\(A=\frac{8}{4+2\sqrt{x}}-\frac{2-\sqrt{x}}{4-x}\)

\(=\frac{8}{2\left(2+\sqrt{x}\right)}-\frac{2-\sqrt{x}}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}\)

\(=\frac{4}{2+\sqrt{x}}-\frac{1}{2+\sqrt{x}}\)

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

\(B=\frac{x\sqrt{y}+y\sqrt{x}}{\sqrt{xy}}:\frac{1}{\sqrt{x}-\sqrt{y}}-x\)

\(=\frac{\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{xy}}.\left(\sqrt{x}-\sqrt{y}\right)-x\)

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

<br class="Apple-interchange-newline"><div id="inner-editor"></div>x>2;y>1

Khi đó Pt ⇔36√x−2 +4√x−2+4√y−1 +√y−1=28

theo BĐT Cô si ta có 36√x−2 +4√x−2≥2.√36√x−2 .4√x−2=24

                                  và 4√y−1 +√y−1≥2√4√y−1 .√y−1=4

Pt đã cho có VT>= 28 Dấu "=" xảy ra ⇔

36√x−2 =4√x−2⇔x=11

và 4√y−1 =√y−1⇔y=5

Đối chiếu với ĐK thì x=11; y=5 là nghiệm của PT

\(A=\dfrac{x^{\dfrac{5}{4}}y+xy^{\dfrac{5}{4}}}{\sqrt[4]{x}+\sqrt[4]{y}}\\ =\dfrac{xy\left(x^{\dfrac{1}{4}}+y^{\dfrac{1}{4}}\right)}{x^{\dfrac{1}{4}}+y^{\dfrac{1}{4}}}\\ =xy\)

\(B=\left(\sqrt[7]{\dfrac{x}{y}\sqrt[5]{\dfrac{y}{x}}}\right)^{\dfrac{35}{4}}\\= \left(\sqrt[7]{\dfrac{x}{y}\cdot\left(\dfrac{x}{y}\right)^{-\dfrac{1}{5}}}\right)^{\dfrac{35}{4}}\\ =\left(\sqrt[7]{\left(\dfrac{x}{y}\right)^{\dfrac{4}{5}}}\right)^{\dfrac{35}{4}}\\ =\left[\left(\dfrac{x}{y}\right)^{\dfrac{4}{35}}\right]^{\dfrac{35}{4}}\\ =\left(\dfrac{x}{y}\right)^{\dfrac{4}{35}\cdot\dfrac{35}{4}}\\ =\left(\dfrac{x}{y}\right)^1\\ =\dfrac{x}{y}\)

ĐKXĐ: \(x\ge1\) 

Ta có: \(\frac{x^2-4}{x}+4+\frac{y^2-4}{y}+4=4\left(\sqrt{x-1}+\sqrt{y-1}\right)\)  

Lại có: \(\frac{x^2-4}{x}+4=x+\frac{4x-4}{x}\ge4\sqrt{x-1}\) 

Tương tự: \(\frac{y^2-4}{y}+4\ge4\sqrt{y-1}\) 

Cộng từng vế: \(\frac{x^2-4}{x}+\frac{y^2-4}{y}+8\ge4\left(\sqrt{x-1}+\sqrt{y-1}\right)\) 

Dấu "=" xảy ra khi: x=y=2 

Vậy (x;y)=(2'2) 

ĐKXĐ: \(x\ge1;y\ge1\)

Ta có: \(\frac{x^2-4}{x}+\frac{y^2-4}{y}+8=4\left(\sqrt{x-1}+\sqrt{y-1}\right)\)

\(\Leftrightarrow\frac{x^2-4}{x}+\frac{y^2-4}{y}=4\left[\left(\sqrt{x-1}-1\right)+\left(\sqrt{y-1}+1\right)\right]\)

