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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/ \(D=\left(2x-1\right)^2-2\left(2x-3\right)^2+4\)
\(D=\left(4x^2-4x+1\right)-2\left(4x^2-12x+9\right)+4\)
\(D=4x^2-4x+1-8x^2+24x-18+4\)
\(D=-4x^2+20x-13\)
Ta có: (2x + 1)2 + (2x- 1)2 - 2(1 + 2x)(2x - 1)
= (2x + 1 - 2x + 1)2
= 22 = 4
1) \(x\left(2^2-3\right)-x^2\left(5x+1\right)+x^2\)
\(=x-5x^3-x^2+x\)
\(=2x-5x^3-x^2\)
2) \(\left(3x-5\right)\left(2x+11\right)-\left(2x+3\right)\left(3x+7\right)\)
\(=\left(6x^2+23x-55\right)-\left(6x^2+23x+21\right)\)
\(=-76\)
Làm lại câu 1
\(x\left(2^2-3\right)-x^2\left(5x+1\right)+x^2\)
\(=x-5x^3-x^2+x^2\)
\(=x-5x^3\)
\(\frac{x^2}{\left(x+y\right)\left(1-y\right)}-\frac{y^2}{\left(x+y\right)\left(1+x\right)}-\frac{x^2y^2}{\left(1+x\right)\left(1-y\right)}.\)
\(=\frac{x^2\left(1+x\right)-y^2\left(1-y\right)-x^2y^2\left(x+y\right)}{\left(x+y\right)\left(1+x\right)\left(1-y\right)}\)
\(=\frac{x^2+x^3-y^2+y^3-x^3y^2-x^2y^3}{\left(x+y\right)\left(1-x\right)\left(1-y\right)}\)
\(=\frac{\left(x^2-y^2\right)+\left(x^3+y^3\right)-x^2y^2\left(x+y\right)}{\left(x+y\right)\left(1+x\right)\left(1-y\right)}\)
\(=\frac{\left(x+y\right)\left(x-y\right)+\left(x+y\right)\left(x^2-xy+y^2\right)-x^2y^2\left(x+y\right)}{\left(x+y\right)\left(1+x\right)\left(1-y\right)}\)
\(=\frac{\left(x+y\right)\left(x-y+x^2-xy+y^2-x^2y^2\right)}{\left(x+y\right)\left(1+x\right)\left(1-y\right)}\)
\(=\frac{2x+x^2+y^2-x^2y^2}{\left(1+x\right)\left(1-y\right)}\)
\(a.\) 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: \(A=\frac{1}{x}-\frac{1}{x+5}=\frac{x+5-x}{x\left(x+5\right)}=\frac{5}{x\left(x+5\right)}\)
Đề sai ! Sửa nhé :
a) \(ĐKXĐ:\hept{\begin{cases}x\ne0\\x\ne\pm2\end{cases}}\)
\(A=\left(\frac{2}{x+2}-\frac{4}{x^2+4x+4}\right):\left(\frac{2}{x^2-4}+\frac{1}{2-x}\right)\)
\(\Leftrightarrow A=\left(\frac{2}{x+2}-\frac{4}{\left(x+2\right)^2}\right):\left(\frac{2}{\left(x-2\right)\left(x+2\right)}-\frac{1}{x-2}\right)\)
\(\Leftrightarrow A=\frac{2\left(x+2\right)-4}{\left(x+2\right)^2}:\frac{2-\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}\)
\(\Leftrightarrow A=\frac{2x+4-4}{\left(x+2\right)^2}.\frac{\left(x+2\right)\left(x-2\right)}{-x}\)
\(\Leftrightarrow A=\frac{2x\left(x-2\right)}{-x\left(x+2\right)}\)
\(\Leftrightarrow A=-\frac{2\left(x-2\right)}{x+2}\)
b) Để \(A\le-2\)
\(\Leftrightarrow-\frac{2\left(x-2\right)}{x+2}\le-2\)
\(\Leftrightarrow\frac{2\left(x-2\right)}{x+2}\ge2\)
\(\Leftrightarrow\frac{x-2}{x+2}\ge1\)
\(\Leftrightarrow x-2\ge x+2\)
\(\Leftrightarrow-2\ge2\)(ktm)
Vậy để \(A\le-2\Leftrightarrow x\in\varnothing\)
a.
