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a) \(\left(x-1\right)^2-\left(x-2\right)\left(x+2\right)=x^2-2x+1-x^2+4=5-2x\)
mình nghĩ là câu b bạn ghi đề sai vì như thế không có hằng đẳng thức nhé
b)\(\left(x^2+\frac{1}{3}x+\frac{1}{9}\right)\left(x-\frac{1}{3}\right)-\left(x-\frac{1}{3}\right)^3=x^3-\frac{1}{27}-x^3+\frac{1}{27}+x^2-\frac{1}{3}x=x^2-\frac{1}{3}x\)
b,\(\left(x^2+\frac{1}{x}+\frac{1}{9}\right)\left(x-\frac{1}{3}\right)-\left(x-\frac{1}{3}\right)^3\)
\(=\)\(\left(x-\frac{1}{3}\right)\left[\left(x^2+\frac{1}{x}+\frac{1}{9}\right)-\left(x-\frac{1}{3}\right)^2\right]\)
\(=\)\(\left(x-\frac{1}{3}\right)\left(x^2+\frac{1}{x}+\frac{1}{9}-x^2+\frac{2}{3}x-\frac{1}{9}\right)\)
\(=\left(x-\frac{1}{3}\right)\left(\frac{1}{x}+\frac{2}{3}x\right)\) \(=1+\frac{2}{3}x^2-\frac{1}{3x}-\frac{2}{9}x\)
\(B=\left(\frac{21}{x^2-9}-\frac{x-4}{3-x}+\frac{x-1}{3+x}\right)\div\left(1-\frac{1}{x+3}\right)\)
\(B=\left(\frac{21}{x^2-9}+\frac{x-4}{x-3}+\frac{x-1}{x+3}\right)\div\left(\frac{x+3}{x+3}-\frac{1}{x+3}\right)\)
\(B=\left(\frac{21}{\left(x+3\right)\left(x-3\right)}+\frac{\left(x-4\right)\left(x+3\right)}{\left(x+3\right)\left(x-3\right)}+\frac{\left(x-1\right)\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}\right)\div\frac{x+2}{x+3}\)
\(B=\left(\frac{21}{\left(x+3\right)\left(x-3\right)}+\frac{x^2-x-12}{\left(x+3\right)\left(x-3\right)}+\frac{x^2-4x+3}{\left(x+3\right)\left(x-3\right)}\right)\cdot\frac{x+3}{x+2}\)
\(B=\left(\frac{21+x^2-x-12+x^2-4x+3}{\left(x+3\right)\left(x-3\right)}\right)\cdot\frac{x+3}{x+2}\)
\(B=\frac{2x^2-5x+12}{\left(x+3\right)\left(x-3\right)}\cdot\frac{x+3}{\left(x+2\right)}\)
\(B=\frac{2x^2-5x+12}{\left(x-3\right)\left(x+2\right)}\)
\(B=\frac{2x^2-5x+12}{x^2-x-6}\)
Đến đây là chịu ạ :(
a) ĐKXĐ: \(x\ne\pm1\)
\(A=\left(\frac{\left(x+1\right)^2}{\left(x-1\right)\left(x+1\right)}-\frac{\left(x-1\right)^2}{\left(x+1\right)\left(x-1\right)}\right):\left(\frac{1-x}{\left(1+x\right)\left(1-x\right)}-\frac{x\left(1+x\right)}{\left(1-x\right)\left(1+x\right)}+\frac{x}{x^2-1}\right)\)
\(=\frac{4x-1}{x^2-1}:\left(\frac{-x^2-2x+1}{1-x^2}-\frac{x}{1-x^2}\right)=\frac{4x-1}{x^2-1}:\frac{-x^2-3x+1}{1-x^2}\)
\(=\frac{1-4x}{1-x^2}:\frac{-x^2-3x+1}{1-x^2}=\frac{\left(1-4x\right)\left(1-x^2\right)}{\left(1-x^2\right)\left(-x^2-3x+1\right)}\)
\(=\frac{1-4x}{-x^2-3x+1}=\frac{4x-1}{x^2+3x-1}\) (chắc hết rút gọn được rồi)
\(A=\left(\frac{3-x}{x+3}.\frac{x^2+6x+9}{x^2-9}+\frac{x}{x+3}\right):\frac{3x^2}{x+3}\)
