(6x2-2x + 1 ) :(3x -1) tính hàng ngang nhé ạ
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1. \(\left(x-y\right)\left(6x^2-4y^2+\dfrac{1}{2}xy\right)\)
\(=6x^3-4xy^2+\dfrac{1}{2}x^2y-6x^2y+4y^3-\dfrac{1}{2}xy^2\)
\(=6x^3+4y^3-\dfrac{11}{2}x^2y-\dfrac{9}{2}xy^2\)
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2. \(\left(6x-1\right)\left(3+x\right)+\left(2x+5\right)\left(-3x\right)\)
\(=18x+6x^2-3-x-6x^2-15x\)
\(=2x-3\)
Chúc bạn học tốt!
`1)(x-y)(6x^2-4y^2+1/2xy)`
`=6x^3-4xy^2+1/2x^2y-6x^2y+4y^3-1/2xy^2`
`=6x^3-11/2x^2y-9/2xy^2+4y^3`
`2)(6x-1)(3+x)+(2x+5)(-3x)`
`=18+6x^2-x-3-6x^2-15x`
`=-15x+15`
1)\(\left(x-y\right)\left(6x^2-4y^2+\dfrac{1}{2}xy\right)=6x^3+4y^3-\dfrac{9}{2}xy^2-\dfrac{11}{2}x^2y\)
2)\(\left(6x-1\right)\left(3+x\right)+\left(2x+5\right)\left(-3x\right)=\left(6x^2+17x-3\right)+\left(-6x^2-15x\right)=2x-3\)
tick mik nha
\(1,=\left(x+3\right)\left(x-2\right):\left(x+3\right)=x-2\\ 2,=\left(x-5\right)\left(x+6\right):\left(x+6\right)=x-5\\ 3,=\left[3x\left(2x-1\right)-5\right]:\left(2x-1\right)=3x.dư.\left(-5\right)\)
1)\(\left(x+x^2-6\right):\left(x+3\right)=\left[x\left(x+3\right)-2\left(x+3\right)\right]:\left(x+3\right)=\left[\left(x+3\right)\left(x-2\right)\right]:\left(x+3\right)=x-2\)
2) \(\left(x+x^2-30\right):\left(x+6\right)=\left[x\left(x+6\right)-5\left(x+6\right)\right]:\left(x+6\right)=\left[\left(x+6\right)\left(x-5\right)\right]:\left(x+6\right)=x-5\)
3) \(\left(5-3x+6x^2\right):\left(2x-1\right)=\left[3x\left(2x-1\right)+5\right]:\left(2x-1\right)=3x+\dfrac{5}{2x-1}\)
\(6x^2-\left(2x-3\right)\left(3x+2\right)-1=0\)
\(\Rightarrow6x^2-6x^2+5x+6-1=0\)
\(\Rightarrow5x=-5\Rightarrow x=-1\)
\(\Rightarrow6x^2-\left(6x^2-5x-6\right)-1=0\\ \Rightarrow5x+5=0\\ \Rightarrow x=-1\)
\(a,6x^2-5x+3=2x-3x\left(3-2x\right)\)
\(\Leftrightarrow6x^2-5x+3=2x-9x+6x^2\)
\(\Leftrightarrow6x^2-6x^2-5x-2x+9x=-3\)
\(\Leftrightarrow2x=-3\)
\(\Leftrightarrow x=-\dfrac{3}{2}\)
\(b,\left(3x-1\right)\left(4x+3\right)=2\left(3x-1\right)\)
\(\Leftrightarrow\left(3x-1\right)\left(4x+3\right)-2\left(3x-1\right)=0\)
\(\Leftrightarrow\left(3x-1\right)\left(4x+3-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-1=0\\4x+1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=-\dfrac{1}{4}\end{matrix}\right.\)
\(6x^2-5x+3=2x-3x\left(3-2x\right)\)
\(\Leftrightarrow6x^2-5x+3=2x-9x+6x^2\)
\(\Leftrightarrow6x^2-5x+3-2x+9x-6x^2=0\)
\(\Leftrightarrow2x+3=0\)
\(\Leftrightarrow2x=-3\)
\(\Leftrightarrow x=\dfrac{-3}{2}\)
\(\text{Vậy phương trình có tập nghiệm là }S=\left\{\dfrac{-3}{2}\right\}\)
\(\left(3x-1\right)\left(4x+3\right)=2\left(3x-1\right)\)
\(\Leftrightarrow\left(3x-1\right)\left(4x+3\right)-2\left(3x-1\right)=0\)
\(\Leftrightarrow\left(3x-1\right)\left(4x+3-2\right)=0\)
\(\Leftrightarrow\left(3x-1\right)\left(4x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-1=0\\4x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=1\\4x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=\dfrac{1}{4}\end{matrix}\right.\)
\(\text{Vậy phương trình có tập nghiệm là }S=\left\{\dfrac{1}{3};\dfrac{1}{4}\right\}\)
`@` `\text {Ans}`
`\downarrow`
`6x^2-2x+1=(6x^2-2x)+1=2x(3x-1)+1`
`->` Thương của phép chia `6x^2-2x+1` cho `3x-1` là `2x` và có số dư là `1`