a: \(A=\dfrac{1}{2a-1}\cdot\sqrt{5a^2}\cdot\left|2a-1\right|\)

\(=\dfrac{2a-1}{2a-1}\cdot a\sqrt{5}=a\sqrt{5}\)(do a>1/2)

b: \(A=\dfrac{\sqrt{x-1-2\sqrt{x-1}+1}}{\sqrt{x-1}-1}+\dfrac{\sqrt{x-1+2\sqrt{x-1}+1}}{\sqrt{x-1}+1}\)

\(=\dfrac{\left|\sqrt{x-1}-1\right|}{\sqrt{x-1}-1}+\dfrac{\sqrt{x-1}+1}{\sqrt{x-1}+1}\)

\(=\dfrac{\sqrt{x-1}-1}{\sqrt{x-1}-1}+1=1+1=2\)

c:

\(=\dfrac{a+b}{b^2}\cdot\dfrac{ab^2}{a+b}=a\)

d: Sửa đề: \(A=\left(\dfrac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right)\left(\dfrac{1-\sqrt{a}}{1-a}\right)^2\)

\(=\left(1+\sqrt{a}+a+\sqrt{a}\right)\cdot\left(\dfrac{1}{1+\sqrt{a}}\right)^2\)

\(=\dfrac{\left(\sqrt{a}+1\right)^2}{\left(\sqrt{a}+1\right)^2}=1\)

e:

\(A=\dfrac{x-1}{\sqrt{y}-1}\cdot\sqrt{\dfrac{\left(\sqrt{y}-1\right)^2}{\left(x-1\right)^4}}\)

\(=\dfrac{x-1}{\sqrt{y}-1}\cdot\dfrac{\sqrt{y}-1}{\left(x-1\right)^2}=\dfrac{1}{x-1}\)

f:

\(A=\sqrt{\dfrac{m}{\left(1-x\right)^2}\cdot\dfrac{4m\left(1-2x+x^2\right)}{81}}\)

\(=\sqrt{\dfrac{m}{\left(x-1\right)^2}\cdot\dfrac{4m\left(x-1\right)^2}{81}}\)

\(=\sqrt{\dfrac{4m^2}{81}}=\dfrac{2m}{9}\)

\(A=\left|a-3\right|-3a=3-a-3a=3-4a\)

\(B=4a+3-\left|2a-1\right|=4a+3-2a+1=2a+4\)

\(C=\dfrac{4}{a^2-4}\left|a-2\right|=\dfrac{-4\left(a-2\right)}{\left(a-2\right)\left(a+2\right)}=\dfrac{-4}{a+2}\)

\(D=\dfrac{a^2-9}{12}:\sqrt{\dfrac{\left(a+3\right)^2}{16}}=\dfrac{a^2-9}{12}:\dfrac{\left|a+3\right|}{4}=\dfrac{\left(a-3\right)\left(a+3\right).4}{-12\left(a+3\right)}=\dfrac{3-a}{3}\)

\(A=\sqrt{\left(a-3\right)^2}-3a\)

=3-a-3a

=3-4a

Sai đề ở vế phải. Cái này tôi làm rồi nên biết:  819598 (học 24)

BDT cần cm tương đương

\(\frac{\left(2+6a+3b+6\sqrt{2bc}\right)\left(\sqrt{2b^2+2\left(a+c\right)^2}+3\right)}{2a+b+2\sqrt{2bc}}\ge16\)

Áp dụng bdt C-S và AM-GM:

\(VT=\frac{\left(2+6a+3b+6\sqrt{2bc}\right)\left(\sqrt{2b^2+2\left(a+c\right)^2}+3\right)}{2a+b+2\sqrt{2bc}}\)

\(=\left(\frac{2}{2a+b+2\sqrt{2bc}}+3\right)\left(\sqrt{2\left(b^2+\left(a+c\right)^2\right)}+3\right)\)

\(\ge\left(\sqrt{2\cdot\frac{\left(a+b+c\right)^2}{2}}+3\right)\left(\frac{2}{2a+b+b+2c}+3\right)\)

\(=\left(a+b+c+3\right)\left(\frac{1}{a+b+c}+3\right)\)

\(\ge\left(3+1\right)^2=16=VP\)

dau '=' khi a+b+c=1, b=a+c, 2c=b bn tự giải not

Chuyên toán Vĩnh Phúc đây mà :) Em chụp lại nha,chớ e mà viết ra nhiều người nhảy vào cà khịa ghê lắm:(

Viết BĐT về dạng \(\frac{2}{2a+b+2\sqrt{2bc}}-\frac{16}{\sqrt{2b^2+2\left(a+c\right)^2}+3}\ge0\)

Ta có: \(\frac{2}{2a+b+2\sqrt{2bc}}\ge\frac{2}{2a+b+b+2c}=\frac{1}{a+b+c}\)

Đẳng thức xảy ra <=> b=2c

Áp dụng BĐT Cauchy-Schwarz ta có:

\(\left(a+b+c\right)^2\le\left(1+1\right)\left[\left(a+c\right)^2+b^2\right]\)

\(\Rightarrow a+b+c\le\sqrt{2\left(a+c\right)^2+2b^2}\)