\(A=\left(\frac{2}{x+2}-\frac{4}{x^2+4+4}\right):\left(\frac{2}{x^2-4}+\frac{1}{2-x}\right)\)
\(A=\left(\frac{2.\left(x^2+8\right)}{\left(x+2\right).\left(x^2+8\right)}-\frac{4\left(x+2\right)}{\left(x+2\right)\left(x^2+8\right)}\right):\left(\frac{2}{\left(x-2\right)\left(x+2\right)}+\frac{1}{2-x}\right)\)
\(A=\left(\frac{2x^2+8-4x+8}{\left(x+2\right)\left(x^2+8\right)}\right):\left(\frac{2}{\left(x-2\right)\left(x+2\right)}+\frac{-1}{x-2}\right)\)
\(A=\left(\frac{2x\left(x-2\right)+16}{\left(x+2\right)\left(x^2+8\right)}\right):\left(\frac{2}{\left(x-2\right)\left(x+2\right)}+\frac{-x-2}{\left(x-2\right)\left(x+2\right)}\right)\)
\(A=\left(\frac{2x\left(x-2\right)+16}{\left(x+2\right)\left(x^2+8\right)}\right):\left(\frac{2-x-2}{\left(x-2\right)\left(x+2\right)}\right)\)
\(A=\left(\frac{\left(2x\left(x-2\right)+16\right)\left(x-2\right)\left(x+2\right)}{\left(x+2\right)\left(x^2+8\right)\left(-x\right)}\right)\)
\(A=\frac{\left(2x\left(x-2\right)+16\right)\left(x-2\right)}{\left(x^2+8\right)\left(-x\right)}\)
\(A=\frac{\left(2x^2-4x+16\right)\left(x-2\right)}{\left(x^2+8\right)\left(-x\right)}\)
\(A=\frac{\left(2x^3-4x-4x-4x^2+8x+16x-32\right)}{-x^3+8}\)
\(A=\frac{2x^3-4x^2+16x-32}{-x^3+8}\)
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) \(=\frac{2\left(x+y\right)+5\left(x+y\right)}{2\left(x+y\right)-5\left(x+y\right)}\)
\(=\frac{7\left(x+y\right)}{-3\left(x+y\right)}=\frac{-7}{3}\)
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^
\(\left(x-1\right)^3-\left(x+2\right)\left(x^2-2x+4\right)+3\left(x-1\right)\left(x+1\right).\)
\(=x^3-3x^2+3x-1-\left(x^3-2^3\right)+3\left(x^2-1\right)\)
\(=x^3-3x^2+3x-1-x^3+8+3x^2-3\)
\(=3x+4\)
\(\left(x-1\right)^3-\left(x+2\right)\left(x^2-2x+4\right)+3\left(x-1\right)\left(x+1\right)\)
\(=\left(x-1\right)^3-\left(x^3+8\right)+3\left(x-1\right)\left(x+1\right)\)
\(=\left(x-1\right)^3-x^3-8+3\left(x-1\right)\left(x+1\right)\)
\(=\left(x-1-x\right)+3x\left(x-1\right)\left(x-1-x\right)-8+3\left(x-1\right)\left(x+1\right)\)(1)
\(=-1-3x\left(x-1\right)-8+3\left(x-1\right)\left(x+1\right)\)
\(=3\left(x-1\right)\left(-x+x+1\right)-9=3\left(x-1\right)-9=3\left(x-4\right)=3x-12\)
(1) là hằng đẳng thức \(x^3-y^3=\left(x-y\right)^3+3xy\left(x-y\right)\)