\(\Leftrightarrow\frac{\left(x-2\right)\left(x+2\right)}{x}+\frac{\left(y-2\right)\left(y+2\right)}{y}=4\left(\frac{x-1-1}{\sqrt{x-1}+1}+\frac{y-1-1}{\sqrt{y-1}+1}\right)\)

\(\Leftrightarrow\left(x-2\right)\left(\frac{x+2}{x}-\frac{4}{\sqrt{x-1}+1}\right)+\left(y-2\right)\left(\frac{y+2}{y}-\frac{4}{\sqrt{y-1}+1}\right)=0\)

\(\Leftrightarrow\left(x-2\right)\frac{x\sqrt{x-1}+2\sqrt{x-1}+2+x-4x}{x\left(\sqrt{x-1}+1\right)}+\left(y-2\right)\frac{y\sqrt{y-1}+2\sqrt{y-1}+y-4y}{y\left(\sqrt{y-1}+1\right)}=0\)

\(\Leftrightarrow\left(x-2\right)\frac{\left( x-1\right)\sqrt{x-1}+3\sqrt{x-1}-3\left(x-1\right)-1}{x\left(\sqrt{x-1}+1\right)}\)

      \(+\left(y-2\right)\frac{\left(y-1\right)\sqrt{y-1}+3\sqrt{y-1}-3\left(y-1\right)-1}{y\left(\sqrt{y-1}+1\right)}=0\)

\(\Leftrightarrow\left(x-2\right)\frac{\left(\sqrt{x-1}-1\right)^3}{x\left(\sqrt{x-1}+1\right)}+\left(y-2\right)\frac{\left(\sqrt{y-1}-1\right)^3}{y\left(\sqrt{y-1}+1\right)}=0\)

\(\Leftrightarrow\left(x-2\right)\frac{\left(\sqrt{x-1}-1\right)^3\left(\sqrt{x-1}+1\right)^3}{x\left(\sqrt{x-1}+1\right)^4}+\left(y-2\right)\frac{\left(\sqrt{y-1}-1\right)^3\left(\sqrt{y-1}+1\right)^3}{y\left(\sqrt{y-1}+1\right)^4}=0\)

\(\Leftrightarrow\frac{\left(x-2\right)^4}{x\left(\sqrt{x-1}+1\right)^4}+\frac{\left(y-2\right)^4}{y\left(\sqrt{y-1}+1\right)^4}=0\)

Vì \(x\ge1;y\ge1\Rightarrow\frac{\left(x-2\right)^4}{x\left(\sqrt{x-1}+1\right)^4}\ge0;\frac{\left(y-2\right)^4}{y\left(\sqrt{y-1}+1\right)^4}\ge0\)\(\Rightarrow\frac{\left(x-2\right)^4}{x\left(\sqrt{x-1}+1\right)^4}+\frac{\left(y-2\right)^4}{y\left(\sqrt{y-1}+1\right)^4}\ge0\)

Do đó dấu ''='' xảy ra khi \(\frac{\left(x-2\right)^4}{x\left(\sqrt{x-1}+1\right)^4}=\frac{\left(y-2\right)^4}{y\left(\sqrt{y-1}+1\right)^4}=0\Leftrightarrow x-2=y-2=0\Leftrightarrow x=y=2\)

Vậy \(x=y=2\).

(14,78-a)/(2,87+a)=4/1

14,78+2,87=17,65

Tổng số phần bằng nhau là 4+1=5

Mỗi phần có giá trị bằng 17,65/5=3,53

=>2,87+a=3,53

=>a=0,66.

a,\(\sqrt{x-4+4\sqrt{x-4}+4}\) +\(\sqrt{x-4-4\sqrt{x-4}+4}\)

=\(\sqrt{x-4}+2+\left|\sqrt{x-4}-2\right|\) (vi x>=8)

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

b, \(\sqrt{x-1+2\sqrt{x\left(x-1\right)}+x}+\sqrt{x-1-2\sqrt{x\left(x-1\right)}+x}\)

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

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

c,d sai dau bai hay sao y