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`#3107.101107`
\(x(x+5)(x-5) - (x+2)(x^2-2x+4)=5\)
`<=> x(x^2 - 25) - (x^3 + 2^3) = 5`
`<=> x^3 - 25x - x^3 - 8 = 5`
`<=> -25x - 8 = 5`
`<=> -25x = 13`
`<=> x = -13/25`
Vậy, `x = -13/25`
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\((x+1)^3 - (x-1)^3 -6(x-1)^2 = -19\)
`<=> x^3 + 3x^2 + 3x + 1 - (x^3 - 3x^2 + 3x - 1) - 6(x^2 - 2x + 1) = -19`
`<=> x^3 + 3x^2 + 3x + 1 - x^3 + 3x^2 - 3x + 1 - 6x^2 + 12x - 6 = -19`
`<=> (x^3 - x^3) + (3x^2 + 3x^2 - 6x^2) + (3x - 3x + 12x) + (1 + 1 - 6) = -19`
`<=> 12x - 4 = -19`
`<=> 12x = -15`
`<=> x = -15/12 = -5/4`
Vậy, `x = -5/4.`
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`@` Sử dụng các hđt:
`1)` `A^2 + B^2 = (A - B)(A + B)`
`2)` `A^3 + B^3 = (A + B)(A^2 - AB + B^2)`
`3)` `(A - B)^3 = A^3 - 3A^2B + 3AB^2 - B^3`
`4)` `(A + B)^3 = A^3 + 3A^2B + 3AB^2 + B^3`
`5)` `(A - B)^2 = A^2 - 2AB + B^2.`
a: \(x\left(x+5\right)\left(x-5\right)-\left(x+2\right)\left(x^2-2x+4\right)=5\)
=>\(x\left(x^2-25\right)-x^3-8=5\)
=>\(x^3-25x-x^3-8=5\)
=>-25x=13
=>\(x=-\dfrac{13}{25}\)
b: \(\left(x+1\right)^3-\left(x-1\right)^3-6\left(x-1\right)^2=-19\)
=>\(x^3+3x^2+3x+1-x^3+3x^2-3x+1-6\left(x^2-2x+1\right)=-19\)
=>\(6x^2+2-6x^2+12x-6=-19\)
=>12x-4=-19
=>12x=-15
=>x=-5/4
1) 4x(x-5)-(x-1)(4x-3)=5
<=>4x2-20x-4x2+3x+4x-3=5
<=>-13x=8
<=>x=-8/13
Thôi mỏi tay quá tìm x luôn nha
2) x=1.875
3) x=17/7
Bài làm
( 2x - 1 )( 3 - x ) + ( x + 4 )( x - 3 ) = -3
( x - 3 )( 2x - 1 + x + 4 ) + 3 = 0
( x - 3 )( 3x + 3 ) = 0
=> x - 3 = 0 hoặc 3x - 3 = 0
=> x = 3 hoặc 3x = 3
=> x = 3 hoặc x = 3 : 3
=> x = 3 hoặc x = 1
Vậy x = 3; x = 1
# Học tốt #
\(\left(2x-1\right)\left(3-x\right)+\left(x+4\right)\left(x-3\right)=-3\)
\(\Leftrightarrow\left(2x-1\right)\left(3-x\right)-\left(x+4\right)\left(3-x\right)=-3\)
\(\Leftrightarrow\left(3-x\right)\left(2x-1-x-4\right)=-3\)
\(\Leftrightarrow\left(3-x\right)\left(x-5\right)=-3\)
\(\Leftrightarrow x^2-8x+15=-3\)
\(\Leftrightarrow x^2-8x+18=0\)
Ta có: \(\Delta=8^2-4.18=-8< 0\)
Vậy pt vô nghiệm
a) x(x+1)+3(x+1)=0
⇌ (x+1)(x+3)=0
\(\Rightarrow\left[{}\begin{matrix}x+1=0\\x+3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-1\\x=-3\end{matrix}\right.\)
b)3x(12x-4)-2x(18x+3)=0
⇒36x2-12x-36x2+6x=0
⇒ -6x = 0
⇒ x=0
3(x+2)2+(2x-1)2-7(x-3)(x+3)=36
<--> 3(x2+4x+4)+(4x2-4x+1)-7(x2-9)=36
<--> 3x2+12x+12+4x2-4x+1-7x2+63=36
<-->8x+76=36
<-->8x= -40
<--> x= -5
(x+1)3-x(x+3)(x-3)=1
<=>x3+3x2+3x+1-x(x2-9)=1
<=>x3+3x2+3x+1-x3+9x=1
<=>3x2+12x+1=1
<=>3x2+12x =1-1=0
<=>3x(x+4)=0
<=>3x=0 hoặc x+4=0
<=>x=0 hoặc x=-4