ĐK: \(-1\le x\le1\)\(;\)\(x\ne0\)

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

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

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

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

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

\(\Leftrightarrow\)\(x^2-\frac{2\sqrt{2}}{3}x=0\)

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

\(\Leftrightarrow\)\(\orbr{\begin{cases}x=0\left(l\right)\\x=\frac{2\sqrt{2}}{3}\left(tm\right)\end{cases}}\)

\(\frac{x-1}{x+1}=\frac{\frac{2\sqrt{2}}{3}-1}{\frac{2\sqrt{2}}{3}+1}=\frac{\frac{2\sqrt{2}-3}{3}}{\frac{2\sqrt{2}+3}{3}}=\frac{2\sqrt{2}-3}{2\sqrt{2}+3}=12\sqrt{2}-17\) ( giống như tìm x ở trên ) 

\(\hept{\begin{cases}|x-2|+2|y-1|=9\\x+|y-1|=-1\end{cases}}\)<=> \(\hept{\begin{cases}\left(x-2\right)+2\left(y-1\right)=9\\x+\left(y-1\right)=-1\end{cases}}\)

<=> \(\hept{\begin{cases}x-2+2y-2=9\\x+y-1=-1\end{cases}}\)<=>\(\hept{\begin{cases}x+2y=13\\x+y=0\end{cases}}\)<=> \(\hept{\begin{cases}x=-13\\y=13\end{cases}}\)

Bình phương đi bạn

TL:

1đk:x<1

.\(1+3x-1=9x^2\) 

     \(3x=9x^2\) 

   x=3x\(^2\) 

 =>x=0(ktm)  hoặc  x= \(\frac{1}{3}\left(tm\right)\) 

vậy x=\(\frac{1}{3}\) 

hc tốt:)

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

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

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

\(\Leftrightarrow\orbr{\begin{cases}x=-1\\\sqrt{x^2-3x+4}=x-4\left(1\right)\end{cases}}\)

PT (1) \(\Leftrightarrow\hept{\begin{cases}x^2-3x+1=\left(x-4\right)^2\\x-4\ge0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x^2-3x+1=x^2-8x+16\\x\ge4\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}5x=15\\x\ge4\end{cases}\Leftrightarrow\hept{\begin{cases}x=5\left(tm\right)\\x\ge4\end{cases}}}\)

Vậy: \(S=\left\{-1;5\right\}\)

=.= hk tốt!!

Đặt \(x+1=a,\sqrt{x^2+x+3}=b\left(b>0\right)\)

=> \(a^2+2b^2=x^2+2x+1+2\left(x^2+x+3\right)=3x^2+4x+7\)

Khi đó PT 

<=> \(a^2+2b^2-3ab=0\)

<=> \(\orbr{\begin{cases}a=b\\a=2b\end{cases}}\)

+ a=b

=> \(x+1=\sqrt{x^2+x+3}\)

<=>\(\hept{\begin{cases}x\ge-1\\x^2+2x+1=x^2+x+3\end{cases}}\)

=> x=2

+ a=2b

=> \(x+1=2\sqrt{x^2+x+3}\)

<=> \(\hept{\begin{cases}x\ge-1\\x^2+2x+1=4\left(x^2+x+3\right)\end{cases}}\)

=> \(\hept{\begin{cases}x\ge-1\\3x^2+2x+11=0\end{cases}}\)(vô nghiệm )

Vậy x=2

cảm ơn bạn nhiều

G/s: Tam giác đều ABC có cạnh bằng a

Đặt AM=x, AN =y, x, y dương và bé hơn a

=> MB=a-x, NC=a-y

Theo bài ra ta có:

\(\frac{x}{a-x}+\frac{y}{a-y}=1\)

\(\Leftrightarrow-\frac{x}{a-x}-\frac{y}{a-y}=-1\)

\(\Leftrightarrow1-\frac{a}{a-x}+1-\frac{a}{a-y}=-1\)

\(\Leftrightarrow\frac{a}{a-x}+\frac{a}{a-y}=3\)

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

\(\Leftrightarrow x+y\le\frac{2a}{3}\)

Diện tích tam giác AMN:

\(S_{\Delta AMN}=\frac{1}{2}AM.AN.\sin\widehat{MAN}=\frac{1}{2}.xy.\frac{\sqrt{3}}{2}\)

\(=\frac{\sqrt{3}}{4}.xy\le\frac{\sqrt{3}}{4}\frac{\left(x+y\right)^2}{4}\le\frac{\sqrt{3}}{16}\frac{4a^2}{9}=\frac{\sqrt{3}a^2}{36}\)

Dấu "=" xảy ra khi và chỉ khi: \(x=y=\frac{a}{3}\)

Vậy AM=1/3AB, AN=1/3AC thì diện tích tam giác AMN lớn nhất bằng \(\frac{\sqrt{3}a^2}{36}\)