Câu hỏi của UZUMAKI NARUTO

Question

Tìm x,biết :a) 2x^2-7x+5=0b) x(2x-5) - 4x+10=0c) (x-5)(x+5) - x(x-2)=15d) x^2(2x-3) - 12+8x=0e) x(x - 1)+5x - 5=0f) (2x-3)^2 - 4x(x - 1)=5g) x(5 - 2x)+2x(x - 1)=13 h)2(x+5)(2x - 5)+(x - 1)(5 - 2x)=0

Phương An · Accepted Answer

$2x^2-7x+5=0$$2x^2-2x-5x+5=0$$2x\left(x-1\right)-5\left(x-1\right)=0$$\left(x-1\right)\left(2x-5\right)=0$$\left[\begin{array}{nghiempt}x-1=0\2x-5=0\end{array}\right.$$\left[\begin{array}{nghiempt}x=1\2x=5\end{array}\right.$$\left[\begin{array}{nghiempt}x=1\x=\frac{5}{2}\end{array}\right.$$x\left(2x-5\right)-4x+10=0$$x\left(2x-5\right)-2\left(2x-5\right)=0$$\left(2x-5\right)\left(x-2\right)=0$$\left[\begin{array}{nghiempt}x-2=0\2x-5=0\end{array}\right.$$\left[\begin{array}{nghiempt}x=2\2x=5\end{array}\right.$$\left[\begin{array}{nghiempt}x=2\x=\frac{5}{2}\end{array}\right.$$\left(x-5\right)\left(x+5\right)-x\left(x-2\right)=15$$x^2-25-x^2+2x=15$$2x=15+25$$2x=40$$x=\frac{40}{2}$$x=20$$x^2\left(2x-3\right)-12+8x=0$$x^2\left(2x-3\right)+4\left(2x-3\right)=0$$\left(2x-3\right)\left(x^2+4\right)=0$$2x-3=0$ (vì $x^2\ge0\Rightarrow x^2+4\ge4>0$)$2x=3$$x=\frac{3}{2}$$x\left(x-1\right)+5x-5=0$$x\left(x-1\right)+5\left(x-1\right)=0$$\left(x-1\right)\left(x+5\right)=0$$\left[\begin{array}{nghiempt}x-1=0\x+5=0\end{array}\right.$$\left[\begin{array}{nghiempt}x=1\x=-5\end{array}\right.$$\left(2x-3\right)^2-4x\left(x-1\right)=5$$4x^2-12x+9-4x^2+4x=5$$-8x=5-9$$-8x=-4$$x=\frac{4}{8}$$x=\frac{1}{2}$$x\left(5-2x\right)+2x\left(x-1\right)=13$$5x-2x^2+2x^2-2x=13$$3x=13$$x=\frac{13}{3}$$2\left(x+5\right)\left(2x-5\right)+\left(x-1\right)\left(5-2x\right)=0$$\left(2x+10\right)\left(2x-5\right)-\left(x-1\right)\left(2x-5\right)=0$$\left(2x-5\right)\left(2x+10-x+1\right)=0$$\left(2x-5\right)\left(x+11\right)=0$$\left[\begin{array}{nghiempt}2x-5=0\x+11=0\end{array}\right.$$\left[\begin{array}{nghiempt}2x=5\x=-11\end{array}\right.$$\left[\begin{array}{nghiempt}x=\frac{5}{2}\x=-11\end{array}\right.$Câu trả lời: $2x^2-7x+5=0$$2x^2-2x-5x+5=0$$2x\left(x-1\right)-5\left(x-1\right)=0$$\left(x-1\right)\left(2x-5\right)=0$$\left[\begin{array}{nghiempt}x-1=0\2x-5=0\end{array}\right.$$\left[\begin{array}{nghiempt}x=1\2x=5\end{array}\right.$$\left[\begin{array}{nghiempt}x=1\x=\frac{5}{2}\end{array}\right.$$x\left(2x-5\right)-4x+10=0$$x\left(2x-5\right)-2\left(2x-5\right)=0$$\left(2x-5\right)\left(x-2\right)=0$$\left[\begin{array}{nghiempt}x-2=0\2x-5=0\end{array}\right.$$\left[\begin{array}{nghiempt}x=2\2x=5\end{array}\right.$$\left[\begin{array}{nghiempt}x=2\x=\frac{5}{2}\end{array}\right.$$\left(x-5\right)\left(x+5\right)-x\left(x-2\right)=15$$x^2-25-x^2+2x=15$$2x=15+25$$2x=40$$x=\frac{40}{2}$$x=20$$x^2\left(2x-3\right)-12+8x=0$$x^2\left(2x-3\right)+4\left(2x-3\right)=0$$\left(2x-3\right)\left(x^2+4\right)=0$$2x-3=0$ (vì $x^2\ge0\Rightarrow x^2+4\ge4>0$)$2x=3$$x=\frac{3}{2}$$x\left(x-1\right)+5x-5=0$$x\left(x-1\right)+5\left(x-1\right)=0$$\left(x-1\right)\left(x+5\right)=0$$\left[\begin{array}{nghiempt}x-1=0\x+5=0\end{array}\right.$$\left[\begin{array}{nghiempt}x=1\x=-5\end{array}\right.$$\left(2x-3\right)^2-4x\left(x-1\right)=5$$4x^2-12x+9-4x^2+4x=5$$-8x=5-9$$-8x=-4$$x=\frac{4}{8}$$x=\frac{1}{2}$$x\left(5-2x\right)+2x\left(x-1\right)=13$$5x-2x^2+2x^2-2x=13$$3x=13$$x=\frac{13}{3}$$2\left(x+5\right)\left(2x-5\right)+\left(x-1\right)\left(5-2x\right)=0$$\left(2x+10\right)\left(2x-5\right)-\left(x-1\right)\left(2x-5\right)=0$$\left(2x-5\right)\left(2x+10-x+1\right)=0$$\left(2x-5\right)\left(x+11\right)=0$$\left[\begin{array}{nghiempt}2x-5=0\x+11=0\end{array}\right.$$\left[\begin{array}{nghiempt}2x=5\x=-11\end{array}\right.$$\left[\begin{array}{nghiempt}x=\frac{5}{2}\x=-11\end{array}\right.$