\(A=\left(\frac{3-x}{x+3}.\frac{\left(x+3\right)^2}{\left(x-3\right)\left(x+3\right)}+\frac{x}{x+3}\right):\frac{3x^2}{x+3}\)
\(A=\left(\frac{-\left(x-3\right)}{x+3}.\frac{\left(x+3^2\right)}{\left(x-3\right)\left(x+3\right)}+\frac{x}{x+3}\right):\frac{3x^2}{x+3}\)
\(A=\left(-1+\frac{x}{x+3}\right):\frac{3x^2}{x+3}\)
\(A=\left(\frac{-x-3+x}{x+3}\right):\frac{3x^2}{x+3}\)
\(A=\left(-\frac{3}{x+3}\right).\frac{x+3}{3x^2}\)
\(A=-x^2\)
\(B=\left(\frac{21}{x^2-9}+\frac{\left(x-4\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}+\frac{\left(x-1\right)\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}\right):\frac{x+2}{x+3}\)
\(B=\frac{2x^2-5x+12}{x^2-9}\cdot\frac{x+3}{x+2}\)
\(B=\frac{2x^2-5x-12}{\left(x-3\right)\left(x+2\right)}\)
\(B=\frac{2x^2-5x+12}{x^2-x-6}\)
Thik thì tách tiếp nha
b. Sử dụng các hằng đẳng thức
\(a^3+b^3+c^2-3abc=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)\)
\(=3\left(a^2+b^2+c^2-ab-bc-ca\right)\)
và \(\left(a-b\right)^3+\left(b-c\right)^3+\left(c-a\right)^3=3\left(a-b\right)\left(b-c\right)\left(c-a\right)\)
nên \(A=\frac{a^2+b^2+c^2-ab-bc-ca}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=\frac{1}{2}.\frac{\left[\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\right]}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)
Do (a - b) + (b - c) + (c - a) = 0 nên áp dụng hđt \(X^2+Y^2+Z^2=-2\left(XY+YZ+ZX\right)\)khi X + Y + Z = 0, ta có:
\(A=-2\left(\frac{1}{a-b}+\frac{1}{b-c}+\frac{1}{c-a}\right).\)
Bài 1 :
\(b,ax^2+3ax+9=a^2\)
\(\Leftrightarrow a^2x+3ax+9-a^2=0\)
\(\Leftrightarrow ax\left(a+3\right)+\left(a+3\right)\left(3-a\right)=0\)
\(\Leftrightarrow\left(a+3\right)\left(ax+3-a\right)=0\)
Vì \(a\ne3\Rightarrow\left(a+3\right)\ne0\Rightarrow ax+3-a=0\)
\(\Leftrightarrow ax=a-3\)
Vì \(a\ne0\Rightarrow x=\frac{a-3}{a}\)
ĐKXĐ \(\hept{\begin{cases}x\ne3\\x\ne-3\\x\ne0\end{cases}}\)
\(A=\left(\frac{x^2-3+x+3}{\left(x-3\right)\left(x+3\right)}\right).\frac{x+3}{x}\)
\(=\frac{x^2+x}{x^2-3x}\)
Ta có: A = \(\left(\frac{x^2-3}{x^2-9}+\frac{1}{x-3}\right):\frac{x}{x+3}\)
\(\Leftrightarrow\) A = \(\left(\frac{x^2-3}{\left(x-3\right)\left(x+3\right)}+\frac{1}{x-3}\right):\frac{x}{x+3}\)
\(\Leftrightarrow\) A = \(\left(\frac{x^2-3+x+3}{\left(x-3\right)\left(x+3\right)}\right).\frac{x+3}{x}\)
\(\Leftrightarrow\) A = \(\frac{x^2+x}{\left(x-3\right)\left(x+3\right)}.\frac{x+3}{x}\)
\(\Leftrightarrow\) A = \(\frac{x^2+x}{x^2-3x}\)