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a, \(4x-3=2\left(x-3\right)\Leftrightarrow4x-3=2x-6\)
\(\Leftrightarrow2x=-3\Leftrightarrow x=-\frac{3}{2}\)
b, \(5x^2+x=0\Leftrightarrow x\left(5x+1\right)=0\Leftrightarrow x=-\frac{1}{5};x=0\)
c, \(\left(3x-5\right)\left(x+7\right)=0\Leftrightarrow x=-7;x=\frac{5}{3}\)
d, \(\frac{2}{x-3}-\frac{3}{x+3}=\frac{7x-1}{x^2-9}\)ĐK : \(x\ne\pm3\)
\(\Leftrightarrow\frac{2\left(x+3\right)-3\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}=\frac{7x-1}{\left(x-3\right)\left(x+3\right)}\)
\(\Rightarrow2x+6-3x+9=7x-1\Leftrightarrow-x+15=7x-1\)
\(\Leftrightarrow-8x=-16\Leftrightarrow x=2\)( tmđk )
e, \(\left(12x-1\right)\left(6x-1\right)\left(4x-1\right)\left(3x-1\right)=330\)
\(\Leftrightarrow\left(12x-1\right)\left(12x-2\right)\left(12x-3\right)\left(12x-4\right)=330.24=7920\)
\(\Leftrightarrow\left(12x-1\right)\left(12x-4\right)\left(12x-2\right)\left(12x-3\right)=7920\)
\(\Leftrightarrow\left(144x^2-60x+4\right)\left(144x^2-60x+6\right)=7920\)
Đặt \(144x^2-60x+4=t\)
\(t\left(t+2\right)=7920\Leftrightarrow t^2+2t-7920=0\)
\(\Leftrightarrow\left(t-88\right)\left(t+90\right)=0\Leftrightarrow t=88;t=-90\)
suy ra :TH1 : \(144x^2-60x+4=88\Leftrightarrow12\left(12x+7\right)\left(x-1\right)=0\Leftrightarrow x=-\frac{7}{12};x=1\)
TH2 : \(144x^2-60x+4=-90\Leftrightarrow144x^2-60x+94=0\)
\(\Leftrightarrow x=\frac{5\pm3\sqrt{39}i}{24}\)
\(a,9\left(2x+1\right)=4\left(x-5\right)^2\)
\(4x^2-40x+100=18x+9\)
\(4x^2-58x+91=0\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{29+3\sqrt{53}}{4}\\x=\frac{29-3\sqrt{53}}{4}\end{cases}}\)
\(b,x^3-4x^2-12x+27=0\)
\(\left(x+3\right)\left(x^2-7x+9\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x+3=0\\x^2-7x+9=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-3\\x=\frac{7\pm\sqrt{13}}{2}\end{cases}}}\)
\(c,x^3+3x^2-6x-8=0\)
\(\left(x+4\right)\left(x-2\right)\left(x+1\right)=0\)
\(Th1:x+4=0\Leftrightarrow x=-4\)
\(Th2:x-2=0\Leftrightarrow x=2\)
\(Th3:x+1=0\Leftrightarrow x=-1\)
\(a,9.\left(2x+1\right)=4.\left(x-5\right)^2\)
\(< =>4x^2-40x+100=18x+9\)
\(< =>4x^2+58x+91=0\)
\(< =>\orbr{\begin{cases}x=\frac{29-3\sqrt{53}}{4}\\x=\frac{29+3\sqrt{53}}{4}\end{cases}}\)
\(b,x^3-4x^2-12x+27=0\)
\(< =>\left(x+3\right)\left(x^2-7x+9\right)=0\)
\(< =>\orbr{\begin{cases}x+3=0\\x^2-7x+9=0\end{cases}}\)
\(< =>\orbr{\begin{cases}x=-3\\x=\frac{7\pm\sqrt{13}}{2}\end{cases}}\)
\(a,PT\Leftrightarrow8x^3-6x^2+4x-3=3x^3-36x^2+x-12\)
\(\Leftrightarrow5x^3+30x^2+3x+9=0\)
\(\Leftrightarrow x=-5,95...\)
\(b,PT\Leftrightarrow2x+22-3x^2-33x=6x-15x^2-4+10x\)
\(\Leftrightarrow12x^2-47x+26=0\)
<=> (3x - 2)(4x - 13) = 0
<=> x = 2/3 hoặc x = 13/4
c, Tách ra <=> (2x - 1)(2x - 5) = 0 <=> ...
a, \(\Leftrightarrow\left(9x^2-4\right)\left(x+1\right)-\left(3x+2\right)\left(x-1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(\left(9x^2-4\right)-\left(\left(3x+2\right)\left(x-1\right)\right)\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(9x^2-4-\left(3x^2-x-2\right)\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(9x^2-4-3x^2+x+2\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(3x^2+x-2\right)=0\)
\(\Leftrightarrow\left(x+1\right)=0;3x^2+x-2=0\)
=> x=-1
với \(3x^2+x-2=0\)
ta sử dụng công thức bậc 2 suy ra : \(x=\dfrac{2}{3};x=-1\)
Vậy ghiệm của pt trên \(S\in\left\{-1;\dfrac{2}{3}\right\}\)
b: \(\Leftrightarrow x^2-2x+1-1+x^2=x+3-x^2-3x\)
\(\Leftrightarrow2x^2-2x=-x^2-2x+3\)
\(\Leftrightarrow3x^2=3\)
hay \(x\in\left\{1;-1\right\}\)
c: \(\Leftrightarrow\left(x-1\right)\left(x+1\right)\left(x+2\right)\left(x-3\right)-\left(x-1\right)\left(x-2\right)\left(x+2\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+2\right)\left[\left(x+1\right)\left(x-3\right)-\left(x-2\right)\left(x+5\right)\right]=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+2\right)\left(x^2-2x-3-x^2-3x+10\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+2\right)\left(-5x+7\right)=0\)
hay \(x\in\left\{1;-2;\dfrac{7}{5}\right\}\)
a) Ta có: \(\left(x-1\right)^2+2=x^2+3x\)
\(\Leftrightarrow x^2-2x+1+2-x^2-3x=0\)
\(\Leftrightarrow-5x=-3\)
hay \(x=\dfrac{3}{5}\)
câu b sai đề bn