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Dung la cai do m** r**!

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

\(\Rightarrow4x^2-12x+9-3\left(-x^2+3x-2\right)=4+12x+9x^2-\left(8-12x\right)\)

\(\Rightarrow4x^2-12x+9+3x^2-9x+6-4-12x-9x^2+8-12x=0\)

\(\Rightarrow-2x^2-45x+19=0\)

Có: \(\Delta=\left(-45\right)^2-4.\left(-2\right).19=2177\)

\(\Rightarrow\sqrt{\Delta}=\sqrt{2177}\)

\(\Rightarrow x=-\frac{45+\sqrt{2177}}{4}\) (nhận) hoặc \(x=-\frac{45-\sqrt{2177}}{4}\) (nhận)

Loc oi gan giong nhu cau hoi tuong tu

\(\Leftrightarrow\)2(9x²+6x+1)=(3x+1)(x-2)

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

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

\(\Leftrightarrow\)(3x+1)(6x+2-x+2)=0

\(\Leftrightarrow\)(3x+1)(5x+4)=0

\(\Leftrightarrow\)\(\orbr{\begin{cases}3x+1=0\Leftrightarrow3x=-1\Leftrightarrow x=\frac{-1}{3}\\5x+4=0\Leftrightarrow5x=-4\Leftrightarrow x=\frac{-4}{5}\end{cases}}\)​

f(x)=9x³-1/3x+3x²-3x+1/3x²-1/9x³-3x²-9x+27+3x

    = 9x³-1/9x³+3x²+1/3x²-3x²-1/3-3x-9x+3x+27

   = 80/9x³+1/3x²-28/3x+27

       \(2\left(9x^2+6x+1\right)=\left(3x+1\right)\left(x-2\right)\)

\(\Leftrightarrow2\left(3x+1\right)^2-\left(3x+1\right)\left(x-2\right)=0\)

\(\Leftrightarrow\left(3x+1\right)\left[2\left(3x+1\right)-\left(x-2\right)\right]=0\)

\(\Leftrightarrow\left(3x+1\right)\left(6x+2-x+2\right)=0\)

\(\Leftrightarrow\left(3x+1\right)\left(5x+4\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}3x+1=0\\5x+4=0\end{cases}\Leftrightarrow\orbr{\begin{cases}3x=-1\\5x=-4\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=-\frac{1}{3}\\x=-\frac{4}{5}\end{cases}}}\)

Vậy phương trình trên có tập nghiệm \(S=\left\{-\frac{1}{3};-\frac{4}{5}\right\}\)

kb vs mk nhoa !!

theo dau bai ta co

2^3x + 2^3x +2^3x +.......+2^3x  ( 16 so hang )

2^3x . 16 = 1024 

2^3x         = 1024 : 16

2^3x          = 64 

2^3x            = 2⁶ 

=> 3x      = 6 

      x        = 6 : 3

      x        =  2

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