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g) \(x^5-3x^4+3x^3-x^2=x^2\left(x^3-3x^2+3x-1\right)=x^2\left(x-1\right)^3\)
f) \(x^2-25-2xy+y^2=\left(x^2-2xy+y^2\right)-25=\left(x-y\right)^2-5^2=\left(x-y-5\right)\left(x-y+5\right)\)
e) \(16x^3+54y^3=2\left(8x^3+27y^3\right)=2\left[\left(2x\right)^3+\left(3y\right)^3\right]=2\left(2x+3y\right)\left(4x^2-6xy+9y^2\right)\)
d) \(3y^2-3z^2+3x^2+6xy=3\left(x^2+2xy+y^2-z^2\right)=3\left[\left(x+y\right)^2-z^2\right]=3\left(x+y+z\right)\left(x+y-z\right)\)
Bài 209 : đăng tách ra cho mn cùng làm nhé
a,sửa đề : \(A=\left(3x+1\right)^2-2\left(3x+1\right)\left(3x+5\right)+\left(3x+5\right)^2\)
\(=\left(3x+1-3x-5\right)^2=\left(-4\right)^2=16\)
b, \(B=\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{32}+1\right)\)
\(2B=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{32}+1\right)=\left(3^{32}-1\right)\left(3^{32}+1\right)\)
\(2B=3^{64}-1\Rightarrow B=\frac{3^{64}-1}{2}\)
c, \(C=\left(a+b-c\right)^2+\left(a-b+c\right)^2-2\left(b-c\right)^2\)
\(=2\left(a-b+c\right)^2-2\left(b-c\right)^2=2\left[\left(a-b+c\right)^2-\left(b-c\right)^2\right]\)
\(=2\left(a-b+c-b+c\right)\left(a-b+c+b-c\right)=2a\left(a-2b+2c\right)\)
Answer:
\(5x^2-10xy+5y^2-20z^2\)
\(=5.\left(x^2-2xy+y^2-4z^2\right)\)
\(=5.[\left(x+y\right)^2-\left(2z\right)^2]\)
\(=5.\left(x+y-2z\right).\left(x+y+2z\right)\)
\(16x-5x^2-3\)
\(=\left(-5x^2+15x\right)+\left(x-3\right)\)
\(=-5x.\left(x-3\right)+\left(x-3\right)\)
\(=\left(1-5x\right).\left(x-3\right)\)
\(x^2-5x+5y-y^2\)
\(=(x-y).(x+y)-5.(x-y)\)
\(=(x-y).(x+y-5)\)
\(3x^2-6xy+3y^2-12z^2\)
\(=3.(x^2-2xy+y^2-4z^2)\)
\(=3[\left(x-y\right)^2-\left(2z\right)^2]\)
\(=3.(x-y-2z).(x-y+2z)\)
\(x^2+4x+3\)
\(=(x^2+x)+(3x+3)\)
\(=x.(x+1)+3.(x+1)\)
\(=(x+1).(x+3)\)
\((x^2+1)^2-4x^2\)
\(=(x^2-2x+1).(x^2+2x+1)\)
\(=(x-1)^2.(x+1)^2\)
\(x^2-4x-5\)
\(=(x^2+x)-(5x+5)\)
\(=x.(x+1)-5.(x+1)\)
\(=(x-5).(x+1)\)
Trả lời:
a, \(-xy.\left(x^2+2xy-3\right)=-x^3y-2x^2y^2+3xy\)
b, \(\left(12x^6y^5-3x^3y^4+4x^2y\right):6x^2y\)
\(=12x^6y^5:6x^2y^2-3x^3y^4:6x^2y+4x^2y+6x^2y\)
\(=2x^4y^3-\frac{1}{2}xy^3+\frac{2}{3}\)
a.\(\left(-xy\right)\left(x^2+2xy-3\right)=-x^3y-2x^2y^2+6xy\)
b.\(\left(12x^6y^5-3x^3y^4+4x^2y\right):6x^2y=2x^4y^4-\frac{1}{2}xy^3+\frac{2}{3}\)
\(4x^2-25+\left(2x+7\right).\left(5-2x\right)\)
\(=\left(2x+5\right).\left(2x-5\right)-\left(2x+7\right).\left(2x-5\right)\)
\(=\left(2x+5-2x-7\right).\left(2x-5\right)\)
\(=-2.\left(2x-5\right)\)
\(a^2x^2-a^2x^2-b^2x^2+b^2y^2\)
\(=a^2.\left(x^2-y^2\right)-b^2.\left(x^2-y^2\right)\)
\(=\left(a^2-b^2\right).\left(x^2-y^2\right)\)
\(=\left(a-b\right).\left(a+b\right).\left(x-y\right).\left(x+y\right)\)
\(x^2-y^2+12y-36\)
\(=x^2-\left(y^2-12y+36\right)\)
\(=x^2-\left(y-6\right)^2\)
\(=\left(x-y+6\right).\left(x+y-6\right)\)
