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a) \(x\left(x-1\right)\left(x+1\right)-\left(x+1\right)\left(x^2-x+1\right)\)
\(=\left(x+1\right)\cdot\left[x\cdot\left(x-1\right)-\left(x^2-x+1\right)\right]\)
\(=\left(x+1\right)\left(x^2-x-x^2+x-1\right)\)
\(=\left(x+1\right)\cdot\left(-1\right)\)
\(=-1\left(x+1\right)\)
b) \(\left(x-1\right)^3-\left(x+2\right)\left(x^2-2x+4\right)+3\left(x+4\right)\left(x-4\right)\)
\(=x^3-3x^2+3x-1-\left(x^3+8\right)+\left(3x+12\right)\left(x-1\right)\)
\(=x^3-3x^2+3x-1-\left(x^3+8\right)+3x^2-3x+12x-12\)
\(=x^3-1-x^3-8+12x-12\)
\(=-21+12x\)
c) \(3x^2\left(x+1\right)\left(x-1\right)+\left(x^2-1\right)^3-\left(x^2-1\right)\left(x^4+x^2+1\right)\)
\(=3x^2\left(x^2-1\right)+x^6-3x^4+3x^2-1-\left(x^6-1\right)\)
\(=3x^4-3x^2+x^6-3x^4+3x^2-1-x^6+1\)
\(=0\)
Chỗ dấu bằng thứ hai sai nên bạn làm cũng chưa đúng
x^6 -y^6 = (x^2-y^2)(x^4 +x^2 .y^2 + y^4)
Bạn hiểu ra chỗ sai của mình chưa.Chúc bạn học tốt.
a) \(\left(x+2\right)\left(x^2-4x+4\right)-\left(x^3+2x^2\right)=5\)
\(\Leftrightarrow\left(x+2\right)\left(x^2-4x+4\right)-x^2\left(x+2\right)=5\)
\(\Leftrightarrow\left(x+2\right)\left(x^2-4x+4-x^2\right)=5\)
\(\Leftrightarrow\left(x+2\right)\left(4-4x\right)=5\)
\(\Leftrightarrow4x-4x^2+8-8x=5\)
\(\Leftrightarrow-4x^2-4x+3=0\)
\(\Leftrightarrow4x^2+4x-3=0\)
\(\Leftrightarrow4x^2-2x+6x-3=0\)
\(\Leftrightarrow2x\left(2x-1\right)+3\left(2x-1\right)=0\)
\(\Leftrightarrow\left(2x-1\right)\left(2x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=0\\2x+3=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{2}\\x=-\frac{3}{2}\end{matrix}\right.\)
Vậy \(x=\left\{\frac{1}{2};-\frac{3}{2}\right\}\)
b) \(6x^2-6x\left(-2+x\right)=36\)
\(\Leftrightarrow6x^2+12x-6x^2=36\)
\(\Leftrightarrow12x=36\)
\(\Leftrightarrow x=3\)
Vậy x = 3
c) \(\left(x+2\right)^2+\left(x-3\right)^2-2\left(x-1\right)\left(x+1\right)=9\)
\(\Leftrightarrow x^2+4x+4+x^2-6x+9-2\left(x^2-1\right)=9\)
\(\Leftrightarrow2x^2-2x+13-2x^2+2=9\)
\(\Leftrightarrow15-2x=9\)
\(\Leftrightarrow2x=6\)
\(\Leftrightarrow x=3\)
Vậy x = 3
d) \(\left(x+5\right)^2-9=0\)
\(\Leftrightarrow\left(x+5\right)^2=9\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(x+5\right)^2=3^2\\\left(x+5\right)^2=\left(-3\right)^2\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x+5=3\\x+5=-3\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=-8\end{matrix}\right.\)
Vậy x ={-2; -8}
e) \(\left(x-2\right)^3=x^3+6x^2=7\) (Câu này sai đề thì phải! Mình sửa lại đề, có gì không giống với đề của bạn thì ib mình sửa nha!)
\(\left(x-2\right)^3-x^3+6x^2=7\)
\(\Leftrightarrow x^3-6x^2+12x-8-x^3+6x^2=7\)
\(\Leftrightarrow12x-8=7\)
\(\Leftrightarrow12x=15\)
\(\Leftrightarrow x=\frac{5}{4}\)
Vậy \(x=\frac{5}{4}\)
#Chúc bạn học tốt!