\(\Rightarrow-\frac{16}{\sqrt{2b^2+2\left(a+c\right)^2}+3}\ge-\frac{16}{a+b+c+3}\)

Đẳng thức xảy ra <=> a+c=b

\(\Rightarrow\frac{2}{2a+b+2\sqrt{bc}}-\frac{16}{\sqrt{2b^2+2\left(a+c\right)^2}+3}+3\ge\frac{1}{a+b+c}-\frac{16}{a+b+c+3}+3\)

\(=\frac{3\left(a+b+c-1\right)^2}{\left(a+b+c\right)\left(a+b+c+3\right)}\ge0\)

Đẳng thức xảy ra \(\Leftrightarrow\hept{\begin{cases}a+b+c-1=0\\b=2c\\a+c=b\end{cases}\Leftrightarrow\hept{\begin{cases}a=c=\frac{1}{4}\\b=\frac{1}{2}\end{cases}}}\)

\(\frac{a-b}{4b^2}\cdot\sqrt{\frac{4a^2b^4}{a^2-2ab+b^2}}\)

\(=\frac{a-b}{4b^2}\cdot\sqrt{\frac{\left(2ab^2\right)^2}{\left(a-b\right)^2}}\)

\(=\frac{a-b}{4b^2}\cdot\frac{2ab}{a-b}\)

\(=\frac{a}{2b}\)

Biểu thức b chắc ghi nhầm, 1 căn dấu trừ thì hợp lý

\(a^3=6+3a.\sqrt[3]{9-4.2}=3a+6\Rightarrow a^3-3a=6\)

\(b^3=34+3b.\sqrt{17^2-12^2.2}=3b+34\Rightarrow b^3-3b=34\)

\(\Rightarrow A=a^3-3a+b^3-3b=6+34=40\)

2/ \(\Leftrightarrow\left\{{}\begin{matrix}2y^2-x^2=1\\2x^3-y^3=1.\left(2y-x\right)\end{matrix}\right.\)

\(\Rightarrow2x^3-y^3=\left(2y^2-x^2\right)\left(2y-x\right)\)

\(\Leftrightarrow x^3+2x^2y+2xy^2-5y^3=0\)

\(\Leftrightarrow\left(x-y\right)\left(x^2+3xy+5y^2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=y\Rightarrow2x^2-x^2=1\Rightarrow...\\x^2+3xy+5y^2=0\left(1\right)\end{matrix}\right.\)

Xét (1): \(\Leftrightarrow\left(x+\frac{3y}{2}\right)^2+\frac{11y^2}{4}=0\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=0\end{matrix}\right.\) thay vào hệ ko thỏa mãn (loại)

\(\frac{1}{m}+\frac{1}{n}=\frac{1}{2}\Leftrightarrow2\left(m+n\right)=mn\)

\(\left\{{}\begin{matrix}\Delta_1=m^2-4n\\\Delta_2=n^2-4m\end{matrix}\right.\)

\(\Rightarrow P=\Delta_1+\Delta_2=m^2+m^2-4\left(m+n\right)\)

\(=m^2+n^2-2mn=\left(m-n\right)^2\ge0\)

\(\Rightarrow\) Luôn có ít nhất 1 trong 2 giá trị \(\Delta_1\) hoặc \(\Delta_2\) không âm nên luôn có ít nhất 1 trong 2 pt trên có nghiệm \(\Rightarrow\) pt luôn luôn có nghiệm

\(P=\sum\frac{\sqrt{3}.a}{\sqrt{3a}.\sqrt{2b+2c-a}}\ge\sum\frac{2\sqrt{3}a}{3a+2b+2c-a}=\sum\frac{2\sqrt{3}a}{2\left(a+b+c\right)}=\sum\frac{\sqrt{3}a}{a+b+c}=\sqrt{3}\)

Dấu "=" xảy ra khi \(a=b=c\) hay tam giác đã cho đều

1)

Ta có: \(M=\Sigma_{cyc}\frac{\sqrt{3}\left(a+b+4c\right)}{\sqrt{3\left(a+b\right)\left(a+b+4c\right)}}\ge\Sigma_{cyc}\frac{\sqrt{3}\left(a+b+4c\right)}{\frac{3\left(a+b\right)+\left(a+b+4c\right)}{2}}=\Sigma_{cyc}\frac{\sqrt{3}\left(a+b+4c\right)}{2\left(a+b+c\right)}=3\sqrt{3}\)

Dấu "=" xảy ra khi a=b=c

2)

\(\Sigma_{cyc}\sqrt[3]{\left(\frac{2a}{ab+1}\right)^2}=\Sigma_{cyc}\frac{2a}{\sqrt[3]{2a\left(ab+1\right)^2}}\ge\Sigma_{cyc}\frac{2a}{\frac{2a+\left(ab+1\right)+\left(ab+1\right)}{3}}=3\Sigma_{cyc}\frac{a}{ab+a+1}\)

Ta có bổ đề: \(\frac{a}{ab+a+1}+\frac{b}{bc+b+1}+\frac{c}{ca+c+1}=1\left(abc=1\right)\)

\(\Rightarrow\Sigma_{cyc}\sqrt[3]{\left(\frac{2a}{ab+1}\right)^2}\ge3\)