\(\left(x+2\right)^2-x^2+2x-1\)
\(=\left(x+2\right)^2-\left(x^2-2x+1\right)\)
\(=\left(x+2\right)^2-\left(x-1\right)^2\)
\(=[x+2-\left(x-1\right)].[x+2+\left(x-1\right)]\)
\(=\left(x+2-x+1\right).\left(x+2+x-1\right)\)
\(=3.\left(2x+1\right)\)
\(16x^2-y^2=\left(4x\right)^2-y^2=\left(4x-y\right).\left(4x+y\right)\)
\(1+27x^3=1^3+\left(3x\right)^3=\left(1+3x\right).\left(1-3x+9x^2\right)\)
a) x2 - 2xy + 5x - 10y
= x(x - 2y) + 5(x - 2y)
= (x + 5)(x - 2y)
b) x(2x - 3y) - 6y2 + 4xy
= x(2x - 3y) + 2y(2x - 3y)
= (x + 2y)(2x - 3y)
c) 8x3 + 4x2 - y2 - y3
= (4x2 - y2) + (8x3 - y3)
= (2x - y)(2x + y) - (2x - y)(4x2 + 2xy + y2)
= (2x - y)(-4x2 - 2xy - y2 + 2x + y)
d) a3 - a2b - ab2 + b3
= a2(a- b) - b2(a - b)
= (a2 - b2)(a - b) = (a - b)2(a + b)
e) ab2c3 + 64ab2
= ab2(c3 + 64)
= ab2(c + 4)(c2 + 4c + 16)
f) 27x3y - a3b3y
= y[27 - (ab)3]
= y(3 - ab)(a2b2 + 3ab + 9)
Answer:
Số lượng bài khá nhiều trong một câu hỏi nên mình sẽ gửi từng bài nhé!
Bài 5:
\(A=9x^2-6x+11\)
\(=9x^2-6x+1+10\)
\(=\left(3x-1\right)^2+10\)
\(\left(3x-1\right)^2\ge0\forall x\)
\(\Rightarrow\left(3x-1\right)^2+10\ge0\forall x\)
Dấu "=" xảy ra khi: \(3x-1=0\Rightarrow x=\frac{1}{3}\)
Vậy giá trị nhỏ nhất của \(A=10\) khi \(x=\frac{1}{3}\)
\(B=4x^2-20x+101\)
\(=4x^2-20x+25+76\)
\(=\left(2x-5\right)^2+76\)
\(\left(2x-5\right)^2\ge0\forall x\)
\(\Rightarrow\left(2x-5\right)^2+76\ge76\forall x\)
Dấu "=" xảy ra khi: \(2x-5=0\Rightarrow x=\frac{5}{2}\)
Vậy giá trị nhỏ nhất của \(B=76\) khi \(x=\frac{5}{2}\)
Bài 6:
\(A=x-x^2\)
\(=-\left(x^2-x\right)\)
\(=-\left(x-\frac{1}{2}\right)^2+\frac{1}{4}\)
\(\left(x-\frac{1}{2}\right)^2\ge0\forall x\)
\(\Rightarrow-\left(x-\frac{1}{2}\right)^2+\frac{1}{4}\le\frac{1}{4}\)
\(\Rightarrow A\le\frac{1}{4}\)
Dấu "=" xảy ra khi: \(x-\frac{1}{2}=0\Rightarrow x=\frac{1}{2}\)
Vậy giá trị lớn nhất của \(A=\frac{1}{4}\) khi \(x=\frac{1}{2}\)
\(B=-x^2+6x-11\)
\(=-\left(x^2-6x\right)-11\)
\(\Rightarrow-\left(x-3\right)^2-2\)
\(\left(x-3\right)^2\ge0\forall x\)
\(\Rightarrow-\left(x-3\right)^2-2\le-2\)
\(\Rightarrow B\le-2\)
Dấu "=" xảy ra khi: \(x-3=0\Rightarrow x=3\)
Vậy giá trị lớn nhất của \(B=-2\) khi \(x=3\)
Answer:
Bài 1:
\(2x\left(3x^2-5x+3\right)\)
\(=2x.3x^2-2x.5x+2x.3\)
\(=6x^3-10x^2+6x\)
\(\left(-2x-1\right)\left(x^2+5x-3\right)-\left(x-1\right)^3\)
\(=\left(-2x^3-10x^2+6x-x^2-5x+3\right)-x^3+3x^2-3x+1\)
\(=-2x^3-11x^2+x+3-x^3+3x^2-3x+1\)
\(=-\left(2x^3+x^3\right)-\left(11x^2-3x^2\right)-\left(3x-x\right)+\left(3+1\right)\)
\(=-3x^3-8x^2-2x+4\)
\(\left(2x-y\right)\left(4x^2+2xy+y^2\right)\)
\(=\left(2x-y\right)[\left(2x\right)^2+2xy+y^2]\)
\(=8x^3-y^3\)
\(\left(6x^5y^2-9x^4y^3+15x^3y^4\right):3x^3y^2\)
\(=(6x^5y^2:3x^3y^2)-(9x^4y^3:3x^3y^2)+(15x^3y^4:3x^3y^2)\)
\(=2x^2-3xy+5y^2\)
