yêu BTS nhiều

Mình cũng ra rồi :

                                                   ( 30+29+41+35 ) : 4 = 33,75 kg

                                                              Đ/S : 33,75 kg

             Nhớ tik nha mấy bồ

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

\(\Leftrightarrow2x^2-x-4x+2=2x^2-3+6x\)

\(\Leftrightarrow2x^2-x-4x-2x^2-6x=-3-2\)

\(\Leftrightarrow-11x=-5\)

\(\Leftrightarrow x=\dfrac{5}{11}\)

\(\Rightarrow S=\left\{\dfrac{5}{11}\right\}\)

So nguoi dan tang la:

          5000:1000x18=90(nguoi)

Nam sau xa co so nguoi la:

          5000+90=5090(nguoi)

                   Ds:...

a, \(\dfrac{2a^2-3a-2}{a^2-4}=2\) ĐKXĐ: \(a\ne2,a\ne-2\)

\(\Rightarrow2a^2-3a-2=2\left(a^2-4\right)\)

\(\Leftrightarrow2a^2-3a-2=2a^2-8\)

\(\Leftrightarrow2a^2-3a-2a^2=-8+2\)

\(\Leftrightarrow-3a=-6\)

\(\Leftrightarrow a=2\) (loại)

Vậy không có giá trị nào của a để biểu thức có giá trị bằng 2.

b, \(\dfrac{3a-1}{3a+1}+\dfrac{a-3}{a+3}=2\) ĐKXĐ: \(a\ne\dfrac{-1}{3};a\ne-3\)

\(\Rightarrow\left(3a-1\right)\left(a+3\right)+\left(a-3\right)\left(3a+1\right)=2\left(a+3\right)\left(3a+1\right)\)

\(\Leftrightarrow3a^2+9a-a-3+3a^2+a-9a-3=6a^2+2a+18a+6\)

\(\Leftrightarrow3a^2+9a-a+3a^2+a-9a-6a^2-2a-18a=6+3+3\)

\(\Leftrightarrow-20a=12\)

\(\Leftrightarrow a=\dfrac{-3}{5}\) (TĐK)

Vậy a = 2 thì biểu thức có giá trị bằng 2.

a, \(\dfrac{2x-1}{x+3}=\dfrac{2x+1}{x-3}\) Điều kiện xác định: \(x\ne-3,x\ne3\)

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

\(\Leftrightarrow2x^2-x-6x+3=2x^2+x+6x+3\)

\(\Leftrightarrow2x^2-x-6x-2x^2-x-6x=3-3\)

\(\Leftrightarrow-14x=0\)

\(\Leftrightarrow x=0\left(TĐK\right)\)

\(\Rightarrow S=\left\{0\right\}\)

b,\(\dfrac{x^2+3}{x-2}=x+5\) Điều kiện xác định:\(x\ne2\)

\(\Rightarrow x^2+3=\left(x+5\right)\left(x-2\right)\)

\(\Leftrightarrow x^2+3=x^2-2x+5x-10\)

\(\Leftrightarrow x^2-x^2+2x-5x=-10-3\)

\(\Leftrightarrow-3x=-13\)

\(\Leftrightarrow x=\dfrac{-13}{-3}=\dfrac{13}{3}\left(TĐK\right)\)

\(\Rightarrow S=\left\{\dfrac{13}{3}\right\}\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x-6=0\\2x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\x=\dfrac{-3}{2}\end{matrix}\right.\)

\(\Rightarrow S=\left\{6;\dfrac{-3}{2}\right\}\)

d, \(\left(x-1\right)\left(2x-4\right)\left(3x-9\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\2x-4=0\\3x-9=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\\x=3\end{matrix}\right.\)

\(\Rightarrow S=\left\{1;2;3\right\}\)

a, \(x^2+\left(x+2\right)\left(11x-7\right)=4\)

\(\Leftrightarrow x^2+11x^2-7x+22x-14=4\)

\(\Leftrightarrow12x^2+15x-14-4=0\)

\(\Leftrightarrow12x^2+15x-18=0\)

\(\Leftrightarrow\left(12x^2+24x\right)-\left(9x+18\right)=0\)

\(\Leftrightarrow12x\left(x+2\right)-9\left(x+2\right)=0\)

\(\Leftrightarrow\left(x+2\right)\left(12x-9\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+2=0\\12x-9=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{3}{4}\end{matrix}\right.\)

b, \(4x^2-12x+5=0\)

\(\Leftrightarrow4x^2-10x-2x+5=0\)

\(\Leftrightarrow\left(4x^2-10x\right)-\left(2x-5\right)=0\)

\(\Leftrightarrow2x\left(2x-5\right)-\left(2x-5\right)=0\)

\(\Leftrightarrow\left(2x-5\right)\left(2x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-5=0\\2x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=\dfrac{1}{2}\end{matrix}\right.\)

\(\dfrac{x+101}{2001}+\dfrac{x+99}{2003}=\dfrac{x+100}{2002}+\dfrac{x+98}{2004}\)

\(\Leftrightarrow\left(\dfrac{x+101}{2001}+1\right)+\left(\dfrac{x+99}{2003}+1\right)=\left(\dfrac{x+100}{2002}+1\right)+\left(\dfrac{x+98}{2004}+1\right)\)

\(\Leftrightarrow\dfrac{x+2102}{2001}+\dfrac{x+2102}{2003}=\dfrac{x+2102}{2002}+\dfrac{x+2102}{2004}\)

\(\Leftrightarrow\dfrac{x+2102}{2001}+\dfrac{x+2102}{2003}-\dfrac{x+2102}{2002}-\dfrac{x+2102}{2004}=0\)

\(\Leftrightarrow\left(x+2102\right)\left(\dfrac{1}{2001}+\dfrac{1}{2003}-\dfrac{1}{2002}-\dfrac{1}{2004}\right)=0\)

Vì \(\dfrac{1}{2002}+\dfrac{1}{2003}-\dfrac{1}{2002}-\dfrac{1}{2004}\ne0\)

\(\Rightarrow x+2102=0\)

\(\Rightarrow x=-2102\)

\(\Rightarrow S=\left\{-2102\right\}\)