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NT
1
4 tháng 12 2021
\(=\dfrac{x^2\left(x-2\right)+5\left(x-2\right)}{x-2}=x^2+5\)
ND
Nguyễn Đức Trí
VIP
8 tháng 9 2023
d) \(2x^2+5x-7=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{7}{2}\end{matrix}\right.\) \(\left(a+b+c=1\right)\)
S
2
28 tháng 2 2021
`(x+1)(x+3)=2x^2-2`
`<=>x^2+x+3x+3=2x^2-2`
`<=>x^2-4x-5=0`
`<=>x^2-5x+x-5=0`
`<=>x(x-5)+(x-5)=0`
`<=>(x-5)(x+1)=0`
`<=>` $\left[ \begin{array}{l}x=5\\x=-1\end{array} \right.$
Vậy `S={5,-1}`
28 tháng 2 2021
Ta có: \(\left(x+1\right)\left(x+3\right)=2x^2-2\)
\(\Leftrightarrow\left(x+1\right)\left(x+3\right)-2x^2+2=0\)
\(\Leftrightarrow\left(x+1\right)\left(x+3\right)-2\left(x^2-1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x+3\right)-2\left(x+1\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left[x+3-2\left(x-1\right)\right]=0\)
\(\Leftrightarrow\left(x+1\right)\left(x+3-2x+2\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(5-x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\5-x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=5\end{matrix}\right.\)
Vậy: S={-3;5}
20 tháng 12 2021
c: \(=\dfrac{x^3+2x+2x^2+2x+x^2-x+1}{\left(x+1\right)\left(x^2-x+1\right)}\)
\(=\dfrac{x^3+3x^2+3x+1}{\left(x+1\right)\left(x^2-x+1\right)}=\dfrac{x^2+2x+1}{x^2-x+1}\)
8 tháng 7 2021
Sửa đề: Biểu thức luôn có giá trị dương
Ta có: \(3x^2+2x-5\)
\(=3\left(x^2+\dfrac{2}{3}x-\dfrac{5}{3}\right)\)
\(=3\left(x^2+2\cdot x\cdot\dfrac{1}{3}+\dfrac{1}{9}-\dfrac{16}{9}\right)\)
\(=3\left(x+\dfrac{1}{3}\right)^2-\dfrac{16}{3}\ge-\dfrac{16}{3}\forall x\)
\(\Leftrightarrow\dfrac{1}{3\left(x+\dfrac{1}{3}\right)^2-\dfrac{16}{3}}\le\dfrac{1}{\dfrac{-16}{3}}=\dfrac{-3}{16}\forall x\)
\(\Leftrightarrow\dfrac{-1}{3\left(x+\dfrac{1}{3}\right)^2-\dfrac{16}{3}}\ge\dfrac{3}{16}>0\forall x\)(đpcm)
PA
13 tháng 10 2016
\(x^2-10x+16=x^2-8x-2x+16=x\left(x-8\right)-2\left(x-8\right)=\left(x-8\right)\left(x-2\right)\)
\(x^2-2x-15=x^2-5x+3x-15=x\left(x-5\right)+3\left(x-5\right)=\left(x-5\right)\left(x+3\right)\)
\(2x^2+7x+3=2x^2+x+6x+3=x\left(2x+1\right)+3\left(2x+1\right)=\left(x+3\right)\left(2x+1\right)\)
13 tháng 10 2016
a) \(x^2-10x+16=x^2-8x-2x+16=\left(x^2-8x\right)-\left(2x-16\right)=x\left(x-8\right)-2\left(x-8\right)=\left(x-8\right)\left(x-2\right)\)b) \(x^2-2x-15=x^2+3x-5x-15=\left(x^2+3x\right)-\left(5x+15\right)=x\left(x+3\right)-5\left(x+3\right)=\left(x+3\right)\left(x-5\right)\)c) \(2x^2+7x+3=2x^2+x+6x+3=\left(2x^2+x\right)+\left(6x+3\right)=x\left(2x+1\right)+3\left(2x+1\right)=\left(2x+1\right)\left(x+3\right)\)
DM
2
17 tháng 8 2017
\(\left(x+3\right)^3-x\left(3x+1\right)^2+\left(2x+1\right)\left(4x^2-2x+1\right)=28\)
\(\Leftrightarrow x^3+9x^2+27x+27-9x^3-6x^2-x+8x^3+1=28\)
\(\Leftrightarrow3x^2+26x+28=28\)
\(\Leftrightarrow3x^2+26x=0\)\(\Leftrightarrow x\left(3x+26\right)=0\)
Suy ra x=0 hoặc x=-26/3
\(\left(3-2x\right)^2=\left(x-2\right)\left(2x-3\right)\)
\(\Leftrightarrow\left(3x-2\right)^2-\left(x-2\right)\left(2x-3\right)=0\)
\(\Leftrightarrow9x^2-12x+4-\left(2x^2-7x+6\right)=0\)
\(\Leftrightarrow9x^2-12x+4-2x^2+7x-6=0\)
\(\Leftrightarrow7x^2-5x-2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{2}{7}\end{matrix}\right.\)
Vậy \(S=\left\{1;-\dfrac{2}{7}\right\}\)
`(3-2x)^2=(x-2)(2x-3)`
`<=>(2x-3)^2 -(x-2)(2x-3)=0`
`<=> (2x-3)(2x-3-x+2)=0`
`<=> (2x-3)(x-1)=0`
\(< =>\left[{}\begin{matrix}2x-3=0\\x-1=0\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=1\end{matrix}\right.\)