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\(C1:=3+1-3y\)
\(=4-3y\)
\(C2:\)
\(a.=3x\left(2y-1\right)\)
\(b.=\left(x-y\right)\left(x+y\right)+4\left(x+y\right)\)
\(=\left(x-y+4\right)\left(x+y\right)\)
\(C3:\)
\(a.6x^2+2x+12x-6x^2=7\)
\(14x=7\)
\(x=\frac{1}{2}\)
\(b.\frac{1}{5}x-2x^2+2x^2+5x=-\frac{13}{2}\)
\(\frac{26}{5}x=-\frac{13}{2}\)
\(x=-\frac{13}{2}\times\frac{5}{26}\)
\(x=-\frac{5}{4}\)
Bạn Moon làm kiểu gì vậy ?
1) \(\left(3x^2y^2+x^2y^2\right):\left(x^2y^2\right)-3y\)
\(=\left[\left(x^2y^2\right)\left(3+1\right)\right]:\left(x^2y^2\right)-3y\)
\(=4-3y\)
2) a, \(6xy-3x=\left(3x\right)\left(2y-1\right)\)
b, \(x^2-y^2+4x+4y=\left(x+y\right)\left(x-y\right)+4\left(x+y\right)\)
\(=\left(x+y\right)\left(x-y+4\right)\)
3) a, \(2x\left(3x+1\right)+\left(4-2x\right)3x=7\)
\(< =>6x^2+2x+12x-6x^2=7\)
\(< =>14x=7< =>x=\frac{7}{14}\)
b, \(\frac{1}{2}x\left(\frac{2}{5}-4x\right)+\left(2x+5\right)x=-6\frac{1}{2}\)
\(< =>\frac{x}{2}.\frac{2}{5}-\frac{x}{2}.4x+2x^2+5x=-\frac{13}{2}\)
\(< =>\frac{x}{5}-2x^2+2x^2+5x=-\frac{13}{2}\)
\(< =>\frac{26x}{5}=\frac{-13}{2}\)
\(< =>26x.2=\left(-13\right).5\)
\(< =>52x=-65< =>x=-\frac{65}{52}=-\frac{5}{4}\)
\(\left(x+5\right)^2-3\left(x+5\right)\)
\(=\left(x+5\right)\left(x+5-3\right)\)
\(=\left(x+5\right)\left(x+2\right)\)
\(2x\left(x-3\right)-\left(x-3\right)^2\)
\(=\left(x-3\right)\left(2x-x+3\right)\)
\(=\left(x-3\right)\left(x+3\right)\)
a) \(4x^2-6x=2x\left(2x-3\right)\)
b) \(9x^4y^3+3x^2y^4=3x^2y^3\left(3x^2+y\right)\)
c) \(3\left(x-y\right)-5x\left(y-x\right)=3\left(x-y\right)+5x\left(x-y\right)\)
\(=\left(5x+3\right)\left(x-y\right)\)
d) \(x^3-2x^2+5x=x\left(x^2-2x+5\right)\)
e) \(5\left(x+3y\right)-15x\left(x+3y\right)=\left(5-15x\right)\left(x+3y\right)\)
\(=5\left(1-3x\right)\left(x+3y\right)\)
f) \(2x^2\left(x+1\right)-4\left(x+1\right)=\left(2x^2-4\right)\left(x+1\right)\)
\(=\left(\sqrt{2}x-2\right)\left(\sqrt{2}x+2\right)\left(x+1\right)\)
\(B=x^2-6x+y^2-2y+12=\left(x^2-6x+9\right)\left(y^2-2y+1\right)+2\)
\(B=\left(x-3\right)^2+\left(y-1\right)^2+2\text{ }\)
Ta thấy B lớn hơn hoặc bằng 2 suy ra GTNN của B là 2
Dấu = xảy ra khi x=3; y=1
\(C=2x^2-6x=\left(2x^2-6x+4,5\right)-4,5=2\left(x^2-3x+2,25\right)-4,5\)
\(C=2\left(x-1,5\right)^2-4,5\)
Ta thấy C luôn luôn lớn hơn hoặc bằng -4,5 nên GTNN của C là -4,5
Dấu = xảy ra khi x=1,5
Tối mình full cho còn giờ mình đi đá bóng đây
1) \(D=\frac{2016}{-4x^2+4x-5}\). Để D đạt giá trị nhỏ nhất suy ra \(-4x^2+4x-5\)đạt giá trị lớn nhất.
Ta có \(-4x^2+4x-5=-4x^2+4x-1-4=\left(-4x^2+4x-1\right)-4\)
\(-4\left(x^2-x+\frac{1}{4}\right)-4=-4\left(x-\frac{1}{2}\right)^2-4\).
