Đặt \(\left\{{}\begin{matrix}\sqrt{1+x}=a\\\sqrt{1-x}=b\end{matrix}\right.\) \(\Rightarrow2=a^2+b^2\)

\(A=\dfrac{\sqrt{1-ab}\left(a^3+b^3\right)}{a^2+b^2-ab}=\dfrac{\sqrt{\dfrac{2}{2}-ab}\left(a+b\right)\left(a^2+b^2-ab\right)}{a^2+b^2-ab}\)

\(=\sqrt{\dfrac{a^2+b^2}{2}-ab}\left(a+b\right)=\left(a+b\right)\sqrt{\dfrac{\left(a-b\right)^2}{2}}=\dfrac{\left|a-b\right|\left(a+b\right)}{\sqrt{2}}\)

\(=\pm\dfrac{a^2-b^2}{\sqrt{2}}=\pm\dfrac{2x}{\sqrt{2}}=\pm\sqrt{2}x\)

b.

\(A\ge\dfrac{1}{2}\Rightarrow\left[{}\begin{matrix}\sqrt{2}x\ge\dfrac{1}{2}\left(x\ge0\right)\\-\sqrt{2}x\ge\dfrac{1}{2}\left(x\le0\right)\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x\ge\dfrac{\sqrt{2}}{4}\\x\le-\dfrac{\sqrt{2}}{4}\end{matrix}\right.\)

Kết hợp ĐKXĐ \(\Rightarrow\left[{}\begin{matrix}\dfrac{\sqrt{2}}{4}\le x\le1\\-1\le x\le-\dfrac{\sqrt{2}}{4}\end{matrix}\right.\)

ĐKXĐ: \(-1\le x\le1\)

Đặt \(\left\{{}\begin{matrix}\sqrt{1-x}=a\\\sqrt{1+x}=b\end{matrix}\right.\) \(\Rightarrow a^2+b^2=2\) ta được:

\(A=\dfrac{\sqrt{1-ab}\left(a^3+b^3\right)}{2-ab}=\dfrac{\sqrt{\dfrac{a^2+b^2}{2}-ab}\left(a+b\right)\left(a^2+b^2-ab\right)}{a^2+b^2-ab}\)

\(=\sqrt{\dfrac{a^2+b^2-2ab}{2}}\left(a+b\right)=\dfrac{\left|a-b\right|\left(a+b\right)}{\sqrt{2}}\)

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

- Với \(-1\le x\le0\Rightarrow A=\dfrac{\left(\sqrt{1-x}-\sqrt{1+x}\right)\left(\sqrt{1-x}+\sqrt{1+x}\right)}{\sqrt{2}}=-\sqrt{2}x\)

- Với \(0\le x\le1\Rightarrow A=\dfrac{\left(\sqrt{1+x}-\sqrt{1-x}\right)\left(\sqrt{1+x}+\sqrt{1-x}\right)}{\sqrt{2}}=\sqrt{2}x\)

b.

TH1: \(\left\{{}\begin{matrix}-1\le x\le0\\-\sqrt{2}x\ge\dfrac{1}{2}\end{matrix}\right.\) \(\Rightarrow-1\le x\le-\dfrac{1}{2\sqrt{2}}\)

TH2: \(\left\{{}\begin{matrix}0\le x\le1\\\sqrt{2}x\ge\dfrac{1}{2}\end{matrix}\right.\) \(\Rightarrow\dfrac{1}{2\sqrt{x}}\le x\le1\)

\(y=2^{\sqrt{\left|x-3\right|-\left|8-x\right|}}+\sqrt{\frac{-\log_{0,5}\left(x-1\right)}{\sqrt{x^2-2x+8}}}\)

Điều kiện : \(\begin{cases}\left|x-3\right|-\left|8-x\right|\ge0\\\frac{-\log_{0,5}\left(x-1\right)}{\sqrt{x^2-2x+8}}\ge0\end{cases}\)

             \(\Leftrightarrow\begin{cases}\left|x-3\right|\ge\left|8-x\right|\\x^2-2x-8>0\\\log_{0,5}\left(x-1\right)\le0\end{cases}\)  \(\Leftrightarrow\begin{cases}\left(x-3\right)^2\ge\left(8-x\right)^2\\x^2-2x-8>0\\x-1\ge1\end{cases}\)

