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a) \(\dfrac{2}{x+3}+\dfrac{1}{x}\) MTC: \(x\left(x+3\right)\)
\(=\dfrac{2x}{x\left(x+3\right)}+\dfrac{x+3}{x\left(x+3\right)}\)
\(=\dfrac{2x+x+3}{x\left(x+3\right)}\)
\(=\dfrac{3x+3}{x\left(x+3\right)}\)
b) \(\dfrac{x+1}{2x-2}+\dfrac{-2x}{x^2-1}\)
\(=\dfrac{x+1}{2\left(x-1\right)}+\dfrac{-2x}{\left(x-1\right)\left(x+1\right)}\) MTC: \(2\left(x-1\right)\left(x+1\right)\)
\(=\dfrac{\left(x+1\right)\left(x+1\right)}{2\left(x-1\right)\left(x+1\right)}+\dfrac{-2x.2}{2\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{\left(x+1\right)^2-4x}{2\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{\left(x+1\right)-4x}{2\left(x-1\right)}\)
\(=\dfrac{x+1-4x}{2\left(x-1\right)}\)
\(=\dfrac{1-3x}{2\left(x-1\right)}\)
c) \(\dfrac{y-12}{6y-36}+\dfrac{6}{y^2-6y}\)
\(=\dfrac{y-12}{6\left(y-6\right)}+\dfrac{6}{y\left(y-6\right)}\) MTC: \(6y\left(y-6\right)\)
\(=\dfrac{y\left(y-12\right)}{6y\left(y-6\right)}+\dfrac{6.6}{6y\left(y-6\right)}\)
\(=\dfrac{y\left(y-12\right)+6^2}{6y\left(y-6\right)}\)
\(=\dfrac{y^2-12y+6^2}{6y\left(y-6\right)}\)
\(=\dfrac{\left(y-6\right)^2}{6y\left(y-6\right)}\)
\(=\dfrac{y-6}{6y}\)
Bạn Nguyễn Nam làm sai câu b rồi , làm lại cho tất nè
a) \(\dfrac{2}{x+3}+\dfrac{1}{x}=\dfrac{2x+x+3}{x\left(x+3\right)}=\dfrac{3x+3}{x\left(x+3\right)}\)
b) \(\dfrac{x+1}{2x-2}+\dfrac{-2x}{x^2-1}=\dfrac{x+1}{2\left(x-1\right)}-\dfrac{2x}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{\left(x+1\right)^2-4x}{2\left(x-1\right)\left(x+1\right)}=\dfrac{x^2+2x+1-4x}{2\left(x-1\right)\left(x+1\right)}=\dfrac{x^2-2x+1}{2\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{\left(x-1\right)^2}{2\left(x-1\right)\left(x+1\right)}=\dfrac{x-1}{2\left(x+1\right)}\)
c) \(\dfrac{y-12}{6y-36}+\dfrac{6}{y^2-6y}=\dfrac{y-12}{6\left(y-6\right)}+\dfrac{6}{y\left(y-6\right)}\)
\(=\dfrac{y^2-12y+36}{6y\left(y-6\right)}=\dfrac{\left(y-6\right)^2}{6y\left(y-6\right)}=\dfrac{y-6}{6y}\)
d) \(\dfrac{6x}{x+3}+\dfrac{3}{2x+6}=\dfrac{6x}{x+3}+\dfrac{3}{2\left(x+3\right)}=\dfrac{12x}{2\left(x+3\right)}\)( sửa đề )
Giúp em vs ạ em cảm ơn nhiều 😄😄😄😄
h) \(=8-12y+6y^2-y^3\)
i) \(=8y^3-125\)
j) \(=27y^3+64\)
k) \(=x^3-9x^2+27x-27+8-12x+6x-x^3=-9x^2+21x-19\)
\(AH=\sqrt{4\cdot9}=6\left(cm\right)\)
BC=BH+CH=13(cm)
\(AM=\dfrac{BC}{2}=6.5\left(cm\right)\)
\(HM=\sqrt{6.5^2-6^2}=2.5\left(cm\right)\)
\(S_{AHM}=\dfrac{2.5\cdot6}{2}=2.5\cdot3=7.5\left(cm^2\right)\)
\(ab\left(b-a\right)-bc\left(b-c\right)-ac\left(c-a\right)\)ư
\(=ab\left(b-a\right)-bc\left[\left(b-a\right)-\left(c-a\right)\right]-ac\left(c-a\right)\)
\(=ab\left(b-a\right)-bc\left(b-a\right)+bc\left(c-a\right)-ac\left(c-a\right)\)
\(=b\left(b-a\right)\left(a-c\right)+c\left(c-a\right)\left(b-a\right)\)
\(=b\left(b-a\right)\left(a-c\right)-c\left(a-c\right)\left(b-a\right)\)
\(=\left(b-c\right)\left(b-a\right)\left(a-c\right)\)
Mình gửi mà nó mất cái ảnh đâu r á, chắc chưa tải lên xong... Sorry bạn nha!
\(\left(4x\right)^2-3^2\left(x-2\right)^2=0\Leftrightarrow\left(4x\right)^2-\left(3x-6\right)^2=0\Leftrightarrow\left(4x-3x+6\right)\left(4x+3x-6\right)=0\Leftrightarrow\left(x+6\right)\left(7x-6\right)=0\Leftrightarrow\left\{{}\begin{matrix}x+6=0\\7x-6=0\end{matrix}\right.\left\{{}\begin{matrix}x=-6\\x=\dfrac{-6}{7}\end{matrix}\right.\)