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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.
\(A=\left(\frac{2}{x+2}-\frac{4}{x^2+4+4}\right):\left(\frac{2}{x^2-4}+\frac{1}{2-x}\right)\)
\(A=\left(\frac{2.\left(x^2+8\right)}{\left(x+2\right).\left(x^2+8\right)}-\frac{4\left(x+2\right)}{\left(x+2\right)\left(x^2+8\right)}\right):\left(\frac{2}{\left(x-2\right)\left(x+2\right)}+\frac{1}{2-x}\right)\)
\(A=\left(\frac{2x^2+8-4x+8}{\left(x+2\right)\left(x^2+8\right)}\right):\left(\frac{2}{\left(x-2\right)\left(x+2\right)}+\frac{-1}{x-2}\right)\)
\(A=\left(\frac{2x\left(x-2\right)+16}{\left(x+2\right)\left(x^2+8\right)}\right):\left(\frac{2}{\left(x-2\right)\left(x+2\right)}+\frac{-x-2}{\left(x-2\right)\left(x+2\right)}\right)\)
\(A=\left(\frac{2x\left(x-2\right)+16}{\left(x+2\right)\left(x^2+8\right)}\right):\left(\frac{2-x-2}{\left(x-2\right)\left(x+2\right)}\right)\)
\(A=\left(\frac{\left(2x\left(x-2\right)+16\right)\left(x-2\right)\left(x+2\right)}{\left(x+2\right)\left(x^2+8\right)\left(-x\right)}\right)\)
\(A=\frac{\left(2x\left(x-2\right)+16\right)\left(x-2\right)}{\left(x^2+8\right)\left(-x\right)}\)
\(A=\frac{\left(2x^2-4x+16\right)\left(x-2\right)}{\left(x^2+8\right)\left(-x\right)}\)
\(A=\frac{\left(2x^3-4x-4x-4x^2+8x+16x-32\right)}{-x^3+8}\)
\(A=\frac{2x^3-4x^2+16x-32}{-x^3+8}\)
3/
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/ \(D=\left(2x-1\right)^2-2\left(2x-3\right)^2+4\)
\(D=\left(4x^2-4x+1\right)-2\left(4x^2-12x+9\right)+4\)
\(D=4x^2-4x+1-8x^2+24x-18+4\)
\(D=-4x^2+20x-13\)
\(a,x^3-x^2-12x+45=0\)
\(\left(x-3\right)\left(x-3\right)\left(x+5\right)=0\)
\(x=3;3;-5\)
\(b,2x^3-5x^2+8x-5=0\)
\(\left(2x^2-3x+5\right)\left(x-1\right)=0\)
\(x=1\)
lm 1 câu đã chán ngắt , giải mấy câu nữa não tớ nổ bùmmm , tớ bt đây là trang web để hc nhưng tạo nên tiếng cười là chính nha ^^
Tại vì nó được đề bài cho nên có nghĩa,k có nghĩa thì lm kiểu đếch j?
\(d,\frac{10x+3}{8}=\frac{7-8x}{12}\)
\(\left(10x+3\right):8=\left(7-8x\right):12\)
\(\left(10x+3\right).\frac{1}{8}=\left(7-8x\right).\frac{1}{12}\)
\(\frac{5}{4}x+\frac{3}{8}=\frac{7}{12}-\frac{8}{12}x\)
\(\frac{5}{4}x+\frac{8}{12}x=\frac{7}{12}-\frac{3}{8}\)
\(\frac{23}{12}x=\frac{5}{24}\)
\(x=\frac{5}{46}\)
E mới lớp 6 nên giải sai thì thông cảm ạ UwU
\(b,\frac{x}{10}-\left(\frac{x}{30}+\frac{2x}{45}\right)=\frac{4}{5}\)
\(< =>\frac{9x}{90}-\frac{7x}{90}=\frac{4}{5}\)
\(< =>\frac{x}{45}=\frac{32}{45}\)
\(< =>x=32\)
\(d,\frac{10x+3}{8}=\frac{7-8x}{12}\)
\(< =>\left(10x+3\right).12=\left(7-8x\right).8\)
\(< =>120x+36=56-64x\)
\(< =>184x=56-36=20\)
\(< =>x=\frac{20}{184}=\frac{5}{46}\)
\(a.\) 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: \(A=\frac{1}{x}-\frac{1}{x+5}=\frac{x+5-x}{x\left(x+5\right)}=\frac{5}{x\left(x+5\right)}\)
a) \(A=\left(\frac{2}{x+2}-\frac{4}{x^2+4x+4}\right):\left(\frac{2}{x^2-4}+\frac{1}{2-x}\right)\)
\(=\left(\frac{2\left(x+2\right)}{\left(x+2\right)^2}-\frac{4}{\left(x+2\right)^2}\right):\left(\frac{2}{x^2-4}-\frac{x+2}{\left(x-2\right)\left(x+2\right)}\right)\)
\(=\left(\frac{2x+4}{\left(x+2\right)^2}-\frac{4}{\left(x+2\right)^2}\right):\left(\frac{2}{x^2-4}-\frac{x+2}{x^2-4}\right)\)
\(=\frac{2x}{\left(x+2\right)^2}:\frac{-x}{x^2-4}=\frac{2x}{\left(x+2\right)^2}.\frac{\left(x+2\right)\left(x-2\right)}{-x}\)
\(=\frac{-2\left(x-2\right)}{\left(x+2\right)}=\frac{-2x+4}{x+2}\)
b) \(x^2-3x=0\Leftrightarrow x\left(x-3\right)=0\Leftrightarrow\orbr{\begin{cases}x=0\\x=3\end{cases}}\)
+) x = 0 \(\Rightarrow A=\frac{-2.0+4}{0+2}=\frac{4}{2}=2\)
+) x = 3 \(\Rightarrow A=\frac{-2.3+4}{3+2}=\frac{-2}{5}\)
bạn giúp mk câu c vs