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a) \(49-\left(3x-1\right)^2=0\)
\(\Leftrightarrow7^2-\left(3x-1\right)^2=0\)
\(\Leftrightarrow\left(7-3x+1\right)\left(7+3x-1\right)=0\)
\(\Leftrightarrow\left(8-3x\right)\left(6+3x\right)=0\)
\(\hept{\begin{cases}8-3x=0\\6+3x=0\end{cases}}\Rightarrow\hept{\begin{cases}x=\frac{8}{3}\\x=-2\end{cases}}\)
Vậy \(x=\frac{8}{3};x=-2\)
b) \(\left(x-1\right)^3-\left(x+2\right)\left(x^2-2x+4\right)=3\left(1-x^2\right)\)
\(\Leftrightarrow\left(x-1\right)^3-\left(x^3+2^3\right)-3\left(1-x^2\right)=0\)
\(\Leftrightarrow x^3-3x^2+3x-1-x^3-8-3+3x^2=0\)
\(\Leftrightarrow3x-12=0\)
\(\Rightarrow x=4\)
Vậy \(x=4\)
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 \(A⋮B\)
Bài 1 :
a) Ta có : x2 - 9x + 8 = x2 - x - 8x + 8 = x(x - 1) - 8(x - 1) = (x - 8)(x - 1)
b) Ta có : x2 + 6x + 8 = x2 + 6x + 9 - 1 = (x + 3)2 - 1 = (x + 3 - 1)(x + 3 + 1) = (x + 2)(x + 4)
Bài 2 :
b) 4x2 - 25 = 0
=> 4x2 = 25
=> (2x)2 = 52
=> 2x = -5;5
=> x = -5/2 ; 5/2
b) = x^2 + 2.x.3 + 3^2 - 1
=(x + 3)^2 - 1
=(x + 3 + 1)(x + 3 - 1)
=(x + 4)(x + 2)
Phần a mk nghĩ bn nên tự lm.
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\(\inƯ\left(-3\right)\)
\(\Rightarrow\begin{cases}x+1=3\\x+1=-3\\x+1=1\\x+1=-1\end{cases}\)\(\Rightarrow\begin{cases}x=2\\x=-4\\x=0\\x=-2\end{cases}\)
không hỉu đề nói rõ đc không