bạn chịu khó đánh máy ra đc không? Chứ khó nhìn đề lắm

=>8(x+1/x)^2+4[(x+1/x)^2-2]^2-4[(x+1/x)^2-2](x+1/x)^2=(x+4)^2

Đặt x+1/x=a(a>=2)

=>8a^2+4[a^2-2]^2-4[a^2-2]*a^2=(x+4)^2

=>8a^2+4a^4-16a^2+16-4a^4+8a^2=(x+4)^2

=>(x+4)^2=16

=>x+4=4 hoặc x+4=-4

=>x=-8;x=0

Điều kiện: \(x\ne0\)

\(\Leftrightarrow8\left(x+\dfrac{1}{x}\right)^2+4\left(x^2+\dfrac{1}{x^2}\right)\left[\left(x^2+\dfrac{1}{x^2}\right)-\left(x+\dfrac{1}{x}\right)^2\right]=\left(x+4\right)^2\)

\(\Leftrightarrow8\left(x+\dfrac{1}{x}\right)^2-8\left(x^2+\dfrac{1}{x^2}\right)=\left(x+4\right)^2\\ \Leftrightarrow\left(x+4\right)^2=16\\ \Rightarrow\left\{{}\begin{matrix}x=0\\x=-8\end{matrix}\right.\)

Vì \(x\ne0\) nên \(S=\left\{-8\right\}\)

ĐK: \(-1\le x\le1\)

Đặt \(\sqrt{1-x}=a;\sqrt{x+1}=b\Rightarrow3-x=2a^2+b^2\)

\(pt\Leftrightarrow2a-b+3ab=2a^2+b^2\)

\(\Leftrightarrow2a^2+b^2-2a+b-3ab=0\)

\(\Leftrightarrow2a^2-a\left(3b+2\right)+b^2+b=0\)

\(\Delta=\left(3b+2\right)^2-4.2.\left(b^2+b\right)=9b^2+12b+4-8b^2-8b\)

\(=b^2+4b+4=\left(b+2\right)^2\)

\(\Rightarrow\left[{}\begin{matrix}a=\dfrac{3b+2-\left(b+2\right)}{4}=\dfrac{2b}{4}=\dfrac{b}{2}\Leftrightarrow2a=b\left(1\right)\\a=\dfrac{3b+2+b+2}{4}=\dfrac{4b+4}{4}=b+1\left(2\right)\end{matrix}\right.\)

pt (1) \(\Leftrightarrow2\sqrt{1-x}=\sqrt{x+1}\)

\(\Leftrightarrow4\left(1-x\right)=x+1\)

\(\Leftrightarrow5x=3\Leftrightarrow x=\dfrac{5}{3}\left(tm\right)\)

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

\(\Leftrightarrow1-x=1+x+1+2\sqrt{x+1}\)

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

\(\Leftrightarrow\left\{{}\begin{matrix}x\le-\dfrac{1}{2}\\4x^2+4x+1=4x+4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\le-\dfrac{1}{2}\\4x^2=3\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\le-\dfrac{1}{2}\\\left[{}\begin{matrix}x=\dfrac{\sqrt{3}}{2}\left(l\right)\\x=-\dfrac{\sqrt{3}}{2}\left(tm\right)\end{matrix}\right.\end{matrix}\right.\)

Vậy, pt có tập nghiệm là: \(S=\left\{-\dfrac{\sqrt{3}}{2};\dfrac{5}{3}\right\}\)

\(\dfrac{x+3}{x-3}-\dfrac{x-3}{x+3}=\dfrac{9}{x^2-9}\left(ĐKXĐ:x\ne\pm3\right)\)

\(\Leftrightarrow\dfrac{\left(x+3\right)^2-\left(x-3\right)^2}{\left(x-3\right)\left(x+3\right)}=\dfrac{9}{\left(x-3\right)\left(x+3\right)}\)

\(\Rightarrow x^2+6x+9-x^2+6x-9=9\)

\(\Leftrightarrow12x=9\)

\(\Leftrightarrow x=\dfrac{3}{4}\) (nhận).

Vậy \(S=\left\{\dfrac{3}{4}\right\}\)

Bài này quá dễ 

x/40 - x/50= 1/3

<=>5x/200 - 4x/200=1/3

<=> x/200= 1/3

<=> x= 200/3.

bạn ê, mik bị ngu toán, lười suy nghĩ ấy mà nên đừng nói dễ hay khó j vs mik

Ờm, k cho mình đi

ĐKXĐ: \(x^2+5x+2>=0\)

=>\(\left[{}\begin{matrix}x>=\dfrac{-5+\sqrt{17}}{2}\\x< =\dfrac{-5-\sqrt{17}}{2}\end{matrix}\right.\)

\(\left(x+1\right)\left(x+4\right)-3\sqrt{x^2+5x+2}=6\)

=>\(x^2+5x+4-3\sqrt{x^2+5x+2}-6=0\)

=>\(x^2+5x+2-3\sqrt{x^2+5x+2}-4=0\)(1)

Đặt \(\sqrt{x^2+5x+2}=a\)(a>=0)

Phương trình (1) trở thành:

\(a^2-3a-4=0\)

=>(a-4)(a+1)=0

=>\(\left[{}\begin{matrix}a=4\left(nhận\right)\\a=-1\left(loại\right)\end{matrix}\right.\)

=>\(x^2+5x+2=4^2=16\)

=>\(x^2+5x-14=0\)

=>\(\left(x+7\right)\left(x-2\right)=0\)

=>\(\left[{}\begin{matrix}x=-7\left(nhận\right)\\x=2\left(nhận\right)\end{matrix}\right.\)

a

\(\left|2x-3\right|=3-2x\)

\(ĐK:x\le\dfrac{3}{2}\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-3=3-2x\\3-2x=3-2x\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\0=0\left(đúng\right)\end{matrix}\right.\)

Vậy \(S=\left\{x\in R;x=\dfrac{3}{2}\right\}\)