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\(B=\left(5x-4y\right)^2-\left(6x+4y\right)\left(5x-4y\right)+\left(3x+2y\right)^2\)
\(B=\left(5x-4y\right)\left(5x-4y-6x-4y\right)+\left(3x+2y\right)^2\)
\(B=\left(5x-4y\right)\left(-x-8y\right)+\left(3x+2y\right)^2\)
\(B=-5x^2-40xy+4xy+32y^2+9x^2+12xy+4y^2\)
\(B=4x^2-24xy+36y^2\)
\(B=x^2-6xy+6y^2\)
Bài chưa đc ktra lại đâu . Có gì sai sót thì bỏ qua
Bài 1 :
a, \(\left(x^2-2x+3\right)\left(x-4\right)=0\)
TH1 : \(x^2-2x+3=0\)
\(\left(-2\right)^2-4.3=4-12< 0\)vô nghiệm
TH2 : \(x-4=0\Leftrightarrow x=4\)
b, \(\left(2x^2-3x-1\right)\left(5x+2\right)=0\)
TH1 : \(\left(-3\right)^2-4.\left(-1\right).2=9+8=17>0\)
\(\Rightarrow x_1=\frac{3-\sqrt{17}}{4};x_2=\frac{3+\sqrt{17}}{4}\)
TH2 ; \(5x+2=0\Leftrightarrow x=-\frac{2}{5}\)
c, đưa về hệ đc ko ?
d, \(\left(5x^3-x^2+2x-3\right)\left(4x^2-x+2\right)=0\)
TH1 : \(x=0,74...\) ( bấm máy cx ra )
TH2 : \(\left(-1\right)^2-4.2.4< 0\)vô nghiệm
KL : vô nghiệm
Bài 2 :
a, \(\left(3x-1\right)\left(2x+7\right)-\left(x+1\right)\left(6x-5\right)-\left(18x-12\right)\)
\(=6x^2+21x-2x-7-6x^2+5x-6x+5-18x+12=10\)
Vậy biểu thức ko phụ thuộc vào biến
b, \(\left(x-y\right)\left(x^3+x^2y+xy^2+y^3\right)-x^4y^4\)
\(=x^4+x^3y+x^2y^2+xy^3-yx^3-y^2x^2-y^3x-y^4-x^4y^4\)
\(=x^4-y^4-x^4y^4\)Vậy biểu thức phụ thuộc vào biến
\(A=3x^2+5x-2\)
\(A=3\left(x^2+\frac{5}{3}x-\frac{2}{3}\right)\)
\(A=3\left(x^2+2.\frac{5}{6}x+\left(\frac{5}{6}\right)^2-\frac{49}{36}\right)\)
\(A=3\left(x^2+2.\frac{5}{6}x+\left(\frac{5}{6}\right)^2\right)-\frac{49}{12}\)
\(A=3\left(x+\frac{5}{6}\right)^2-\frac{49}{12}\)
Vì \(3\left(x+\frac{5}{6}\right)^2\ge0\)
Do đó \(3\left(x+\frac{5}{6}\right)^2-\frac{49}{12}\ge-\frac{49}{12}\)
Dấu = xảy ra khi \(x+\frac{5}{6}=0\Rightarrow x=-\frac{5}{6}\)
Vậy Min A=\(-\frac{49}{12}\) khi x=\(-\frac{5}{6}\)
mk làm ý a thôi, mấy ý sau dựa vào mà làm.
A = \(3x^2+5x-2\)
=> \(\frac{A}{3}=x^2+\frac{5}{3}x-\frac{2}{3}\)(chia cả 2 vế cho 3)
\(\Leftrightarrow\frac{A}{3}=x^2+2.x.\frac{5}{6}+\left(\frac{5}{6}\right)^2-\frac{49}{36}\)
\(\Leftrightarrow\frac{A}{3}=\left(x+\frac{5}{6}\right)^2-\frac{49}{36}\)
\(\Rightarrow A=3\left(x+\frac{5}{6}\right)^2-\frac{49}{12}\ge-\frac{49}{12}\)
Đẳng thức xảy ra <=> x = - 5/6.
