x^4+2x^3+3x^2+2x+1
x^4+2x^3+3x^2+2x+1
\(x^4+2x^3+3x^2+2x+1\)
\(=\left(x^4+x^3+x^2\right)+\left(x^3+x^2+x\right)+\left(x^2+x+1\right)\)
\(=x^2\left(x^2+x+1\right)+x\left(x^2+x+1\right)+1\left(x^2+x+1\right)\)
\(=\left(x^2+x+1\right)\left(x^2+x+1\right)\)
\(=\left(x^2+x+1\right)^2\)
\(x^4+2x^3+3x^2+2x+1\)
\(=x^4+x^3+x^2+x^3+x^2+x+x^2+x+1\)
\(=x^2\left(x^2+x+1\right)+x\left(x^2+x+1\right)+\left(x^2+x+1\right)\)
\(=\left(x^2+x+1\right)^2\)
x mũ 2 (x-1)-16 (x-1)= 0 giúp vs ạ
\(x^2\left(x-1\right)-16\left(x-1\right)=0\\ \left(x^2-16\right)\left(x-1\right)=0\\ \left(x-4\right)\left(x+4\right)\left(x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-4=0\\x+4=0\\x-1=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=4\\x=-4\\x=1\end{matrix}\right.\)
Vậy..........
Ta có: \(x^2\left(x-1\right)-16\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-4\right)\left(x+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=4\\x=-4\end{matrix}\right.\)
x mũ 2 - 3y mũ 2 giúp vs ạ cần gấp
\(x^2-\left(3y\right)^2=\left(x-3y\right)\left(x+3y\right)\)
\(x^2-3y^2=\left(x-y\sqrt{3}\right)\left(x+y\sqrt{3}\right)\)
tìm x 4x mũ 2 - 49 = 0 câu thứ 2 x mũ 2 +36 =12x câu thứ 3 10 (x-5) -8x (5-x0 =0
1. \(4x^2-49=0\)
\(\Leftrightarrow\left(2x+7\right)\left(2x-7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+7=0\Leftrightarrow x=-\dfrac{7}{2}\\2x-7=0\Leftrightarrow x=\dfrac{7}{2}\end{matrix}\right.\)
Vậy: \(x=-\dfrac{7}{2}\) hoặc \(x=\dfrac{7}{2}\)
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2. \(x^2+36=12x\)
\(\Leftrightarrow x^2-12x+36=0\)
\(\Leftrightarrow\left(x-6\right)^2=0\)
\(\Leftrightarrow x=6\)
Vậy: \(x=6\)
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3. \(10\left(x-5\right)-8x\left(5-x\right)=0\)
\(\Leftrightarrow10\left(x-5\right)+8x\left(x-5\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left(10+8x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-5=0\Leftrightarrow x=5\\10+8x=0\Leftrightarrow x=-\dfrac{5}{4}\end{matrix}\right.\)
Vậy: \(x=5\) hoặc \(x=-\dfrac{5}{4}\)
1: Ta có: \(4x^2-49=0\)
\(\Leftrightarrow\left(2x-7\right)\left(2x+7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}\\x=-\dfrac{7}{2}\end{matrix}\right.\)
2: Ta có: \(x^2+36=12x\)
\(\Leftrightarrow x^2-12x+36=0\)
\(\Leftrightarrow\left(x-6\right)^2=0\)
\(\Leftrightarrow x-6=0\)
hay x=6
a: Ta có: \(x^2+10x+25=0\)
\(\Leftrightarrow\left(x+5\right)^2=0\)
\(\Leftrightarrow x+5=0\)
hay x=-5
b: Ta có: \(x^2+\dfrac{2}{5}x+\dfrac{1}{25}=\dfrac{4}{25}\)
\(\Leftrightarrow\left(x+\dfrac{1}{5}\right)^2=\dfrac{4}{25}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{5}=\dfrac{2}{5}\\x+\dfrac{1}{5}=-\dfrac{2}{5}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{5}\\x=-\dfrac{3}{5}\end{matrix}\right.\)
d: Ta có: \(x^2-49=0\)
\(\Leftrightarrow\left(x-7\right)\left(x+7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=7\\x=-7\end{matrix}\right.\)
a,\(x^2+10x+25=0\)
\(\Leftrightarrow\left(x+5\right)^2=0\)
\(\Leftrightarrow x+5=0\)
\(\Leftrightarrow x=-5\)
d,\(x^2-49=0\)
\(\Leftrightarrow x^2-7^2=0\)
\(\Leftrightarrow\left(x+7\right)\left(x-7\right)=0\)
TH1: x+7=0 =>x=-7
TH2: x-7=0=>x=7
a) \(x^2+10x+25=0\Rightarrow\left(x+5\right)^2=0\Rightarrow x+5=0\Rightarrow x=-5\)
