A) \(\sqrt{25x-25}-\dfrac{15}{2}\sqrt{\dfrac{x-1}{9}}=6+\sqrt{x-1}\)

\(\Leftrightarrow5\sqrt{x-1}-\dfrac{15}{2}\dfrac{\sqrt{x-1}}{3}-\sqrt{x-1}=6\)

\(\Leftrightarrow5\sqrt{x-1}-\dfrac{5}{2}\sqrt{x-1}-\sqrt{x-1}=6\)

\(\Leftrightarrow\dfrac{3}{2}\sqrt{x-1}=6\)

\(\Leftrightarrow\sqrt{x-1}=4\Leftrightarrow x-1=16\)

\(\Leftrightarrow x=17\)

Vậy, x=17

A: \(\Leftrightarrow5\sqrt{x-1}-\dfrac{15}{2}\cdot\dfrac{\sqrt{x-1}}{3}=6+\sqrt{x-1}\)

=>5/2*căn x-1-căn x-1=6

=>3/2*căn x-1=6

=>căn x-1=4

=>x-1=16

=>x=17

B:

a: ĐKXĐ: x>=0; x<>1

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

=căn x-1+x-căn x+1

=x

B) a) \(ĐK:\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)

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

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

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

a.

\(y=\sqrt{x+2}\Rightarrow y^2=\left(\sqrt{x+2}\right)^2\)

                    \(\Rightarrow y^2=x+2\)

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

thay vào A ta có:\(A=x-2\sqrt{x+2}\)

\(\Rightarrow A=y^2-2y=y^2-2y-2\)

b.

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

Điều kiện:x+2≥0⇔x>-2

ta có:\(A=x-2\sqrt{x+2}\)

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

            \(=\left(\sqrt{x+12}-1\right)^2-3\)

vì \(\left(\sqrt{x+2}-1\right)^2\ge0\forall x\)

\(\Rightarrow\left(\sqrt{x+2}-1\right)^2-3\ge-3\forall x\)

vậy GTNN của A là-3

a/ y=\(\sqrt{x+2}\)→\(y^2-2=x\)

⇒A=\(y^2-2-2y\)

b/ A=\(y^2-2y-2\)=\(\left(y^2-2y+1\right)-3\)=\(\left(y-1\right)^2-3\)≥ -3

⇒\(A_{min}=-3\)

dấu = xảy ra khi y=1⇒x= -1

Ta có : a= \(\sqrt[3]{2-\sqrt{3}}\)  + \(\sqrt[3]{2+\sqrt{3}}\)

Suy ra a^3 = 3a +4  => (a^2 -3)a=4  

<=> \(\left(\frac{4}{a^2-3}\right)^3\)= a^3  <=>\(\frac{64}{\left(a^2-a\right)^3}\) -3a = 4   

mà 4 nguyên suy ra đpcm

Ta có \(a=3\sqrt{2-\sqrt{3}}+\sqrt{3}^32_{\sqrt{3}}\)

Suy ra ta được 3= 3a + 4 => (a ngũ 2 - 3)a =4

Vậy kết quả khi tính đ là

=> (4 trên a2 - 3) trên 3 =a ngũ 3 <=> 64 trên a 2 - a3 - 3a =4

a: Ta có: \(B=\left(\dfrac{6}{a-1}+\dfrac{10-2\sqrt{a}}{a\sqrt{a}-a-\sqrt{a}+1}\right)\cdot\dfrac{\left(\sqrt{a}-1\right)^2}{4\sqrt{a}}\)

\(=\dfrac{6\sqrt{a}-6+10-2\sqrt{a}}{\left(\sqrt{a}-1\right)^2\cdot\left(\sqrt{a}+1\right)}\cdot\dfrac{\left(\sqrt{a}-1\right)^2}{4\sqrt{a}}\)

\(=\dfrac{4\left(\sqrt{a}+1\right)}{\sqrt{a}+1}\cdot\dfrac{1}{4\sqrt{a}}\)

