Tìm x: (x+2)^2-(x+2)(x-2)=0 x^2-7x+10=0 x^3+3x=3x^2+1 7x^2-28=0 2/3x(x^2-4)=0 2x.(3x-5)-(5-3x)=0

2 bình luận về “Tìm x: (x+2)^2-(x+2)(x-2)=0 x^2-7x+10=0 x^3+3x=3x^2+1 7x^2-28=0 2/3x(x^2-4)=0 2x.(3x-5)-(5-3x)=0”

1. Giải đáp: $\begin{array}{l}a)x=–2\\ b)x=2;x=5\\ c)x=1\\ d)x=2;x=–2\\ e)x=0;x=–2;x=2\\ f)x={\displaystyle \frac{5}{3}};x=–{\displaystyle \frac{1}{2}}\end{array}$

    

   Lời giải và giải thích chi tiết:

   $\begin{array}{l}a){\left(x+2\right)}^2–\left(x+2\right)\left(x–2\right)=0\\ \iff \left(x+2\right)\left(x+2–x+2\right)=0\\ \iff \left(x+2\right).4=0\\ \iff x+2=0\\ \iff x=–2\\ Vậyx=–2\\ b)x^2–7x+10=0\\ \iff x^2–2x–5x+10=0\\ \iff \left(x–2\right)\left(x–5\right)=0\\ \iff [\begin{array}{c}x–2=0\\ x–5=0\end{array}\\ \iff [\begin{array}{c}x=2\\ x=5\end{array}\\ Vậyx=2;x=5\\ c)x^3+3x=3x^2+1\\ \iff x^3–3x^2+3x–1=0\\ \iff {\left(x–1\right)}^3=0\\ \iff x–1=0\\ \iff x=1\\ Vậyx=1\\ d)7x^2–28=0\\ \iff 7x^2=28\\ \iff x^2=4\\ \iff x=2;x=–2\\ Vậyx=2;x=–2\\ e){\displaystyle \frac{2}{3}}x\left(x^2–4\right)=0\\ \iff {\displaystyle \frac{2}{3}}x\left(x–2\right)\left(x+2\right)=0\\ \iff [\begin{array}{c}x=0\\ x–2=0\\ x+2=0\end{array}\\ \iff [\begin{array}{c}x=0\\ x=2\\ x=–2\end{array}\\ Vậyx=0;x=–2;x=2\\ f)2x\left(3x–5\right)–\left(5–3x\right)=0\\ \iff 2x\left(3x–5\right)+\left(3x–5\right)=0\\ \iff \left(3x–5\right)\left(2x+1\right)=0\\ \iff [\begin{array}{c}3x–5=0\\ 2x+1=0\end{array}\\ \iff [\begin{array}{c}x={\displaystyle \frac{5}{3}}\\ x={\displaystyle \frac{–1}{2}}\end{array}\\ Vậyx={\displaystyle \frac{5}{3}};x=–{\displaystyle \frac{1}{2}}\end{array}$

   Trả lời
2. 1.(x+2)^2-(x+2)(x-2)=0

   =>(x+2)[(x+2)-(x-2)]=0

   =>(x+2)(x+2-x+2)=0

   =>4(x+2)=0

   =>x+2=0

   =>x=-2

   2.x^2-7x+10=0

   =>x^2-2x-5x+10=0

   =>x(x-2)-5(x-2)=0

   =>(x-5)(x-2)=0

   => $[\begin{array}{l}x-5=0\\ x-2=0\end{array}$ 

   => $[\begin{array}{l}x=5\\ x=2\end{array}$ 

   3.x^3+3x=3x^2+1

   =>x^3-3x^2+3x-1=0

   =>(x-1)^3=0

   =>x-1=0

   =>x=1

   4.7x^2-28=0

   =>7(x^2-4)=0

   =>7(x-2)(x+2)=0

   => $[\begin{array}{l}x-2=0\\ x+2=0\end{array}$ 

   => $[\begin{array}{l}x=2\\ x=-2\end{array}$ 

   5. 2/3 x(x^2-4)=0

   =>2/3 x(x-2)(x+2)=0

   => $[\begin{array}{l}x=0\\ x-2=0\\ x+2=0\end{array}$ 

   => $[\begin{array}{l}x=0\\ x=2\\ x=-2\end{array}$ 

   6.2x(3x-5)-(5-3x)=0

   =>2x(3x-5)+(3x-5)=0

   =>(2x+1)(3x-5)=0

   => $[\begin{array}{l}2x+1=0\\ 3x-5=0\end{array}$ 

   => $[\begin{array}{l}2x=-1\\ 3x=5\end{array}$ 

   => $[\begin{array}{l}x=-{\displaystyle \frac{1}{2}}\\ x={\displaystyle \frac{5}{3}}\end{array}$ 

   Trả lời