Vậy Min A = - 49/12 khi và chỉ khi x = - 5/6.
a) \(5x^2-12xy+9y^2-4x+4=\left(4x^2-12xy+9y^2\right)+x^2-4x+4=\left(2x-3y\right)^2+\left(x-2\right)^2\ge0\)
b) \(-x^2-2y^2+12x-4y+7=-\left(x^2-12x+36\right)-2\left(y^2+2y+1\right)+45=-\left(x-6\right)^2-2\left(y+1\right)^2+45\le45\)
c)\(4y^2+10x^2+12xy+6x+7=\left(4y^2+12xy+9x^2\right)+x^2+6x+9-2=\left(2y+3x\right)^2+\left(x+3\right)^2-2\ge-2\)
d) \(3-10x^2-4xy-4y^2=3-\left(4y^2+4xy+x^2\right)-9x^2=-\left(2y+x\right)^2-9x^2+3\le3\)
e)\(x^2-5x+y^2-xy-4y+16=\left(\frac{1}{2}x^2-xy+\frac{1}{2}y^2\right)+\frac{1}{2}\left(x^2-10x+25\right)+\frac{1}{2}\left(y^2-8y+16\right)-\frac{9}{2}=\frac{1}{2}\left(x-y\right)^2+\frac{1}{2}\left(x-5\right)^2+\frac{1}{2}\left(y-4\right)^2-\frac{9}{2}\ge-\frac{9}{2}\)Phần e) mới nghĩ đk v, tui biết đáp án sao do k xảy ra dấu bằng
a: \(\dfrac{-6x^3y^4+4x^4y^3}{2x^3y^3}\)
\(=\dfrac{-6x^3y^4}{2x^3y^3}+\dfrac{4x^4y^3}{2x^3y^3}\)
\(=-3y+2x\)
b: \(\dfrac{5x^4y^2-x^3y^2}{x^3y^2}=\dfrac{5x^4y^2}{x^3y^2}-\dfrac{x^3y^2}{x^3y^2}\)
\(=5x-1\)
c: \(\dfrac{27x^3y^5+9x^2y^4-6x^3y^3}{-3x^2y^3}\)
\(=-\dfrac{27x^3y^5}{3x^2y^3}-\dfrac{9x^2y^4}{3x^2y^3}+\dfrac{6x^3y^3}{3x^2y^3}\)
\(=-9xy^2-3y+2x\)
a) \(\dfrac{-6x^3y^4+4x^4y^3}{2x^3y^3}\)
\(=\dfrac{2x^3y^3\cdot\left(-3y+2x\right)}{2x^3y^3}\)
\(=-3y+2x\)
\(=2x-3y\)
b) \(\dfrac{5x^4y^2-x^3y^2}{x^3y^2}\)
\(=\dfrac{5x\cdot x^3y^2-x^3y^2\cdot1}{x^3y^2}\)
\(=\dfrac{x^3y^2\cdot\left(5x-1\right)}{x^3y^2}\)
\(=5x-1\)
c) \(\dfrac{27x^3y^5+9x^2y^4-6x^3y^3}{-3x^2y^3}\)
\(=\dfrac{-3x^2y^3\cdot-9xy^2+-3x^2y^3\cdot-3y+-3x^2y^3\cdot2x}{-3x^2y^3}\)
\(=\dfrac{-3x^2y^3\cdot\left(-9xy^2-3y+2x\right)}{-3x^2y^3}\)
\(=-9xy^2-3x+2x\)
\(\left(5x-4y\right)^2-\left(6x+4y\right).\left(5x-4y\right)+\left(3x+2y\right)^2\)
\(=\left(5x-4y\right)^2-2.\left(5x-4y\right).\left(3x+2y\right)+\left(3x+2y\right)^2\)
\(=\left(5x-4y-3x-2y\right)^2\)
\(=\left(2x-6y\right)^2\)
\(=4x^2-24xy+36y^2\)