b) \(x^2+\dfrac{2}{5}x+\dfrac{1}{25}=\dfrac{4}{25}\Rightarrow\left(x+\dfrac{1}{5}\right)^2=\dfrac{4}{25}\Rightarrow\left[{}\begin{matrix}x+\dfrac{1}{5}=\dfrac{2}{5}\\x+\dfrac{1}{5}=-\dfrac{2}{5}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{5}\\x=-\dfrac{3}{5}\end{matrix}\right.\)
d) \(x^2-49=0\Rightarrow\left(x-7\right)\left(x+7\right)=0\Rightarrow\left[{}\begin{matrix}x=7\\x=-7\end{matrix}\right.\)
e) \(16x^2-1=0\Rightarrow\left(4x-1\right)\left(4x+1\right)=0\Rightarrow\left[{}\begin{matrix}4x-1=0\\4x+1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{4}\\x=-\dfrac{1}{4}\end{matrix}\right.\)
Phân tích thành nhân tử:
A = (6x - 3y) + (4x2 - 4xy + y2)
B= 9x2 - (y2 - 4y + 4)
C= -25x2 + y2 - 6y + 9
D= x2 - 4x - y2 - 8y -12
\(A=\left(6x-3y\right)+\left(4x^2-4xy+y^2\right)=3\left(2x-y\right)+\left(2x-y\right)^2=\left(2x-y\right)\left(2+2x-y\right)\)
\(B=9x^2-\left(y^2-4y+4\right)=9x^2-\left(y-2\right)^2=\left(3x-y+2\right)\left(3x+y-2\right)\)
\(C=-25x^2+y^2-6y+9=\left(y^2-6y+9\right)-25x^2=\left(y-3\right)^2-\left(5x\right)^2=\left(y-3-5x\right)\left(y-3+5x\right)\)\(D=x^2-4x-y^2-8y-12=\left(x^2-4x+4\right)-\left(y^2+8y+16\right)=\left(x-2\right)^2-\left(y+4\right)^2=\left(x-2-y-4\right)\left(x-2+y+4\right)=\left(x-y-6\right)\left(x+y+2\right)\)
a: Ta có: \(A=\left(6x-3y\right)+\left(4x^2-4xy+y^2\right)\)
\(=3\left(2x-y\right)+\left(2x-y\right)^2\)
\(=\left(2x-y\right)\left(2x-y+3\right)\)
b: Ta có: \(B=9x^2-\left(y^2-4y+4\right)\)
\(=9x^2-\left(y-2\right)^2\)
\(=\left(3x-y+2\right)\left(3x+y-2\right)\)
Bài 1. Phân tích đa thức bằng nhân tử.
a) 2x3 - 3x2
b) 3x4 - 24x
c) x3y + 5x2y
d) 7x2 + 14xy
a. \(2x^3-3x^2=x^2\left(2x-3\right)\)
b. \(3x^4-24x=3x\left(x^3-8\right)=3x\left(x-2\right)\left(x^2+2x+4\right)\)
c. \(x^3y+5x^2y=x^2y\left(x+5\right)\)
d. \(7x^2+14xy=7x\left(x+2y\right)\)
a)2x^3-3x^2=x^2(2x-3)
b)3x^4-24x
=3x(x^3-8x)
=3x(x-2)(x^2+2x+4)
c)x^3y+5x^2y
=x^2y(x+5)
d)7x^2+14xy
=7x(x+2y)
3A:
a: Ta có: \(75\cdot20.9+5^2\cdot20.9\)
\(=20.9\cdot100\)
=2090
b: Ta có: \(86\cdot15+150\cdot1.4\)
\(=86\cdot15+15\cdot14\)
\(=15\cdot100=1500\)
Giúp mình vs ạ mai mình học rùi
So sánh 2 số sau bằng cách vận dụng hằng đẳng thức :
a) A = 1999.2001 và B = 20002
b) A = 2^16 và B = (2 + 1)(2^2 + 1)(2^4 + 1)(2^8 + 1)
c) A = 2011.2013 và B = 2012^2
d) A = 4(3^2 + 1)(3^4 + 1)....(3^64 + 1) và B = 3^128 - 1
Bài 12A:
a: Ta có: \(x^2+2x-8\)
\(=x^2+4x-2x-8\)
\(=x\left(x+4\right)-2\left(x+4\right)\)
\(=\left(x+4\right)\left(x-2\right)\)
b: Ta có: \(x^2+5x+6\)
\(=x^2+2x+3x+6\)
\(=x\left(x+2\right)+3\left(x+2\right)\)
\(=\left(x+2\right)\left(x+3\right)\)
c: Ta có: \(4x^2-12x+8\)
\(=4x^2-4x-8x+8\)
\(=4x\left(x-1\right)-8\left(x-1\right)\)
\(=4\left(x-1\right)\left(x-2\right)\)
d: Ta có: \(x^2+8xy-9y^2\)
\(=x^2+9xy-xy-9y^2\)
\(=x\left(x+9y\right)-y\left(x+9y\right)\)
\(=\left(x+9y\right)\left(x-y\right)\)
Bài 13A:
a: Ta có: \(x^2+6x+8\)
\(=x^2+2x+4x+8\)
\(=x\left(x+2\right)+4\left(x+2\right)\)
\(=\left(x+2\right)\left(x+4\right)\)
b: Ta có: \(x^2+14x+13\)
\(=x^2+x+13x+13\)
\(=x\left(x+1\right)+13\left(x+1\right)\)
\(=\left(x+1\right)\left(x+13\right)\)
c: Ta có: \(9x^2+24x+15\)
\(=9x^2+9x+15x+15\)
\(=9x\left(x+1\right)+15\left(x+1\right)\)
\(=3\left(x+1\right)\left(3x+5\right)\)
d: Ta có: \(6x^2-xy-7y^2\)
\(=6x^2-7xy+6xy-7y^2\)
\(=x\left(6x-7y\right)+y\left(6x-7y\right)\)
\(=\left(6x-7y\right)\left(x+y\right)\)