Tìm STN x biết:(9x-18)(x+5)>0
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Ta có 5(x+x+1+x+2)=1018÷218
=>5(3x+3)=(10÷2)18
=>5(3x+3)=518
=>3x+3=18
=>3x=18-3
=>3x=15
=>x=15÷3
=>x=5
Vậy với x=5 thì 5x×5x+1×5x+2=100....0:218(18 c/s 0)
\(a,8x-75=5x+21\)
\(8x-5x=21+75\)
\(3x=96\)
\(x=32\)
\(b,9x+25=-\left(2x-58\right)\)
\(9x+25=-2x+58\)
\(9x+2x=58-25\)
\(11x=33\)
\(x=3\)
\(c,\left(5-x\right).\left(x+2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}5-x=0\\x+2=0\end{cases}\Rightarrow\orbr{\begin{cases}x=5\\x=-2\end{cases}}}\)
Ta có : (x3 - 2x2) - 9x + 18 = 0
<=> x2(x - 2) - (9x - 18) = 0
<=> x2(x - 2) - 9(x - 2) = 0
=> (x2 - 9) (x - 2) = 0
\(\Leftrightarrow\orbr{\begin{cases}x^2-9=0\\x-2=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x^2=9\\x=2\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=3;-3\\x=2\end{cases}}\)
2) \(\frac{1}{5}\sqrt{25x+50}-5\sqrt{x+2}+\sqrt{9x+18}+9=0\)
\(\frac{1}{5}\sqrt{25\left(x+2\right)}-5\sqrt{x+2}+\sqrt{9x+18}+9=0\)
\(\frac{1}{5}.\sqrt{25}.\sqrt{x+2}-5\sqrt{x+2}+\sqrt{9x+18}+9=0\)
\(\frac{1}{5}.5\sqrt{x+2}-5\sqrt{x+2}+\sqrt{9x+18}+9=0\)
\(\frac{1}{5}.5\sqrt{x+2}-5\sqrt{x+2}+\sqrt{9\left(x+2\right)}+9=0\)
\(\frac{1}{5}.5\sqrt{x+2}-5\sqrt{x+2}+\sqrt{9}.\sqrt{x+2}+9=0\)
\(\frac{1}{5}.5\sqrt{x+2}-5\sqrt{x+2}+3\sqrt{x+2}+9=0\)
\(\sqrt{x+2}-5\sqrt{x+2}+3\sqrt{x+2}+9=0\)
\(-\sqrt{x+2}=-9\)
\(x+2=81\)
\(\Rightarrow x=79\)
3) \(\sqrt{x^2-4x+4}=7x-1\)
\(\sqrt{x^2-2.x.2+2^2}=7x-1\)
\(\sqrt{\left(x-2\right)^2}=7x-1\)
\(x-2=7x-1\)
\(-2=7x-1-x\)
\(-2+1=7x-x\)
\(-1=6x\)
\(-\frac{1}{6}=x\)
\(\Rightarrow x=-\frac{1}{6}\)
\(\frac{2}{9}.5x+\frac{1}{2}-\frac{1}{18}=\frac{5}{36}\)
\(\frac{2}{9}.5x=\frac{5}{36}+\frac{1}{18}-\frac{1}{2}\)
\(\frac{2}{9}.5x=\frac{-11}{36}\)
\(5x=\frac{-11}{36}:\frac{2}{9}\)
\(5x=\frac{-11}{8}\)
\(x=\frac{-11}{8}:5\)
\(x=\frac{-11}{40}\)
Chú ý dấu chấm là dâu nhân nha
\(\left(5-x\right)\left(9x^2-4\right)=0\)
=>\(\left(x-5\right)\left(3x-2\right)\left(3x+2\right)=0\)
=>\(\left[{}\begin{matrix}x-5=0\\3x-2=0\\3x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=\dfrac{2}{3}\\x=-\dfrac{2}{3}\end{matrix}\right.\)
\(\left(5-x\right)\left(9x^2-4\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}5-x=0\\9x^2-4=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=5\\x^2=\dfrac{4}{9}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=5\\x=\dfrac{2}{3}\\x=-\dfrac{2}{3}\end{matrix}\right.\)
\(5\sqrt{x-2}=10+3\sqrt{x+2}\)
\(\Leftrightarrow25x-50=100+60\sqrt{x+2}+9x+18\)
\(\Leftrightarrow25x-9x=168+60\sqrt{x+2}\)
\(\Leftrightarrow16x-168=60\sqrt{x+2}\)
\(\Leftrightarrow256x^2-5376x+28224=3600x+7200\)
\(\Leftrightarrow256x^2-8976x+21024=0\)
....................
\(5x^2-9x+18=0\)
Ta có: \(\Delta=b^2-4ac=\left(-9\right)^2-4\cdot5\cdot18=-279< 0\)
Vậy phương trình vô nghiệm
\(x\ge3\text{ với mọi x}\in N\text{ thì thỏa mãn pt:}\left(9x-18\right)\left(x+5\right)>0\)