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DP
0
2
26 tháng 9 2023
`4)x^2-5x+6`
`=x^2-2x-3x+6`
`=x(x-2)-3(x-2)=(x-2)(x-3)`
`5)x^2+7x+10`
`=x^2+5x+2x+10`
`=x(x+5)+2(x+5)=(x+5)(x+2)`
`6)x+7\sqrt{x}+10` `ĐK: x >= 0`
`=(\sqrt{x})^2+5\sqrt{x}+2\sqrt{x}+10`
`=\sqrt{x}(\sqrt{x}+5)+2(\sqrt{x}+5)=(\sqrt{x}+5)(\sqrt{x}+2)`
`7)3x^4+7x^2+4`
`=3x^4+3x^2+4x^2+4`
`=3x^2(x^2+1)+4(x^2+1)=(x^2+1)(3x^2+4)`
`8)x^2-x-2`
`=x^2-2x+x-2`
`=x(x-2)+(x-2)=(x-2)(x+1)`
`9)x^6-x^3-2`
`=x^6+x^3-2x^3-2`
`=x^3(x^3+1)-2(x^3+1)`
`=(x^3+1)(x^3-2)`.
20 tháng 6 2018
Đặt \(\hept{\begin{cases}\sqrt[3]{65+x}=a\\\sqrt[3]{65-x}=b\end{cases}}\)
\(\Rightarrow a^2+4b^2=5ab\)
\(\Leftrightarrow\left(b-a\right)\left(4b-a\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}a=b\\a=4b\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}\sqrt[3]{65+x}=\sqrt[3]{65-x}\\\sqrt[3]{65+x}=4\sqrt[3]{65-x}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}65+x=65-x\\65+x=4\left(65-x\right)\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=39\end{cases}}\)
DQ
26 tháng 8 2018
a) \(\dfrac{x+43}{57}+\dfrac{x+46}{54}=\dfrac{x+49}{51}+\dfrac{x+52}{48}\)
\(\left(\dfrac{x+43}{57}+1\right)+\left(\dfrac{x+46}{54}+1\right)=\left(\dfrac{x+49}{51}+1\right)+\left(\dfrac{x+52}{48}\right)\)
\(\dfrac{x+43+57}{57}+\dfrac{x+46+54}{54}-\dfrac{x+49+51}{51}-\dfrac{x+52+48}{48}=0\)
\(\dfrac{x+100}{57}+\dfrac{x+100}{54}-\dfrac{x+100}{51}-\dfrac{x+100}{48}=0\)
\(\left(x+100\right)\left(\dfrac{1}{57}+\dfrac{1}{54}-\dfrac{1}{51}-\dfrac{1}{48}\right)=0\)
Mà \(\dfrac{1}{57}+\dfrac{1}{54}-\dfrac{1}{51}-\dfrac{1}{48}\ne0\)
Nên: \(x+100=0\)
\(x=-100\)
4 tháng 11 2017
1) \(x^2+x-6=x\left(x-2\right)+3\left(x-2\right)=\left(x+3\right)\left(x-2\right)\)
2) \(x^2-x-6=\left(x-3\right)\left(x+2\right)\)
3) \(x^2+2x-48=\left(x-6\right)\left(x+8\right)\)
4) \(x^2-2x-48=\left(x-8\right)\left(x+6\right)\)
5) \(x^2+x-42=\left(x-6\right)\left(x+7\right)\)
6) \(x^2-x-42=\left(x-7\right)\left(x+6\right).\)
KB
16 tháng 2 2019
ĐKXĐ : \(\forall x\)
Ta có : \(\dfrac{x^2}{x^2+2x+2}+\dfrac{x^2}{x^2-2x+2}-\dfrac{4\left(x^2-5\right)}{x^4+4}=\dfrac{322}{65}\)
\(\Leftrightarrow\dfrac{x^2\left(x^2-2x+2\right)+x^2\left(x^2+2x+2\right)-4\left(x^2-5\right)}{x^4+4}=\dfrac{322}{65}\)
\(\Leftrightarrow\dfrac{x^4-2x^3+2x^2+x^4+2x^3+2x^2-4x^2+20}{x^4+4}=\dfrac{322}{65}\)
\(\Leftrightarrow\dfrac{2x^4+20}{x^4+4}=\dfrac{322}{65}\)
\(\Leftrightarrow65\left(2x^4+20\right)=322\left(x^4+4\right)\)
\(\Leftrightarrow130x^4+1300=322x^4+1288\)
\(\Leftrightarrow192x^4-12=0\)
\(\Leftrightarrow x^4=\dfrac{12}{192}\)
\(\Leftrightarrow x^4=\dfrac{1}{16}\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=\dfrac{1}{2}\end{matrix}\right.\)
Vậy ...
NV
Nguyễn Việt Lâm
Giáo viên
16 tháng 2 2019
\(\dfrac{x^2}{x^2+2x+2}+\dfrac{x^2}{x^2-2x+2}-\dfrac{4\left(x^2-5\right)}{x^4+4}=\dfrac{322}{65}\)
\(\Leftrightarrow x^2\left(\dfrac{1}{x^2+2+2x}+\dfrac{1}{x^2+2-2x}\right)-\dfrac{4\left(x^2-5\right)}{x^4+4}=\dfrac{322}{65}\)
\(\Leftrightarrow x^2\left(\dfrac{2x^2+4}{x^4+4}\right)-\dfrac{4\left(x^2-5\right)}{x^4+4}=\dfrac{322}{65}\)
\(\Leftrightarrow\dfrac{2x^4+4x^2-4x^2+20}{x^4+4}=\dfrac{322}{65}\)
\(\Leftrightarrow\dfrac{2x^4+20}{x^4+4}=\dfrac{322}{65}\)
\(\Leftrightarrow65x^4+650=161x^4+644\)
\(\Leftrightarrow96x^4=6\)
\(\Leftrightarrow x^4=\dfrac{1}{16}\)
\(\Rightarrow x=\pm\dfrac{1}{2}\)