Các bạn ơi giúp mình với

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 5^x×5^x+1×5^x+2=100....0:2¹⁸(18 c/s 0)

Thiếu 0;1;2;3;4

\(a,8x-75=5x+21\)

    \(8x-5x=21+75\)

               \(3x=96\)

                  \(x=32\)

\(b,9x+25=-\left(2x-58\right)\)

    \(9x+25=-2x+58\)

    \(9x+2x=58-25\)

            \(11x=33\)

                 \(x=3\)

\(c,\left(5-x\right).\left(x+2\right)=0\)

\(\Rightarrow\orbr{\begin{cases}5-x=0\\x+2=0\end{cases}\Rightarrow\orbr{\begin{cases}x=5\\x=-2\end{cases}}}\)

a) \(8x-75=5x+21\)

\(8x-5x=21+75\)

\(3x=96\)

\(x=32\)

vậy \(x=32\)

b) \(9x+25=-\left(2x-58\right)\)

\(9x+25=-2x+58\)

\(9x+2x=58-25\)

\(11x=33\)

\(x=3\)

vậy \(x=3\)

c) \(\left(5-x\right)\left(x+2\right)=0\)

\(\Rightarrow\orbr{\begin{cases}5-x=0\\x+2=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=5\\x=-2\end{cases}}\)

vậy....

d) \(3.\left|x+3\right|-4=-7-\left(-18\right)\)

\(3.\left|x+3\right|-4=-7+18\)

\(3.\left|x+3\right|=11+4\)

\(3.\left|x+3\right|=15\)

\(\left|x+3\right|=5\)

\(\Rightarrow\orbr{\begin{cases}x+3=5\\x+3=-5\end{cases}}\Rightarrow\orbr{\begin{cases}x=2\\x=-8\end{cases}}\)

vậy..

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

nhầm rồi 5x+1/2

\(\left(5-x\right)\left(9x^2-4\right)=0\)

=>\(\left(x-5\right)\left(3x-2\right)\left(3x+2\right)=0\)

=>\(\left[{}\begin{matrix}x-5=0\\3x-2=0\\3x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=\dfrac{2}{3}\\x=-\dfrac{2}{3}\end{matrix}\right.\)

\(\left(5-x\right)\left(9x^2-4\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}5-x=0\\9x^2-4=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=5\\x^2=\dfrac{4}{9}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=5\\x=\dfrac{2}{3}\\x=-\dfrac{2}{3}\end{matrix}\right.\)

\(5\sqrt{x-2}=10+3\sqrt{x+2}\)

\(\Leftrightarrow25x-50=100+60\sqrt{x+2}+9x+18\)

\(\Leftrightarrow25x-9x=168+60\sqrt{x+2}\)

\(\Leftrightarrow16x-168=60\sqrt{x+2}\)

\(\Leftrightarrow256x^2-5376x+28224=3600x+7200\)

\(\Leftrightarrow256x^2-8976x+21024=0\)

....................

\(5x^2-9x+18=0\)

Ta có: \(\Delta=b^2-4ac=\left(-9\right)^2-4\cdot5\cdot18=-279< 0\)

Vậy phương trình vô nghiệm