              \(\Leftrightarrow\begin{cases}x\ge\frac{11}{2}\\x< -2;x>4\\x\ge2\end{cases}\)

              \(\Leftrightarrow x\ge\frac{11}{2}\) là tập xác định của hàm số

6:ĐKXĐ: x>=0; x<>1/25

BPT=>\(\dfrac{3\sqrt{x}}{5\sqrt{x}-1}+3< =0\)

=>\(\dfrac{3\sqrt{x}+15\sqrt{x}-5}{5\sqrt{x}-1}< =0\)

=>\(\dfrac{18\sqrt{x}-5}{5\sqrt{x}-1}< =0\)

=>\(\dfrac{1}{5}< \sqrt{x}< =\dfrac{5}{18}\)

=>\(\dfrac{1}{25}< x< =\dfrac{25}{324}\)

7:

ĐKXĐ: x>=0

BPT \(\Leftrightarrow\dfrac{\sqrt{x}+1}{2\sqrt{x}+3}>\dfrac{8}{3}:\dfrac{8}{3}=1\)

=>\(\dfrac{\sqrt{x}+1}{2\sqrt{x}+3}-1>=0\)

=>\(\dfrac{\sqrt{x}+1-2\sqrt{x}-3}{2\sqrt{x}+3}>=0\)

=>\(-\sqrt{x}-2>=0\)(vô lý)

8:

ĐKXĐ: x>=0; x<>9/4

BPT \(\Leftrightarrow\dfrac{\sqrt{x}-2}{2\sqrt{x}-3}+4< 0\)

=>\(\dfrac{\sqrt{x}-2+8\sqrt{x}-12}{2\sqrt{x}-3}< 0\)

=>\(\dfrac{9\sqrt{x}-14}{2\sqrt{x}-3}< 0\)

TH1: 9căn x-14>0 và 2căn x-3<0

=>căn x>14/9 và căn x<3/2

=>14/9<căn x<3/2

=>196/81<x<9/4

TH2: 9căn x-14<0 và 2căn x-3>0

=>căn x>3/2 hoặc căn x<14/9

mà 3/2<14/9

nên trường hợp này Loại

9: 

ĐKXĐ: x>=0

\(BPT\Leftrightarrow\dfrac{2\sqrt{x}+3}{5\sqrt{x}+7}< =-\dfrac{1}{3}\)

=>\(\dfrac{2\sqrt{x}+3}{5\sqrt{x}+7}+\dfrac{1}{3}< =0\)

=>\(\dfrac{6\sqrt{x}+9+5\sqrt{x}+7}{3\left(5\sqrt{x}+7\right)}< =0\)

=>\(\dfrac{11\sqrt{x}+16}{3\left(5\sqrt{x}+7\right)}< =0\)(vô lý)

10: 

ĐKXĐ: x>=0; x<>1/49

\(BPT\Leftrightarrow\dfrac{6\sqrt{x}-2}{7\sqrt{x}-1}+6>0\)

=>\(\dfrac{6\sqrt{x}-2+42\sqrt{x}-6}{7\sqrt{x}-1}>0\)

=>\(\dfrac{48\sqrt{x}-8}{7\sqrt{x}-1}>0\)

=>\(\dfrac{6\sqrt{x}-1}{7\sqrt{x}-1}>0\)

TH1: 6căn x-1>0 và 7căn x-1>0

=>căn x>1/6 và căn x>1/7

=>căn x>1/6

=>x>1/36

TH2: 6căn x-1<0 và 7căn x-1<0

=>căn x<1/6 và căn x<1/7

=>căn x<1/7

=>0<=x<1/49

Mn ơi giúp mk với , cảm ơn nhiều !!

1) (x−1):0,16=−9:(1−x)

\(\Rightarrow\)(x-1):0,16= 9:(-1):(x-1)

\(\Rightarrow\)(x-1):0,16=9:(x-1)

\(\Rightarrow\)(x-1).(x-1)= 9. 0,16

\(\Rightarrow\)(x-1)\(^2\)= 1,44=1,2\(^2\)=(-1,2)\(^2\)

\(\Rightarrow\)x-1=1,2\(\Rightarrow\)x=2,2

hoặc x-1= -1,2\(\Rightarrow\)x= -0,2

Vậy x =2,2 ; x=0,2

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