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C1
a) -7x(3x-2)=-21x^2+14x
b) 87^2+26.87+13^2=87^2+2.13.87+13^2=(87+13)^2=100^2
C2
a) (x-5)(x+5)
b)3x(x+5)-2(x+5)=(3x-2)(x+5)=0
\(\Rightarrow\left[\begin{array}{nghiempt}3x-2=0\\x+5=0\end{array}\right.\)
\(\Rightarrow\left[\begin{array}{nghiempt}x=\frac{2}{3}\\x=-5\end{array}\right.\)
Vậy S={-5;2/3}
C3:
a)3x^3-2x^2+2=(x+1)(3x^2-5x-5)-3
b) Để A chia hết cho B=> x+1\(\inƯ\left(-3\right)\)
\(\Rightarrow\begin{cases}x+1=3\\x+1=-3\\x+1=1\\x+1=-1\end{cases}\)\(\Rightarrow\begin{cases}x=2\\x=-4\\x=0\\x=-2\end{cases}\)
a) 3x3-2x2+2 chia x+1= 3x2-5x+5 dư -3 b) -3 chia hết x+1 vậy chon x =2
1)
a) \(-7x\left(3x-2\right)\)
\(=-21x^2+14x\)
b) \(87^2+26.87+13^2\)
\(=87^2+2.87.13+13^2\)
\(=\left(87+13\right)^2\)
\(=100^2\)
\(=10000\)
2)
a) \(x^2-25\)
\(=x^2-5^2\)
\(=\left(x-5\right)\left(x+5\right)\)
b) \(3x\left(x+5\right)-2x-10=0\)
\(\Leftrightarrow3x\left(x+5\right)-\left(2x-10\right)=0\)
\(\Leftrightarrow3x\left(x+5\right)-2\left(x-5\right)=0\)
\(\Leftrightarrow\left(x+5\right)\left(3x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+5=0\\3x-2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-5\\3x=2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=\dfrac{2}{3}\end{matrix}\right.\)
Vậy..........
3)
a) \(A:B=\left(3x^3-2x^2+2\right):\left(x+1\right)\)
Vậy \(\left(3x^3-2x^2+2\right):\left(x+1\right)=\left(3x^2-5x-5\right)+7\)
b)
Để \(A⋮B\Rightarrow7⋮\left(x+1\right)\)
\(\Rightarrow\left(x+1\right)\in U\left(7\right)=\left\{-1;1-7;7\right\}\)
Vì x là số nguyên nên x=0 ; x=6 thì \(A⋮B\)
ta có: A=x3+x2-x+a=(x2-x)(x+2)+(x+a)
Để A chia hết cho x+2<=>(x+a)chia hết cho x+2<=>a=2
Vậy a=2
Trả lời:
1, \(\left(x-2\right)^3-\left(x+1\right)\left(x^2-x+1\right)+6\left(x-1\right)^2\)
\(=x^3-6x^2+12x-8-\left(x^3+1\right)+6\left(x^2-2x+1\right)\)
\(=x^3-6x^2+12x-8-x^3-1+6x^2-12x+6\)
\(=-2\)
2, \(-x\left(x+2\right)^2+\left(2x+1\right)^2+\left(x+3\right)\left(x^2-3x+9\right)-1\)
\(=\)\(-x\left(x^2+4x+4\right)+4x^2+4x+1+x^3+27-1\)
\(=-x^3-4x^2-4x+4x^2+4x+1+x^3+27-1\)
\(=27\)
a ) \(\left(x-1\right)^3+\left(2-x\right)\left(4+2x+x^2\right)+3x\left(x+2\right)=17\)
\(\Leftrightarrow x^3-3x^2+3x-1+8-x^3+3x^2+6x=17\)
\(\Leftrightarrow\left(x^3-x^3\right)+\left(3x^2-3x^2\right)+\left(6x+3x\right)+\left(8-1\right)=17\)
\(\Leftrightarrow9x+7=17\)
\(\Leftrightarrow9x=10\)
\(\Leftrightarrow x=\dfrac{10}{9}\)
Vậy nghiệm của p/t là : \(\dfrac{10}{9}\)
b ) \(x\left(x-5\right)\left(x+5\right)-\left(x+2\right)\left(x^2-2x+4\right)=3\)
\(\Leftrightarrow x\left(x^2-25\right)-\left(x^3+8\right)=3\)
\(\Leftrightarrow x^3-25x-x^3-8=3\)
\(\Leftrightarrow-25x-8=3\)
\(\Leftrightarrow-25x=11\)
\(\Leftrightarrow x=-\dfrac{11}{25}\)
Vậy nghiệm của p/t là : \(-\dfrac{11}{25}\)
a) \(\left(x+3\right)^3-x.\left(3x+1\right)^2+\left(2x+1\right).\left(4x^2-2x+1\right)-3x^2=54\)
\(\Leftrightarrow x^3+9x^2+27x+27-x.\left(9x^2+6x+1\right)+8x^3+1-3x^2=54\)
\(\Leftrightarrow x^3+9x^2+27x+27-9x^3-6x^2-x+8x^3+1-3x^2=54\)
\(\Leftrightarrow26x+28=54\Leftrightarrow26x=54-28\Leftrightarrow26x=26\Leftrightarrow x=1\)
Vậy nghiệm của phương trình là x=1
b) \(\left(x-3\right)^3-\left(x-3\right).\left(x^2+3x+9\right)+6.\left(x+1\right)^2+3x^2=-33\)
\(\Leftrightarrow x^3-9x^2+27x-27-\left(x^3-27\right)+6.\left(x^2+2x+1\right)+3x^2=-33\)
\(\Leftrightarrow x^3-9x^2+27x-27-x^3+27+6x^2+12x+6+3x^2=-33\)
\(\Leftrightarrow27x+12x+6=-33\Leftrightarrow39x=-33-6\Leftrightarrow39x=-39\Leftrightarrow x=-1\)
Vậy nghiệm của phương trình là x = -1
Trần Anh: Hí hí =)) ÀI LỚP DIU CHIU CHIU CHÍU :3 CẢM ƠN PẠN NHIỀU NHÁ ;) ;) ;)
12
2x+2x-6=42
=>4x=48
=>x=12 nhé