T.Thùy Ninh · Answer

$a,2x^2-7x+5=0\Leftrightarrow2x^2-2x-5x+5=0\Leftrightarrow2x\left(x-1\right)-5\left(x-1\right)=0\Leftrightarrow\left(x-1\right)\left(2x-5\right)=0\Rightarrow\left[{}\begin{matrix}x-1=0\2x-5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\2x=5\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\x=2,5\end{matrix}\right.$$b,x\left(2x-5\right)-4x+10=0\Rightarrow x\left(2x-5\right)-2\left(2x-5\right)=0\Leftrightarrow\left(x-2\right)\left(2x-5\right)=0\Rightarrow\left[{}\begin{matrix}x-2=0\2x-5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2\2x=5\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2\x=2,5\end{matrix}\right.$$c,\left(x-5\right)\left(x+5\right)-x\left(x-2\right)=15\Leftrightarrow x^2-25-x^2+2x-15=0\Leftrightarrow2x-40=0\Rightarrow2x=40\Rightarrow x=20$$d,x^2\left(2x-3\right)-12+8x=0\Rightarrow2x^3-3x^2-12+8x=0\Leftrightarrow2x^3+8x-3x^2-12=0\Leftrightarrow2x\left(x^2+4\right)-2\left(x^2+4\right)=0\Leftrightarrow\left(2x-2\right)\left(x^2+4\right)=0\Rightarrow\left[{}\begin{matrix}2x-2=0\x^2+4=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x=2\x^2=-4\end{matrix}\right.\Rightarrow x=1$Câu trả lời: $a,2x^2-7x+5=0\Leftrightarrow2x^2-2x-5x+5=0\Leftrightarrow2x\left(x-1\right)-5\left(x-1\right)=0\Leftrightarrow\left(x-1\right)\left(2x-5\right)=0\Rightarrow\left[{}\begin{matrix}x-1=0\2x-5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\2x=5\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\x=2,5\end{matrix}\right.$$b,x\left(2x-5\right)-4x+10=0\Rightarrow x\left(2x-5\right)-2\left(2x-5\right)=0\Leftrightarrow\left(x-2\right)\left(2x-5\right)=0\Rightarrow\left[{}\begin{matrix}x-2=0\2x-5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2\2x=5\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2\x=2,5\end{matrix}\right.$$c,\left(x-5\right)\left(x+5\right)-x\left(x-2\right)=15\Leftrightarrow x^2-25-x^2+2x-15=0\Leftrightarrow2x-40=0\Rightarrow2x=40\Rightarrow x=20$$d,x^2\left(2x-3\right)-12+8x=0\Rightarrow2x^3-3x^2-12+8x=0\Leftrightarrow2x^3+8x-3x^2-12=0\Leftrightarrow2x\left(x^2+4\right)-2\left(x^2+4\right)=0\Leftrightarrow\left(2x-2\right)\left(x^2+4\right)=0\Rightarrow\left[{}\begin{matrix}2x-2=0\x^2+4=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x=2\x^2=-4\end{matrix}\right.\Rightarrow x=1$

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