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* 4x - 1 = 3x - 2
⇔ 4x - 3x = -2 + 1
⇔ x = -1
Vậy tập nghiệm của pt là S = {-1}
* \(\frac{3}{4}-3x=0\)
⇔ \(\frac{3}{4}-\frac{3x.4}{4}=0\)
⇒ 3 - 12x = 0
⇔ 12x = 3
⇔ x = \(\frac{3}{12}=\frac{1}{4}\)
Vậy tập nghiệm của pt là S = \(\left\{\frac{1}{4}\right\}\)
* 3x - 2 = 2x + 3
⇔ 3x - 2x = 3 + 2
⇔ x = 5
Vậy tập nghiệm của pt là S = {5}
* 2(x - 3) = 5(x + 4)
⇔ 2x - 6 = 5x + 20
⇔ 2x - 5x = 20 + 6
⇔ -3x = 26
⇔ x = \(\frac{-26}{3}\)
Vậy tập nghiệm của pt là S = \(\left\{\frac{-26}{3}\right\}\)
\(A,5x-25=0\)
\(\Leftrightarrow5x-5^2=0\)
\(\Leftrightarrow5\left(x-1\right)=0\)
\(\Leftrightarrow x-1=0\)
\(\Rightarrow x=1\)
Chúc bạn học tốt !
\(a.\left(3x+2\right)\left(x^2-1\right)=\left(9x^2-4\right)\left(x+1\right)\\ \left(3x+2\right)\left(x^2-1\right)-\left(9x^2-4\right)\left(x+1\right)=0\\ \left(3x+2\right)\left(x+1\right)\left(x-1\right)-\left(3x-2\right)\left(3x+2\right)\left(x+1\right)=0\\ \left(3x+2\right)\left(x+1\right)\left[\left(x-1\right)-\left(3x-2\right)\right]=0\\ \left(3x+2\right)\left(x+1\right)\left(x-1-3x+2\right)=0\\ \left(3x+2\right)\left(x+1\right)\left(1-2x\right)=0\\ \left[{}\begin{matrix}3x+2=0\\x+1=0\\1-2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{-2}{3}\\x=-1\\x=\frac{1}{2}\end{matrix}\right.\)
\(b.x\left(x+3\right)\left(x-3\right)-\left(x+2\right)\left(x^2-2x+4\right)=0\\ x\left(x^2-9\right)-\left(x^3+8\right)=0\\ x^3-9x-x^3-8=0\\ -9x-8=0\\ -9x=8\\ x=\frac{-8}{9}\)
\(c.2x\left(x-3\right)+5\left(x-3\right)=0\\ \left(x-3\right)\left(2x+5\right)=0\\ \left[{}\begin{matrix}x-3=0\\2x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\frac{-5}{2}\end{matrix}\right.\)
\(d.\left(3x-1\right)\left(x^2+2\right)=\left(3x-1\right)\left(7x-10\right)\\ \left(3x-1\right)\left(x^2+2\right)-\left(3x-1\right)\left(7x-10\right)=0\\ \left(3x-1\right)\left[\left(x^2+2\right)-\left(7x-10\right)\right]=0\\ \left(3x-1\right)\left(x^2+2-7x+10\right)=0\\ \left(3x-1\right)\left(x^2-7x+12\right)=0\\ \left(3x-1\right)\left(x^2-4x-3x+12\right)=0\\ \left(3x-1\right)\left[\left(x^2-4x\right)+\left(-3x+12\right)\right]=0\\ \left(3x-1\right)\left[x\left(x-4\right)-3\left(x-4\right)\right]=0\\ \left(3x-1\right)\left(x-4\right)\left(x-3\right)=0\\ \left[{}\begin{matrix}3x-1=0\\x-4=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{3}\\x=4\\x=3\end{matrix}\right.\)
\(e.\left(x+2\right)\left(3-4x\right)=x^2+4x+4\\ \left(x+2\right)\left(3-4x\right)=\left(x+2\right)^2\\ \left(x+2\right)\left(3-4x\right)-\left(x+2\right)^2=0\\ \left(x+2\right)\left[\left(3-4x\right)-\left(x+2\right)\right]=0\\ \left(x+2\right)\left(3-4x-x-2\right)=0\\ \left(x+2\right)\left(1-5x\right)=0\left[{}\begin{matrix}x+2=0\\1-5x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\frac{1}{5}\end{matrix}\right.\)
\(f.x\left(2x-7\right)-4x+14=0\\ x\left(2x-7\right)-2\left(2x-7\right)=0\\ \left(2x-7\right)\left(x-2\right)=0\\ \left[{}\begin{matrix}2x-7=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{7}{2}\\x=2\end{matrix}\right.\)
\(g.3x-15=2x\left(x-5\right)\\ 3\left(x-5\right)=2x\left(x-5\right)\\ 3\left(x-5\right)-2x\left(x-5\right)=0\\ \left(x-5\right)\left(3-2x\right)=0\\ \left[{}\begin{matrix}x-5=0\\3-2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=\frac{3}{2}\end{matrix}\right.\)
