QA

Tìm x, biết :

a) (x-2)3 +6(x+1)2-x3+12=0

b) (x-5) (x+5) - (x+3)2+3(x-2)2=(x+1)2- (x+4)(x-4)+3x2

c) (2x+3)+(x-1)(x+1)=5(x+2)2-(x-5)(x+1)+(x+4)

d) (1-3x)2-(x-2)(9x+1)=(3x-4)(3x+4)-9(x+3)2

Giúp mk với ạ, mk cảm ơn !

HN
25 tháng 7 2021 lúc 14:59

a) (x-2)3+6(x+1)2-x3+12=0

\(\Rightarrow\)x3-6x2+12x-8+6(x2+2x+1)-x3+12=0

\(\Rightarrow\)x3-6x2+12x-8+6x2+12x+6-x3+12=0

\(\Rightarrow\)24x+10=0

\(\Rightarrow\)24x=-10

\(\Rightarrow\)x=\(\dfrac{-10}{24}=\dfrac{-5}{12}\)

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HN
25 tháng 7 2021 lúc 15:09

b)(x-5)(x+5)-(x+3)2+3(x-2)2=(x+1)2-(x-4)(x+4)+3x2

\(\Rightarrow\)x2-25-(x2+6x+9)+3(x2-4x+4)=x2+2x+1-(x2-16)+3x2

\(\Rightarrow\)x2​-25-x2-6x-9+3x2-12x+12=x2+2x+1-x2+16+3x2

\(\Rightarrow\)3x2-18x-22=3x2+2x+17

\(\Rightarrow\)3x2-18x-22-3x2-2x-17=0

\(\Rightarrow\)-20x-39=0

\(\Rightarrow\)-20x=39

\(\Rightarrow\)x=\(-\dfrac{39}{20}\)

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HN
25 tháng 7 2021 lúc 15:16

c) (2x+3)+(x-1)(x+1)=5(x+2)2-(x-5)(x+1)+(x+4)

⇒4x2+12x+9+x2-1=5(x2+4x+4)-(x2+x-5x-5)+x+4

⇒5x2+12x+8=5x2+20x+20-x2-x+5x+5+x+4

⇒5x2+12x+8-5x2-20x-20+x2+x-5x-5-x-4=0

⇒x2-13x-21=0

 

 

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NT
25 tháng 7 2021 lúc 22:58

a) Ta có: \(\left(x-2\right)^3+6\left(x+1\right)^2-x^3+12=0\)

\(\Leftrightarrow x^3-6x^2+12x-8+6\left(x^2+2x+1\right)-x^3+12=0\)

\(\Leftrightarrow-6x^2+12x+4+6x^2+12x+6=0\)

\(\Leftrightarrow24x=-10\)

hay \(x=-\dfrac{5}{12}\)

b) Ta có: \(\left(x-5\right)\left(x+5\right)-\left(x+3\right)^2+3\left(x-2\right)^2=\left(x+1\right)^2-\left(x+4\right)\left(x-4\right)+3x^2\)

\(\Leftrightarrow x^2-25-x^2-6x-9+3\left(x^2-4x+4\right)=x^2+2x+1-\left(x^2-16\right)+3x^2\)

\(\Leftrightarrow-6x-34+3x^2-12x+12=x^2+2x+1-x^2+16+3x^2\)

\(\Leftrightarrow3x^2-18x-22=3x^2+2x+17\)

\(\Leftrightarrow-18x-2x=17+22\)

\(\Leftrightarrow-20x=39\)

hay \(x=-\dfrac{39}{20}\)

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NT
25 tháng 7 2021 lúc 23:01

c) Ta có: \(\left(2x+3\right)^2+\left(x-1\right)\left(x+1\right)=5\left(x+2\right)^2-\left(x-5\right)\left(x+1\right)+\left(x+4\right)\)

\(\Leftrightarrow4x^2+12x+9+x^2-1=5\left(x+2\right)^2-\left(x-5\right)\left(x+1\right)+x+4\)

\(\Leftrightarrow5x^2+12x+8=5x^2+20x+20-\left(x^2+x-5x-5\right)+x+4\)

\(\Leftrightarrow5x^2+12x+8-5x^2-20x-20-x-4-\left(x^2-4x-5\right)=0\)

\(\Leftrightarrow-8x-16-x^2+4x+5=0\)

\(\Leftrightarrow-x^2-4x-11=0\)(Vô lý)

d) Ta có: \(\left(1-3x\right)^2-\left(x-2\right)\left(9x+1\right)=\left(3x-4\right)\left(3x+4\right)-9\left(x+3\right)^2\)

\(\Leftrightarrow9x^2-6x+1-\left(9x^2+x-18x-2\right)=9x^2-16-9\left(x^2+6x+9\right)\)

\(\Leftrightarrow9x^2-6x+1-9x^2+17x+2=9x^2-16-9x^2-54x-81\)

\(\Leftrightarrow11x+3=-54x-97\)

\(\Leftrightarrow11x+54x=-97-3\)

\(\Leftrightarrow65x=-100\)

hay \(x=-\dfrac{20}{13}\)

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