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a/ \(A=\left(x-y\right)^2+\left(x+y\right)^2.\)
\(A=\left(x^2-2xy+y^2\right)+\left(x^2+2xy+y^2\right)\)
\(A=x^2-2xy+y^2+x^2+2xy+y^2\)
\(A=2x^2+2y^2\)
b/ \(B=\left(2a+b\right)^2-\left(2a-b\right)^2\)
\(B=\left(4a^2+4ab+b^2\right)-\left(4a^2-4ab+b^2\right)\)
\(B=4a^2+4ab+b^2-4a^2+4ab-b^2\)
\(B=8ab\)
c/ \(C=\left(x+y\right)^2-\left(x-y\right)^2\)
\(C=\left(x^2+2xy+y^2\right)-\left(x^2-2xy+y^2\right)\)
\(C=x^2+2xy+y^2-x^2+2xy-y^2\)
\(C=4xy\)
d/ \(D=\left(2x-1\right)^2-2\left(2x-3\right)^2+4\)
\(D=\left(4x^2-4x+1\right)-2\left(4x^2-12x+9\right)+4\)
\(D=4x^2-4x+1-8x^2+24x-18+4\)
\(D=-4x^2+20x-13\)
a. \(8x\left(x-2017\right)-2x+4034=0\)
\(8x\left(x-2017\right)-2\left(x-2017\right)=0\)
\(\left(8x-2\right)\left(x-2017\right)=0\)
\(\Rightarrow TH1:8x-2=0\)
\(8x=2\)
\(x=\frac{1}{4}\)
\(TH2:x-2017=0\)
\(x=2017\)
Vậy \(x\in\left\{\frac{1}{4};2017\right\}\)
Bài 1
a) \(8x\left(x-2017\right)-2x+4034=0\)
\(\Rightarrow8x\left(x-2017\right)-2\left(x-2017\right)=0\)
\(\Rightarrow\left(x-2017\right)\left(4x-1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=2017\\x=\frac{1}{4}\end{cases}}\)
\(B=x^2-6x+y^2-2y+12=\left(x^2-6x+9\right)\left(y^2-2y+1\right)+2\)
\(B=\left(x-3\right)^2+\left(y-1\right)^2+2\text{ }\)
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
\(a.\) Với \(a+b+c=0\) thì \(\frac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{abc}=\frac{\left(-c\right).\left(-a\right).\left(-b\right)}{abc}=\frac{-abc}{abc}=-1\)
\(b.\) Công thức tổng quát: \(\frac{1}{n\left(n+1\right)}=\frac{1}{n}-\frac{1}{n+1}\)
Ta có:
\(\frac{1}{x\left(x+1\right)}=\frac{1}{x}-\frac{1}{x+1}\)
\(\frac{1}{\left(x+1\right)\left(x+2\right)}=\frac{1}{x+1}-\frac{1}{x+2}\)
\(\frac{1}{\left(x+2\right)\left(x+3\right)}=\frac{1}{x+2}-\frac{1}{x+3}\)
\(\frac{1}{\left(x+3\right)\left(x+4\right)}=\frac{1}{x+3}-\frac{1}{x-4}\)
\(\frac{1}{\left(x+4\right)\left(x+5\right)}=\frac{1}{x+4}-\frac{1}{x+5}\)
Do đó, suy ra được: \(A=\frac{1}{x}-\frac{1}{x+5}=\frac{x+5-x}{x\left(x+5\right)}=\frac{5}{x\left(x+5\right)}\)
Bài 6
\(\left(a-b\right)^2=a^2-2ab+b^2\)
\(=\left(a^2+2ab+b^2\right)-4ab\)
\(=\left(a+b\right)^2-4ab\)
Bài 5 :
\(a,16x^2-\left(4x-5\right)^2=15\)
\(16x^2-16x^2+40x-25-15=0\)
\(40x-40=0\)
\(40x=40\)
\(x=1\)
\(b,\left(2x+3\right)^2-4\left(x-1\right)\left(x+1\right)=49\)
\(4x^2+12x+9-4x^2+4=49\)
\(12x=36\)
\(x=3\)
\(c,\left(2x+1\right)\left(2x-1\right)+\left(1-2x\right)^2=18\)
\(4x^2-1+1-4x+4x^2=18\)
\(8x^2-4x-18=0\)
\(2\left(4x^2-2x-9\right)=0\)
\(x=\frac{1-\sqrt{37}}{4}\)
\(d,2\left(x+1\right)^2-\left(x-3\right)\left(x+3\right)-\left(x-4\right)^2=0\)
\(2x^2+4x+2-x^2+9-x^2+8x-16=0\)
\(12x=4\)
\(x=\frac{1}{3}\)
\(a,\frac{1}{2}x+\frac{1}{2}+\frac{1}{4}x+\frac{3}{4}=3-\frac{1}{3}x-\frac{2}{3}\)
\(\frac{13}{12}x=\frac{13}{12}\Rightarrow x=1\)
a) \(=\frac{2\left(x+y\right)+5\left(x+y\right)}{2\left(x+y\right)-5\left(x+y\right)}\)
\(=\frac{7\left(x+y\right)}{-3\left(x+y\right)}=\frac{-7}{3}\)
b)\(=\frac{3x\left(x+y\right)}{y}\)
c) \(\frac{5\left(x-y\right)+3\left(x-y\right)}{10\left(x-y\right)}\)
\(=\frac{8\left(x-y\right)}{10\left(x-y\right)}=\frac{4}{5}\)
a) \(\frac{2x+2y+5x+5y}{2x+2y-5x-5y}=\frac{7x+7y}{-3x-3y}=\frac{7\left(x+y\right)}{-3\left(x+y\right)}=-\frac{7}{3}.\)
b) \(\frac{15x\left(x+y\right)^3}{5y\left(x+y\right)^2}=\frac{3x\left(x+y\right)}{y}=\frac{3x^2+3xy}{y}\)
c) \(\frac{5\left(x-y\right)-3\left(y-x\right)}{10\left(x-y\right)}=\frac{5\left(x-y\right)+3\left(x-y\right)}{10\left(x-y\right)}=\frac{8\left(x-y\right)}{10\left(x-y\right)}=\frac{4}{5}\)
d) \(\frac{3\left(x-y\right)\left(x-z\right)^2}{6\left(x-y\right)\left(x-z\right)}=\frac{x-z}{2}\)
h) \(\frac{3x\left(1-x\right)}{2\left(x-1\right)}=-\frac{3x\left(x-1\right)}{2\left(x-1\right)}=\frac{-3x}{2}\)
j) \(\frac{6x^2y^2}{8xy^5}=\frac{3x}{4y^3}\)
Câu b) bạn xem lại nhé.
Học tốt ^3^
Khó vl , dẹp mẹ điiii
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 !