\(h.\left(2x+1\right)\left(3x-2\right)=\left(5x-8\right)\left(2x+1\right)\\ \left(2x+1\right)\left(3x-2\right)-\left(5x-8\right)\left(2x+1\right)=0\\ \left(2x+1\right)\left[\left(3x-2\right)-\left(5x-8\right)\right]=0\\ \left(2x+1\right)\left(3x-2-5x+8\right)=0\\ \left(2x+1\right)\left(6-2x\right)=0\\ \left[{}\begin{matrix}2x+1=0\\6-2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{-1}{2}\\x=3\end{matrix}\right.\)
\(o,x^2-9x+20=0\)
\(\Leftrightarrow x^2-4x-5x+20=0\)
\(\Leftrightarrow x\left(x-4\right)-5\left(x-4\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(x-5\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-4=0\\x-5=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=4\\x=5\end{cases}}\)
\(n,3x^3-3x^2-6x=0\)
\(\Leftrightarrow3x\left(x^2-x-2\right)=0\)
\(\Leftrightarrow3x\left(x^2+x-2x-2\right)=0\)
\(\Leftrightarrow3x\left[x\left(x+1\right)-2\left(x+1\right)\right]=0\)
\(\Leftrightarrow3x\left(x+1\right)\left(x-2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}\orbr{\begin{cases}3x=0\\x+1=0\end{cases}}\\x-2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}\orbr{\begin{cases}x=0\\x=-1\end{cases}}\\x=2\end{cases}}\)
\(\begin{array}{l} a)3x - 1 = x + 3\\ \Leftrightarrow 3x - x = 3 + 1\\ \Leftrightarrow 2x = 4\\ \Leftrightarrow x = 2\\ b)15 - 7x = 9 - 3x\\ \Leftrightarrow - 7x + 3x = 9 - 15\\ \Leftrightarrow - 4x = - 6\\ \Leftrightarrow x = \dfrac{3}{2}\\ c)x - 3 = 18\\ \Leftrightarrow x = 18 + 3\\ \Leftrightarrow x = 21 \end{array}\)
\(\begin{array}{l} d)2x + 1 = 15 - 5x\\ \Leftrightarrow 2x + 5x = 15 - 1\\ \Leftrightarrow 7x = 14\\ \Leftrightarrow x = 2\\ e)3x - 2 = 2x + 5\\ \Leftrightarrow 3x - 2x = 5 + 2\\ \Leftrightarrow x = 7\\ f) - 4x + 8 = 0\\ \Leftrightarrow - 4x = - 8\\ \Leftrightarrow x = 2 \end{array}\)
Anh sửa đề câu cuối.
\(\begin{array}{l} g)4x + 5 = 3x\\ \Leftrightarrow 4x - 3x = - 5\\ \Leftrightarrow x = - 5\\ h)3 + 2.25 + 2.6 = 2x + 5.0,4x\\ \Leftrightarrow 3 + 50 + 12 = 2x + 2x\\ \Leftrightarrow 47 = 4x\\ \Leftrightarrow x = \dfrac{{47}}{4}\\ i)2x{\left( {x + 2} \right)^2} - 8{x^2} = 2\left( {{x^3} - 8} \right)\\ \Leftrightarrow 2{x^3} + 8{x^2} + 8x - 8{x^2} = 2{x^3} - 16\\ \Leftrightarrow 8x = - 16\\ \Leftrightarrow x = - 2 \end{array}\)
\(a.3x-1=x+3\\\Leftrightarrow 3x-x=1+3\\ \Leftrightarrow2x=4\\ \Leftrightarrow x=2\)
\(b.15-7x=9-3x\\ \Leftrightarrow-7x+3x=-15+9\\\Leftrightarrow -4x=-6\\\Leftrightarrow x=\frac{3}{2}\)
\(c.x-3=18\\\Leftrightarrow x=21\)
\(d.2x+1=15-5x\\ \Leftrightarrow2x+5x=-1+15\\ \Leftrightarrow7x=14\\ \Leftrightarrow x=2\)
\(e.3x-2=2x+5\\ \Leftrightarrow3x-2x=2+5\\ \Leftrightarrow x=7\)
\(f.-4x+8=0\\\Leftrightarrow -4x=-8\\\Leftrightarrow x=2\)
\(g.4x+5=3x\\ \Leftrightarrow4x-3x=5\\\Leftrightarrow x=5\)
\(h.3+2,25+2,6=2x+5+0,4x\\ \Leftrightarrow3+2,25+2,6-5=2,4x\\\Leftrightarrow 2,85=2,4x\\ \Leftrightarrow x=1,1875\)
\(i.2x\left(x+2\right)^2-8x^2=2\left(x-2\right)\left(x^2+2x+4\right)\\ \Leftrightarrow2x\left(x^2+4x+4\right)-8x^2=2\left(x^3-8\right)\\ \Leftrightarrow2x^3+8x^2+8x-8x^2-2x^3=-16\\\Leftrightarrow 8x=-16\\\Leftrightarrow x=-2\)
Mạn phép sửa đề câu \(i\) nhé tiểu cô nương :D
Dạ em cảm ơn chị nhiều nha