\(\Leftrightarrow\dfrac{4\cdot90\cdot\left(x+5\right)-4\cdot90\cdot x}{4x\left(x+5\right)}=\dfrac{x\left(x+5\right)}{4x\left(x+5\right)}\)

\(\Leftrightarrow x^2+5x-1800=0\)

\(\text{Δ}=5^2-4\cdot1\cdot\left(-1800\right)=7225>0\)

Vì Δ>0 nên phương trình có hai nghiệm phân biệt là:

\(\left\{{}\begin{matrix}x_1=\dfrac{-5-85}{2}=\dfrac{-90}{2}=-45\left(nhận\right)\\x_2=\dfrac{-5+85}{2}=40\left(nhận\right)\end{matrix}\right.\)

`1/9(x-3)^2-1/25(x+5)^2=0`

`<=>(1/3x-1)^2-(1/5x+1)^2=0`

`<=>(1/3x-1-1/5x-1)(1/3x-1+1/5x+1)=0`

`<=>(2/15x-2). 8/15x=0`

`<=>2/15x-2=0` hoặc `8/15x=0`

`<=>x=15`         hoặc `x=0`

Vậy `S=`{`15;0`}

y x 25 - 107 = 3968

y x 25          = 3968 + 107

y x 25          = 4075

y                  = 4075 : 25

y                 = 163

\(y\times25-107=3968\)

\(y\times25=3968+107\)

\(y\times25=4075\)

\(y=4075:25\)

\(y=163\)

\(3\left(x-2\right)^2-25=2=>3\left(x-2\right)^2=27=>\left(x-2\right)^2=\frac{27}{3}=9=3^2=\left(-3\right)^2\)

\(=>\orbr{\begin{cases}x-2=3\\x-2=-3\end{cases}=>\orbr{\begin{cases}x=5\\x=-1\end{cases}}}\)

\(3\left(x-2\right)^2-25=2\)

\(=>3\left(x^2-4x+4\right)-25=2\)

\(=>3x^2-12x+12-25=2\)

\(=>3x^2-12x-13=2\)

\(=>3x^2-12x-15=0\)

\(=>\left(x-5\right)\left(x+1\right)=0\)

\(=>\orbr{\begin{cases}x-5=0\\x+1=0\end{cases}=>\orbr{\begin{cases}x=5\\x=-1\end{cases}}}\)

\(\left\{{}\begin{matrix}x-2y=3\left(1\right)\\x^2+xy-5y=25\left(2\right)\end{matrix}\right.\)

\(\left(2\right)\Leftrightarrow x^2+xy-5y-25=0\) 

\(\Delta=b^2-4ac=\left(y+10\right)^2\ge0\)

=> phương trình (2) có 2 nghiệm \(\left\{{}\begin{matrix}x_1=\dfrac{-b+\sqrt{\Delta}}{2a}=5\\x_2=\dfrac{-b-\sqrt{\Delta}}{2a}=-y-5\end{matrix}\right.\)

- với x=5 thì y=1

-với x=-y-5 thay vào (1)=> y=\(-\dfrac{8}{3}\);\(x=-\dfrac{7}{3}\)

\(\frac{80}{x+4}+\frac{80}{x-4}=\frac{25}{3}\left(ĐkXĐ:x\ne\pm4\right)\\ \Leftrightarrow\frac{3.80\left(x-4\right)+3.80\left(x+4\right)-25.\left(x^2-16\right)}{\left(x^2-16\right).3}=0\\ \Leftrightarrow\frac{240x-960+240x+960-25x^2+400}{3.\left(x^2-16\right)}=0\\ \Leftrightarrow\frac{-25x^2+480x+400}{3.\left(x^2-16\right)}=0\\ \Leftrightarrow\frac{-25.\left(x+\frac{4}{5}\right)\left(x-20\right)}{3.\left(x^2-16\right)}=0\\ \Leftrightarrow\left[{}\begin{matrix}x+\frac{4}{5}=0\\x-20=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-\frac{4}{5}\left(Nhận\right)\\x=20\left(Nhận\right)\end{matrix}\right.\\ \Rightarrow S=\left\{-\frac{4}{5};20\right\}\)

\(3.\left(x-2\right)^5+19=115\)

\(3.\left(x-2\right)^5=96\)

\(\left(x-2\right)^5=32\)

\(\left(x-2\right)^5=2^5\)

\(\Rightarrow x-2=2\)

\(\Rightarrow x=4\)

vậy \(x=4\)

b) \(5.3^{x+1}-45=90\)

\(5.3^{x+1}=90+45\)

\(5.3^{x+1}=135\)

\(3^{x+1}=27\)

\(3^{x+1}=3^3\)

\(\Rightarrow x+1=3\)

\(\Rightarrow x=2\)

vậy \(x=2\)

a,        3(x-2)⁵ + 19=115

                    3(x-2)⁵=115-19

                    3(x-2)⁵=96

                      (x-2)⁵=96:3

                      (x-2)⁵=32

           \(\Rightarrow\)(x-2)⁵=2⁵      mà số mũ 5 > 1

           \(\Rightarrow\)x-2=2

                       x=2+2

                       x=4

                        Vậy x=4

b,    5.3^x+1-45=90

            5.3^x+1=90-45

            5.3^x+1=45

               3^x+1=45:5

               3^x+1=9       mà 9=3³

        \(\Rightarrow\)3^x+1=3³    mà cơ số 3 > 1

       \(\Rightarrow\)x+1=3

                     x=3-1

                     x=2

                      Vậy x=2.