a-b-c+a+b+c+\sqrt{abc}=2a+\sqrt{abc}>0\))

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