C. =8+2*(12-6/2)^2

C

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\(\Leftrightarrow x\left(3x+2\right)+\left(x+1\right)^2-\left(2x-5\right)\left(2x+5\right)=-12\)

\(\Leftrightarrow3x^2+2x+x^2+2x+1-4x^2+25=-12\)

\(\Leftrightarrow4x+26=-12\)

\(\Leftrightarrow4x=-38\)

\(\Leftrightarrow x=-\frac{19}{2}\)

Vậy tập nghiệm của phương trình là \(S=\left\{-\frac{19}{2}\right\}\)

SAI GẦN HẾT

\(A=\left(\dfrac{2x+1}{2\left(x+2\right)}-\dfrac{x}{3\left(x-2\right)}-\dfrac{2x^2}{3\left(x-2\right)\left(x+2\right)}\right)\cdot\dfrac{24-12x}{13x+6}\)

\(=\dfrac{3\left(2x+1\right)\left(x-2\right)-2x\left(x+2\right)-4x^2}{6\left(x-2\right)\left(x+2\right)}\cdot\dfrac{12\left(2-x\right)}{13x+6}\)

\(=\dfrac{3\left(2x^2-3x-2\right)-2x^2-4x-4x^2}{x+2}\cdot\dfrac{-2}{13x+6}\)

\(=\dfrac{6x^2-9x-6-6x^2-4x}{x+2}\cdot\dfrac{-2}{13x+6}=\dfrac{2}{x+2}\)

\(=\dfrac{x^3+2x^2+2x^2-72+108-6x}{2x\left(x+6\right)}\\ =\dfrac{x^3+6x^2-2x^2-12x+6x+36}{2x\left(x+6\right)}\\ =\dfrac{\left(x+6\right)\left(x^2-2x+6\right)}{2x\left(x+6\right)}=\dfrac{x^2-2x+6}{2x}\)

\(\dfrac{x^2+2x}{2x+12}+\dfrac{x-6}{x}+\dfrac{108-6x}{2x\left(x+6\right)}\)

\(=\dfrac{x^2+2x}{2\left(x+6\right)}+\dfrac{x-6}{x}+\dfrac{108-6x}{2x\left(x+6\right)}\)

\(=\dfrac{x\left(x^2+2x\right)+2\left(x+6\right)\left(x-6\right)+108-6x}{2x\left(x+6\right)}\)

\(=\dfrac{x^3+2x^2+2\left(x^2-36\right)+108-6x}{2x\left(x+6\right)}\)

\(=\dfrac{x^3+2x^2+2x^2-72+108-6x}{2x\left(x+6\right)}\)

\(=\dfrac{x^3+4x^2+36-6x}{2x\left(x+6\right)}\)

\(=\dfrac{x^3+6x^2-2x^2-12x+6x+36}{2x\left(x+6\right)}\)

\(=\dfrac{\left(x^3+6x^2\right)+\left(-2x^2-12x\right)+\left(6x+36\right)}{2x\left(x+6\right)}\)

\(=\dfrac{x^2\left(x+6\right)-2x\left(x+6\right)+6\left(x+6\right)}{2x\left(x+6\right)}\)

\(=\dfrac{\left(x+6\right)\left(x^2-2x+6\right)}{2x\left(x+6\right)}\)

\(=\dfrac{x^2-2x+6}{2x}\)

\(a,\frac{7}{3}x+5=\frac{9}{6}x-7\)

\(\frac{7}{3}x-\frac{9}{6}x=-7-5\)

  \(\left(\frac{7}{3}-\frac{9}{6}\right).x=-12\)

                  \(\frac{5}{6}.x=-12\)

                        \(x=\left(-12\right):\frac{5}{6}\)

                        \(x=-\frac{72}{5}\)

\(b,3x+\frac{6}{4}=5-\frac{7x}{6}\)

\(3x+\frac{7x}{6}=5-\frac{6}{4}\)

\(\left(3+\frac{7}{6}\right).x=\frac{7}{2}\)

\(\frac{25}{6}.x=\frac{7}{2}\)

\(x=\frac{7}{2}:\frac{25}{6}\)

\(x=\frac{21}{25}\)

\(c,\frac{12}{5}x+6=\frac{17}{3}x+12\)

\(\frac{12}{5}x-\frac{17}{3}x=12-6\)

            \(-\frac{49}{15}x=6\)

                       \(x=6:\left(-\frac{49}{15}\right)\)

                       \(x=-\frac{90}{49}\)

\(d,6\left(3x+7\right)=12\left(2x-4\right)\)

       \(18x+42=24x-48\)

     \(18x-24x=-48-42\)

                 \(-6x=-90\)

                        \(x=15\)

\(e,13\left(2x-9\right)=8\left(3x-13\right)\)

      \(26x-117=24x-104\)

      \(26x-24x=-104+117\)

                      \(2x=13\)

                        \(x=\frac{13}{2}\)

1: Ta có: \(\dfrac{x+4}{4}+\dfrac{3x-7}{5}=\dfrac{7x+2}{20}\)

\(\Leftrightarrow5x+20+12x-28=7x+2\)

\(\Leftrightarrow17x-7x=2+8=10\)

hay x=1

2: Ta có: \(\dfrac{x}{6}+\dfrac{1-3x}{9}=\dfrac{-x+1}{12}\)

\(\Leftrightarrow\dfrac{6x}{36}+\dfrac{4\left(1-3x\right)}{36}=\dfrac{3\left(-x+1\right)}{36}\)

\(\Leftrightarrow6x+4-12x=-3x+3\)

\(\Leftrightarrow-6x+3x=3-4\)

hay \(x=\dfrac{1}{3}\)

3: Ta có: \(\dfrac{x-3}{3}-\dfrac{x+2}{12}=\dfrac{2x-1}{4}\)

\(\Leftrightarrow4x-12-x-2=6x-3\)

\(\Leftrightarrow3x-14-6x+3=0\)

\(\Leftrightarrow-3x=11\)

hay \(x=-\dfrac{11}{3}\)

4: Ta có: \(\dfrac{x-2}{4}-\dfrac{2x+3}{3}=\dfrac{x+6}{12}\)

\(\Leftrightarrow3x-6-8x-12=x+6\)

\(\Leftrightarrow-5x-x=6+18\)

hay x=-4

5: Ta có: \(\dfrac{2x-1}{12}-\dfrac{3-x}{18}=\dfrac{-1}{36}\)

\(\Leftrightarrow6x-3+2x-6=-1\)

\(\Leftrightarrow8x=8\)

hay x=1

- Mình có thể giúp bài 1 :)

 5^x +  5^ ( x + 2 ) = 650

5^x +  5^x . 5² = 650

 5^x .( 1 + 25 ) = 650

 5^x . 26 = 650 

5^x = 650 : 26

 5^x = 25

 5^x = 5² 

=> x = 2 

Vậy x = 2 

The Reflection Of Light~ Ủa đề  là 5^x-5^x+2= 650 sao ông lại đổi thành 5^x-5^(x+2) zậy