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`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)`.
Đặt \(\hept{\begin{cases}\sqrt[3]{65+x}=a\\\sqrt[3]{65-x}=b\end{cases}}\)
\(\Rightarrow a^2+4b^2=5ab\)
\(\Leftrightarrow\left(b-a\right)\left(4b-a\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}a=b\\a=4b\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}\sqrt[3]{65+x}=\sqrt[3]{65-x}\\\sqrt[3]{65+x}=4\sqrt[3]{65-x}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}65+x=65-x\\65+x=4\left(65-x\right)\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=39\end{cases}}\)
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: \(x+100=0\)
\(x=-100\)
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) \(x^2-x-42=\left(x-7\right)\left(x+6\right).\)
Ta có : \(x^2+2x+y^2-2y-2xy+65\)
\(=\left(x-y\right)^2+2\left(x-y\right)+65\)
Mà \(x=y+5\)
\(\Rightarrow x-y=5\)
- Thay x - y = 5 vào đa thức trên ta được :
\(=\left(x-y\right)^2+2\left(x-y\right)+65=100\)
Vậy ...
\(\dfrac{x+1}{65}+\dfrac{x+3}{63}\) = \(\dfrac{x+5}{61}\) + \(\dfrac{x+7}{59}\)
<=> \(\dfrac{x+1}{65}+1+\dfrac{x+3}{63}+1\) = \(\dfrac{x+5}{61}\) + 1 + \(\dfrac{x+7}{59}\) + 1
<=> \(\dfrac{x+66}{65}+\dfrac{x+66}{63}\) = \(\dfrac{x+66}{61}\) + \(\dfrac{x+66}{59}\)
<=> \(\dfrac{x+66}{65}+\dfrac{x+66}{63}\) - \(\dfrac{x+66}{61}\) - \(\dfrac{x+66}{59}\) = 0
<=> (x + 66) . (\(\dfrac{1}{65}+\dfrac{1}{63}+\dfrac{1}{61}+\dfrac{1}{59}\)) = 0
<=> x + 66 = 0
<=> x = -66