(x+7).(x-9)=0
Cần giải gấp!!!!!!!Thanks
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\(\left(x+7\right)\left(9-3x\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x+7=0\\9-3x=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=-7\\x=2\end{cases}}\)
Ta có : \(x-\frac{3}{1.5}-\frac{3}{5.9}-\frac{3}{9.13}-\frac{3}{13.17}-\frac{3}{17.21}=\frac{2}{7}\)
=> \(x-\left(\frac{3}{1.5}+\frac{3}{5.9}+\frac{3}{9.13}+\frac{3}{13.17}+\frac{3}{17.21}\right)=\frac{2}{7}\)
=> \(x-\frac{3}{4}\left(\frac{4}{1.5}+\frac{4}{5.9}+\frac{4}{9.13}+\frac{4}{13.17}+\frac{4}{17.21}\right)=\frac{2}{7}\)
=> \(x-\frac{3}{4}\left(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+\frac{1}{3}-\frac{1}{17}+\frac{1}{17}-\frac{1}{21}\right)=\frac{2}{7}\)
=> \(x-\frac{3}{4}\left(1-\frac{1}{21}\right)=\frac{2}{7}\)
=> \(x-\frac{3}{4}.\frac{20}{21}=\frac{2}{7}\)
=> \(x-\frac{5}{7}=\frac{2}{7}\)
=> x = 1
Vậy x = 1
3 . (-x + 1) - 2.(x + 7) - 32 = -(-2x - 9) - 158
=> -3x + 3 - 2x - 14 - 9 = 2x + 9 - 1
=> -3x - 2x - 2x = 9 - 1 + 9 + 14 - 3
=> -7x = 28
=> x = 28 : (-7)
=> x = -4
Vậy ...
=>(-3x+3)-(2x+14)-9=2x+9-1
=>(-3x-2x)-(14-3+9)=2x+8
=>-5x-20=2x+8
=>2x+5x=-8-20
=>-7x=-28
=>x=-28/7
ĐK: \(-\frac{7}{2}\le x\le5.\)
pt <=> \(x+25-4\sqrt{5-x}-6\sqrt{2x-7}=0\)
<=> \(5-x-4\sqrt{5-x}+4+2x+7-6\sqrt{2x+7}+9=0\)
<=> \(\left(\sqrt{5-x}-2\right)^2+\left(\sqrt{2x+7}-3\right)^2=0\)
<=> \(\hept{\begin{cases}\sqrt{5-x}-2=0\\\sqrt{2x+7}-3=0\end{cases}\Leftrightarrow}\hept{\begin{cases}5-x=4\\2x+7=9\end{cases}\Leftrightarrow}x=1\left(tmdk\right)\)
\(\Rightarrow\orbr{\begin{cases}x+7=0\Rightarrow x=-7\\x-9=0\Rightarrow x=9\end{cases}}\)
Vậy \(x\in\left\{-7;9\right\}\)
(x+7).(x-9)=0
* x+7=0 * x-9=0
x=0-7 x=0+9
x=-7 x=9
vậy x=-7 hoặc x= 9