\(\left(x^3-3x^2+x-3\right):\left(x-3\right)\)
\(=[\left(x^3-3x^2\right)+\left(x-3\right)]:\left(x-3\right)\)
\(=\left(x-3\right)\left(x^2+1\right):\left(x-3\right)\)
\(=x^2+1\)
Bài 2:
\(5x\left(x-1\right)=10\left(x-1\right)\)
\(\Rightarrow5x\left(x-1\right)-10\left(x-1\right)=0\)
\(\Rightarrow\left(x-1\right)\left(5x-10\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-1=0\\5x-10=0\end{cases}\Rightarrow\orbr{\begin{cases}x=1\\x=2\end{cases}}}\)
\(2\left(x+5\right)-x^2-5x=0\)
\(\Rightarrow2\left(x+5\right)-x\left(x+5\right)=0\)
\(\Rightarrow\left(x+5\right)\left(2-x\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x+5=0\\2-x=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-5\\x=2\end{cases}}}\)
\(x^3-x=0\)
\(\Rightarrow x\left(x^2-1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x^2-1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=\pm1\end{cases}}\)
\(\left(2x-1\right)^2-\left(4x-3\right)^2=0\)
\(\Rightarrow[\left(2x-1\right)-\left(4x-3\right)][\left(2x-1\right)+\left(4x-3\right)]=0\)
\(\Rightarrow\left(-2x+2\right)\left(6x-4\right)=0\)
\(\Rightarrow\orbr{\begin{cases}-2x+2=0\\6x-4=0\end{cases}}\Rightarrow\orbr{\begin{cases}-2x=-2\\6x=4\end{cases}}\Rightarrow\orbr{\begin{cases}x=1\\x=\frac{2}{3}\end{cases}}\)
\(\left(5x+3\right)\left(x-4\right)-\left(x-5\right)x=\left(2x-5\right)\left(5+2x\right)\)
\(\Rightarrow\left(5x^2-20x+3x-12\right)-x^2+5x=\left(2x-5\right)\left(2x+5\right)\)
\(\Rightarrow\left(5x^2-17x-12\right)-x^2+5x=\left(2x\right)^2-5^2\)
\(\Rightarrow\left(5x^2-x^2\right)-\left(17x-5x\right)-12-\left(4x^2-25\right)=0\)
\(\Rightarrow\left(4x^2-4x^2\right)-12x+\left(25-12\right)=0\)
\(\Rightarrow12x=-13\)
\(\Rightarrow x=\frac{-13}{12}\)
Bài 3:
\(x\left(3x+12\right)-\left(7x-20\right)+x^2\left(2x-3\right)-x\left(2x^2+5\right)\)
\(=3x^2+12x-7x+20+2x^3-3x^2-2x^3-5x\)
\(=20\)
Vậy giá trị của biểu thức không phụ thuộc vào biến
\(3\left(2x-1\right)-5\left(x-3\right)+6\left(3x-4\right)-19x\)
\(=6x-3-5x+15+18x-24-19x\)
\(=-12\)
Vậy giá trị của biểu thức không phụ thuộc vào biến
Bài 4:
\(10x\left(x-y\right)-8\left(y-x\right)\)
\(=10x\left(x-y\right)+8\left(x-y\right)\)
\(=\left(10x+8\right)\left(x-y\right)\)
\(=2\left(5x+4\right)\left(x-y\right)\)
\(\left(3x+1\right)^2-\left(2x+1\right)^2\)
\(=\left(3x+1-2x-1\right)\left(3x+1+2x+1\right)\)
\(=x\left(5x+2\right)\)
\(-5x^2+10xy-5y^2+20z^2\)
\(=-5\left(x^2-2xy+y^2-4z^2\right)\)
\(=-5\left(\left(x-y\right)^2-4z^2\right)\)
\(=-5\left(x-y-2z\right)\left(x-y+2z\right)\)
\(2x^2-9xy-5y^2\)
\(=2x^2-10xy+xy-5y^2\)
\(=2x\left(x-5y\right)+y\left(x-5y\right)\)
\(=\left(x-5y\right)\left(2x+y\right)\)
\(x^3-4x^2+4x-xy^2\)
\(=x[\left(x^2-4x+4\right)-y^2]\)
\(=x[\left(x-2\right)^2-y^2]\)
\(=x\left(x-y-2\right)\left(x+y-2\right)\)