Ta Thấy:\(-4\left(x-\frac{1}{2}\right)^2\) bé hơn hoặc bằng 0 nên \(-4\left(x-\frac{1}{2}\right)^2-4\)bé hơn hoặc bằng -4
nên ..... bạn tự kết luận
\(x^3+y^3+z^3-3xyz\)
\(=\left(x+y\right)^3-3xy\left(x+y\right)+z^3-3xyz\)
\(=\left(x+y+z\right)\left[\left(x+y\right)^2-\left(x+y\right)z+z^2\right]-3xy\left(x+y+z\right)\)
\(=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)\)
BÀI 1:
a) \(x^4+2x^2y+y^2=\left(x^2+y\right)^2\)
b) \(\left(2a+b\right)^2-\left(2b+a\right)^2=\left(2a+b+2b+a\right)\left(2a+b-2b-a\right)\)
\(=\left(3a+3b\right)\left(a-b\right)=3\left(a+b\right)\left(a-b\right)\)
c) \(\left(a^3-b^3\right)+\left(a-b\right)^2=\left(a-b\right)\left(a^2+ab+b^2\right)+\left(a-b\right)^2\)
\(=\left(a-b\right)\left[a^2+ab+b^2+\left(a-b\right)\right]=\left(a-b\right)\left(a^2+ab+b^2+a-b\right)\)
d) \(\left(x^2+1\right)^2-4x^2=\left(x^2+1-2x\right)\left(x^2+1+2x\right)=\left(x-1\right)^2\left(x+1\right)^2\)
e) \(\left(y^3+8\right)+\left(y^2-4\right)=\left(y+2\right)\left(y^2-y+2\right)\)
f) \(1-\left(x^2-2xy+y^2\right)=1-\left(x-y\right)^2=\left(1-x+y\right)\left(1+x-y\right)\)
g) \(x^4-1=\left(x-1\right)\left(x+1\right)\left(x^2+1\right)\)
h) ktra lại đề
m) \(\left(x-a\right)^4-\left(x+a\right)^4=-8ax\left(a^2+x^2\right)\)
a) \(A=\left(\frac{1}{4}x-y\right)\left(x^2+4xy+16y^2\right)+4\left(4y^3-\frac{1}{16}x^3+1\right)\)
\(\Leftrightarrow A=\frac{1}{4}\left(x-4y\right)\left(x^2+4xy+16y^2\right)+16y^3-\frac{1}{4}x^3+4\)
\(\Leftrightarrow A=\frac{1}{4}\left(x^3-64y^3\right)+16y^3-\frac{1}{4}x^3+4\)
\(\Leftrightarrow A=\frac{1}{4}x^3-16y^3+16y^3-\frac{1}{4}x^3+4\)
\(\Leftrightarrow A=4\)
b) \(B=2x\left(x-4\right)^2-\left(x+5\right)\left(x-2\right)\left(x+2\right)+2\left(x-5\right)^2-\left(x-1\right)^2\)
\(\Leftrightarrow B=2x\left(x^2-8x+16\right)-\left(x+5\right)\left(x^2-4\right)+2\left(x^2-10x+25\right)-\left(x^2-2x+1\right)\)
\(\Leftrightarrow B=2x^3-16x^2+32x-x^3-5x^2+4x+20+2x^2-20x+50-x^2+2x-1\)
\(\Leftrightarrow B=x^3-20x^2+18x+69\)
c) \(C=\frac{80x^3-125x}{3\left(x-3\right)-\left(x-3\right)\left(8-4x\right)}\)
\(\Leftrightarrow C=\frac{5x\left(16x^2-25\right)}{\left(x-3\right)\left(3-8+4x\right)}\)
\(\Leftrightarrow C=\frac{5x\left(4x-5\right)\left(4x+5\right)}{\left(x-3\right)\left(4x-5\right)}\)
\(\Leftrightarrow C=\frac{5x\left(4x+5\right)}{x-3}\)
\(\Leftrightarrow C=\frac{20x^2+25x}{x-3}\)
d) \(D=\frac{\left(a-b\right)\left(c-d\right)}{\left(b^2-a^2\right)\left(d^2-c^2\right)}\)
\(\Leftrightarrow D=\frac{\left(a-b\right)\left(c-d\right)}{\left(a^2-b^2\right)\left(c^2-d^2\right)}\)
\(\Leftrightarrow D=\frac{\left(a-b\right)\left(c-d\right)}{\left(a-b\right)\left(a+b\right)\left(c-d\right)\left(c+d\right)}\)
\(\Leftrightarrow D=\frac{1}{\left(a+b\right)\left(c+d\right)}\)
Chúc bạn học tốt !