vô nghiệm

Mình thiếu điều kiện xác định ^_^

Cho mình bổ xung thêm

\(ĐKXĐ:x\ne\pm1\)

và mình sửa lại nữa là: \(\orbr{\begin{cases}x=-1\left(L\right)\\x=-3\left(TM\right)\end{cases}}\)

Vậy \(S=\left\{-3\right\}\)

\(\frac{x+1}{x-1}-\frac{x-1}{x+1}=\frac{x^2+3}{1-x^2}\) đkxđ \(x\ne\pm1\)

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

\(\Leftrightarrow x^2+2x+1-x^2-2x-1+x^2+3=0\)

\(\Leftrightarrow x^2+3=0\)

\(\Leftrightarrow x^2=-3\)

\(\Leftrightarrow x\in\varnothing\)

\(ĐKXĐ:x\ne\pm1\)

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

\(\Leftrightarrow\left(x+1\right)^2-\left(x-1\right)^2=-x^2-3\)

\(\Leftrightarrow x^2+2x+1-x^2+2x-1=-x^2-3\)

\(\Leftrightarrow x^2+4x+3=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=-3\\x=-1\end{cases}}\)

Đkxđ: \(\hept{\begin{cases}x\ne2\\x\ne0\end{cases}}\)

\(\frac{x+3}{x-2}+\frac{x+2}{x}=2\) 

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

\(\Rightarrow x\left(x+3\right)+\left(x-2\right)\left(x+2\right)=2x\left(x-2\right)\)

\(\Leftrightarrow x^2+3x+x^2-4=2x^2-4x\)

\(\Leftrightarrow x^2+3x+x^2-2x^2+4x=4\)

\(\Leftrightarrow7x=4\)

\(\Leftrightarrow x=\frac{4}{7}\)

\(\frac{3-x}{5-x}=\left(\frac{3}{5}\right)^2\)

\(\Rightarrow\frac{3-x}{5-x}=\frac{9}{25}\)

\(\Rightarrow\left(3-x\right)25=9\left(5-x\right)\)

\(\Rightarrow75-25x=45-9x\)

\(\Rightarrow-25x+9x=45-75\)

\(\Rightarrow-16x=-30\)

\(\Rightarrow x=\frac{15}{8}\)

cảm ơn bạn nha

Ta có:

\(\frac{x-1}{2}\) =\(\frac{y-2}{3}\)=\(\frac{z-3}{4}\)=k =>x=2k+1

                                          y=3k+2

                                          z=4k+3

                     Thay vào: x  -  2y  + 3z  =  -10

                              (2k+1)-2x(3k+2)+3x(4k+3)= -10

                              (2k+1)-(6k+4)+(12k+9)= -10

                               (2k-6k+12k)+(1-4+9) = -10

                                      8k    +  6             = -10  

                                              8k               = -16

                                                k               = -2

                                   =>    x = 2x(-2)+1 = -3

                                           y = 3x(-2)+2 = -4

                                           z =4x(-2)+3 =  -5

                                                Vậy .............

                          Nếu đúng nhớ **** cho mk nha!

Ta có: \(\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}=\frac{x-2y+z}{2-3+4}=\frac{-10}{3}\)

Mặt khác: \(\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}=\frac{x+y+z-6}{9}\)

=> \(\frac{x+y+z-6}{9}=\frac{-10}{3}\)

=> x + y + z - 6 = -10.9 : 3 = -30

=> x + y + z = -24

\(\frac{x+3}{-4}=-\frac{9}{x+3}\)

\(\Leftrightarrow\left(x+3\right)\left(x+3\right)=-4\cdot\left(-9\right)\)

\(\Leftrightarrow\left(x+3\right)^2=36\)

\(\Leftrightarrow\orbr{\begin{cases}\left(x+3\right)^2=6^2\\\left(x+3\right)^2=\left(-6\right)^2\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x+3=6\\x+3=-6\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=3\\x=-9\end{cases}}\)

Vậy ....

quy đồng

\(\left(x+3\right)^2=36\)

\(\left(x+3\right)^2-6^2=0\)

áp dụng định lí "  \(a^2-b^2=\left(a+b\right)\left(a-b\right)\) ta được

\(\left(x+3-6\right)\left(x+3+6\right)=0\)

\(x=3,x=-9\)

nhân chéo: 

(x+3)(x+3)=-4.(-9)

(x+3)^2   =36

Hai TH: 

TH1: x+3=6

=>   x=3

TH2: x+3=-6

=>  x=-9

Vậy x=3 hoặc x=-9

a) 4/3 - x = 3/5 + 1/2

=> 4/3 - x= 0,8

=> x = 4/3 + 0/8 

=> x = 5/8

b) ( 1/2 + 1/3 +1/6)x = 2/9

= 1x = 2/9 

=> x = 2/9

không biết

\(\frac{x+5}{x-5}+\frac{x-5}{x+5}=\frac{2\left(x^2+25\right)}{x^2-25}\left(x\ne\pm5\right)\)

\(\Leftrightarrow\frac{x+5}{x-5}+\frac{x-5}{x+5}-\frac{2\left(x^2+25\right)}{\left(x-5\right)\left(x+5\right)}=0\)

\(\Leftrightarrow\frac{\left(x+5\right)^2}{\left(x-5\right)\left(x+5\right)}+\frac{\left(x-5\right)^2}{\left(x-5\right)\left(x+5\right)}-\frac{2x^2+50}{\left(x-5\right)\left(x+5\right)}=0\)

\(\Leftrightarrow\frac{x^2+10x+25}{\left(x-5\right)\left(x+5\right)}+\frac{x^2-10x+25}{\left(x-5\right)\left(x+5\right)}-\frac{2x^2+50}{\left(x-5\right)\left(x+5\right)}=0\)

\(\Leftrightarrow\frac{x^2+10x+25+x^2-10x+25-2x^2-50}{\left(x-5\right)\left(x+5\right)}=0\)

\(\Rightarrow\frac{0}{\left(x-5\right)\left(x+5\right)}=0\)

=> PT đúng với mọi x khác \(\pm5\)

Refund QB nhìn logic :V 

\(\frac{x+5}{x-5}+\frac{x-5}{x+5}=\frac{2\left(x^2+25\right)}{x^2-25}\)

\(\frac{x+5}{x-5}+\frac{x-5}{x+5}=\frac{2\left(x^2+25\right)}{\left(x+5\right)\left(x-5\right)}\)

\(\left(x+5\right)^2-\left(x-5\right)^2=2\left(x^2+25\right)\)

\(20x=2x^2+50\)

\(20x-2x^2-50=0\)

\(2\left(10x-x^2-25\right)=0\)

\(-x^2+10x+25=0\)

\(x^2-10x+25=0\)

\(x^2-2\left(x\right)\left(5\right)+5^2=0\)

\(\left(x-5\right)^2=0\)

\(x-5=0\Leftrightarrow x=5\)

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