b) ĐK: \(a\ge0,a\ne6\)

\(\frac{\left(3-\sqrt{a}\right)\left(3+\sqrt{a}\right)}{\sqrt{a}+3}-\frac{\left(\sqrt{a}-3\right)^2}{\sqrt{a}-3}-6\)

=\(\left(3-\sqrt{a}\right)-\left(\sqrt{a}-3\right)-6=3-\sqrt{a}-\sqrt{a}+3-6\)

\(=-2\sqrt{a}\)

\(a+b+c=3\)

\(\Leftrightarrow\left(a+b+c\right)^2=9\)

\(\Leftrightarrow a^2+b^2+c^2+2\left(ab+bc+ca\right)=9\)

\(\Rightarrow a^2+c^2+b^2=9-2\left(ab+bc+ca\right)\)

Nếu \(ab+bc+ca>3\) thì \(a^2+b^2+c^2< 3\left(vl\right)\)

\(\Rightarrow ab+bc+ca\le3\)

\(3^{2012}-1=\left(4-1\right)^{2012}-1=BS4^{2012}+1-1\)

\(=BS4^{2012}=BS2^{2014}⋮2^{2014}\)

ĐPCM

\(a,\)\(\left(2-\frac{\sqrt{a}\left(\sqrt{a}-3\right)}{\sqrt{a}-3}\right).\left(2-\frac{\sqrt{a}\left(b+5\right)}{\sqrt{b}-5}\right).\)

\(=\left(2-\sqrt{a}\right)\left(\frac{2\sqrt{b}-10-\sqrt{ab}-5\sqrt{a}}{\sqrt{b}-5}\right)\)

\(=\left(2-\sqrt{a}\right)\left(\frac{2\left(\sqrt{b}-5\right)-\sqrt{a}\left(\sqrt{b}-5\right)}{\sqrt{b}-5}\right)\)

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

\(=a-4\sqrt{a}+4\)

\(b,\frac{9-a}{\sqrt{a}+3}-\frac{9-6\sqrt{a}+a}{\sqrt{a}-3}-6\)

\(=\frac{\left(3-\sqrt{a}\right)\left(3+\sqrt{a}\right)}{\sqrt{a}+3}-\frac{\left(\sqrt{a}-3\right)^2}{\sqrt{a}-3}-6\)

\(=3-\sqrt{a}-\left(\sqrt{a}-3\right)-6\)

\(=-2\sqrt{a}\)

Lời giải :

a) \(A=3\sqrt{x-1}+7\ge7\forall x\)

Dấu "=" xảy ra \(\Leftrightarrow x=1\)

b) \(B=\frac{4}{\sqrt{x}+3}\le\frac{4}{3}\forall x\)

Dấu "=" xảy ra \(\Leftrightarrow x=0\)

c) \(C=\frac{3\sqrt{x}+8}{\sqrt{x}+3}=\frac{3\left(\sqrt{x}+3\right)-1}{\sqrt{x}+3}=3-\frac{1}{\sqrt{x}+3}\)

Có \(\frac{1}{\sqrt{x}+3}\le\frac{1}{3}\forall x\)

\(\Leftrightarrow-\frac{1}{\sqrt{x}+3}\ge\frac{-1}{3}\)

\(\Leftrightarrow3-\frac{1}{\sqrt{x}+3}\ge3-\frac{1}{3}=\frac{8}{3}\)

\(\Leftrightarrow C\ge\frac{8}{3}\forall x\)

Dấu "=" xảy ra \(\Leftrightarrow x=0\)

d) \(D=x-3\sqrt{x}+2\)

\(D=\left(\sqrt{x}\right)^2-2\cdot\sqrt{x}\cdot\frac{3}{2}+\frac{9}{4}-\frac{1}{4}\)

\(D=\left(\sqrt{x}-\frac{3}{2}\right)^2-\frac{1}{4}\ge\frac{-1}{4}\forall x\)

