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e) = \(\dfrac{3}{2\left(x+3\right)}\) - \(\dfrac{x-6}{2x\left(x+3\right)}\)
= \(\dfrac{3x}{2x\left(x+3\right)}\) - \(\dfrac{x-6}{2x\left(x+3\right)}\) = \(\dfrac{3x-x+6}{2x\left(x+3\right)}\)
= \(\dfrac{2x-6}{2x\left(x+3\right)}\)
= \(\dfrac{2\left(x-3\right)}{2x\left(x+3\right)}\)
c) = \(\dfrac{2\left(a^3-b^3\right)}{3\left(a+b\right)}\) . \(\dfrac{6\left(a+b\right)}{a^2-2ab+b^2}\)
= \(\dfrac{-2\left(a+b\right)\left(a^2-2ab+b^2\right)}{3\left(a+b\right)}\) . \(\dfrac{6\left(a+b\right)}{a^2-2ab+b^2}\)
= \(\dfrac{-2\left(a+b\right)}{1}\) . \(\dfrac{2}{1}\) = -4 (a+b)
a, \(\dfrac{x^2-x}{x-2}+\dfrac{4-3x}{x-2}\)
\(=\dfrac{x^2-x+4-3x}{x-2}=\dfrac{x^2-4x+4}{x-2}\)
c) \(\dfrac{2}{x^2-9}+\dfrac{1}{x+3}\)
Ta có: \(\dfrac{1}{x+3}=\dfrac{1\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}=\dfrac{x-3}{x^2-9}\)
\(\Rightarrow\dfrac{2}{x^2-9}+\dfrac{1}{x+3}=\dfrac{2}{x^2-9}+\dfrac{x-3}{x^2-9}=\dfrac{2+x-3}{x^2-9}=\dfrac{x-1}{x^2-9}\)
1)
\(\Leftrightarrow\left(x^2-2+\dfrac{1}{x^2}\right)+\left(y^2-2+\dfrac{1}{y^2}\right)+z^2=0\)
\(\Leftrightarrow\left(x-\dfrac{1}{x}\right)^2+\left(y-\dfrac{1}{y}\right)^2+z^2=0\)
\(\left\{{}\begin{matrix}x-\dfrac{1}{x}=0\Rightarrow\left|x\right|=1\\y-\dfrac{1}{y}=0\Rightarrow\left|y\right|=1\\z=0\end{matrix}\right.\)
dk\(x,y,z,a,b,c\ne0\)\(\left\{{}\begin{matrix}\dfrac{a}{x}=A\\\dfrac{b}{y}=B\\\dfrac{c}{z}=C\end{matrix}\right.\) \(\Rightarrow A,B,C\ne0\)
\(\left\{{}\begin{matrix}A+B+C=2\\\dfrac{1}{A}+\dfrac{1}{B}+\dfrac{1}{C}=0\end{matrix}\right.\)
\(\left\{{}\begin{matrix}A^2+B^2+C^2+2\left(AB+BC+AC\right)=4\\\dfrac{ABC}{A}+\dfrac{ABC}{B}+\dfrac{ABC}{C}=0\end{matrix}\right.\)
\(\left\{{}\begin{matrix}AB+BC+AC=0\\A^2+B^2+C^2=4\end{matrix}\right.\)
\(\left(\dfrac{a}{x}\right)^2+\left(\dfrac{b}{y}\right)^2+\left(\dfrac{c}{z}\right)^2=4\)
Ta có \(7a^2-15ab+2b^2=0\Leftrightarrow7a^2-14ab-ab+2b^2=0\Leftrightarrow7a\left(a-2b\right)-b\left(a-2b\right)=0\Leftrightarrow\left(a-2b\right)\left(7a-b\right)=0\Leftrightarrow\)\(\left[{}\begin{matrix}7a-b=0\\a-2b=0\end{matrix}\right.\)(*)
Vì a-2b\(\ne0\)(Để E xác định)
Vậy (*)\(\Leftrightarrow7a-b=0\Leftrightarrow7a=b\)
Thay vào E ta có \(E=\dfrac{a-7a}{2a+7a}-\dfrac{3a-7a}{a-14a}=\dfrac{-6a}{9a}-\dfrac{-4a}{-13a}=\dfrac{-6}{9}-\dfrac{4}{13}=-\dfrac{38}{39}\)
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giúp mk vs
a) \(\dfrac{x^3}{x+1}+\dfrac{x^2}{x-1}+\dfrac{1}{x+1}+\dfrac{1}{1-x}\)
\(=\dfrac{x^3}{x+1}+\dfrac{x^2}{x-1}+\dfrac{1}{x+1}+\dfrac{-1}{x-1}\)
\(=\dfrac{x^3\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}+\dfrac{x^2\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}+\dfrac{1\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}+\dfrac{-1\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}\)
\(=\dfrac{x^4-x+x^3+x+x-1-x+1}{\left(x+1\right)\left(x-1\right)}\)
\(=\dfrac{x^4+x^3}{\left(x+1\right)\left(x-1\right)}\)
\(=\dfrac{x^3\left(x+1\right)}{\left(x+1\right)\left(x-1\right)}=\dfrac{x^3}{x-1}\)
b) \(\dfrac{x^3}{x-1}-\dfrac{x^2}{x+1}-\dfrac{1}{x-1}+\dfrac{1}{x+1}\)
\(=\dfrac{x^3\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}-\dfrac{x^2\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}-\dfrac{1\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}+\dfrac{1\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}\)
\(=\dfrac{x^3\left(x+1\right)-x^2\left(x-1\right)-1\left(x+1\right)+1\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{x^4+x^3-x^3+x^2-x-1+x-1}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{x^4+x^2-2}{\left(x-1\right)\left(x+1\right)}\)
c) \(\dfrac{4-2x+x^2}{2+x}-2-x\)
\(=\dfrac{4-2x+x^2}{2+x}-\dfrac{2\left(2+x\right)}{2+x}-\dfrac{x\left(2+x\right)}{2+x}\)
\(=\dfrac{4-2x+x^2-4-2x-2x-x^2}{2+x}\)
\(=\dfrac{-6x}{2+x}\)
Còn lại thì dễ rồi, bạn tự làm nha ^^
Lời giải:
Vì \(2a-b=5\Rightarrow b=2a-5\Rightarrow 2b=4a-10\)
\(\Rightarrow 7a-2b=7a-(4a-10)=3a+10\)
\(\Rightarrow \frac{7a-2b}{3a+10}=\frac{3a+10}{3a+10}=1\)
Lại có:
\(2a-b=5\Rightarrow 2a=b+5\Rightarrow 4a=2b+10\)
\(\Rightarrow 7b-4a=7b-(2b+10)=5b-10\)
\(\Rightarrow \frac{7b-4a}{15b-30}=\frac{5b-10}{15b-30}=\frac{5b-10}{3(5b-10)}=\frac{1}{3}\)
Vậy: \(A=1-\frac{1}{3}=\frac{2}{3}\)
de dung ko vay ban