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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) \(6\text{x}\left(x-y\right)-3\left(x-y\right)^2\)
\(=\left(x-y\right)\left(6\text{x}-3\text{x}+3y\right)\)
\(=\left(x-y\right)\left(3\text{x}+3y\right)\)
\(=3\left(x-y\right)\left(x+y\right)\)
\(=3\left(x^2-y^2\right)\)
b) \(2\text{a}\left(x-1\right)+b\left(x-1\right)-\left(1-x\right)\)
\(=2\text{a}\left(x-1\right)+b\left(x-1\right)+\left(x-1\right)\)
\(=\left(x-1\right)\left(2\text{a}+b+1\right)\)
c) \(2\text{x}\left(x-2\right)-\left(x-2\right)^2+5\text{x}\left(2-x\right)\)
\(=2\text{x}\left(x-2\right)-\left(x-2\right)^2-5\text{x}\left(x-2\right)\)
\(=\left(x-2\right)\left(2\text{x}-x+2-5\text{x}\right)\)
\(=\left(x-2\right)\left(2-4\text{x}\right)\)
a) \(6x\left(x-y\right)-3\left(x-y\right)^2\)
\(=3\left(x-y\right)\left(2x-x+y\right)\)
b) \(2a\left(x-1\right)+b\left(x-1\right)-\left(1-x\right)\)
\(=2a\left(x-1\right)+b\left(x-1\right)+\left(x-1\right)\)
\(=\left(x-1\right)\left(2a+b+1\right)\)
c) \(2x\left(x-2\right)-\left(x-2\right)^2+5x\left(2-x\right)\)
\(=2x\left(x-2\right)-\left(x-2\right)^2-5x\left(x-2\right)\)
\(=\left(x-2\right)\left(2x-x+2-5x\right)\)
\(=\left(x-2\right)\left(-4x+2\right)\)
1.
a) \(\left(-2x^3\right)\)\(\left(x^2+5x-\frac{1}{2}\right)\) = \(-2x^5\)\(-10x^4\) \(+x^3\)
b) (\(6x^3-7x^2\)\(-x+2\))\(:\left(2x+1\right)\)=\(3x^2-5x+2\)
2.
a) 9x(3x-y) + 3y (y-3x)=9x(3x-y)-3y(3x-y)
= (9x-3y)(3x-y)
= 3(3x-y)(3x-y)
= 3(3x-y)^2
b) \(x^3-3x^2\)\(-9x+27\)= \(\left(x^3-3x^2\right)\)\(-\left(9x-27\right)\)
= \(x^2\left(x-3\right)\)\(-9\left(x-3\right)\)
= \(\left(x^2-9\right)\left(x-3\right)\)
= \(\left(x+3\right)\left(x-3\right)\left(x-3\right)\)
= \(\left(x+3\right)\left(x-3\right)^2\)
Bài 1 ) a ) \(\left(-2x^3\right)\left(x^2+5x-\frac{1}{2}\right)\)
\(=-2x^5-10x^4+x^3\)
b ) \(\left(6x^3-7x^2+x+2\right):\left(2x+1\right)\)
\(=3x^2-5x+2\)
2 ) a ) \(9x\left(3x-y\right)+3y\left(y-3x\right)\)
\(=9x\left(3x-y\right)-3y\left(3x-y\right)\)
\(=\left(3x-y\right)\left(9x-3y\right)\)
\(=3\left(3x-y\right)\left(x-y\right)\)
b ) \(x^3-3x^2-9x+27\)
\(=\left(x^3-3x^2\right)-\left(9x-27\right)\)
\(=x^2\left(x-3\right)-9\left(x-3\right)\)
\(=\left(x^2-9\right)\left(x-3\right)\)
\(=\left(x-3\right)\left(x+3\right)\left(x-3\right)\)
a, Dùng phương pháp đổi biến (đầu tiên ghép cặp (x+2) với (x+5) và cặp còn lại, rồi đổi biến)
b, Dùng phương pháp thêm bớt cùng 1 hạng tử
c, Dùng phương pháp nhóm hang tử
Mọi người giúp mình trả lời nhé, nay mình kiểm tra 1 tiết toán nên cần gấp đáp án ạ!
Ta có \(a^2-5ax+6x^2\)
\(=a^2-4ax+\left(2x\right)^2-ãx+2x^2\)
\(=\left(a^2+4ax+\left(2x\right)^2\right)-\left(ãx-2x^2\right)\)
\(=\left(a-2x\right)^2-x.\left(a-2x\right)\)
\(=\left(a-2x\right).\left(\left(a-2x\right)-x\right)\)
\(a^2-5ax+6x^2=a^2-2ax-3ax+6x^2=a\left(a-2x\right)-3x\left(a-2x\right)=\left(a-2x\right)\left(a-3x\right)\)