Dấu "=" xảy ra \(\Leftrightarrow\sqrt{x}=\frac{3}{2}\Leftrightarrow x=\frac{9}{4}\)

e) \(E=\frac{4}{x-2\sqrt{x}+3}=\frac{4}{\left(\sqrt{x}-1\right)^2+2}\le\frac{4}{2}=2\forall x\)

Dấu "=" xảy ra \(\Leftrightarrow\sqrt{x}=1\Leftrightarrow x=1\)

a) Vì \(3\sqrt{x-1}\ge0\forall x\ge1\) 

 \(\Rightarrow3\sqrt{x-1}+7\ge7\forall x\ge1\) 

Dấu "=" xảy ra <=>\(3\sqrt{x-1}=0\Leftrightarrow\sqrt{x-1}=0\Leftrightarrow x-1=0\Leftrightarrow x=1\) 

Vậy A_min =7 tại x=1

Thuy Duong Nguyen đánh đề cẩn thận hơn bạn nhé

Lời giải :

a) ĐKXĐ : \(x\ne1\)

 \(A=\frac{15\sqrt{x}-11}{x+2\sqrt{x}-3}+\frac{3\sqrt{x}-2}{1-\sqrt{x}}-\frac{2\sqrt{x}+3}{\sqrt{x}+3}\)

\(A=\frac{15\sqrt{x}-11}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}+\frac{\left(\sqrt{x}+3\right)\left(2-3\sqrt{x}\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}-\frac{\left(2\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)

\(A=\frac{15\sqrt{x}-11-3x+6-7\sqrt{x}-2x-\sqrt{x}+3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)

\(A=\frac{-5x+7\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)

\(A=\frac{\left(\sqrt{x}-1\right)\left(-5\sqrt{x}+2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)

\(A=\frac{2-5\sqrt{x}}{\sqrt{x}+3}\)

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

\(\Leftrightarrow\sqrt{x}=\sqrt{2}-1\)

Khi đó \(A=\frac{2-5\left(\sqrt{2}-1\right)}{\sqrt{2}-1+3}\)

\(A=\frac{2-5\sqrt{2}+5}{\sqrt{2}+2}=\frac{7-5\sqrt{2}}{\sqrt{2}+2}\)

c) \(A=\frac{1}{2}\)

\(\Leftrightarrow\frac{2-5\sqrt{x}}{\sqrt{x}+3}=\frac{1}{2}\)

\(\Leftrightarrow2\left(2-5\sqrt{x}\right)=\sqrt{x}+3\)

\(\Leftrightarrow4-10\sqrt{x}-\sqrt{x}-3=0\)

\(\Leftrightarrow1-11\sqrt{x}=0\)

\(\Leftrightarrow11\sqrt{x}=1\)

\(\Leftrightarrow\sqrt{x}=\frac{1}{11}\)

\(\Leftrightarrow x=\frac{1}{121}\)( thỏa )

d) A nguyên \(\Leftrightarrow2-5\sqrt{x}⋮\sqrt{x}+3\)

\(\Leftrightarrow-5\left(\sqrt{x}+3\right)+17⋮\sqrt{x}+3\)

Vì \(-5\left(\sqrt{x}+3\right)⋮\sqrt{x}+3\)

\(\Rightarrow17⋮\sqrt{x}+3\)

\(\Rightarrow\sqrt{x}+3\inƯ\left(17\right)=\left\{17\right\}\)( vì \(\sqrt{x}+3\ge3\))

\(\Leftrightarrow\sqrt{x}=14\)

\(\Leftrightarrow x=196\)( thỏa )

Vậy....

\(a,ĐKXĐ:\orbr{\begin{cases}x+2\sqrt{x}+3\ne0\\\sqrt{x}+3\ne0\end{cases}}\)

\(\Leftrightarrow\orbr{ }\sqrt{x}\ne-3\)

Rút gọn: p/s: sau phân số thứ 2 ở mẫu ko có x à? Bạn chép đề sai?