1: \(\Leftrightarrow2x^2-10x-3x-2x^2=0\)
=>-13x=0
=>x=0
2: \(\Leftrightarrow5x-2x^2+2x^2-2x=13\)
=>3x=13
=>x=13/3
3: \(\Leftrightarrow4x^4-6x^3-4x^3+6x^3-2x^2=0\)
=>-2x^2=0
=>x=0
4: \(\Leftrightarrow5x^2-5x-5x^2+7x-10x+14=6\)
=>-8x=6-14=-8
=>x=1
`1)2x(x-5)-(3x+2x^2)=0`
`<=>2x^2-10x-3x-2x^2=0`
`<=>-13x=0`
`<=>x=0`
___________________________________________________
`2)x(5-2x)+2x(x-1)=13`
`<=>5x-2x^2+2x^2-2x=13`
`<=>3x=13<=>x=13/3`
___________________________________________________
`3)2x^3(2x-3)-x^2(4x^2-6x+2)=0`
`<=>4x^4-6x^3-4x^4+6x^3-2x^2=0`
`<=>x=0`
___________________________________________________
`4)5x(x-1)-(x+2)(5x-7)=0`
`<=>5x^2-5x-5x^2+7x-10x+14=0`
`<=>-8x=-14`
`<=>x=7/4`
___________________________________________________
`5)6x^2-(2x-3)(3x+2)=1`
`<=>6x^2-6x^2-4x+9x+6=1`
`<=>5x=-5<=>x=-1`
___________________________________________________
`6)2x(1-x)+5=9-2x^2`
`<=>2x-2x^2+5=9-2x^2`
`<=>2x=4<=>x=2`
a) \(\left(3x-1\right)^2-3x\left(x-5\right)=21\)
\(\Leftrightarrow9x^2-6x+1-3x^2+15x=21\)
\(\Leftrightarrow6x^2+9x-20=0\)
\(\Leftrightarrow x\in\left\{-\sqrt{\frac{\sqrt{561}+9}{12}};\sqrt{\frac{\sqrt{561}-9}{12}}\right\}\)
b) \(3\left(x+2\right)^2+\left(2x-1\right)^2-7\left(x+3\right)\left(x-2\right)=36\)
\(\Leftrightarrow3x^2+12x+12+4x^2-4x+1-7x^2+63=36\)
\(\Leftrightarrow8x+76=36\)
\(\Leftrightarrow8x=-40\)
\(\Leftrightarrow x=-5\)
\(=3x^2\left(x^2-1\right)+\left(x^8-3x^4+3x^2-1\right)-\left(x^8-1\right)\)
\(=3x^4-3x^2+x^8-3x^4+3x^2+1-x^8+1\)
\(=2\)
=2 nha ban
(con cach lam ban nhan dang thuc len rui rut gon lai)
Cm: Ta có:
a) A = x2 - 8x + 20 = (x2 - 8x + 16) + 4 = (x - 4)2 + 4 > 0 \(\forall\) x(vì (x - 4)2 \(\ge\)0 \(\forall\)x ; 4 > 0)
=> A luôn dương với mọi x
b) B = 4x2 - 12x + 11 = [(2x)2 - 12x + 9] + 2 = (2x - 3)2 + 2 > 0 \(\forall\)x (vì (2x - 3)2 \(\ge\)0 \(\forall\)x; 2 > 0)
=> B luôn dương với mọi x
c) C = x2 - x + 1 = (x2 - x + 1/4) + 3/4 = (x - 1/2)2 + 3/4 > 0 \(\forall\)x (vì (x - 1/2)2 \(\ge\)0 \(\forall\)x; 3/4 > 0)
=> C luôn dương với mọi x
* Tìm x
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 = 36 - 76
=> 8x = -40
=> x = -40 : 8 = -5
d) \(\left(4x^2-2x+1\right)\left(2x+1\right)\)
\(=\left(2x+1\right)\left[\left(2x\right)^2-2x.1+1^2\right]\)
\(=\left(2x\right)^3+1\)
\(=8x^3+1\)
a) \(\left(x+2y\right)^3=x^3+3.x^2.2y+3.x.\left(2y\right)^2+\left(2y\right)^3\)
\(=x^3+6x^2y+12xy^2+8y^3\)
b) \(\left(2x-y\right)^3=\left(2x\right)^3-3.\left(2x\right)^2.y+3.2x.y^2-y^3\)
\(=8x^3-12x^2y+6xy^2-y^3\)
c) \(\left(x^2+x+1\right).\left(x-1\right)=x^3-x^2+x^2-x+x-1\)
\(=\left(x^3-1\right)\)
#Câu này mình làm chi tiết 1 tí :) Bạn có thể tự làm gọn cho lẹ luôn nha :)
d) \(\left(4x^2-2x+1\right)\left(2x+1\right)=\left(2x+1\right)\left(4x^2-2x+1\right)\)
\(=\left(2x+1\right)\left(\left(2x\right)^2-2x.1+1^2\right)\)
\(=\left(2x\right)^3+1\)
\(=8x^3+1\)
3(x + 2)^2 + (2x - 1)^2 - 7(x + 3)(x - 3) = 36
=> 3(x^2 + 4x + 4) + 4x^2 - 4x + 1 - 7(x^2 - 9) = 36
=> 3x^2 + 12x + 12 + 4x^2 - 4x + 1 - 7x^2 + 63 = 36
=> 8x + 76 = 36
=> 8x = -40
=> x = -5