\(=\dfrac{1}{\sqrt{a}}\)

a) \(B=\left(\dfrac{6}{a-1}+\dfrac{10-2\sqrt{a}}{a\sqrt{a}-a-\sqrt{a}+1}\right).\dfrac{\left(\sqrt{a}-1\right)^2}{4\sqrt{a}}=\left(\dfrac{6}{a-1}+\dfrac{10-2\sqrt{a}}{\left(a-1\right)\left(\sqrt{a}-1\right)}\right).\dfrac{\left(\sqrt{a}-1\right)^2}{4\sqrt{a}}=\dfrac{6\left(\sqrt{a}-1\right)+10-2\sqrt{a}}{\left(a-1\right)\left(\sqrt{a}-1\right)}.\dfrac{\left(\sqrt{a}-1\right)^2}{4\sqrt{a}}=\dfrac{4\left(\sqrt{a}+1\right)}{\left(\sqrt{a}-1\right)^2\left(\sqrt{a}+1\right)}.\dfrac{\left(\sqrt{a}-1\right)^2}{4\sqrt{a}}=\dfrac{1}{\sqrt{a}}\)

b) \(C=B.\left(a-\sqrt{a}+1\right)=\dfrac{a-\sqrt{a}+1}{\sqrt{a}}=\sqrt{a}-1+\dfrac{1}{\sqrt{a}}\ge2\sqrt{\sqrt{a}.\dfrac{1}{\sqrt{a}}}-1=1\)(bất đẳng thức Cauchy cho 2 số dương)

\(a=\lim\dfrac{\sqrt{2n+1}}{\sqrt{n}+1}=\lim\dfrac{\sqrt{2+\dfrac{1}{n}}}{1+\dfrac{1}{\sqrt{n}}}=\sqrt{2}\)

\(\Rightarrow\lim\dfrac{3-4\sqrt{2}n^2}{\left(\sqrt{2}n-2\right)^2}=\lim\dfrac{\dfrac{3}{n^2}-4\sqrt{2}}{\left(\sqrt{2}-\dfrac{2}{n}\right)^2}=\dfrac{-4\sqrt{2}}{2}=-2\sqrt{2}\)

1: \(B=\dfrac{2\sqrt{x}-x-2\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}}\)

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

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

Đặt x=a-2,ta có : \(P=\frac{\sqrt{x}-2}{3}.\left(\frac{\sqrt{x}}{3+\sqrt{x}}+\frac{x+9}{9-x}\right):\left(\frac{3\sqrt{x}+1}{x-3\sqrt{x}}-\frac{1}{\sqrt{x}}\right)\)

\(=\frac{\sqrt{x}-2}{3}.\left(\frac{\sqrt{x}\left(3-\sqrt{x}\right)+x+9}{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)}\right):\left(\frac{3\sqrt{x}+1-\left(\sqrt{x}-3\right)}{\sqrt{x}\left(\sqrt{x}-3\right)}\right)\)

\(=\frac{\sqrt{x}-2}{3}.\left(\frac{3\left(\sqrt{x}+3\right)}{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)}\right):\left(\frac{2\left(\sqrt{x}+2\right)}{\sqrt{x}\left(\sqrt{x}-3\right)}\right)\)

\(=\frac{\sqrt{x}-2}{3}.\frac{3}{3-\sqrt{x}}.\frac{\sqrt{x}\left(\sqrt{x}-3\right)}{2\left(\sqrt{x}+2\right)}\)

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

Chú ý tới đẳng thức : \(\left(x+y\right)^3=x^3+y^3+3xy\left(x+y\right)\)

\(a=\sqrt[3]{2-\sqrt{3}}+\sqrt[3]{2+\sqrt{3}}\)

\(\Leftrightarrow a^3=2-\sqrt{3}+2+\sqrt{3}+3\sqrt[3]{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}\cdot a\)

\(\Leftrightarrow a^3=4+3\sqrt[3]{4-3}\cdot a\)

\(\Leftrightarrow a^3=4+3a\)

\(\Leftrightarrow a^3-3a=4\)

Khi đó: \(\frac{64}{\left(a^3-3a\right)^3}-3=\frac{64}{4^3}-3=1-3=-2\)

Ta có đpcm.

p/s: Mình nghĩ đề sai và sửa luôn rồi, có gì bạn ib lại.

Lập phương lên bạn!