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a. 3x2+3xy-x-y
= (3x2+3xy)-(x-y)
= 3x(x+y)-(x+y) (sau dấu trừ đổi dấu)
=(x+y)(3x-1)
(x+1)2+2x(x-2)=3(x+4)(x+1)
<= >(x2+2x+1)+2x2-4x=(3x+12)(x+1)
<=>x2+2x+1+2x2-4x=3x2+3x+12x+12
<=>x2+2x+2x2-4x-3x2-3x-12x=12-1
<=>-17x=11
<=>x=\(\frac{-11}{17}\)
4x2-1+x(2x-1)=0
<=>4x2-1+2x2-x=0
<=> 6x2-x-1=0
<=>(6x2-x)-1=0
<=>x (6x-1) =1
<=>\(6x=2\)
<=>x=\(\frac{2}{6}=\frac{1}{3}\)
^-^
\(-x^4-x^3-2x^2+x-3\)
\(=-x^4-2x^3-3x^2+x^3+2x^2+3x-x^2-2x-3\)
\(=-x^2\left(x^2+2x+3\right)+x\left(x^2+2x+3\right)-\left(x^2+2x+3\right)\)
\(=\left(-x^2+x-1\right)\left(x^2+2x+3\right)\)
2. Đặt c + d = x
Ta có: \(a+b+c+d=0\Rightarrow a+b+x=0\Rightarrow a^3+b^3+c^3+d^3=3abx\)
\(\Rightarrow a^3+b^3+c^3+d^3+3cd\left(c+d\right)=3ab\left(c+d\right)\)
\(\Rightarrow a^3+b^3+c^3+d^3=3ab\left(c+d\right)-3cd\left(c+d\right)=3\left(ab-cd\right)\left(c+d\right)\)
Câu 4:
\(a^{2016}+b^{2016}+c^{2016}=a^{1008}b^{1008}+b^{1008}c^{1008}+c^{1008}+a^{1008}\)
\(\Rightarrow2a^{2016}+2b^{2016}+2c^{2016}-2a^{1008}b^{1008}-2b^{1008}c^{1008}-2c^{1008}a^{1008}=0\)
\(\Rightarrow\left(a^{1008}-b^{1008}\right)^2+\left(b^{1008}-c^{1008}\right)^2+\left(c^{1008}-a^{1008}\right)^2=0\)
\(\Rightarrow a^{1008}=b^{1008},b^{1008}=c^{1008},c^{1008}=a^{1008}\)
\(\Rightarrow a=b,b=c,c=a\) (vì a,b,c > 0 nên \(a\ne-b,b\ne-c,c\ne-a\) )
\(\Rightarrow a-b=0,b-c=0,a-c=0\)
Thay vào A ta tính được A = 0
Ta có:
x2+y2+z2=12;x+y+z=6⇒3(x2+y2+z2)−(x+y+z)2=0⇔3(x2+y2+z2)−(x2+y2+z2+2xy+2xz+2yz)=0⇔2x2+2y2+2z2−2xy−2xz−2yz=0⇔(x−y)2+(y−z)2+(z−x)2=0(1)x2+y2+z2=12;x+y+z=6⇒3(x2+y2+z2)−(x+y+z)2=0⇔3(x2+y2+z2)−(x2+y2+z2+2xy+2xz+2yz)=0⇔2x2+2y2+2z2−2xy−2xz−2yz=0⇔(x−y)2+(y−z)2+(z−x)2=0(1)
Mà (x−y)2,(y−z)2,(z−x)2≥0,∀x,y,z(x−y)2,(y−z)2,(z−x)2≥0,∀x,y,z
⇒(x−y)2+(y−z)2+(z−x)2≥0,∀x,y,z⇒(x−y)2+(y−z)2+(z−x)2≥0,∀x,y,z
→(1)→(1) đúng chỉ khi dấu bằng xảy ra
(x−y)2=(y−z)2=(z−x)2=0⇔x−y=y−z=z−x=0⇔x=y=z(x−y)2=(y−z)2=(z−x)2=0⇔x−y=y−z=z−x=0⇔x=y=z
Mà x+y+z=6x+y+z=6⇒x=y=z=2⇒x=y=z=2\, suy ra A=10
a) 2x2 + 5x - 3 = (2x - 1).(x + 3)
b) x - 2√xy +5√x - 10y = [(√x)2 – 2 y√x] + (5√x - 10y)
c. 5x(y + 1) – 2(y + 1) = (y + 1)(5x - 2)
d. 3x + 12√xy = 3√x(√x + 4y)
e. x2 – 2xy + 5x – 10y = (x2 – 2xy) + (5x – 10y)
f. x - 3√x + √xy – 3y = (x - 3√x) + (√xy – 3y)
\(x.\left(x+1\right)+3.\left(x+1\right)=0\)
\(\left(x+1\right)\left(x+3\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x+1=0\\x+3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-1\\x=-3\end{cases}}}\)
Vậy \(\orbr{\begin{cases}x=-1\\x=-3\end{cases}}\)
\(3x\left(12x-4\right)-2x\left(18x+3\right)=36\)
\(36x^2-12x-36x^2-6x=36\)
\(-18x=36\)
\(\Rightarrow x=-2\)
Vậy \(x=-2\)
Tham khảo nhé~